Talk:Monty Hall problem/Arguments

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It is all just terminology

Perhaps all this argument is about nothing but terminology. Does anyone disagree with any of the following statements? It is assumed throughout that the car is placed randomly, the players initial pick is random, and that the host always opens an unchosen door to reveal a goat.

1) The probability of winning by switching, given that the player has picked door 1 and the host has opened door 3, is between 1/2 and 1 depending on the hosts door preference. (This is clearly a conditional probability)

2) The probability of winning by switching, given that the player has picked door 1 and the host has opened door 2, is between 1/2 and 1 depending on the hosts door preference. (This is clearly a conditional probability)

3) The probability of winning by switching, given that the player has picked door 1 and the host has opened one of the unchosen doors, is 2/3 for all host preferences. (Morgan call this the unconditional case, maybe this is where the confusion arises).

4) The probability of winning by switching given that the player has picked a door and the host has opened one of the unchosen doors is 2/3 for all host preferences.

5) The probability of winning by switching on average for player who switch is 2/3 for all host preferences.

Martin Hogbin (talk) 10:26, 21 March 2009 (UTC)

You're right. And notice that in case 3, opening one of the unchosen doors is not a condition, because it is part of the rules of the game. The only condition, if you would consider the full model, there is the initial picking of door 1. In case 4 there is no condition at all in the full model and hence it is the same as case 5. But ... notice that in the MHP a specific door has been chosen and a door opened.Nijdam (talk) 10:43, 21 March 2009 (UTC)

I continue here. The difficulty is what we do mean by probability. If you mean an average over all cases in which any door has been opened, there is, as I mentioned, no condition. If you mean: a door has been opened, but I do not know which one, and I want to average over all cases concerning the same door, the answer depends on the door and is a number between 1/2 and 1. Nijdam (talk) 17:55, 23 March 2009 (UTC)

I do not quite agree with you. If we average over all cases where a door has been opened, but you do not know which one, the answer is 2/3 if we include all possible host strategies in our average. The only way to get an answer other than 2/3 is to average over cases where the host has a fixed strategy and the same door is opened every time. Martin Hogbin (talk) 20:50, 23 March 2009 (UTC)

Nijdam, do you see the connection with my simple problem in which a prize is placed behind one of two doors and the player then picks a door? If the player picks randomly the probability of winning is 1/2. If the player has picked just one specific door then the probability of winning depends on the initial prize placement. Martin Hogbin (talk) 09:07, 24 March 2009 (UTC)

Who else agrees? Martin Hogbin (talk) 18:43, 21 March 2009 (UTC)

I agree with #1 and #2. I find #3, #4 and #5 somewhat ambiguous without further clarification, for example I don't know precisely what you mean by "for all host preferences". I would agree with the following revised versions:

By 'for all host preferences' I mean regardless of whether the host policiy is to: always open door 2 where possible, always open door 3 where possible, choose randomly where there is a choice, or employ any other method of deciding which door to open within the stated rules. Martin Hogbin (talk) 21:25, 21 March 2009 (UTC)

3) The probability of winning by switching, given that the player has picked door 1, is 2/3.

4) The probability of winning by switching, given that the player has picked a door, is 2/3.

5) The probability of winning by switching on average for players who switch is 2/3.

These are clearly "unconditional" statements that say nothing about the player deciding after the host opens a door. You've skipped one, perhaps 2.5, which I think is how most people actually interpret the question, specifically:

2.5) The probability of winning by switching, given that the player has picked some specific door and the host has opened a specific different door, is between 1/2 and 1 depending on the host's door preference.

What is a specific door? Do you mean a specified door (ie 1 or 2 or the rightmost door) or something else, in which case what. The host must always, according to the above rules, open an unchosen door, what is required to make this door a this a 'specific door'? Martin Hogbin (talk) 21:25, 21 March 2009 (UTC)

I mean the probability given the player has picked door i and the host has opened door j, for 1<=i<=3, 1<=j<=3, and i not equal j. -- Rick Block (talk) 22:30, 21 March 2009 (UTC)

This is also (clearly!) a conditional probability. -- Rick Block (talk) 20:24, 21 March 2009 (UTC)

What is your answer to 3, exactly as I asked it ('for all host preferences' is defined above)?

3) The probability of winning by switching, given that the player has picked door 1 and the host has opened one of the unchosen doors. Martin Hogbin (talk) 21:25, 21 March 2009 (UTC)

I still can't tell what you mean. When does the player decide, before or after the host opens a door? If before, 2/3. If after, I'd say it depends on the host preference. I think what you actually mean is "considering all cases where the player picks door 1", which is effectively the "before" case. -- Rick Block (talk) 22:30, 21 March 2009 (UTC)

I mean exactly what I say, 'given that the host has opened one of the unchosen doors. I mean that the host has already opened one of the unchosen doors to reveal a goat but we are given no information as to which door it is. Martin Hogbin (talk) 23:26, 21 March 2009 (UTC)

The host has opened a door and you're just not saying which one it is? The the probability of winning by switching is between 1/2 and 1, specifically 1/(1+p) where p is the host's preference for the door he's opened (the probability of the host opening that door when the player initially picks the car). I think this is equivalent to "what is the probability if the player picks door 1 and the host opens door j, for some j in {2,3} (right?). I believe the question must refer either to all cases where the player picks door 1 (probability is 2/3, equivalent to deciding to switch before the host opens a door) or to a specific case where the player has picked door 1 and the host has opened some door j. The question that asks both of these simultaneously is the first one, which is apparently not what you're going for. What is your answer for 2.5? -- Rick Block (talk) 03:26, 22 March 2009 (UTC)

The door that the host has opened is not identified thus there is no way of quantifying the host's door preference. If the host has opened door 3 then it could be the case that the host always opens door 3 if possible. If the host has opened the rightmost door it could be that the host preference is always to open the rightmost door. On the other hand, if all we are told is that the host has opened one of the unchosen doors there is no way to specify a host preference that could have resulted in the actual choice. The choice is made random by this lack of information.

This is a very important point to resolve, do you have any suggestions on ways to do so? Martin Hogbin (talk) 10:17, 22 March 2009 (UTC)

Resolve in what sense? In a Wikipedia sense there is no issue, since you've still never offered up a single source supporting anything you've said. My suggestion is for you to drop this ridiculous foray into WP:OR and pay attention to what the math sources say.

If you mean resolve in the "understanding" sense, I think (my opinion) if the player picks a door, and then the host opens a door, the player can clearly see the door she picked and the door the host opened. From the structure of the problem, which door the host opens is plainly obvious. I think this means we have three choices - consider all cases (the fully unconditional problem), consider all cases where the player picks door 1 (or door i for some i if you must, unconditional with regard to the specific door the host opens), or consider one case where the player has picked door 1 (or door i) and the host has opened door 3 (or door j if you must - fully conditional). There is no middle ground here. You can't turn this into an urn problem by coyly refusing to identify which door the host opens since the doors are distinguishable. The way (the only way) to make it equivalent to an urn problem is to force the host to pick randomly between two goats. In urn problems the balls are indistinguishable, so any choice among them must be random - in terms of MHP this means the p we keep talking about has to be 1/2. -- Rick Block (talk) 17:03, 22 March 2009 (UTC)

You state your view as if it is supported by some evidence. That facts are simple, if the door opened by the host is not identified in the statement of the question then it is not possible to propose a host strategy that makes the chances of winning by switching anything other than 2/3. There is nothing in Morgan which says otherwise. Morgan clearly and consistently describes a situation in which one specific and identified door has been opened by the host. This is what the words words say and what the maths in the paper describes. There is absolutely nothing in the paper that says the host door preference is relevant to the case where one unknown door has been opened.

There clearly is a middle ground here, which is the case that an unchosen and unspecified door has been opened. Morgan call this the unconditional case and express it as 'a mixture of the two conditional cases' and proceed to show that probability of winning by switching in the scenario we are discussing is always 2/3. I am happy to refer to this particular case as unconditional if you like but it is clear from the maths that Morgan are referring to the case where either door 2 or door 3 has been opened by the host but we do not know which.

So my assertion is not OR, it is based on reading a reliable source on the subject. You insist on interpreting it in a way that is not supported by the wording or the maths in the paper itself, if you disagree then please show me where Morgan make clear that the host opening door 3 is intended to represent the more general case where the host has opened either door 2 or door three. I believe that Nijdam agrees with on this particular point, so I suggest that we do need to resolve it in some way. Martin Hogbin (talk) 18:05, 22 March 2009 (UTC)

Martin - your argument here boils down to claiming that a paper co-authored by 4 statistics professors published in a peer reviewed math journal applies to the one and only case where the player picks door 1 and the host opens door 3, and that their solution is not meant to be taken as a solution for the general case of player picks door i and host opens door j (for all i and j, 1<=i<=3, 1<=j<=3, and i not equal j), and that their solution doesn't apply if we don't know the specific number of the door the host opens.

In the section you're reading, where they talk about the unconditional problem, they say "The unconditional problem is of interest, too, for it evaluates the proportion of winners out of all games with the player following a switch strategy." This is almost exactly the same as my wording above ("all cases where the player picks door 1"). It is the probability given only that the player picks door 1 (which might as well be door i - or are you still arguing this one as well?). In their phrasing of it (quoted in the section above), the unconditional problem has a major difference from the MHP which is that the decision to switch is made before, not after, the host opens a door.

Earlier, talking about F1, they say (I've also quoted this above) "The distinction between the conditional and unconditional situations here seems to confound many, ...". You wouldn't by any chance be confounded about this, would you? -- Rick Block (talk) 21:07, 22 March 2009 (UTC)

Yes, what I am saying is that in a a paper co-authored by 4 statistics professors published in a peer reviewed math journal you can take it that what is written is what is intended, and what is written is, 'To avoid any confusion, here is the situation: The player has chosen door 1, the host has then revealed a goat behind door 3...'. That seems pretty clear to me. Are you saying that the learned gentlemen actually meant something else? If so, they make no mention of this fact anywhere in the paper. Also the maths which follows is based literally on the quoted statement. Surely, if the authors had intended the calculation to be representative of a more general case (the host has then revealed a goat behind one of the unchosen doors) they would have said so.

Regarding the section where they talk about the unconditional problem, I agree that the wording is not entirely clear, but the maths is. They simply sum the two probabilities for the door 3 and the door 2 cases (weighted according to the chances that the particular door would be chosen). This is exactly what you would do to get the probability of winning given that that either door was opened. Nijdam says that the case where one (unspecified) of the doors was opened should not be called conditional because there is no condition other than the game rules. Perhaps he would like to comment. I would also say that there is a good case for calling this unconditional as none of the probabilities for the event in the sample space are changed by the host action, but I am no expert.

All I was saying when I talked about resolving the issue was that, rather than fight this out, perhaps we could get some other opinions on the matter. Martin Hogbin (talk) 22:23, 22 March 2009 (UTC)

I'll wait for Nijdam to reply, but can you clarify what you mean by "the probability of winning given that either door was opened"? Your use of "either" here makes this somewhat ambiguous. Do you mean the probability of winning given the player picks door i and the rule that the host will open a door revealing a goat (which is the unconditional probability of winning given the rules and that the player picks door i)? Or, do you mean the general solution for the probability of winning given the player picked door i and the host has opened door j (for some i, j - which I think makes this the conditional probability given the player picked door i and the host opened door j). Do you see the difference between these, and (if so) which one do you mean?

I mean the latter, the the host has already opened an unspecified door consistent with the game rules as described above. I do see the difference but in this particular case it does not affect the probability of winning by switching. Martin Hogbin (talk) 09:54, 23 March 2009 (UTC)

BTW - I think it's obvious Morgan et al. intend the solution for the specific case of door 1 and door 3 to be taken as the general solution for player picks door i and host opens door j. One thing that makes this obvious is that their solution ends up expressed as a formula, 1/(1+q), not involving door numbers (and, yes, q is literally defined as probabilities involving the host's preference for door 3). It is also common when discussing conditional probabilities to use a specific case as representative of all similar cases - consistent with the "say No. x" wording in the problem statement. -- Rick Block (talk) 23:45, 22 March 2009 (UTC)

You are entitled to your opinion as to what Morgan mean as I am to mine but they are both opinions, let us see what others think. You point out the flaw in your own argument. If we do not specify which door has been opened then we cannot define what q means. Once we specify a door we can give some meaning to q. Note that q (or p) must be the probability that the host will open a specified door.

Let me ask you this question. In the case where the host has opened an unspecified door, what is the probability that the host will win by switching. You say 1/(1+q), where q=1-p. Where p is what? You said earlier that p in that case would be the probability that the host would open the door he did. After he has opened a door that probability is clearly 1. Martin Hogbin (talk) 09:54, 23 March 2009 (UTC)

p is the host's preference for the unopened door (door 2 in the problem as stated). If the probability of winning by switching is 2/3 for all i,j (where the player picks door i and host opens door j), how can it be something different for i=1 and j=3? If you want the general probability for any i,j take out Morgan's paper, and replace "door 1" with "door i" and "door 3" with "door j" anywhere you see door 1 and door 3. q becomes the host's preference for door j. You've previously said you're a physicist - so surely you understand the meaning of using a variable. If you object to naming the door the host opens "door j" then you're not talking about a specific case where the host has opened a particular door - you're instead talking about all cases where the player picks door i. You can't be both specific and unspecific at the same time. -- Rick Block (talk) 13:48, 23 March 2009 (UTC)

Yes I do understand what variable is, which is why I can say that the host cannot have a preference for door j, where j is a variable. This is a statement with no meaning. If the host actually has a preference, then it must be for one identifiable door or another, it cannot be for door j. Remember we are referring to the host's strategy, not the host's action, so we could ask the host before the game, what their strategy would be should the player choose the door that the car is behind. The host could answer any of, 'I will choose randomly from the other two doors', 'I will always open door 3 if I can', 'I will always open the center door if I can', or I will always open door 3 if the car is behind door 1', but the host cannot answer, 'I will alway open door i'. Martin Hogbin (talk) 19:01, 23 March 2009 (UTC)

(outindent) I think we need to stop trying to describe this in words. Are you talking about:

1) $P({\text{win by switching}})\,$

or

2) $P({\text{win by switching}}|{\text{player picks door i}})\,$

or

3) $P({\text{win by switching}}|{\text{player picks door i and host opens door j}})\,$

or (perhaps)

4) $P({\text{win by switching}}|{\text{player picks door i and (host opens door j or host opens door k)}})\,$

I am talking about #4. Martin Hogbin (talk) 09:34, 25 March 2009 (UTC)

where in the Parade version of the problem i=1 and j=3 (and k=2). What Nijdam is saying above is that #4 is equivalent to #2. What I'm saying is that if the player has made an initial pick and is now standing in front of two closed doors and one door the host has opened, we can call the player's pick door i and the door the host opened door j, meaning we're talking about #3. Morgan et al. show the probability for #3 (specifically in the case i=1 and j=3) is 1/(1+p) where p is the host's preference for door 3.

Agreed. Martin Hogbin (talk) 09:34, 25 March 2009 (UTC)

This obviously generalizes to player pick i and host door j where the probability p refers to the host's preference for door j. -- Rick Block (talk) 03:30, 24 March 2009 (UTC)

No it does not, but I am having difficulty at present explaining why. Martin Hogbin (talk) 09:34, 25 March 2009 (UTC)

Do we mean something different by generalize? What I mean by #3 (player pick i and host opens j) is simply shorthand for each of the following cases.

• Player picks door 1 and host opens door 3 (and, in this case, q is the host's preference for door 3) - this is the specific Morgan et al. case
• Player picks door 1 and host opens door 2 (and, in this case, q is the host's preference for door 2)
• Player picks door 2 and host opens door 3 (and, in this case, q is the host's preference for door 3)
• Player picks door 2 and host opens door 1 (and, in this case, q is the host's preference for door 1)
• Player picks door 3 and host opens door 2 (and, in this case, q is the host's preference for door 2)
• Player picks door 3 and host opens door 1 (and, in this case, q is the host's preference for door 1)

What q is in each case may vary. We could talk about q₁₃ to mean (for example) the q in the first case above and have 6 different q's - but they're all variables between 0 and 1. In their paper Morgan et al. literally address the first case above, but by substituting "door 2" everywhere they say "door 3" (and vice versa) doesn't this turn their paper into an analysis of the second case above? My assertion is this means Morgan et al.'s analysis generalizes to show the answer in each case is 1/(1+q), for some q between 0 and 1. This is what I mean by generalize. -- Rick Block (talk) 14:07, 25 March 2009 (UTC)

I rest my case until Martin and I have finished our analysis in mathematical terms on the Analysis page. Nijdam (talk) 10:17, 25 March 2009 (UTC)

Rick your analysis makes no sense. The parameter q represents the host's preference for a particular door. This is a fixed preference, representing the host strategy. To take your first two examples:

• Player picks door 1 and host opens door 3 (and, in this case, q is the host's preference for door 3)
• Player picks door 1 and host opens door 2 (and, in this case, q is the host's preference for door 2)

The host cannot have a preference of q for door 3 and a preference of q for door 2. This is a pre-existing preference, which exists before the player chooses a door and certainly before the host opens a door. After the host has opened a door his probability of opening that particular door is clearly 1.

To show this, we could consider the example of the malevolent host. With the normal rules (including player has chosen randomly), let us say that the player picks door 1 and the car is behind door 1 (the only case where the host has a choice). The host wants the player to do as badly as possible, what action does he take? There is nothing he can do to change the players chance of winning by switching. Martin Hogbin (talk) 09:24, 27 March 2009 (UTC)

Martin - did you read the part about specifying these as individually named parameters, e.g. q₁₃? If you'd like we can say the probability of winning by switching is one of 1/(1+q₁₃), 1/(1+q₁₂), etc. for the six different host preferences. Every one of these is of the general form 1/(1+q) where q is a variable that has a value between 0 and 1. The answer is always some unknown value between 1/2 and 1.

Yes I did read that bit, note that p=1-q in Morgan's example. These are fixed probabilities for the case where the player initially picks door 1 and the host could write them down on a card at the start of the game (the player could have cards for the cases that the player pick door 2 or door 3 but I agree that this is not important). After the player has chosen a door the host can then look to see where the car is and then using the probability written on his card open a door.

In your malevolent host example P(switching wins|player picked the car) is 0, regardless of what the host does. I think what you're really suggesting is that the host try to minimize P(switching wins|player picked door i) for some i (like, say, door 1). This is always 2/3, regardless of the host's preference so a player who decides to switch before the host opens a door wins 2/3 of the time. However, if the host has a known strategy this can help a player who gets to decide after the host opens a door, i.e. affects P(switching wins|player picked door i and host opened door j) - which I think is the situation described in the problem. If you think nothing the host does can change the player's 2/3 chance of winning by switching, the real money offer I made to Glkanter above still stands and I'd be happy to play this game with you as well if you're willing. -- Rick Block (talk) 14:02, 27 March 2009 (UTC)

What you say is true but it changes nothing. One time in three the player will have chosen the car and will lose by switching, whatever the host does and whenever the player makes the choice of what do do. Two times out of three the host will only have one door to choose and the player will win by switching to the only remaining door. The host can do nothing to change this.

I am not sure what you offered Glkanter but here is what I would offer you. The 'car' is placed randomly behind one of three doors. I will pick a door (I will always pick door 1, unless you have a problem with that, in which case I will pick randomly). You must then always tell me the number of a loosing door that I have not picked (2 or 3 if I have picked 1). You can use any strategy you like to do this. I have the option of switching to the remaining door or sticking to my original door. If I win I will pay you 6 units, if I loose you pay me 5 units. If you want to play this for real money I am game. Martin Hogbin (talk) 20:29, 27 March 2009 (UTC)

Again, what the host is not changing is P(win by switching|player picks door i), which is not the question we're trying to answer. Your offer also addresses this probability. This probability is 2/3. We all agree about that. The other probability is the one where the player can decide after the host opens a door - with knowledge of what door the host has opened. My offer to Glkanter is described in the section immediately above, I'll repeat it here - see section 'Game challenge'.

Game challenge

Start with an ace and two jokers. You're the dealer.

• Shuffle the 3 cards. Deal them face down left to right. The leftmost card corresponds to "door 1", the middle one "door 2" and the rightmost one "door 3".
• initial pick is random - I'll pick the card on the left (door 1). I now have a 1/3 chance of having the ace.
• Look at the card in the middle and then the card on the right (the two I didn't pick). Immediately turn over the other one if either is the ace. If you haven't done this, now you have a choice to make. You can comply with the MHP rules by turning either card over. It doesn't matter if I know how you're deciding, what matters is whether you decide (with preference p) or whether you're forced to take a random action (i.e. your preference p=1/2). To make the effect of this the most obvious (p=1), if you have a choice please turn over the rightmost card (door 3). If it doesn't matter how the host decides in this case and can't affect my initial 1/3 chance of having the ace, you should be willing to accept this method of deciding.
• host makes the offer to switch
• Player decides after you flip a card - I'll never switch, but simply vary my bet that I have the ace.

We've scrupulously followed the rules of the MHP (right?), so your claim is I have a 1/3 chance of having the ace and you have a 2/3 chance regardless of which card you turned over - i.e. you have a 2-1 advantage so should be willing to give me 1.5 to 1 odds if I never switch. I'll stay every time, but I'll bet a penny if you turn over the middle card (door 2) and $100 if you turn over the card on the right (door 3). Repeat at least 60 times and I'll do this for real money. My guess is I'll end up roughly $500 up even though we both know you'll win 2/3 of the time. My chances of having the ace change depending on which card you turn up, and by varying my bet I can make money if you give me anything better than even (1-1) odds. -- Rick Block (talk) 22:08, 27 March 2009 (UTC)

That is not an accurate representation of the MH scenario. You want to be the player in a case where the player knows the host strategy. In the MHP the host strategy is unknown. If we agree that the host strategy should be known only to the host (as is the case in the stated problem) then I am happy to play you as either the host or the player. Martin Hogbin (talk) 11:23, 28 March 2009 (UTC)

Where does it say the probability of interest is the probability from the player's perspective? As I read it, the question is about the probability given everything we know from the problem statement - which presumably the player knows as well, but in my view this is actually irrelevant (perhaps the player has no probabilistic skill whatsoever and can only guess). For example, if the problem statement said the host rolled a die and put the car behind door 1 if die=1, and door 2 if die=2, and door 3 otherwise and we asked what is the player's chance of winning by switching given which door the player initially picked and which door the host opens, we wouldn't "solve" this by assuming the player doesn't know how the host picked which door to put the car behind and start by assuming a 1/3:1/3:1/3 distribution.

I agree that there is nothing to say that the probability must be from the player's perspective but it must be based on the information given in the question and the host's strategy is not given. Thus in any game based on the problem the host strategy cannot be defined.

My view is that by saying the probability for the "MH scenario" is 2/3, we're saying it must be 2/3 for any variant that meets the stated conditions. This variant complies. The nature of a counter-example is to create an example that meets the stated conditions but doesn't have the claimed results. Since I can create a variant where the probability is not 2/3, but 1/2 or 1 depending on which door the host opens, either

• the problem is underspecified (this can be fixed by clarifying that the host must pick randomly or, where you seem to be going [a valid, but perhaps unusual direction], by clarifying that the player doesn't know the host's strategy) - this is exactly the same kind of quibble as the quibble that the host must make the offer to switch and must show a goat, or
• the answer that the probability of winning by switching given knowledge of which door the host opens is 2/3 is incorrect (in the sense that it is not always the case given only what is in the problem statement), or
• the problem is misstated and should be changed to clarify that the probability of interest is the unconditional probability (one way to do this is to force the player to decide before the host opens a door).

Pick one. -- Rick Block (talk) 18:05, 28 March 2009 (UTC)

It looks like neither of us will be winning any money just yet. I understand the argument the conditional probability of winning by switching is between 1/2 and 1. I cannot say that this is incorrect; it as the correct answer, but not to a question that anybody cares about. Martin Hogbin (talk) 22:23, 28 March 2009 (UTC)

Let's Clarify Morgan's Points

According to Morgan, does anybody's solution give a numeric result without the 'equal goat door' constraint?

According to Morgan, why would the 'Combining Doors' solution be a 'false' unconditional solution?

As there is no mention whatsoever of any 'host behaviour', or door numbers in Rick's 5 bullet points, why do we accept Morgan's criticisms that rely on them? Glkanter (talk) 23:34, 22 March 2009 (UTC)

From their conclusion:

In general, we cannot answer the question "What is the probability of winning if I switch, given that I have been shown a goat behind door 3?" unless we either know the host's strategy or are Bayesians with a specified prior."

We can get into the Bayesian thing if you'd like (my guess is you really don't want to do this), but what they're saying is to state a specific number you have to make an assumption about the host's strategy.

The combining doors solution is equivalent to their F5 that I quoted above.

Morgan's criticisms rely on interpreting the problem as a conditional probability problem. The fact that the doors are numbered is part of the problem statement, as is the potential for the host to have a preference for one door over another. Their criticisms match criticisms published in other sources as well, such as the Gillman paper referenced in the article. Here's a quote from the Gillman paper:

What is the probability P that you win if you switch, given that the host has opened door #3? This is a conditional probability, which takes account of this extra condition. When the car is actually at #2, the host will open #3. But when it is at #1, he may open either #2 or #3. The answer to the question just asked depends on his selection strategy when he has this choice—on the probability q that he will then open door #3. (Marilyn did not address this question.)

Two different math papers. Both say almost exactly the same thing. We accept both as reliable sources. -- Rick Block (talk) 00:19, 23 March 2009 (UTC)

Thanks Rick. I admire your patience and willingness to share your knowledge.

What if this is the question: "0. What is the aggregate probability of winning by switching - meaning across all players regardless of which specific door they pick and which door the host opens?"

What if we add 'the equal goat door' constraint to MvS' solution? Does an unconditional solution work then? If so, why is the Morgan conditional solution, which also requires the 'equal goat door' constraint, preferable to an unconditional solution?

Has Morgan really dis-proven the 'Combined Doors' solution? Or just 'dismissed' it?

Isn't there a 'rule' in Probability that says 'whatever is not constrained can (must?) be assumed to be random'?

This may be my last word on the topic: Given that there is no mention of any 'host behaviour' in any description of the problem (not the solutions), how does Monty's inevitable choice of losing doors affect the original 'chosen door = 1/3 probability'? Glkanter (talk) 01:54, 23 March 2009 (UTC)

Question 0 is the unconditional question, and the answer is 2/3 chance of winning by switching (no host preference required, any unconditional solution is just fine).

Adding the equal door constraint enables the unconditional solution to arrive at the correct answer and, therefore, to be used but it requires an argument of some kind like "Due to the natural symmetry of the situation, the conditional probability of interest must be the same as the unconditional probability. We can see the unconditional probability is 2/3 because ...". I'll note (my opinion) that this is a dangerous approach since people generally suck at recognizing when "natural symmetry" actually applies.

Morgan et al. don't directly say anything about the combined doors solution.

The rule in Probability is that whatever is not constrained may have any distribution (random is just one option).

Regarding your last question, let's try this one more time. The player's 1/3 chance at the beginning splits into the player's chances for the case where the host opens door 2 and the case where the host opens door 3. The 1/3 at the end (which is a conditional probability) is only a piece of the original 1/3 (half in the case where p=1/2). p (and 1-p) are the fractions of this 1/3 that end up in each case.

                   1/3           +          1/3         +         1/3    = 1
                   /\                       /\                    /\
                  /  \                     /  \                  /  \
                 /    \                   /    \                /    \
                /      \                 /      \              /      \
host opens:  door 2   door 3          door 2   door 3       door 2   door 3
              /          \             /          \          /          \
             /            \           /            \        /            \
          (1/3)p    +  1/3(1-p)  +   0       +     1/3     1/3     +      0 = 1

If p is 1/2 the two terms at the left are each 1/6, and when the host opens door 3 the (unconditional) chances are (1/6,1/3,0). To express these as conditional probabilities we divide by the sum, i.e. divide each by 1/2, which makes these (1/3, 2/3, 0). If the player is looking at two closed doors and one open door, this split has happened (in accordance with whatever p is). The upshot is that the original (unconditional) 1/3 and the resultant (conditional) 1/3 are never the same 1/3 - and the unconditional solution is talking about the 1/3 on the top line, not the conditional probability you can compute from the bottom line by dividing (1/3)p by ( (1/3)p + 1/3 ). Note that this turns out to be 1/3 if p is 1/2, but depending on p it might be anything between 0 and 1/2. -- Rick Block (talk) 14:55, 23 March 2009 (UTC)

Do you have a reference for your response to my 'Random rule' question? Or is that your opinion?

I've quoted Morgan on this (search for "natural symmetry" on this page). -- Rick Block (talk) 19:13, 23 March 2009 (UTC)

In the article, why not 'define the problem' as question 0 (above)? That's how I've always interpreted it, anyways. Who's to say? Then use the best unconditional solution.

Question 0 doesn't match any description of the MHP I've ever seen published anywhere. The whole point is to put the player in front of two closed doors and one open door and ask what the probabilities are - doing this is what makes it a conditional problem. -- Rick Block (talk) 19:13, 23 March 2009 (UTC)

If Morgan doesn't address the 'Combining Doors' solution, and 3 out of 4 sources do not include the 'equal goat door' constraint, then why not use this for the Solution, without any qualification? Glkanter (talk) 16:18, 23 March 2009 (UTC)

Why not say it's a conditional probability problem which most people answer assuming p=1/2? If we do this, then if someone asks about some stupid variant, like the "host forgets" one (which vos Savant has asked about) someone coming to the article here will see that it is a conditional probability problem and be able to figure out the correct answer. The combining doors solution says the answer to the host forgets variant is 2/3 (but it's actually 1/2) - vos Savant's comments about this are kind of interesting [1]:

Back in 1990, everyone was convinced that it didn’t help to switch, whether the host opened a losing door on purpose or not. Assuming a knowledgeable host who would always open a losing door, that was incorrect. (A knowledgeable host who opened a winning door on purpose wouldn’t have much of a show, would he?!)

Now everyone is convinced that it always helps to switch, regardless of what the host knows. But this is just as incorrect!

As I said above someplace - fuzzy thinking (specifically, treating a conditional problem as if it's unconditional) contributes to this. -- Rick Block (talk) 19:13, 23 March 2009 (UTC)

Without assuming any 'host behaviour' whatsoever, how is the 'Combining Doors' solution inadequate for the MHP problem via MvS as you understand it?

Here's a new can of worms...since 'host behaviour' is never mentioned in any 'non-variant' MHP problem statements (except occasionally to say he chooses goat doors equally), why can't we treat each instance of the game the same as flipping a coin? You know, the past results don't influence the probabilities of the next playing of the game? Glkanter (talk) 20:24, 23 March 2009 (UTC)

If we don't assume any host behavior we don't know how the original (unconditional) 1/3 splits when the player opens a door, so the chances of winning by switching are between 1/2 and 1, not 2/3 - so the combining doors solution is not correct. The issue is the question that is being asked. The question is not "what is the probability of winning by switching for all players who switch", but "given you're looking at two closed doors and one the host has opened, what is the probability of winning by switching". It's a conditional question.

Re worms - I may not be following what you mean. Do you mean treat the host's preference as 1/2?

If you want the problem to be unconditional, I think you have to change the question. Gillman's version is basically "decide before the host opens a door". Morgan's version is "host will open a door showing a goat, what is your chance of winning if your strategy is to switch" (implying a decision before the host opens the door). Grinstead and Snell's version compares players who unconditionally (i.e. decide before the host opens a door) switch vs. players who unconditionally stay. -- Rick Block (talk) 04:04, 24 March 2009 (UTC)

I just read MvS' response to 'the host forgets'. It violates one of the bullet points (premises), and needs no further discussion here. So Morgan doesn't address the 'Combining Doors' solution. I don't see how how it relies on any 'host behaviour'. The fact is, after revealing a goat the formula reads either 1/3 + (2/3 + 0) or 1/3 + (0 + 2/3) where the 1/3 on the left of the '+' is the chosen door and the numbers inside the parenthesis are the left and right unchosen doors. As both formulas are exactly equal to 1/3 + 2/3, I do not see this as conditional, regardless of any 'host behaviour'. Can you please provide a published source which either disproves the 'Combining Doors' solution, or explains why the 'Combining Doors' solution must be solved as a conditional problem? Glkanter (talk) 14:08, 26 March 2009 (UTC)

If you're so sure it's always 2/3 regardless of the host behavior why won't you put real money on it? -- Rick Block (talk) 14:04, 27 March 2009 (UTC)

We're playing cards with an Ace and 2 Jokers? We bet the same amount before each hand? I pick, you discard a Joker, and I get to switch every time? Sure, I'll play. I anxiously await your reply to my previous questions. Glkanter (talk) 14:37, 27 March 2009 (UTC)

No - your claim is the host never affects the odds so it doesn't matter if we bet before or after the host discards, or are you agreeing that the player's odds might change from 2/3 depending on some host behavior? I want to bet after as I described above. Regarding combining doors - it's clearly an unconditional solution and we've already talked about the references saying the problem is a conditional problem (Morgan, Gillman, Grinstead and Snell - there are plenty more if these aren't enough). -- Rick Block (talk) 15:28, 27 March 2009 (UTC)

I find this thread unproductive at this point. I will return to the main talk page and propose moving the 'Combining Doors' solution to a prominant and non-discreditted placement in the article. Glkanter (talk) 15:48, 27 March 2009 (UTC)

Words Have Meanings

I'd like to see the word 'variant' expunged from these discussions.

As it is used, it represents a new premise (or constraint) to the MHP. Any new premise represents a different problem than the specific MHP, with the 5 agreed upon bullet points.

I think it would be more precise to call these 'different problems', which intentionally share characteristics of the MHP. But, by definition, they are no longer the MHP. Glkanter (talk) 19:51, 28 March 2009 (UTC)

I agree that the wording 'Monty Hall Problem' is better used to refer to the simple (unconditional if you like) case. Martin Hogbin (talk) 21:25, 29 March 2009 (UTC)

Can we get this straight

I think that most people believe that Morgan's solution to the MHP applies only to the specific case where the player has chosen door 1 and the host has opened door 3. Who agrees with this and who disagrees? Martin Hogbin (talk) 21:28, 29 March 2009 (UTC)

Clarification Morgan's solution would also apply to a case where we were told that player has chosen door 1 and the host has opened door 2 but it does not apply if we are told that the player has chosen door 1 and the host has opened one of the other two doors. Martin Hogbin (talk) 08:53, 30 March 2009 (UTC)

Martin - The distinction between the conditional and unconditional situations here seems to confound you. Morgan et al.'s solution addresses the conditional case which is, as you say, player chooses door 1 and host opens door 3. Or, player chooses door 1 and host opens door 2. Or player chooses door 2 and host opens door 1. Or player chooses door 2 and host opens door 3. Or player chooses door 3 and host opens door 1. Or player chooses door 3 and host opens door 2. These are the conditional cases. Read the problem description. At the point the player decides, she will be in one and only one of these cases (not two, not all six - this is not a problem in quantum physics). Evaluating the probabilities in effect in one, and only one, of these cases is exactly what it means to evaluate the conditional probability. -- Rick Block (talk) 13:24, 30 March 2009 (UTC)

So let me get this quite straight. You agree that if we are told the player has chosen door 1 and the host has opened one of the other two doors the problem is not conditional and the probability of winning by switching is always 2/3? Martin Hogbin (talk) 18:18, 30 March 2009 (UTC)

Yes. The question as you phrase it is asking about the chances of winning given the host rules and that the player has picked door 1 (not also given which door the host has opened). However that's not how the problem is structured. The player can always see exactly which of the two doors the host opens. At the decision point, the player is standing in front of two closed doors and one open door. The player can therefore see which specific one of the two doors the host opened. There is no uncertainty about which door it is. If the player picks door 1 the question is NOT what are her chances of winning by switching given the host WILL OPEN (or has opened, if you must) an unknown one of door 2 or door 3, but what are her chances of winning given which specific door (one of door 2 or door 3) she has seen the host open with her own two eyes. Again, the distinction between the conditional and unconditional situations confounds many. -- Rick Block (talk) 18:51, 30 March 2009 (UTC)

I have no difficulty in distinguishing conditional from unconditional it is just that, from an earlier conversation with you, I got the impression that you believed that the answer to the question where the host had opened one (unspecified) of the two unchosen doors should be treated conditionally. Obviously that was a misunderstanding on my part.

It is interesting to consider what 'the probability of winning by switching' means in the Morgan context. It is a measure of what proportion of the games the player would win by switching if we repeat the game with the car being placed randomly each time but with the host having exactly the same door opening strategy and having opened the same door each time. Martin Hogbin (talk) 19:05, 30 March 2009 (UTC)

The earlier issue may have been around your (continued?) notion that it is possible to ask the "conditional" question about both doors the host might open simultaneously. Probabilistically, you're either asking an unconditional question (without being given the specific door the host opened - so might as well ask about the situation before the host opens the door) or a conditional question (given the specific door). You are correct about what Morgan et al. mean by "the probability of winning by switching". Again, this is exactly what is meant by the conditional probability (the probability given which door the player picked and which door the host opens). Do you see (yet) how this is the precise question that is asked? -- Rick Block (talk) 00:25, 31 March 2009 (UTC)

I think we have finally reached agreement. Morgan's solution is correct iff you take the question and the answer to have the meanings that I have given above. I am glad we have got round to the question that was asked. Morgan's assumption is certainly a possible and reasonable interpretation of the question but it is far from being the only one. This has been my main point all along. As I have said before, I am happy to accept that the Morgan paper as a reliable source on the subject of mathematics and probability; where I think they do not qualify so highly is in the subject of interpreting the question that Parade reader intended to ask and understanding how they might interpret the answer. As we have reached agreement on some things I will stop this discussion here and start a new section on how I think Morgan have got the question wrong. Martin Hogbin (talk) 08:38, 31 March 2009 (UTC)

Morgan's interpretation

Many people, including myself get the feeling that something is wrong with the Morgan solution to the MHP but find it hard to say exactly where the problem lies. None of Morgan's assumptions is unreasonable but in nearly every case there are others that could have been made. As usual, I am considering only the case that the host always opens an unchosen door to reveal a goat and always offers the swap. So how do Morgan et al manage to conjure up such a unintuitive answer? They do this by the way they interpret the question.

If we state the question in an impersonal form such as:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No 1 then another door opens, say No 3 which has a goat. After the door has opened you then have to decide whether you want to pick door No. 2. What is your probability of winning by switching?

This might well be interpreted to mean that the car is placed a random and the door is opened at random. With that assumption the answer is simply 2/3. There is nothing unreasonable about this assumption, as vos Savant said, we may well take the host to be acting simply as the agent of chance, and anyway the producer and the host have little to gain by adopting anything other than a random strategy.

Now consider the question in a more personal form:

Suppose you're on a game show, and you're given the choice of three doors: The producer has arranged for a car to be placed behind one door and goats behind the others. You pick a door, say No 1 then the host opens another, say No 3 which has a goat. He then says to you, "Do you want to pick door No. 2?" What is your probability of winning by switching?

Here we have two events that depend on the strategy of an individual. The answer is not 2/3 or even that given by Morgan but it depends on the probability that the producer has arranged for the car to be behind door 1. There is nothing unreasonable about this interpretation, we know that one person or another arranged for the car to be behind one of the doors and we have no idea what their method or motives were in this respect. The only real problem with this answer is that the solution is any number between 0 and 1 depending on the producer's car placing strategy - not very interesting.

What Morgan do is take the question and personalise the host, he has a strategy or a probability of taking certain actions if certain events occur. On the other hand Morgan tacitly take it that the car has been placed randomly. So hey presto! we now have an 'elegant' solution that gives a definite answer to 'the question that was asked' and proves everyone else wrong. Easy when you know how.

Now I agree that there is some justification for Morgan's assumption in that a host is mentioned in the question and a producer or stage hand is not, however, I suspect their their decision to take the question literally like this was motivated more by a desire to conjure up an 'elegant' solution than to answer the question that Whitaker probably intended to ask.

The further problem with Morgan's solution, which is relevant here, is the matter of what our readers will take the probability to meant. As we have agreed above it only refers to a case where the car is placed randomly but the host has a fixed strategy. Martin Hogbin (talk) 17:38, 31 March 2009 (UTC)

Martin - If the player's initial choice is random, it's equivalent to the producer (or whoever) randomly placing the car in the first place. If either one of these is random, then the probability of initially selecting the car is 1/3 regardless of which door the player picks (I've offered before, but would you like to see the math for this?). The "tacit" assumption that the car was randomly placed is an explicit assumption, clearly intended to match vos Savant's interpretation - as clarified by the explicit instructions she provided for the simulation she suggested (see [2]). Morgan et al.'s solution applies to a case where either the car was initially placed randomly or the player's choice was random and any case where the host is not constrained to pick randomly if the player initially picks the car (not just for a "fixed host strategy"). In particular, q (the host's preference for one door over another) might change every day, or might be constant - if the host is not constrained to pick randomly we simply don't know what q might be (or what it might depend on - perhaps the host opens the leftmost door on Mondays and the rightmost door on Fridays). There's absolutely no requirement that the player (or even the host!) actually know what value the preference parameter takes. From a problem definition viewpoint, not specifying this allows any number of (what Glkanter doesn't like to call) variants. If we don't specify this and then go on to say the answer is 2/3, the variants where q is not 1/2 are legitimate counterexamples that prove us wrong.

I think what you're basically saying is that it makes no sense not to assume q is 1/2. Sensible or not, it's a math problem - so if it's not given in the problem statement assuming it is wrong. This is important to point out for this specific problem since it is fundamentally a conditional probability problem. How the host decides which door to open in the case the player initially selects the car always affects the answer, whether you want it to or not. Saying the answer is 2/3 is saying p is 1/2. If this is not given then it is an unjustified assumption. Also note that it's not just Morgan et al. who have this viewpoint. Gillman as well. And Grinstead and Snell. And lots of others. It is in some sense a trivial oversight not to say the host picks randomly in this case. However, since it changes the answer from 2/3 to [0.5,1] it is probabilistically significant. -- Rick Block (talk) 19:09, 31 March 2009 (UTC)

Your first statement is wrong because it is given in the question that the player has picked door 1. There is no maths required. Given that the player has picked door 1, suppose that we have a producer who always puts the car behind door 1. You tell me what the probability of winning by switching is in that case? Now suppose that we have a producer who places the car behind door 1 with probability c, what is the probability of winning by switching then?

With regard to Morgan's assumption about random car placement can you tell me where in the paper they state that they have taken this to be the case?

I agree that there is no requirement for the player to know the host door selection parameter, that is because we are told that the host has opened door 3 In exactly the same way because we are told that the player has picked door 1 the manner by which the player picked that door is not important. We have been here before. Suppose the game were very simple and the player simply wins what is behind the door. Suppose that we are told that the player has picked door 1. What is the probability of winning. In that case it is clearly c (where c is the probability that the producer has put the car behind door 1). If we are told only that the player has randomly picked a door then the probability is, as you say, 1/3.

I am not saying that 'it makes no sense not to assume q is 1/2' at all, I am saying that there are four options. We can assume c is 1/3 (where c is again the probability that the producer has put the car behind door 1), or we can assume q is 1/2, or we can assume neither, or we can assume both. Morgan pick one of these possibilities as if it were the only one. How the host picks a door is completely unimportant in answering the question as given as we are told which was the door picked. Think of it as a conditional problem if you like, what is the probability of the player winning given that she picks door 1. It depends on the value of c.Martin Hogbin (talk) 21:57, 31 March 2009 (UTC)

You seem to be missing the main point of the paper, which (IMO) is that the problem is one involving conditional probabilities. Precisely what they assume for their solution is far less important than the basic approach - although (and I've said this numerous times) they assume exactly what vos Savant assumes, nothing more and nothing less (the probability of the initial pick being 1/3 is mentioned at the top of the first column on page 286 - and the bit about generalizations on p 287 clearly shows this to be a conscious assumption). The reason it's important to approach the problem conditionally is so that any assumptions are explicitly exposed. I believe it is the case that before their paper was published many solutions using an unconditional approach were touted as "the solution" with with most people not understanding that these solutions 1) directly address a slightly different question than the one that was asked (e.g. deciding to switch before the host opens a door - or any of the other unconditional versions we've discussed), and 2) if applied to the conditional question carried along an implicit assumption affecting the result.

Regarding your specific objections - if you're taking "say no 1" with regard to the player pick to mean that the only outcome of interest is when the player picks door 1 then why have you been arguing for weeks that the door the host opens is unspecified? Clearly door 1 and door 3 are meant to be illustrative choices, mentioned merely to put us into a conditional setting (so we can imagine the player standing in front of two closed doors and one open door and deciding whether to switch). Is this related to your continued claim that the Morgan et al. solution applies only if the player picks door 1 and the host opens door 3? Do you still not agree that the solution applies in any of the 6 conditional cases? I think virtually anyone reading the problem takes "door 1" and "door 3" to be a randomly selected case from among these 6 cases, with the expectation that the same results pertain to the other 5. It seems to me that this might be the point you're missing - that analyzing one of the 6 conditional cases and taking this to be representative of the answer in all 6 cases is entirely different than coming up with an "overall average" sort of answer. Any unconditional solution does the latter. Only a conditional solution does the former. -- Rick Block (talk) 00:18, 1 April 2009 (UTC)

Let me deal with your second point first. It is not totally clear whether the original questioner intended the question to refer to specific doors. I certainly believe that there is some doubt.

If we take it that the original question was meant to say something along the lines of, 'The player opens a door then the host opens another door', then we both agree that the simple unconditional solution applies and the Morgan paper is irrelevant.

If, on the other hand, we take it, as Morgan do, that the doors are to be specified, as in, 'The player has chosen door 1, the host has then revealed a goat behind door 3...' then we have agreed above that the probability of winning by switching depends on the hosts door preference. Do you also agree that in this case the probability of winning by switching also depends on the producer's initial placement of the car? If you agree this last point then we can continue to discuss the effect of Morgan's choices on the problem solution. Martin Hogbin (talk) 08:01, 1 April 2009 (UTC)

The point is: do we consider the average player of the game, who may decide before anything has happened that his chance of winning the car is 2/3 when switching. Or do we consider a player who made his initial choice and actually sees an open door with a goat. Not only do I think the latter is meant, but it is also the only interesting case.Nijdam (talk) 10:04, 1 April 2009 (UTC)

I am not sure why you keep asking me that question. I have made it quite clear that, if we take it that specific door numbers are given in the problem statement, then I am considering the probability after the host has opened a specified door. I agree that this makes the problem conditional and understand why it does so. As we have all agreed that if specific door numbers are not mentioned in the problem statement the problem is non-conditional, I am now talking only about Morgan's conditional interpretation of the question.

Not exactly. What we have all agreed is that is is possible to carefully craft a wording that logically asks an unconditional question. Where we disagree is whether this is even remotely reasonable and whether by doing so you've somehow made the Morgan paper irrelevant. Even if the doors are not numbered they have a physical location (left, right, and middle - for example), so they can always be distinguished and not numbering them achieves nothing. It is not the door numbers being in the problem statement that makes the problem conditional, it is letting the player decide to switch after a door has been opened. If the player is deciding after a door has been opened the player can tell which one she picked, which one the host opened, and which one is the other one. The question Nijdam asks is precisely the right question. Are we considering an "average" player of some type, or a player who can see two closed doors and a goat? I agree with Nijdam that the latter case is the only interesting one - and moreover would claim it's the only reasonable way to read the question. -- Rick Block (talk) 14:16, 1 April 2009 (UTC)

I am not sure what you mean by craftily worded. We either mention door numbers or we do not, thus:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick door 1, and the host, who knows what's behind the doors, opens door 3, which has a goat. He then says to you, "Do you want to pick door 2?" Is it to your advantage to switch your choice?

is conditional and the answer depends on the host's door opening preference, but in this case:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, "Do you want to pick the remaining door?" Is it to your advantage to switch your choice?

the answer does not depend on the host's door opening preference. Whether you call it conditional or not I do not mind, I would call the 'condition' a null condition.

This has the form of a conditional question, and (as such) the answer does depend on the host's door opening preference. We don't know what it is, so the answer here is something between .5 and 1 (inclusive). -- Rick Block (talk) 01:57, 2 April 2009 (UTC)

However, none of the above affects what I have said. I agree that in the conditional case the probability of winning by switching depends on the host door opening strategy. Now the important point. Do you agree that the probability of winning by switching also depends on the probability with which the car was placed behind the chosen door? Martin Hogbin (talk) 11:12, 1 April 2009 (UTC)

Is the player pick random or not? This is the essential difference between the host preference and the car placement. The host preference affects the theoretical probability whether the player knows it or not. The car placement affects the probability computation only if the initial player pick is also not random (essentially meaning the player knows about it). You keep insisting that this is a problem of some sort with the Morgan et al. paper, but it's a completely different topic. They acknowledge it is possible to pursue this. They chose not to. Solutions that misread the question and present unconditional solutions are not choosing to ignore the potential host preference, they are (apparently unknowingly) answering a different question. -- Rick Block (talk) 14:16, 1 April 2009 (UTC)

No Rick you are wrong. If the car placement is not random it makes no difference whether the player knows about this or not, or whether their pick is random provided that the door that the player picks is specified. If the producer alway puts the car behind door 1 and we are simply told that the player has picked randomly, the probability that the player has picked the car is 1/3, but, if we are told in the problem statement that the player has picked door 1, what is the probability that she has picked the car? You tell me. It is not a hard question. Martin Hogbin (talk) 19:31, 1 April 2009 (UTC)

Martin - if the car is not placed randomly but the player picks randomly, the question "given the player has picked door 1 what is the probability the player has picked the car" basically negates the randomness of the player's pick. You can certainly ask this question but it has no relevance to the situation. It's like asking what the probability is of the car being behind door 1 if it's placed randomly but given the producers put it behind door 1 (??!!). Of course this probability is 1. But, it's a silly question. Reasonable questions here might be "what is the chance the player wins by switching given a non-random placement and a random initial choice?" (the answer is 2/3 if the host chooses between two goats randomly) or "what is the maximum chance the player wins by switching given a non-random placement and a knowledgeable initial choice?" (the answer to this involves three probabilities and examining how the player might maximize her chances of winning). Any question phrased "if the player picks, say door 1 ..." carries a VERY strong implication that this should be taken as a representative case (i.e. either the initial placement was random or the player's choice was random). -- Rick Block (talk) 01:57, 2 April 2009 (UTC)

We either take statements such as "if the player picks, say door 1 ..." to mean literally that the player opens door 1 or we take it to indicate a random choice. If we take it to indicate a random choice then the same applies to "if the host opens a door, say door 3 ...", we take this to mean that the host opens one of the unchosen goat doors randomly (unless there is only one). Alternatively we take the mention of a door number to mean literally that the mentioned numbered door is in fact opened, in which case the chances of winning by switching depend on the host's door opening preference and the initial car placement probability. You cannot have it both ways. Martin Hogbin (talk) 08:29, 2 April 2009 (UTC)

We take it to be a specific choice, but which one randomly selected. -- Rick Block (talk) 13:48, 2 April 2009 (UTC)

Rick, what is your answer to this very simple question? We have a producer who always puts the car behind door 1. A player chooses door 1. What is the probability that the car is behind the door that they have chosen? If you insist that it is important to know how the player chose door 1 then let is say that he rolled a die and divided the result by 2. Martin Hogbin (talk) 18:43, 2 April 2009 (UTC)

How do you interprete this game. You throw a die with a cup. And before you look at the result I have a peek and tell you whether it is below 3 or above 2 , say above 2, and let you guess, say 5. What will be your probability of guessing right? Nijdam (talk) 12:45, 2 April 2009 (UTC)

I interpret this to be asking P(correct guess|above 2). Possible outcomes are 3,4,5, or 6, so 5 is the correct guess with probability 1/4. -- Rick Block (talk) 13:48, 2 April 2009 (UTC)

I'm not surprised about your answer, Rick. The question is meant for Martin. Nijdam (talk) 17:21, 2 April 2009 (UTC)

I am not sure why you are still asking me questions about conditional probability. Perhaps you could explain what it is that you think I do not understand about it. I agree with Rick's answer.

We are now arguing about a very simple question. What is your answer to the question that I ask Rick above. Martin Hogbin (talk) 18:43, 2 April 2009 (UTC)

The conditional probability, given the player has picked door 1 is 1. Simple as that. Nijdam (talk) 09:32, 3 April 2009 (UTC)

(outindent) Martin - can you please explain what you're trying to accomplish here? -- Rick Block (talk) 13:38, 3 April 2009 (UTC)

Yes. I am trying to reach an agreement on what the mathematical facts are. I do not think that, in this respect, I am suggesting anything unusual or contrary to normal teaching in the subject. In particular I am trying to reach agreement on the player's probability of winning by switching in the following circumstances. If we take it that the player has opened door 1 and the host has opened door 3, with the usual rules, then:

If the producer places the car randomly and the host picks a door to open randomly (when he has a choice) the the probability of winning by switching is 2/3.

If the producer does not place the car randomly and the host does not open a door randomly then the probability of winning by switching depends on both the producer's car placement and the host's door opening strategy.

Do you agree? Martin Hogbin (talk) 17:39, 3 April 2009 (UTC)

I take it you wrote by accident "... the player has opened door 1 ...". The random car etc. implies that the conditional probability of winning by switching is 2/3. Let us not speculate about other situations. I specially put emphasis on "conditional", because the whole too lengthy discussion is about this, and you seemingly have forgotten to write this. Nijdam (talk) 18:59, 3 April 2009 (UTC)

Sorry, I meant the player has chosen door 1.

In both cases I meant the conditional probability given that the player has chosen door 1 and the host has opened door 3. In the first case the car was placed randomly and the host door opening strategy is random. In other words the host's strategy is to choose randomly between doors when he has a choice. In this case the conditional probability (that is to say, given that the host has opened door 3) of winning by switching is 2/3.

In the second case I mean that the car has not been placed randomly but according to the preference of the producer and the host does not choose randomly when he has a choice of doors but chooses which door to open according to his preference. In this case the conditional probability (that is to say, given that the host has opened door 3) of winning by switching is dependent on the producer's preference and the host's preference. Do you agree?Martin Hogbin (talk) 19:53, 3 April 2009 (UTC)

Of course, it's what Rick and I have written over and over. Nijdam (talk) 11:29, 4 April 2009 (UTC)

Neither of you has mentioned the effect of car placement before. Martin Hogbin (talk) 20:22, 4 April 2009 (UTC)

And, skipping ahead, do you Martin agree Morgan et al.'s solution is sensitive to both the car placement and the host preference, and that they explicitly, deliberately, and consciously assume the car placement is random, and that this (and their other assumptions, like the host always makes the offer to switch) exactly match vos Savant's published experimental procedure? -- Rick Block (talk) 12:03, 4 April 2009 (UTC)

In Morgan's paper the effect of non-random car placement on the probability of winning by switching is not mentioned anywhere. It is not explicitly stated to be random - if you believe this to be the case, please tell me where they say this. What the authors were conscious of when they wrote the paper I cannot say. I accept that on page 287 they mention generalizations which 'appear to be of less interest' including the 'possibility of nonuniform probabilities of assignment of the car to the three doors'. Considering that non-random car placement has just as much an effect on the result as non-random host door preference it seems odd to leave it as an afterthought.

What vos Savant says is irrelevant. Morgan start their calculation on page 285 where they discuss the more general scenario. There is no mention whatever at that stage that the car placement is taken to be random although the calculation has obviously assumes this fact with no justification or explanation whatever. Later, on page 286, the vos Savant constraints are added. It really is unfair to attempt to blame Morgan's failings on vos Savant. Vos Savant wrote a column in a general interest magazine and made reasonable assumptions in the circumstances, although these were not all made clear initially. Morgan et al were writing a paper for a learned journal and attempted a comprehensive solution to the question.

It does seem that you now agree that the Morgan paper only applies to the specific case where the producer has placed the car randomly but the host chooses with some not-necessarily-random probability which door to open. All other combinations produce different results. The only question that remains is how reasonable are the Morgan assumptions, considering all they achieve is to obfuscate a simple but very unintuitive problem, and how relevant they are to a WP article on the subject. Martin Hogbin (talk) 20:22, 4 April 2009 (UTC)

It is complete unclear to me what you're aiming at. Why bother about the placement of the car, as with the random placement it turned out to be difficult enough to understand. I thought we reached a point where you understood the "conditional nature" of the problem. It is not my intention to complicate the article with such considerations. It is about time the article gets a make-over. Nijdam (talk) 20:36, 4 April 2009 (UTC)

I have always understood the conditional nature of the problem although we still seem not to have reached agreement on exactly what formulations of the problem must be treated as conditional. My point is that the article does indeed need a makeover and the Morgan paper is a bad place to start as it addresses only one specific formulation of the problem.

In fact, can I just confirm that we agree on one other thing which is that the Morgan solution only applies if specific door numbers are mentioned in the problem formulation? Martin Hogbin (talk) 21:06, 4 April 2009 (UTC)

No. As we've discussed before lack of door numbers doesn't change anything since the doors can always be uniquely identified by their position on the stage. -- Rick Block (talk) 22:39, 4 April 2009 (UTC)

Perhaps I should have said that the question must identify the doors in some way, as in the player chooses the centre door and the host opens the leftmost door (but not as in the player chooses a door and the host opens another one) Do you accept that this is a necessary condition for the Morgan solution to apply. Martin Hogbin (talk) 23:16, 4 April 2009 (UTC)

No. And please stop with the word games. The only condition necessary is that the player be given the opportunity to switch after the host opens a door. This is what makes it a problem about conditional probabilities, which is what the Morgan paper addresses. You keep trying to imply it's an urn problem if we don't identify the doors (or something) but it's not an urn problem. The doors are inherently different, so unless you depersonalize the question (e.g. ask about all players who switch) if the player is choosing after the host opens a door it's a conditional probability problem. I think everyone agrees the player gets to pick after, and we want to imagine a player standing in front of two closed doors looking at a goat - this makes it a conditional probability problem. -- Rick Block (talk) 00:20, 5 April 2009 (UTC)

I am not playing word games I am making clear, as you once did, that a probability problem must be solved on the information given in the question. If information allowing us to identify the doors is not given in the question then we cannot use that information in the solution. The facts that the doors may have numbers on or that the player will know which door was opened are irrelevant as they are not given in the statement of the question. It is, if you like, still a conditional problem but the condition is that one of the host opened one of the two possible doors, we cannot say which. In this case the answer is 2/3.

If you want to discuss the relevance of Morgan's solution when the problem is viewed from from the player's perspective then I will be pleased to do so. 86.132.253.23 (talk) 08:46, 5 April 2009 (UTC)

Not everyone agrees with your conclusion that it's necessarily a conditional puzzle. And the player is not just "standing in front of two closed doors looking at a goat". Take a look at the Combining Doors solution. He's looking at the door he selected, the remaining door he did not select, and a goat. There's no condition. Glkanter (talk) 01:45, 5 April 2009 (UTC) Glkanter (talk) 01:58, 5 April 2009 (UTC)

Here's a quote from Feb 15, 2009 from your buddy, the subject matter expert: "By the way, I hope I haven't given the impression that I think the problem must be understood as conditional. As for Marilyn's exact statement, I think it's ambiguous enough on several fronts, and there's no reason to state with absolute confidence it should be conditional." Despite my request, he never explained this any further, and I'd be the last guy to twist his words, but this is what he wrote. Glkanter (talk) 02:09, 5 April 2009 (UTC)

That's a quote from user:C S, archived in Talk:Monty Hall problem/Archive 9. -- Rick Block (talk) 03:51, 5 April 2009 (UTC)

I had to re-read my posting 3 times before I realized the missing 'r', which changes 'you' to 'your'. Sorry for any confusion I caused. Glkanter (talk) 05:01, 5 April 2009 (UTC)

Martin - please explain to me why the Gillman paper says pretty much exactly the same thing as the Morgan et al. paper. They both conclude the chances of winning by switching for the Parade version of the problem (with the vos Savant assumptions) is not 2/3 but 1/(1+q) where they both say q is the unspecified host preference for one of the doors. -- Rick Block (talk) 22:39, 4 April 2009 (UTC)

Because that is correct. Provided of course that the host does have a preference and does not choose randomly and that the producer places the car randomly and does not have a preference. Martin Hogbin (talk) 23:15, 4 April 2009 (UTC)

Correction - provided only that the producer places the car randomly. 1/(1+q) is the solution whether or not the host has a preference. If the host chooses randomly (has "no preference") then q=1/2 and the answer simplifies to 2/3, but the answer is still 1/(1+q). This is a simplified form of (1/3)/(1/3 + q(1/3)) which directly matches the probability of interest (win 1/3 in the case the car is behind the "other" door when the only possibilities are this or the losing case with probability p(1/3)). Did you ever read my explanation to the anon poster at Talk:Monty Hall problem#How complicated is it???? -- Rick Block (talk) 00:20, 5 April 2009 (UTC)

Further correction. If you really want to be pointlessly pedantic then the probability of winning by switching always depends on the producer's car placement. If this is random then, of course, the probability is 1/3 that the car is behind each door and its effect can be ignored. The situation is exactly the same for the producer and the host, you can ignore the effect that their strategy may have or not. 86.132.253.23 (talk) 09:05, 5 April 2009 (UTC)

It's not pointlessly pedantic. I think you're continuing to miss the significance of this. If (assuming random initial placement) you say the answer is 2/3, you're saying q is 1/2 because the answer to the conditional question is always 1/(1+q). Combining this with the other conversation going on in this thread just above, saying the answer is either 2/3 or 1/(1+q) is also saying the initial car placement (or player's initial choice) is random - but this assumption matches how nearly everyone interprets the problem (and is not something Morgan et al. simply overlooked as you keep trying to suggest). Most people are unaware the host preference (which is usually unconstrained by the problem statement) directly influences the probability of interest. Saying I have a 2/3 probability of winning by switching is different than saying all players who switch have a 2/3 probability of winning (you do see the difference, don't you?). This is the point of the card game I proposed that you're unwilling to play with me where I'll have an overall 2/3 probability winning by switching but 1/2 or 1 any given time we play. What I keep hearing you say is my probability of winning, every time I play, is 2/3 but you'll only put money on it if you don't let me use the information about which card you flip (door the host opens) to adjust my bet. You're not stupid - you can see the probability will be 1/2 or 1 and will "average out" to 2/3. What I don't understand is why you then have such a hard time agreeing with me that the probability, per game, can vary between 1/2 and 1. -- Rick Block (talk) 16:20, 5 April 2009 (UTC)

I fully understand why the probability of winning is between 1/2 and 1. The point that I have been trying to make for some time now is that this solution only applies in certain formulations of the problem. Let us consider the problem with the usual rules and the conditional case - we must assess the probability of winning by switching after a door has been opened.

Firstly, it is trivially true to say the required probability is between 0 and 1 inclusive. We want to do better than that, in fact we want to give as narrow a range of probabilities as possible.

If we start with the case where the producer has not placed the car randomly and the player has not chosen randomly then the best we can do is still to say that the answer is somewhere from 0 to 1.

Now add the constraint that the producer has placed the car randomly but the host does not open a door randomly. If we now formulate the problem with identified doors then we both agree that the best that we can do is to say that answer is between 1/2 and 1 depending on what the host's (unknown) door opening strategy happens to be.

Now do the same but do not identify doors. We do not yet agree on this one. I say the probability is still, of course, between 1/2 and 1 but it is in fact 2/3. I am not sure if you still disagree.

Now add the constrain that the host must choose the door to open randomly. The probability is still, of course, between 1/2 and 1 but it is in fact 2/3. We agree on this.

Finally let us make the problem unconditional, the player decides at the start what they will do. Once again we agree that the answer is 2/3.

My point is that the answer of between 1/2 and 1 is only the best that we can do in some of the many possible formulations of the problem. It is easier to see just how restrictive the Morgan scenario is by considering the average results obtained by repeating the game many times. This has nothing to do with the conditional/unconditional issue, we can repeat the game with the relevant condition applied every time. Martin Hogbin (talk) 17:17, 5 April 2009 (UTC)

I think we're on the same page except for the bit about not identifying doors. My view is that the problem is either conditional, i.e. applies to a specific player standing in front of two closed doors and a goat (in which case the doors are inherently identified) or unconditional (applies only in some "average" sort of way). You seem to be thinking there's a third option - "conditional" but without specifying which door. This option is mathematically the same as the unconditional case - and seems pointlessly contrived. The Morgan et al. result can be easily observed, but requires keeping track of which door the player picks and which door the host opens per the #Excel simulation of difference between "random goat" and "leftmost goat" variants above. This is the criticism of the usual experimental validation as well. If you lump all 6 combinations of player pick and host door together, you're not keeping track of the data that might expose a host preference. This is fine if all you're interested in is the overall chance of winning by switching - but not if you want to know the probability of winning by switching given which door the host opens (i.e. this is not good enough if you want to decide whether to switch after the host opens a door). -- Rick Block (talk) 18:55, 5 April 2009 (UTC)

Maybe we are arguing about very little. The case when an unidentified door has been opened is the case that I have been referring to as having a null condition. You say it is mathematically equivalent to the unconditional case. It looks like we agree.

However, that brings us back to Morgan's interpretation of the problem as necessarily conditional. After misquoting the original question, Morgan state that '...the host has revealed a goat behind door 3', clearly conditional, but the question that Whitaker actually asked said, '... the host ... opens another door, say No. 3...'. We might well take the question to mean '...the host has revealed a goat behind another door', which we have just agreed makes the problem unconditional.

As the reliable source Seymann points out, we need to consider what Whitaker actually meant by the question. I very much doubt that he meant the number of the door that was opened by the host to have any significance. Martin Hogbin (talk) 21:31, 5 April 2009 (UTC)

Martin - Is the question what is your, Martin's, chance of winning by switching if you find yourself in this position (and you're allowed to switch after the host opens the door) or is it some overall average? When you, Martin, are in this situation you know which door you originally picked and which door the host opened. And, in this case, Morgan's solution applies. IMO, the whole point of the MHP is to put you in the player's shoes, at the point of the decision (i.e. after the host has opened a door). The absolutely standard way to make the odds 2/3 by switching is to include the random host pick constraint. This fixes everything. Why are you so reluctant to include this as a constraint? -- Rick Block (talk) 23:12, 5 April 2009 (UTC)

I have no objection to using the random goat door constraint as one of the justifications for the 'popular' solution.

I am sure that you agree that Morgan's paper is based on treating the question as a formal probability problem which must answered using only information given in the question. Taking this approach means that whether the player knows the host's strategy or not is unimportant, all that matters is what the host's strategy actually is, which we do not know. I agree that another, and possibly better, approach would be to put ourselves in the place of the player and answer from that perspective. My guess is that this is more like the question that Whitaker intended to ask. If we answer the question from the player's perspective then we can and must ask ourselves what we take the player's exact state of knowledge to be. This changes everything and essentially makes the Morgan paper irrelevant. Let us do it! Martin Hogbin (talk) 09:04, 6 April 2009 (UTC)

I've been waiting for (dreading) the return of the 'whose (who's?) point of view' question. I'm not sure who you would choose, if it was not the contestant. Why not choose the stage hand who parked the car who knows with 100% certainty when to switch? No, it can only be the contestant, and he has no way of knowing of any 'host behaviour'. Did you know there are actually laws in the United States that prevent the producers from sharing such knowledge with the contestant? Back in the 1950s, they did give contestants the answers. It became a huge scandal. To summarize, according to Morgan and Rick, the 1/3 vs 2/3 only holds true after Monty has revealed a goat with the 'equal goat door constraint'. Without it, it becomes, I guess, what, unknowable?. Take a look at the Combining Doors solution. Then extrapolate that Monty could have opened the other door. Still looks like 1/3 vs 2/3 to me.

You guys are missing the most important element in a proof: Common Sense. Glkanter (talk) 13:10, 6 April 2009 (UTC)

From the player's perspective, if she doesn't know the host's preference her chance of winning by switching is perfectly knowable - it's between 1/2 and 1 (and will average 2/3), and she should therefore switch. -- Rick Block (talk) 13:43, 6 April 2009 (UTC)

No, from the player's point of view, the probability of winning by switching is from 0 to 1, averaging out to 2/3. Or as we say in the worlds of statistics and common sense, simply 2/3. Martin Hogbin (talk) 12:58, 7 June 2009 (UTC)

From the player's point of view

As always, the solution depends on exactly what the question is. Let us assume the normal rules (the host always offers the swap and always opens an unchosen door to reveal a goat) and let us consider the particular case where the producer has placed the car randomly.

Let us also take the case that we (the player) have chosen door 1 and just seen the host open door 3 to reveal a goat, one possibility of many.

What is the probability of winning by switching. Before we can answer that we have to clarify one further point about the precise problem formulation. It is easiest to make the necessary distinction by considering the average result of a player who chooses to swap after seeing the door opened in different circumstances.

If we repeat the game, with the car being placed randomly each time and with the same host who has a (unknown) goat door opening strategy, then the answer could be anywhere from 1/2 to 1, that is to say the average result could be, for example, that we would win by switching only 1/2 of the time. (In that case, if the same host were to open door 2 and we swapped we would win always)

If, on the other hand, we repeat the game with a new host each time, each with an different and unknown goat door opening strategy then the answer will also be between 1/2 and 1, but in just the same way that it will be between 0 and 1, it will be 2/3.

I am not sure how we phrase this distinction from the point of view of a single game, but it seems clear to me that we must answer from the point of view of a player who plays the game for the first time, thus we take it to be a new host. In other words, as this is our first game ever, we start with a new universe each time. Martin Hogbin (talk) 19:52, 6 April 2009 (UTC)

Let me ask this question, suppose the prize were $1200 cash, with nothing behind the other doors. Suppose you have chosen door 1 and just seen the host open door 3 and you make your choice of whether to stick or swap. Before the result is revealed the host offers you $700 for the prize that you have now chosen (the game rules are that he always makes this same offer whether you choose to swap or stick). What would you do: swap doors and take the prize, stick and take the prize, or take the cash? I know what I would do. Answer on the basis that you have never seen the game played before but know the rules. Martin Hogbin (talk) 20:02, 6 April 2009 (UTC)

I see that I have no takers on this one. I would swap and take the prize. Is there anyone who would do different? Martin Hogbin (talk) 09:15, 8 April 2009 (UTC)

No answers. Is that because people do not know the answer or dare not give one or just that this bit has now been forgotten? Martin Hogbin (talk) 17:14, 9 April 2009 (UTC)

Since The Contestant Can't Know Of Any Host Behaviour...

In MvS' version of the MHP, there is no reference to a Host Behaviour.

Any inclusion of a Host Behaviour as a premise creates a new puzzle, not the MHP.

It would be illogical for the Host to share this info with the Contestant.

In the United States, it would be illegal for the Host to share any such info.

The only logical 'point of view' for the problem is the Contestant's.

Any discussion of the Contestant's probabilities other than 1/3 vs 2/3 requires knowledge of where the car is. The contestant won't know this until after he has made his switch or no-switch decision.

So when Rick writes:

"From the player's perspective, if she doesn't know the host's preference her chance of winning by switching is perfectly knowable - it's between 1/2 and 1 (and will average 2/3), and she should therefore switch. -- Rick Block (talk) 13:43, 6 April 2009 (UTC)"

The fact that it will average 2/3 is the only salient point. And we already know this from solutions like the Combining Doors. Glkanter (talk) 11:23, 7 April 2009 (UTC)

Why does the forgetful host version (Monty opens a door without knowing there's a goat behind it, and it luckily turns out to be a goat) require a different solution method, or does combining doors work for this different problem as well, in which case the answer is the same? -- Rick Block (talk) 13:28, 7 April 2009 (UTC)

This is a new constraint, making it a different problem. It's a premise that Monty always shows a goat, so if his 'forgetfulness' ever led him to reveal a car, that's a different puzzle. But I'm curious, how would this 'forgetfulness' be evident to the contestant? He was told Monty would reveal a goat, and Monty revealed a goat. Glkanter (talk) 14:01, 7 April 2009 (UTC)

Yes, this is a different problem that vos Savant has discussed in other Parade columns. The contestant doesn't necessarily know (hmmm, just like the contestant doesn't necessarily know about a host preference), but the question is what is the probability of winning by switching given that the host randomly opened one of the two doors but (luckily) showed a goat. She says in this case it's 50/50. My question for you is why doesn't the combining doors solution work here. If it works sometimes but not others, I'm asking how we know when it works and when it doesn't. -- Rick Block (talk) 15:46, 7 April 2009 (UTC)

Rick, this is one of your favorite techniques for prolonging the discussion in order to avoid improving the article. Bring up something different, whether it's a card game, a forgetful host, an Excel simulation, anything but the MHP. I choose not to engage in such unproductive tangents. But I'm curious, how would this 'forgetfulness' be evident to the contestant? He was told Monty would reveal a goat, and Monty revealed a goat. Glkanter (talk) 17:44, 7 April 2009 (UTC)

I bring this up not to prolong discussion but to point out the problem with the approach. The approach is at best incomplete - so basing the article on this approach would make the article incomplete as well. Your failure to understand this is the only thing prolonging anything, and your favorite technique for prolonging the discussion is to simply ignore any and all arguments that expose the fallacy of your thinking. I'm trying to get you to see why and how the approach is incomplete, but I can't force you to understand it. On the other hand, as long as you don't I think your opinion about "improving" the article is essentially worthless.

Whether the forgetfulness is evident to the contestant is irrelevant. The question is what are the odds of winning by switching (not what does the contestant think the odds are). This is not a version I'm making up - it's a version vos Savant has discussed in her column (see [3]). It's directly relevant to the MHP, not a tangent. So, again, why does the combining doors solution work for the standard MHP but not this version? -- Rick Block (talk) 18:54, 7 April 2009 (UTC)

You wrote:

"The question is what are the odds of winning by switching (not what does the contestant think the odds are)."

The odds of switching are either 0 or 1. If the Contestant originally selected the car, then his odds of winning by switching are 0. On the other hand, if he has not chosen the car, the odds are 1. Because there are people who know exactly where the car is. But this is becoming increasingly ridiculous. If you're not solving the problem through the Contestant's POV, whose are you using? Glkanter (talk) 19:19, 7 April 2009 (UTC)

The odds according to what's known from the problem statement of course. You can pursue as many diversionary tangents as you'd like, but the question remains: why does the combining doors solution work for the standard MHP but not the forgetful host version? -- Rick Block (talk) 00:34, 8 April 2009 (UTC)

From the previously agreed bullet points:

• car is randomly placed
• initial pick is random
• host must make the offer to switch
• host must show a goat (and, hence, knows where the car is)
• player decides after the host opens a door

No one not associated with the show knows where the car was placed

There is no host behaviour other than to reveal a goat

After the goat has been revealed all one can do is generalize about the overall probabilities of 1/3 vs 2/3

A valid conclusion is reached: my chances of winning increase from 1/3 to 2/3 if I switch. Glkanter (talk) 02:14, 8 April 2009 (UTC)

And why does this solution work for the standard MHP but not the forgetful host version? -- Rick Block (talk) 04:03, 8 April 2009 (UTC)

Are you in agreement with what I wrote?

When the host forgets, does he reveal a goat? Glkanter (talk) 04:19, 8 April 2009 (UTC)

The only difference is the 4th bullet (host must show a goat). In this version the host happens to show a goat (and doesn't know where the car is). A goat is revealed. Why does the combining doors solution not apply? -- Rick Block (talk) 04:30, 8 April 2009 (UTC)

I'm not sure that still qualifies as a Probability question. How is this new bullet #4 explained to the Contestant? Does the Contestant still expect Monty to reveal a goat after he has chosen a door? Are you saying this happens one time? Does he sometimes forget and show a car? Does he always forget? Glkanter (talk) 04:46, 8 April 2009 (UTC)

This is Marilyn's answer from your link:

"Here’s one way to look at it. A third of the time, the clueless host will choose the door with the prize, and the game will be over immediately. In our puzzle, that didn’t occur. So we’re considering the two-thirds of the time when either: 1) You have chosen the door with the prize; or 2) The prize is behind the unopened door. Each of these two events will occur one-third of the time, so you don’t gain by switching."

So it's a completely different problem! Like I said, it has nothing to do with the MHP. Just another time waster. And you mis-stated the problem anyway. He doesn't 'happen' to show a goat. He randomly showed a goat. So bullet #4 becomes "Monty will reveal either a car or a goat". And if Monty reveals a car, bullet #5 doesn't even take place! What happens to bullet #3? How this has anything to do with a proof of the MHP is unknown. Glkanter (talk) 05:04, 8 April 2009 (UTC) Glkanter (talk) 09:48, 8 April 2009 (UTC)

And since the door is being revealed randomly, it doesn't even need to be the host selecting the door. It could be anybody. It could even be the Contestant. Hey, I know, let's have the Contestant reveal the 2nd door! Why, that would be just like a popular game called "Deal or No Deal"! You've heard of that, right Rick? And only a completely clueless person with no common sense understanding of Probability would be ignoramus enough to try to use Deal or No Deal in an effort to prove or disprove anything related to the MHP, right, Rick? Glkanter (talk) 09:44, 8 April 2009 (UTC)

Yes, it's a different but related problem. The problem statement is here. The reference to the standard MHP is unmistakable. The only difference is the host forgot which door hides the car, but does open a door showing a goat. You have an initial 1/3 chance of having selected the car. The other two doors have a 2/3 chance. At the point you're deciding whether to switch you're standing in front of two closed doors looking at a third open door. Surely the chance the car is behind your door is 1/3 and the chance it's behind the other two doors is 2/3. And, after the host opens a door (as you keep saying) 1/3 + 2/3 = 1. I understand Marilyn's explanation, but why doesn't the combining doors solution work? -- Rick Block (talk) 13:48, 8 April 2009 (UTC)

This is my final response ever to one of your edits. You have now clearly demonstrated, again, that you have no idea what you are talking about. Let's start with the bullets:

• car is randomly placed

    no change

• initial pick is random

    no change

• host must make the offer to switch

    not if he already revealed the goat, he won't

• host must show a goat (and, hence, knows where the car is)

    he will randomly reveal either a goat or a car (and, hence, he doesn't know where the car is)

• player decides after the host opens a door

    1/3 of the time (according to Marilyn), the player will not even be offered a switch

There are no teachings in Probability in which these would be considered the same problem, or where it would be considered appropriate to substitute the results of one for the other.

I've explained how the Contestant, or ANYONE NOT ASSOCIATED WITH THE PRODUCTION OF THE SHOW, will not know of any Host Behaviour, or the original placement of the car.

I've offered up the Combining Doors solution, which is in the article, and has apparently passed 2 (maybe 3) FA reviews as a source for the unconditional solution.

I've shown that the best the Contestant, or ANYONE NOT ASSOCIATED WITH THE PRODUCTION OF THE SHOW, can do is to generalize the 1/3 vs 2/3 to his single opportunity.

And it's been shown (by you!) that without the 'equal goat door constraint', the so-called conditional solution offers up something like "it's between 1/2 and 1, and averages 2/3" as it's best answer. Which, after all that horse hockey and complications, etc., is no better than the Combining Doors solution.

For all these reasons, I am again going to return to the MHP talk page and propose that the focus of the article be turned towards solutions along the lines of the Combining Doors solution. Glkanter (talk) 14:26, 8 April 2009 (UTC)

I'm sorry if this makes you angry. I'm not suggesting substituting the results, just the solution method. It seems like a simple enough question. I take it from your refusal to address it that you cannot explain why the combining doors solution works for the standard MHP but not for the host forgets version (and SHOUTING ABOUT IT doesn't make anything more clear for anyone). Since the question is the same in both cases (what is the probability of winning by switching) and the problems are so similar (only one detail different) it seems like the same method should be able to be used for both. Marilyn's approach for the forgetful host version is a conditional solution. She says "we’re considering the two-thirds of the time when either ...". The reason she does this is because the unconditional solution (2/3) is the wrong answer in this case. The answer to the question I'm asking (why doesn't the combining doors solution work) is because the question is a conditional question and in the host forgets version the conditional answer is different from the unconditional answer.

This finally brings us to the point I'm making about the standard MHP. Since the question in the standard MHP and this forgetful host version is the same (what is the probability of winning by switching), my claim is that the standard MHP is also a conditional question (if it's a conditional question when the host forgets it must also be a conditional question when the host remembers). The combining doors solution (or any other unconditional solution) is not directly answering the question. You can say it ends up with the right answer so who cares, but unless you know why it applies in one case and not another you're not understanding the problem. In fact, it only applies in the standard MHP if the host is also constrained to pick between two goats equally.

A solution to the standard MHP in the same style as Marilyn's solution for the forgetful host version would be something like:

Here’s one way to look at it. Half the time, the host will open door 2. In our puzzle, that didn’t occur. So we’re considering the half of the time when either: 1) You have chosen the door with the prize (and since we're considering half the time, your original 1/3 chance is divided in half and is now 1/6); or 2) The prize is behind the unopened door (in which case the host is forced to open door 3). The car is behind the unopened door one-third of the time but behind your door only 1/6 of the time, so your chances of getting the car double by switching.

This is a conditional solution as well and depends on the host opening door 2 and door 3 exactly half the time when the car is behind door 1. This is either assumed or must be given in the problem statement. If it isn't, then (just like the host forgets version) the conditional and unconditional solutions are different so none of the unconditional solutions (like the combining doors solution) work. -- Rick Block (talk) 19:18, 8 April 2009 (UTC)

Probabilities from the player's perspective

Strictly from the players perspective, what is the probability that she has originally chosen the car? Martin Hogbin (talk) 18:54, 9 April 2009 (UTC)

Formulation of the problem

I think we've reached a point where Martin tries to save the "unconditional" solution, by stating that the formulation of the problem permits an unconditional explanation.

Named door numbers

If in the formulation of the problem explicitly the numbers of the doors picked and opened are mentioned, the problem needs a conditional approach.

More general formulation

The problem may be formulated more general, without mentioning the door numbers explicitly, but indicating that a specific door is picked and a specific door is opened. Again conditonal.

The important distinction is whether the doors mentioned in the problem are identifiable. This could be done by means other that that of giving the door number, for example we could say the host opens the leftmost door or the player chooses the centre door but if we are going to do this we might as well give the doors numbers. The possibility of the answer not being 2/3 depends on the possibility that the host might have a non-uniform probability of opening doors when he has a choice. For it to have any meaning this probability must be expressed in terms of identifiable doors, as Morgan do.

We have all agreed that if the problem is stated in a way in which the doors are not identified, as in, the player has chosen a door and the host has opened another one, the condition is unimportant. Whether we refer to this case as conditional, unconditional, or having a null condition I do not mind, the important fact is that in this case the fact that the host has opened a door has no effect on the answer. Martin Hogbin (talk) 17:28, 7 April 2009 (UTC)

General formulation

It is possible to formulate a related, similar problem in such a way that the unconditional approach is valid. In that case the problem should address any player, regardless which door is picked and which is opened. Like: describe how the game is played and ask what the probability is for an average player to win the car (when switching).

Which one is "our" formulation? I'd say it is the first one, as door numbers are mentioned. If we interpret the naming of the door numbers as a way of expressing that any door numbers may be used, it is the second formulation. A strong argument in favour for either is that the problem says: suppose you are ..., meaning a specific player is intended and not the average player in general. Besides it is actually only an interesting problem if finally a door is opened and the player is confronted with this specific situation. Nijdam (talk) 13:05, 7 April 2009 (UTC)

The un-numbered door interpretation is all that is required for the Contestant to benefit from the 1/3 vs 2/3 advantage. All of your discussion about where the car is, and what to do when there are two goats is immaterial to the Contestant. He simply does not have any idea where the car actually is when he must make his decision. And Monty revealing a door does not change that. Because the Contestant can not know Monty's method. So whether he generalizes that it's 1/3 vs 2/3 overall, and makes his decision, or whether the Contestant for some reason says to himself "I don't know the host's preference, so my chances of winning by switching is perfectly knowable - it's between 1/2 and 1 (and I will average 2/3), I will therefore switch" (paraphrasing Rick Block, above), it doesn't make any difference whatsoever. Glkanter (talk) 14:19, 7 April 2009 (UTC)

What is the un-numbered door interpretation?Nijdam (talk) 16:30, 7 April 2009 (UTC)

Something along the lines of this:

"After the player has chosen a door, the probability it hides the car is 1/3. This probability is not influenced by the opening of a door with a goat by Monty, hence after Monty has opened a door with a goat, the probability the original chosen door hides the car is also 1/3. Because clearly the open door does not show the car, the remaining closed door must hide the car with probability 2/3. Hence switching increases the probability of winning the car from 1/3 to 2/3. Better? Nijdam (talk) —Preceding undated comment added 15:21, 6 April 2009 (UTC)."

But let's focus on the Contestant. He can not be aware of any Host Behaviour. Nor can he have any knowledge of where the car is located. So it makes no sense, and adds no value, to parse his decision beyond the generalized 1/3 vs 2/3. Glkanter (talk) 17:37, 7 April 2009 (UTC)

I believe that when the problem is viewed strictly from the player's perspective and state of knowledge, Morgan's 'goat door choice parameter' - p will evaporate. Martin Hogbin (talk) 18:23, 7 April 2009 (UTC)

I think that it is important to point out that there are at least two different and distinct interpretations in which the unconditional solution applies. The first is that we take the question to be asking about the overall probability of winning by switching, the totally unconditional case in which the questioner wants to know whether, in general it is best to swap or not, the decision being made at the start of the game. This is almost certainly what Whitaker actually wanted to know. When you think about it, most players must have a good idea whether they plan to switch or not at the start of the game and I cannot imagine a player saying to herself, 'Oh no, door 3 has been opened, I had better not switch now I have seen that'.

The other case is the one that, at first sight, appears conditional in that we consider that the decision will be made after a door has been opened but the doors are not identified. As we have all agreed (I hope) this has the effect of making the condition have no effect whatsoever, it is a null condition and the problem can be treated unconditionally.

You have agreed with yourself about this. I have not agreed, but have said repeatedly that if the decision is made after the door has been opened the player must know which door she picked and which door the host opened so this "formulation" of the problem is nonsensical. -- Rick Block (talk) 01:04, 8 April 2009 (UTC)

Rick, it was you who, quite rightly, told me that a probability problem must be solved on the information given in the question. If the information as to which door the host has opened is not given then it cannot be used or affect the answer, regardless of whether the player knows it or not. I thought we had agreed on this. Martin Hogbin (talk) 08:54, 8 April 2009 (UTC)

A third case, which I would also claim can be properly treated unconditionally, is where the host opens a goat door randomly. This again has the effect of making the 'condition' insignificant; no information can be revealed by this random act.

So, Morgan's version of the conditional solution applies only in the case that Whitaker is taken to ask the apparently conditional question, the car is placed randomly, the host has not chosen a goat door randomly, the doors are identified in the question, and, I believe, the problem is considered as a formal probability problem rather that strictly from the players perspective. As it has such limited applicability it should not dominate the article. Martin Hogbin (talk) 18:19, 7 April 2009 (UTC)

Doors not identified

I thought that we had all agreed on this but it would seem not. What are your answers to the following questions, assuming the usual rules that the host always offers the swap and always opens an unchosen door to reveal a goat and also assuming that the initial car placement is random.

We agree that if the problem states that, "the player has chosen door 1 and the host has opened door 3" then it should be treated conditionally and the answer (the probability of winning by switching) could be anywhere between 1/2 and 1 depending on the host's door opening preference.

What is your answer if the problem states, "the player has chosen door 1 and the host has opened one of the other two doors"? Note that the problem statement still clearly refers to the situation after the host has opened a door and we assume that the player can see which door this is. Who agrees that the probability of winning by switching is 2/3 in this case. Can we start with a simple agree/disagree response before we start any discussion on why. Martin Hogbin (talk) 09:10, 8 April 2009 (UTC)

Agree. Martin Hogbin (talk) 09:10, 8 April 2009 (UTC)

Mu (the case as described is nonsensical). -- Rick Block (talk) 13:54, 8 April 2009 (UTC)

It is perfectly simple English, what is nonsensical about it? Martin Hogbin (talk) 22:43, 8 April 2009 (UTC)

Nijdam, what is your answer? Martin Hogbin (talk) 17:15, 9 April 2009 (UTC)

I would have no answer, but a question: which door? Nijdam (talk) 19:54, 10 April 2009 (UTC)

But if you force me to answer, I give a conditional answer: If the host opened door 2, then ..., and if he opened door 3, then ... Nijdam (talk) 19:59, 10 April 2009 (UTC)

If you further insist, I might say: in both cases you win when switching with 2/3 chance.

One number please, you say. Me: 2/3 is the only one I can think off. Would you be satisfied? Nijdam (talk) 20:08, 10 April 2009 (UTC)

Nijdam, are you really trying to tell me that you would be completely unable to answer the question if it were posed in that way? The statement I give makes perfect sense. In fact in Krauss and Wang on page 7 they say something very similar, 'The corresponding formulation would be " Monty now opens another door and reveals a goat"'. It is clear from the context that they are talking about the player making a decision after the door has been opened. Martin Hogbin (talk) 20:12, 10 April 2009 (UTC)

There was simultaneous editing here, my reply only applies to your first answer.

So, Nijdam, do you agree that, if the door opened by the host is not identified, the probability of winning by switching is 2/3 and does not depend on the goat door preference of the host? Martin Hogbin (talk) 20:46, 10 April 2009 (UTC)

It would be pure theoretically. It would mean conditioning on the event "the door opened is not the door chosen", which is just the phrasing of one of the rules of the game. Because this event has probability 1, it is not really a condition. In fact it means the opening of the door didn't happen. Something like a player who is blindfolded. The theory permits to formulate such things, but the problem shows a different situation. In the show as we all know, and as a spectator may see with his own eyes, a door is opened and the player sees which one. That we have to calculate the conditional probabilities, is a reflection of the conditional answer I gave above. If you don't mention the door number, I'll give conditional answers, depending on the door numbers, but nevertheless conditional. It all seems to me a rather artificial effort of giving right of existence to the "unconditional solution", may be because people has strongly supported it, and are not willing to "give in". So I can say: before a door is opened by the host, the chance is 1/3 the player picks the door with the car. This wording has the same meaning as your formulation. Is it your opinion that the MHP asks for this probability? Nijdam (talk) 08:00, 12 April 2009 (UTC)

I do not seriously disagree with anything that you have said. The fact that the host is simply doing what he is bound to do by the rules does not change anything is agreed, that is why I have been consistently referring to this 'condition' as a null condition. I do not know if this is the correct technical term but it seems entirely appropriate in the circumstances. The fact that the player can see the doors is irrelevant. We must answer the question on the information that we are given, not on information that one of the characters in the problem may or may not have, see my comments on problem style.

I agree that what I am doing is an attempt to give a right of existence to the unconditional solution but ask you to consider that it is those who adhere too strongly to the paper by Morgan who are not willing to give in. Morgan is the only 'reliable source' to take such a strong line against the unconditional/simple solution. Many others treat the problem in an unconditional way but mention the issue of conditionality. The unconditional problem does have real existence and it can be extended, with a little intuition, to cover special conditional cases.

Your last question is a good one and it is the one we should all be asking. What does the MHP really ask? It is my opinion that the MHP does ask about the unconditional case. If we start from the most notable problem statement, the letter by Whitaker, then we should ask, 'What is it that he really wants to know?'. It would be a very good idea for us all to discuss that question and I would be happy to start a new section on this very important topic. Martin Hogbin (talk) 23:11, 12 April 2009 (UTC) [Re-signed]

Let me at least add to this that a lot of the strong adversaries of the simple solution, fail or refuse to understand that opening a door changes the initial probabilities into conditional ones, be it, in the usual case, with (only) for the chosen door the same values. Agree?Nijdam (talk) 11:16, 12 April 2009 (UTC)

I am not sure what you mean Nijdam, could you clarify please. Martin Hogbin (talk) 23:11, 12 April 2009 (UTC)

Devlin http://en.wikipedia.org/wiki/Keith_Devlin and many others, subsequent to Morgan's paper, have put forth the unconditional interpretation and solution as appropriate. http://www.maa.org/devlin/devlin_07_03.html Would you say he "fails or refuses to understand that opening a door changes the initial probabilities into conditional ones"? Glkanter (talk) 11:56, 12 April 2009 (UTC)

Well he is called Devlin, not God. The referred article is a column in a mathematical magazine, where mister Devlin gives his popular opinion. Not peer reviewed, so why bother with his ideas. Nijdam (talk) 17:11, 12 April 2009 (UTC)

Actually, 'mister Devlin' is most often referred to as Dr. Devlin. http://www.stanford.edu/~kdevlin/ Glkanter (talk) 19:15, 12 April 2009 (UTC)

His PhD certainly wasn't on a probabilistic subject. Nijdam (talk) 21:16, 12 April 2009 (UTC)

Whilst we are talking about the status of various papers we should see Morgan for what it is. Although it is published in a peer-reviewed journal it is not really a piece of current statistical research but more something of a novelty paper. The paper is of such a poor quality that I am surprised that it got past the referees. It also carries a 'health warning' in the form of a polite but significant critical commentary by Seymann at the end. Martin Hogbin (talk) 23:11, 12 April 2009 (UTC)

Why Morgan are right.

This is for newcomers to this discussion and anyone else who may find it interesting.

In the Monty Hall problem with the usual rules, that the host always offers the swap and always opens an unchosen door to reveal a goat, if we assume that the car was placed randomly then the probability that the player will initially choose the car is 1/3. As a player who swaps must always get the opposite of their original choice their probability of winning by switching must be 2/3.

A paper by Morgan et al states that, because the player decides whether to switch or not after the host has opened a door, the problem is one of conditional probability. The problem is described by Morgan in this way, 'To avoid any confusion, here is the situation: The player has chosen door 1, the host has then revealed a goat behind door 3, and the player is now offered the option to switch'. They then proceed to show that the probability of winning by switching is given by 1/(1+q) where q is the probability that the host will open door 3 if the car is behind door 1. How can the answer be different from that given by the simple argument at the start?

The argument at the start clearly shows that 2/3 of players who switch will get the car, how can it be that the player in the case Morgan describe does not have a 2/3 chance of winning the car? What if we do not consider all players who enter but only a sample of them? If we take a representative (or random) sample then the result must still be 2/3. However if we take a sample that is not properly representative of an average player who decides to swap then the answer may be different.

Although the player may choose any door, Morgan only consider players who have chosen door 1. If, as we have assumed throughout, the car is placed randomly this sample is representative of an average player and the result is still 2/3 for all players who choose door 1.

After the player has chosen door 1, the host may open door 2 or 3 (unless this reveals a car) however, Morgan only consider the case where the host has opened door 3. This makes the sample not representative of the average and makes the result no longer 2/3. Here is why. If the car is behind door 1 then the host can open either door 2 or door 3. We are not told how he decides which to open in this case and it could be, for example, that the host for some reason never opens door 3 if he has the option not to. This means that if the host has opened door 3 it must be because the car is behind door 2. If this is the case then the player will win with certainty if she swaps.

To sum up, because they only consider the case where a particular door has been opened Morgan do not take a representative sample of the of the overall or average case and they therefore get a different result. Martin Hogbin (talk) 20:05, 8 April 2009 (UTC)

Sorry Martin, may be you think: the last words will be the right ones, but this whole summing-up is fallacious. Nijdam (talk) 06:51, 9 April 2009 (UTC)

In what way? Martin Hogbin (talk) 07:20, 9 April 2009 (UTC)

Read what I wrote on the discussion page of MHP. Nijdam (talk) 15:18, 9 April 2009 (UTC)

I see nothing of relevance there. Surely the process of conditioning a sample space is precisely the same as taking a non-representative sample from it? Martin Hogbin (talk) 17:11, 9 April 2009 (UTC)

As no one else seem to want to contribute any intelligent argument to this subject, I will continue my explanation.

The sample taken from the from all possible the events in the unconditional case under the condition that the player chooses door 1 and the host opens door 3 is not always unrepresentative. It depend on the hosts door preference.

Let us take Morgan's case that q=0, which means that the host never opens door 3 unless they have to. Here, the fact that door 3 has been opened means that the car is behind door 2. So, to make that quite clear, if we consider the conditional case and the hosts door preference parameter q happens to be 0, we only consider cases where the car is behind door 2. Hardly a representative sample.

Now consider the case that q=1, the host always opens door 3 if he can. This means that the car is equally likely to be behind doors 1 and door 2 but not 3, again not a representative sample of the general case.

It the host happens to have an equal preference for both doors (q=1/2) then the car has a 1/3 probability of being behind any door and the sample is one that is representative of the overall or unconditional case. As we know, if q=1/2 the conditional and unconditional probabilities are the same. Martin Hogbin (talk) 13:07, 11 April 2009 (UTC)

Where to start? How about, they claim that the host's behaviour may provide information on where the car is. Then they give him a behaviour. Does the Contestant now know this? If so, hasn't this behaviour just become a premise of the puzzle? Well, that changes it so that it's not the MHP any longer. Talk about 'false solutions'! Glkanter (talk) 13:29, 11 April 2009 (UTC)

What you say is really what my section below is all about. In a formal probability problem, which is what Morgan have perversely taken Whitaker's question to be, anything not defined as random must be taken to be possibly non-random. Thus, as the host's behaviour is not specified in the problem statement to be random, we must consider the possibility that it is not random and that there may be some other has some process, such as the host having a preference for one door involved. In the style in which Morgan understand the question, what the player knows is not important, it is what we (as the answerers of the question) know, or do not know, that forms the basis on which the problem must be solved. What Morgan fail to do (amongst other things) is to consider the equally likely possibility that the producer has not placed the car randomly.

Some problems are often considered from the point of view or state of knowledge of one of the characters in the problem. It seems clear to me that Whitaker intended the question to be taken that way. In other words he was asking, from the point of view (state of knowledge) of a player on the show what would be their best estimate of the probability of winning by switching, which is 2/3 plain and simple. Martin Hogbin (talk) 17:04, 11 April 2009 (UTC)

Yes, 2/3, but the conditional probability, and not the unconditional one if that is what you mean by "plain and simple"!! —Preceding unsigned comment added by Nijdam (talk • contribs) 08:06, 12 April 2009 (UTC)

If we take a traditional view from the player's perspective in which we assume that the player has no information about the producer's or the host's behaviour then the 2/3 solution applies to both cases, which are indistinguishable. If we ask the question as to what the players best estimate of her probability of winning by switching is in the above case then all she can do is to assume the producer and host act randomly. In that case, even after an identified door, for example door 1, has been opened, her estimate of her probability of winning by switching remains at 2/3.

Before you tell me, I appreciate that taking the problem from the player's perspective and assuming she has no information about the game is not 'correct', it might be that she has some historical knowledge of where the producer usually places the car, but it is is traditional in these problems to assume no prior knowledge for the characters involved. Look at the Three_Prisoners_problem where it says 'Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3'. This estimate is not strictly justified as it could be that prisoner A has some prior knowledge of his likelihood of being executed but we traditionally take this not to be the case. The only way to resolve these ambiguities is to do what Seymann suggests and consider what Whitaker actually intended by his question. That would be a worthwhile discussion. Martin Hogbin (talk) 09:52, 12 April 2009 (UTC)

You know, this has bothered me all along. As I understand it, without assigning a host behaviour, Morgan says the probability of winning by switching is between 1/2 and 1, and that it averages 2/3. Or, is that only when the Contestant selects the car? No, that probability would be 0. So now we have Morgan saying 'it averages 2/3', and the unconditional solution saying 'it's 2/3'. I guess there's a difference from some standpoint, from a practical standpoint I don't see it. So, without his specious claim that the unconditional solution is false, (and someone better tell Devlin, http://www.maa.org/devlin/devlin_07_03.html as he seems oblivious to this), Morgan has accomplished nothing.

But wait. Don't ask me why, but he decides to ascribe a host behaviour. Now, the host is more likely to choose one door over another. Look, I'm a simple guy. I went to a state school. A land grant college at that. But they 'learned us' that a new premise describes a new problem. Was that true only on my campus, and not at Old Dominion University in Virginia? So how is it relevant, and not a Source of Confusion itself, to discuss these other problems when trying to describe the MHP paradox? Glkanter (talk) 23:35, 11 April 2009 (UTC)

I am not sure why you are arguing with me. I agree that from any real world standpoint the Morgan paper and its conclusion is irrelevant. I have demonstrated that on this page and on my analysis page, where at least one other editor seems to agree with the maths. There have been no significant arguments against my criticism of the Morgan paper on this page just a repeat of the, obviously untrue, claim that I do not understand conditional probability on the main talk page.

Morgan is right from a particular formal standpoint and given a set of arbitrary assumptions. I agree with you that the paper is awful and irrelevant to the real MH problem but is is not actually wrong. Unfortunately it is a paper published in a peer-reviewed journal and we therefore cannot completely ignore it although I wish we could. Martin Hogbin (talk) 01:15, 12 April 2009 (UTC)

My last edits were not directed at you. It just happens that you created a section called 'Why Morgan are right', and it seems like a good place to discuss why Morgan are NOT right. And like I said, this has bothered me all along. If it's a false solution, then why does someone like Devlin use it? And do I mis-understand the principal that new premises make new problems? I really want to know. So, I think Morgan is flat out wrong about the MHP. Published, to be sure. Peer-reviewed, maybe. Worthy of prominence in the MHP article? No way. On this, we agree. Glkanter (talk) 02:30, 12 April 2009 (UTC)

Yes, we do agree that Morgan's paper should not set the tone for this article. I also agree that does not even qualify as a reliable source in many ways. However where we may have to disagree is that the Morgan paper is technically correct in a limited formulation and style of the problem. My point was just this Morgan are right but in a way that is irrelevant. 86.132.253.23 (talk) 10:59, 12 April 2009 (UTC)

Well if you feel better in writing such comments down, ok. But remember: Morgan's is an article in a peer reviewed magazine, and Devlin just wrote in his popular column. Nijdam (talk) 20:33, 13 April 2009 (UTC)

See my comments about the Morgan paper at the end of the section above. It is essentially a 'novelty article'. Martin Hogbin (talk) 08:30, 14 April 2009 (UTC)

Problem style

This is a subject that I believe is central to many of the arguments on this page. Perhaps somebody who is an expert in the subject can clarify something. It seems to me that there are (at least) two styles of problem in probability questions.

Formal style

The formal style is to present a question that must be answered only on the information given in the question; what the characters in the problem may or may not know is irrelevant, as is any information that we may deduce from real life. In such problems it is necessary to specify whether that a certain selection or distribution is random or not. If such information is not explicitly given in the question then we must take it as unknown rather than random.

As an example, suppose this were the question:

A car is placed behind one of three doors. A person picks a door, what is the probability that they have picked a door hiding the car?

If we take this as a formal probability problem we should first observe that no information is given as to how the car was placed or as to how the player picks the door. We cannot assume that this is random, it could be that the car is always placed behind door 2 and the player always picks door 3, for example. If we take this problem to be a formal probability problem, the best we can say is the probability is from 0 to 1.

On the other if the question were to say:

A car is placed randomly behind one of three doors. A person picks a door, what is the probability that they have picked a door hiding the car?

or

A car is placed behind one of three doors. A person randomly picks a door, what is the probability that they have picked a door hiding the car?

The probability of picking the car is 1/3.

If in the case of solving a formal problem if we are to make assumptions, we must state them explicitly, we should not assume motives or strategies of characters in the problem (that is not to say that characters might not have strategise just that w should not assume what they are). So even if the problem were to be:

A car is placed behind one of three glass doors. A person picks a door, what is the probability that they have picked a door hiding the car?

The answer would not be 1 but would still be from 0 to 1 unless it is made clear that the person can actually see the car and will certainly choose it.

Traditional style

Traditionally probability problems have been presented from the point of view or state of knowledge of one of the characters in the problem. Normal human motives are also often traditionally assumed for these characters. For example in the 'Three prisoners problem' we have: 'Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3'.

If we have the same problem as before:

A car is placed behind one of three doors. A person picks a door, what is the probability that they have picked a door hiding the car?

and we treat it more traditionally we might say something like, 'It is natural to assume that the person was not aware of the way the cars were placed behind the door so their estimate of the probability that they have chosen the car would be 1/3'.

Here we put ourselves in the place of the player and assume the player's state of knowledge. Anything that we do not know e take to be random.

Can anybody confirm or expand on what I have said on problem style? Who disagrees? Martin Hogbin (talk) 14:50, 10 April 2009 (UTC)

Maybe the Contestant gains knowledge from the host's behaviour by accident.

Clearly, there is no explanation for how the Contestant gains knowledge of the location of the car when Monty opens a door revealing a goat. His behaviour could never intentionally be transmitted to the Contestant. I mean, that's got to be in the definition of a Game Show, right? In the US it sure is. http://en.wikipedia.org/wiki/Quiz_show_scandal

So, the only remaining way for Monty's action to give useful info is inadvertently. Maybe he walks past a goat, and somehow causes the goat to make a noise, and then opens the other goat door. Or maybe he picks up some goat crap on his shoe, and then opens the other goat door.

So my question is, is it common practice in Probability theory (or puzzle formulation) to factor in inadvertent acts? How about illegal ones? How does one assign a probability to this? Glkanter (talk) 02:28, 13 April 2009 (UTC)

This is the very first sentence of the article.

"The Monty Hall problem is a probability puzzle based on the American television game show Let's Make a Deal." (emphasis mine)

This is the very first sentence of the Wikipedia article "Quiz show scandals":

"The American quiz show scandals of the 1950s were the result of the revelation that contestants of several popular television quiz shows were secretly given assistance by the producers to arrange the outcome of a supposedly fair competition." (emphasis mine)

http://en.wikipedia.org/wiki/Quiz_show_scandals

So, does Morgan perhaps not understand the rules of a game show? Glkanter (talk) 02:45, 13 April 2009 (UTC)

What is interesting about this fact is that, if it were the intention to manipulate the results of the show, this could only be done by means of fixing the car placement, there is no strategy that the host can adopt to change the player's chances of winning. Martin Hogbin (talk) 09:16, 13 April 2009 (UTC)

What is your response to the various questions I ask above? Would you agree that Morgan's claim that the host's behaviour could provide information as to the whereabouts of the car is contrary to the normal (and legal) understanding of how game shows work? Glkanter (talk) 11:03, 13 April 2009 (UTC)

You need to start with the way Morgan set up the question, 'The player has chosen door 1 and the host has then revealed a goat behind door 3...'. Note that it is asked from the point of view of a third party observer. We are to answer the question on the information given. What the player knows in not important. It is only by setting up the problem in this way that Morgan can introduce the host preference into the problem at all. Morgan supporters need to remember this, you cannot say for example that the player must know which door has been opened so we must take that into account. Martin Hogbin (talk) 21:33, 13 April 2009 (UTC)

I should have included this, from the Wikipedia article. It's the problem statement in the first paragraph:

"A well-known statement of the problem was published in Parade magazine:

Suppose you're on a game show, and you're given the choice of three doors..."

So, it's stated that it's a game show, which means the Contestant can not be given any indication where the car is. And it says 'Suppose you're on a game show', which clearly states that this is from the Contestant's point of view. Why is Morgan still given any credibility at this point? Glkanter (talk) 22:42, 13 April 2009 (UTC)

I agree with you. Morgan massage the question (note that they even misquote the original question from Whitaker in their paper to make it closer th their own problem statement) to get an 'elegant' solution. Actually, I now think that Morgan are even more wrong than that. I will explain later. Martin Hogbin (talk) 08:37, 14 April 2009 (UTC)

Well, I love surprises.

But it's more than they're wrong. To me that implies their conclusion is wrong. Their entire premise is false. They completely disregard this simple statement: "Suppose you're on a game show". By definition then, the pov is the Contestant, and all the things that define a game show are (unstated) premises. Like, the host is not going to indicate to you (the Contestant) in any way where the car is. So, like I've been saying, the minute they begin describing host behaviours they're on to a different problem. And the Morganians ALWAYS invent a host behaviour when explaining their pov. Look at the article. Without host behaviours, Morgan would be about 3 sentences long. This is all they can say, "It's between 1/2 and 1, and averages 2/3". That, and 'The unconditional solution is false'. But that is clearly just an opinion that is not popularly held. See Devlin, and many others. Glkanter (talk) 12:09, 14 April 2009 (UTC)

I find it interesting that after 2 1/2 days, none of the Morganians have been motivated to respond to this section. And there's no requirement that they do so. Maybe they could just explain to me how I should interpret these 6 words (from Whitaker?), "Suppose you're on a game show...". Glkanter (talk) 15:43, 15 April 2009 (UTC)

A question

Starting with the standard MHP game rules (host always offers the swap and always opens an unchosen door to reveal a goat). Suppose that we take Morgan's situation, player chooses door 1 and host opens door 3. Now let us suppose that there is a population of equal numbers of hosts of two types: those where p=1 (always open door 3 where possible) and those where p=0 (never open door 3 unless forced to do so). Let us assume that our host was taken randomly from this population and is thus equally likely to be either type.

Here is the question. Given the above, what is the probability that given the host has opened door 3, the player will win by switching? I have already put my answer on another page. Who will give this conditional probability problem a go? Martin Hogbin (talk) 18:34, 14 April 2009 (UTC)

Conditional on the choice of door 1:

p=1
agG* 
gaG*
gGa
p=0
aGg
gaG*
gGa

hence: P(car=1|opened=3)= agG*/[agG* + gaG* + gaG*] = 1/3

Nijdam (talk) 19:54, 14 April 2009 (UTC)

I do not understand your answer. I presume you are saying that the probability that the player will win by sticking is 1/3? Thus the probability of winning by switching is 2/3.

Do you not find this answer interesting? Although we do not know the host behaviour, the probability works out to 2/3. Martin Hogbin (talk) 22:19, 14 April 2009 (UTC)

You did understand it well. I could have written it more extensively of course. What is so peculiar about it? We do know the host's behavior. Instead of a host with random behavior, it is a random host with a fixed behavior, equivalent more or less like to the flipping coin strategy. Nijdam (talk) 06:16, 15 April 2009 (UTC)

I agree with your answer but I think it is more general with that. I believe that the answer (probability of winning by switching) is 2/3 for any distribution of host behaviour that is symmetrical about 1/2. This includes a linear distribution of q, which I think is what Morgan get the answer ln(2) for. Do you agree? Martin Hogbin (talk) 10:18, 15 April 2009 (UTC)

No, ln(2) is the answer where 1/(1+q) is uniformly distributed, not q (Nijdam - this is in the Morgan et al. paper where they compute a specific number using a Bayesian method assuming an uninformative prior). -- Rick Block (talk) 13:35, 15 April 2009 (UTC)

Are you sure, Rick? Can you explain further please? Martin Hogbin (talk) 17:24, 15 April 2009 (UTC)

(outindented) The noninformative distribution is the uniform distribution for q. With this distribution the average probability of winning by switching is the average of 1/(1+q):

$\int _{0}^{1}{\tfrac {1}{1+q'}}f_{q}(q')dq'=\int _{0}^{1}{\tfrac {1}{1+q'}}dq'=ln(2).$ —Preceding unsigned comment added by Nijdam (talk • contribs) 20:32, 16 April 2009 (UTC)

Is that not what I said? We take q to have an equal probability (density) of being any number from 0 to 1 and integrate the probability of winning by switching (1/(1+q)) with respect to q. Martin Hogbin (talk) 00:27, 17 April 2009 (UTC)

Yes, you're right, I wrote it down explicitly, because it seems Rick has doubts.Nijdam (talk) 08:15, 17 April 2009 (UTC)

If we take it that Morgan took the unconditional distribution of q to be uniform then it seems to me that they fail to properly apply the condition that the host has opened door 3 in their calculation. Martin Hogbin (talk) 18:52, 17 April 2009 (UTC)

The prior distribution of q has nothing to do with the terms 'unconditional' and 'conditional', so no accusations of failure. Nijdam (talk) 23:24, 19 April 2009 (UTC)

That was not my point. First we assume a uniform prior distribution of q, we then apply the condition that the host has opened door 3. After we have applied this condition the distribution of q is no longer uniform. Do you not agree? 86.132.253.23 (talk) 22:17, 20 April 2009 (UTC)

You may mean the posterior distribution of q after the host has opened door 3. You may have a point there. The posterior is ${\tfrac {2}{3}}(2-t)$, and puts more weight, of course, on door 3. Nijdam (talk) 09:57, 21 April 2009 (UTC)

Yes, that is exactly my point. My original question of this section was a very simple example. Martin Hogbin (talk) 22:19, 21 April 2009 (UTC)

Well that's the problem with Bayesian approach, it doesn't mean it's wrong. Nijdam (talk) 08:56, 25 April 2009 (UTC)

I have not studied the Bayesian approach but I do not think that it is relevant to this issue. Morgan make an error in their calculation in that they fail to use the information in the door number opened by the host. They perform a calculation using the prior distribution of q whereas they should have used the posterior (after the host has opened a door) distribution. This is an error plain and simple however you look at it. Martin Hogbin (talk) 09:09, 25 April 2009 (UTC)

I think that I now know the calculation that Morgan should have done to calculate the player's probability of winning by switching if the prior distribution of the host's door opening parameter q is uniform. They should have had:

$1/(1+\int _{0}^{1}\ {f_{q}(q)}dq)=1/(1+\int _{0}^{1}\ {q}dq)=2/3$

I can show how I derived this if anyone is interested. Martin Hogbin (talk) 21:19, 29 April 2009 (UTC)

Another question

This is a question the first part of which I will answer myself.

A car is placed (not necessarily randomly) behind one of three doors, which are numbered 1,2, and 3. I propose to throw a fair die and if I throw 1 or 4 I will pick door 1, 2 or 5 door 2, 3 or 6 door 3. What is the probability that the door I am going to pick will have the car behind it?

My answer: 1/3.

Now I have thrown the die it has come up with 4 (door 1). What is now the probability that I have the car?

My answer: From 0 to 1 depending on the probability the car is behind door 1.

The question. How does getting more information give me a less good measure of the probability of winning the car?

Any answers? Martin Hogbin (talk) 19:13, 15 April 2009 (UTC)

Note that you implicitly assume the independence of car placing and throw. Now about you question. Why do you consider it a lesser good measure? It is a perfect measure, but the object of the measurement is unknown. The precise answer would be: the probability with which the car is placed behind door 1.Nijdam (talk) 20:40, 16 April 2009 (UTC)

Yes I do assume independence of the car placing and the throw of the die, it is hard to see how things could be otherwise but I guess I should have made this clear.

What puzzles me is that we go from being able to say that the probability of winning the car is 1/3 (a definite answer) to only being able to say that the answer is given by the probability that the car was placed behind door 1, which we only know is a number from 0 to 1 (an extremely unhelpful answer) but we have been given more information. This just seems wrong somehow. Martin Hogbin (talk) 00:36, 17 April 2009 (UTC)

When All You Have Is A Hammer, Everything Looks Like A Nail

Explain to me again how the original 1/3 probability of the Contestant choosing the car changes based on which goat Monty reveals?

The course titles for the Probability classes I took were 'Intro to Logic' and 'Symbolic Logic'. That's what is missing from the Morganian's arguments: Logic.

"Suppose you're on a game show..."

The MHP is a story problem. It can be solved using whatever technique is best. And being a story problem about a game show, the rules of game shows apply.

Since it's illegal in the US, there can be no 'host behaviour' known to the Contestant, which as per the article, quoted above, is how the problem is laid out. And the Combining Doors solution is not a 'false (unconditional) solution' as the Morganians interpret Morgan's paper. As there is no possibility of any 'host behaviour' known to the Contestant, there is no difference between deciding to switch before or after the goat is revealed. As if that mattered anyways. Glkanter (talk) 12:12, 19 April 2009 (UTC)

It is a conjuring trick and here is how it is done.

Firstly you misquote the original question, making, 'a door, say No 1' into, 'door No 1'.

You then rephrase the question so that it is not from the point of view of the player but is a formal probability probability problem. This means that we no longer consider the state of knowledge of the player but have to consider what actions are stated to be random in the question. If something is not stated in the question to be random then we must assume that it may not be. There are three choices in the problem, none of which is actually stated to be random in the Whitaker statement.

Next you quietly ignore two of the non-random choices, that of the producer placing the car and the player choosing the door. The reason for this is that otherwise the solution becomes rather dull. The probability of winning by sticking would then depend on the probability that the producer placed the car behind the door that the player chose. Of course, if the problem were considered from the players state of knowledge this would not be the case as the player does not know where the car was placed but taken as a formal probability problem where we are told that the player has picked door 1, the solution is obviously dependent on the probability with which the producer has placed the car behind door 1. He might always place the car behind door 1 for example. This effect is so obvious and boring, however, that it is discreetly ignored.

Finally we leave the host choice as non-random. Hey presto, rather than an obvious result like that of the producer placing the car non-randomly we can do a little maths and come up with an 'elegant solution'. Martin Hogbin (talk) 21:44, 19 April 2009 (UTC)

Here is the same trick without the props.

'Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1'. What is the probability that you pick the car. Most people, Morgan included, would give the answer 1/3.

Now misquote.'Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick door No. 1'

Rephrase, 'On a game show a car is placed behind one of three doors. Given that door 1 has been chosen, what is the probability the car is behind the chosen door?' Answer to this, now conditional, question: anything from 0 to 1 depending on the probability the car was placed behind door 1. Martin Hogbin (talk) 22:52, 19 April 2009 (UTC)

What is this about? Do I notice some frustration? Nijdam (talk) 23:25, 19 April 2009 (UTC)

There is some frustration with several things, mainly by lack of ability to improve the article. Because of the way that Morgan's solution and ethos is incorporated throughout the article it is hard to make small improvements that stand alone. The article really need rewriting to make a better and more convincing job of explaining the basic, unconditional if you prefer, solution. Even thinking of a suitable description for this solution that properly describes its status is not easy. The article is currently featured and I am reluctant to make major changes on my own, I would much prefer to convince other editors that such changes are necessary so that we can work together to improve the article.

I am also frustrated that several important issues that I have raised above on the subject of probability and that are very relevant to the MHP have been ignored. The above explanation is intended to convince people how unhelpful and irrelevant the Morgan paper is to the real MHP. Martin Hogbin (talk) 08:50, 20 April 2009 (UTC)

Well we do disagree than on many points. From the discussions with you it seems you got an understanding of the nature of conditional probability. This should lead to the notion that the MHP is inherently of a conditional nature. If the article is to be improved, it should mention the right solution first. Then it may pay some attention to the often heard "unconditional solution", and mention the flaw in it. These considerations are in my opinion the same as Morgan's, and so Morgan's paper is very relevant and helpful. And it is a shame hat people like Devlin have written cheap columns on subject they hardly understand. Nijdam (talk) 22:59, 22 April 2009 (UTC)

You say the MHP is inherently of a conditional nature; this is where we disagree. I accept that it can be made into a conditional problem by ignoring the question that Whitaker undoubtedly wanted the answer to, which is, 'Is it generally better to switch?' and choosing to answer the specific question where the player has chosen door 1 and the host has opened door 3 (or other specified doors). Even when you have made the problem conditional, if you take it that the host chooses randomly, the condition makes no difference, yes, it is a condition, but who cares?

I have to keep coming back to the point that the simple problem, regardless of whether it is treated conditionally or not, is the one that most people get wrong most of the time. Martin Hogbin (talk) 17:58, 24 April 2009 (UTC)

Martin - do you agree that the problem consists of a first step where the player selects a door and a second step where the host opens a door and that the point of decision is after the second step? If so, you're agreeing it's a conditional problem. One can argue that the solution to the conditional problem must be the same as the solution to the unconditional problem (which, if the player doesn't switch, is mathematically equivalent to a problem where the player selects a door and the host does nothing), but the fact that the conditional problem is equivalent to this intuitively obvious unconditional problem (iff the host opens a random door if the player initially selects the car) is NOT obvious, and should have some reasoning behind it. That's all we're saying here. The point is not to complicate matters, but to address the actual problem in a way that's sound and won't lead readers astray if they encounter slightly different problems (like the "host forgets" version). Given a firm grasp of the unconditional solution to the standard version, why doesn't this same approach work for the "host forgets" version? As Glkanter says, it's a different problem, but how should the reader know when an unconditional approach works and when it doesn't? This question apparently completely stumped Glkanter above (I pushed it to the point where he apparently got very angry). Teaching people shortcuts without teaching them when the shortcuts are applicable seems like a bad idea to me. -- Rick Block (talk) 19:24, 24 April 2009 (UTC)

I do not agree the point of decision is necessarily after the host has opened a door. This is clearly the case in Morgan's reformulation of the problem but it may well not be what Whitaker wanted to know. He may well have been asking the question of what the best policy would be for a contestant on the show. You keep talking about 'the actual problem' as though this is Morgan's restatement of Whitaker's original question. The actual problem, if there is one, is what Whitaker actually wanted to know. This may or may not be conditional problem, we have no way of knowing.

You still do not seem to have understood my point about formal versus informal problem treatments. In the first case we must take any information that is not provided as indeterminate in the second it is common to treat it as random.

Regarding your point about addressing the problem in a way that is sound, most of mathematics consists of shortcuts. If we were to address every mathematical problem completely rigorously (and mathematicians can get very rigorous) we would never get off the ground with most problems. In order to make any headway with a problem you have to ignore some of the details initially. These can be considered, if desired, later. Martin Hogbin (talk) 00:18, 25 April 2009 (UTC)

Rick made it completely clear. And even if Whitaker had another problem in mind - and we may never know - the way he formulated the problem, with the stages Rick mentioned, forces it to be answered with conditional probabilities. And I have the same concern as Rick, and even more, as I notice that teachers and students often make use of the MHP and just use the wrong simple solution to explain the problem, supported, I hope only until recently, by Wikipedia. Nijdam (talk) 08:51, 25 April 2009 (UTC)

Rick made his version of the question clear but nobody can claim that Morgan's (or Rick's) formulation is the 'real' MHP. If we take the Parade statement as the definitive problem statement then the definitive formulation is given by 'the precise intent of the questioner'. This fact cannot be disputed. It is standard practice in any discussion of probability that before a question can be answered it must be decided what it means, particularly if much detail is missing from the question statement itself. This view is made quite clear by Seymann at the end of the Morgan paper. Are you disputing that this is the correct way to address a question from a non-expert?

For educational purposes my suggestion is to first formulate the problem in a way that the simple solution applies, if you consider it essential, formulate it unconditionally for this purpose. After that apparently simple but intuitively exceptionally difficult problem has been solved and understood, the conditional problem and could be introduced. Treating the two together at the start only serves to obfuscate both issues. Martin Hogbin (talk) 09:48, 25 April 2009 (UTC)

I'm curious. If Whitaker had asked vos Savant about Deal or No Deal, and had casually referred to unique suit case numbers as in "say, suitcase 13", would the equivalent of Morgan's literal interpretation regarding "say, door 3" still be considered the 'last word' by the Morganians? Glkanter (talk) 02:35, 25 April 2009 (UTC)

Of course, because that way they can get an 'elegant solution'. Martin Hogbin (talk) 18:20, 26 April 2009 (UTC)

The question that no one will address

I have raised this issue before as I believe that it is central to this MHP argument. Replies to date have simple been along the lines of 'No it is not'. Firstly let me make clear this point has nothing to do with whether the problem is conditional or whether the doors are distinguishable. It is about problem style.

Let me start by asking two similar questions that have two completely different answers. I have given my answers. Does anyone disagree with them? I ask those involved in discussion on this subject to add their answers below or to state whether they agree or disagree. Please do not make any assumptions about the way this relates to the MHP at this stage, I just want to get absolutely clear that we agree before proceeding. Martin Hogbin (talk) 10:02, 25 April 2009 (UTC)

1) A car is placed behind one of 3 doors. What is the probability that it is behind door 1?

• Indeterminate. The answer is simply the probability with which it was initially placed behind door 1.Martin Hogbin (talk) 10:02, 25 April 2009 (UTC)

2) A car is placed behind one of 3 doors. What is the probability that a person who chooses randomly will choose the door with the car?

• 1/3 The door choice is defined to be random son the initial car placement probability is important.Martin Hogbin (talk) 10:02, 25 April 2009 (UTC)

To #1 my answer is: If I am to "assume you are on a game show", my answer is 1/3, otherwise "between 0 and 1, averaging 1/3". To #2, my answer is 1/3. I tried my best to follow your instructions. Glkanter (talk) 13:18, 25 April 2009 (UTC)

I wanted the questions to be treated like questions in a test on probability, where you would be expected to answer the questions based only on the information contained within them. You may like to think that question 1 refers to a game show but you are not expected to use any knowledge that you think you might have about game shows to answer it. Martin Hogbin (talk) 18:46, 25 April 2009 (UTC)

@2): Formally the relation between the choice of the person and the placemant of the car must be specified. If they are independent, the answer is 1/3. BTW: what do you mean by: son the initial car placement probability is important? Nijdam (talk) 15:47, 25 April 2009 (UTC)

Sorry, that was a typo, it was meant to be "so the initial car placement probability is unimportant ". Surely a statement that the person chooses randomly is sufficient because a random choice is by definition independent of all else. If you think that it is necessary to add that the person's choice is independent of the car placement to my question 2 I will do so. Apart from that do you agree with my two answers? Martin Hogbin (talk) 18:37, 25 April 2009 (UTC)

I suspect Martin means "so the initial car placement probability is unimportant". Martin - I find your approach here extremely annoying. Please just say what you mean and drop the little puzzles. -- Rick Block (talk) 17:59, 25 April 2009 (UTC)

I do not understand what you find annoying. This is the arguments page, the place for discussion about the MHP. My questions are not meant to be puzzles, I am trying to ensure that we all start on the same wavelength, if we all disagree fundamentally on some aspects of probability it makes it difficult to move forwards. The way a probability problem is worded is extremely important, pretty much all the argument concerning the MHP is about interpretation of the question rather than the maths which is fairly straightforward. Nijdam and Glkanter have given their answers perhaps you could do the same then I will explain my point. Martin Hogbin (talk) 18:37, 25 April 2009 (UTC)

What is annoying is your metering out of your point. Probability theory allows for extremely precise problem statements and there really aren't that many choices for the MHP. It's either talking about:

1. P(win by switching)

2. P(win by switching|player picks door i) (where i is, say, 1)

3. P(win by switching|player picks door i and host opens door j) (where i is, say, 1 and j is, say, 3)

Various wordings in English map to one of these. The wording should make it obvious which one of the above probabilities is being discussed. If it doesn't, it's simply poorly worded. Nijdam and I are both saying the "standard" MHP is meant to map to the third one (and this is backed up by numerous math sources that know the difference). This is the notable problem that most people get wrong. Yes, the answer to #2 happens to be a shortcut to the same answer (under certain assumptions) - but this is not the problem that is asked. Your continued insistence that Whitaker's words actually mean something different is simply absurd. -- Rick Block (talk) 00:52, 26 April 2009 (UTC)

I do not understand the term, 'metering out of your point'. If you mean that I am going through it rather tediously that is because I want to ensure that we all agree at the start. There is little point going through a long argument only to find at the end that you disagree because you had not agreed on one of the starting premises.

You give three possible questions that might be asked but these are clearly not complete problem statements. A problem statement must make clear the setup of the problem. In the case of the MHP that would include the game rules (which are not in dispute) and the basis on which selections are made (are they random, unknown, based on an informed choice, independent of one another etc?) and possibly other things. Without this information the questions that you ask are meaningless.

Regarding Whitaker's words, do you really think that the question he intended to ask was,' Given that a player has picked door 1 and the host has opened door 3, is it best to swap?'? Most people are not aware of the conditional issue (which given reasonable assumptions does not make a blind bit of difference). The event that made the problem notable was the response to vos Savant's answer (in which the conditional issue was ignored).

Finally you have chosen not to confirm your agreement (or otherwise) with my answers above. I am not sure what you think you have to lose by answering unless you fear that it will lead to a discussion that will show up the weaknesses in the current treatment of the problem. I and several other editors (probably the majority of those currently active) believe, for various reasons, that the simple/unconditional problem and solution should be given greater prominence within the article, but rather than rush in and make major changes we have decided to discuss the issues on this page; one specifically set up for that purpose. As you know, I am trying to show up weaknesses in the current MHP treatment and these are concerned with problem style. In order to explain exactly what I mean, it is easiest to give examples to see where any disagreement lies. There is nothing more sinister than that to my 'little puzzles'. Martin Hogbin (talk) 09:47, 26 April 2009 (UTC)

By "metering out your point" I mean going through it slowly, without divulging where you're headed. I find it annoying not because I think I have anything to lose but because it's inefficient. If we knew where you were going with this, Nijdam and I could probably help point out specifically where you're going wrong. If the only way you'll proceed is by gaining consensus at each step then fine - lacking any other information, the answer to your first question is p where 0<=p<=1. The second one is 1/3.

The probability statements I offer are the only different probabilities involving the problem setup we're talking about (#2 and #3 cover the cases of the the specific individual doors, if these probabilities might vary by specific door number). Formally, they depend on the background information typically denoted I, which includes the games rules and the other considerations you mention.

Regarding Whitaker's words, I think he intended to ask a question that maps to #3 above. If you're not still confounded by the distinction between the unconditional and conditional situations here, you're giving a very good impression of one who is. The question is not restricted to door 1 and door 3, but involves a specific player pick and some specific door opened by the host. This problem did not originate with Whitaker, so your seemingly intense focus on analyzing precisely what he meant is entirely misplaced. The problem is a reformulation of other, classic problems in conditional probability (specifically the Three Prisoners problem which itself is essentially a reformulation of Bertrand's box paradox). The entire point of the problem is to create a situation with two unknowns that have identical unconditional probabilities but different conditional probabilities. In the MHP it turns out the conditional probability of the player's door remains the same as the original unconditional probability, but the question is most assuredly asking about the conditional probabilities in effect given the host has opened a specific (known) door after the player's initial (known) selection. Do you really think the question is meant to map to #2 (or #1) above and that it's nothing but an elaborate trick question meant to hide your second question (sort of like those primary school problems like "assume a plane crashes exactly at the North Pole - where are the survivors buried?").

The response to vos Savant's columns included the two math papers we're talking about (Morgan et al. and Gillman) who both clarify that the problem is a conditional probability problem (i.e. maps to #3 above). The reason people typically get this wrong (and to be clear, this is simply my opinion) is because people are unaccustomed to dealing with conditional probabilities. You and several other editors have argued with expert opinions about this so long that most experts here (pretty much anyone involved in wp:WikiProject Mathematics) simply ignore you. This definitely does not mean anyone is agreeing with you. -- Rick Block (talk) 17:05, 26 April 2009 (UTC)

Rick, you seem fixated on conditional probability. I made quiet clear at the start that that was not what this section was about. Anyway, you seem to agree with my answers. Martin Hogbin (talk) 19:48, 26 April 2009 (UTC)

The point that I am trying to make is that, in between the two problem examples that a gave above is a range of possible questions and circumstances where the answer is not quite so clear. Take for example:

3) A car is placed behind one of 3 doors. What is the probability that a person who chooses a door will choose the door with the car? (The circumstances are such that we would not expect this person to have any knowledge of where the car had been placed.)

Who would say that the probability is 1/3?

Could we also say that the car placement was effectively random because the chooser had no knowledge of it?

Now what about if we apply the same logic to the host's choice of door in the MHP. The player has no knowledge of the host's door choice policy, thus it is effectively random. Martin Hogbin (talk) 19:48, 26 April 2009 (UTC)

This is the point where you start to go astray. Problem statements are meant to be clear and this one is not. If the initial placement is random then the answer is 1/3. If the player's choice is random then the answer is 1/3. If by the wording of the problem you mean both of these may be taken as not random (and you either mean these to be taken as random or you don't, or the problem is ambiguous) then this problem is the same as your first problem and the answer is anything between 0 and 1. What about if we pay attention to what the most reliable sources say about the MHP (and I'm including Gillman here, not just Morgan et al.)? Either the problem statement specifies the host's door choice policy includes the constraint that the host choose randomly if the player initially selects the car or it doesn't. If it does, then the answer is 2/3 chance of winning by switching. If it doesn't, then the answer is 1/(1+q) where q reflects the host's preference (which the player may or may not know). -- Rick Block (talk) 20:22, 26 April 2009 (UTC)

Yes, I understand that the right and proper thing to do is to state whether a particular choice is random or not but there is common informal style of probability problem where we assume something to be random if we are given no information. For example in this quote from the Three Prisoners Problem, 'Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3' it is assumed that the decision of whom to pardon is random because the prisoners do not have this information, but nowhere in the question is this stated. Before you rush off to fix the article please bear in mind that this approach is quite common in informal probability puzzles, what is not known is taken to be random. Martin Hogbin (talk) 23:30, 26 April 2009 (UTC)

@Martin: I'm not sure where you're heading, but whatever is said or assumed about all these distributions, the solution is based on CONDITIONAL probabilities. Nijdam (talk) 20:50, 26 April 2009 (UTC)

Yes I know. Martin Hogbin (talk) 23:30, 26 April 2009 (UTC)

Rick and Nijdam, what is your response my assertion that it is common in informal probability problems to assume that, where no information is given regarding a method of choice, that choice is to be taken as random. This is clearly the case in the three prisoners problem. Martin Hogbin (talk) 08:25, 28 April 2009 (UTC)

What is common is to assume the Principle of indifference - which is not quite the same as what we're talking about here. Assuming the prisoner's initial chance of being pardoned is 1/3 is like assuming the car is initially randomly placed. In the Three Prisoner's problem (at least in Gardner's version) the warden flips a coin do decide which name to give in the case prisoner A is the one being pardoned (it's not unspecified and assumed to be random). This is what's similar to the point under dispute in the MHP. The reason it's specified is because the 3PP is also a conditional probability problem and the answer depends on this constraint (and Gardner certainly knew this). -- Rick Block (talk) 01:56, 29 April 2009 (UTC)

Fine, so why do we not apply the principle of indifference to the host's door choice and take it to be random also? It is exactly the same as the car placement choice. Both choices are not defined to be random, both were made by people (the car placement choice by the producer or his agent and the door choice by the host) and the player has no knowledge of either. Martin Hogbin (talk) 17:09, 29 April 2009 (UTC)

Are we talking about probability problems in general, or the MHP as it is generally understood, or specifically the Parade version, or specifically Morgan et al.'s interpretation of the Parade version given vos Savant's subsequent clarifications? Certainly as the MHP is generally understood and given vos Savant's clarifications of the Parade version, the initial distribution of the car is meant to be taken as random. In a game show situation, even if the producers don't randomize the initial placement, the player's initial choice pretty much has to be a random guess so the easiest way to analyze the situation is to assume an initial random distribution (since a random guess from among three alternatives regardless of the actual distribution has a 1/3 chance of being correct). On the other hand the player's choice has no such randomizing effect on the host's preference. -- Rick Block (talk) 19:41, 29 April 2009 (UTC)

I guess that I am talking about a sensible interpretation of the parade statement and how Morgan should have interpreted it.

There are three people who make choices, the player, one I will call the producer, and the host. None of the choices made is specified to be random in the Whitaker statement. The principle of indifference should logically be applied to all of those choices or none of them. There is no logic to saying the producer acts non-randomly, the player chooses randomly and the host acts non-randomly. Martin Hogbin (talk) 21:29, 29 April 2009 (UTC)

As it happens, I believe that Morgan's analysis does not apply if he producer places the car non-randomly and the player picks randomly. See my analysis page Martin Hogbin (talk) 21:39, 29 April 2009 (UTC)

This is Morgan's Entire Argument Against the "Combining Doors" Solution

'F1' refers to the strategy to always switch, as the original choice has a 1/3 chance of being the car.

"...It just does not solve the problem at hand. F1 is a solution to the unconditional problem which may be stated as follows: "You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?" the distinction between the conditional and unconditional situations here seem to confound many, from whence much of the pedagogic and entertainment value is derived"

It's a funny thing. This is exactly how I interpret the problem. I believe that the unconditional solution DOES properly "solve the problem at hand." I reject the notion that the interpretations that are door number defendant are the only "true" interpretations. And the above paragraph in no way explains how the unconditional interpretation is insufficient.

There are countless equally reliable published sources that subsequent to Morgan have put forth the unconditional solutions such as Combining Doors as their only solution.

It's time to bring clarity and common sense to the Monty Hall problem on Wikipedia. Glkanter (talk) 16:05, 26 April 2009 (UTC)

From Wikipedia:Reliable sources# Scholarship: Academic and peer-reviewed publications are usually the most reliable sources when available. However, some scholarly material may be outdated, superseded by more recent research, in competition with alternate theories, or controversial within the relevant field. Reliable non-academic sources may also be used, particularly material from reputable mainstream publications. Wikipedia articles should cover all significant views, doing so in proportion to their published prominence among the most reliable sources.

The Morgan et al. paper is in an academic and peer-reviewed publication. As far as I know, it is the only such source specifically about the precise issue we're spent so much time talking about (I'm not sure Gillman's paper was peer reviewed - and the other sources in the article itself are either non-academic, not peer reviewed, or not about the issue we're talking about). Yes, there are countless published sources that put forth unconditional solutions, but these are NOT "equally reliable". Per the quote above from WP:RS, the article should cover all significant views in proportion to their published prominence among the most reliable sources. Again as far as I know, your view that the unconditional solution solves the problem at hand is supported only by sources that are less reliable (if not FAR less reliable) than Morgan et al. Taking the RS guideline literally, this view doesn't actually need to be mentioned in the article at all.

It's time to stop arguing against what the most reliable sources say. -- Rick Block (talk) 17:36, 26 April 2009 (UTC)

Morgan's paper is an arrogant, patronising, incomplete, inaccurate, sloppy (being somewhat charitable), novelty article that somehow managed to get into in a peer-reviewed publication. It even carries its own 'health warning' at the and in the form of a comment by Seymann. Academic and peer-reviewed publications are usually the most reliable sources... sure, but not this one.Martin Hogbin (talk) 19:21, 26 April 2009 (UTC)

You are entitled to your own opinion, but unless you can cite an academic, peer reviewed source that says something different then this one is the best source we have. -- Rick Block (talk) 19:54, 26 April 2009 (UTC)

@Martin: I'm quite shocked by your burst-out. It even puzzles me where it originates from. From the fact that you doesn't like the outcome? That's quite unscientific. Because the content is wrong? Then come with proof. But I can assure you, in such simple math an error would have been detected ages ago. Because it prevents you from using the simple solution? Well, you should not use it, as it is wrong. So?Nijdam (talk) 20:23, 26 April 2009 (UTC)

I think that all that I say is justified. Let me give some examples:

Incomplete - The article fails to make clear that its analysis is based on the assumption that the car is initially randomly placed. (It mentions in passing after the analysis that people might like to consider non-random car placement)

Inaccurate - I thought you agreed that Morgan's calculation regarding a uniform distribution of q was wrong in that they used the prior distribution.

Sloppy - They misquote the original question by Whitaker (let us assume by accident).

'Novelty article' - This is hardly an paper pushing forward the boundaries of knowledge of probability theory.

'Health warning' - Why was there a comment by Seymann immediately after the paper stressing the importance of problem definition? Martin Hogbin (talk) 23:09, 26 April 2009 (UTC)

Which of the following statements is more accurate?

Morgan has expressed his opinion that the unconditional interpretation of the MHP is not what Whitaker intended to ask.

or

Morgan has demonstrated, using a scientific method, that the Combining Doors solution to the MHP is incorrect.

Glkanter (talk) 15:56, 27 April 2009 (UTC)

Morgan and three other math professors expressed their collective opinion in a peer reviewed math journal that an unconditional solution of the MHP does not solve the problem that Whitaker asked, and this opinion is reflected by another independently published paper (Gillman) and in the Grinstead and Snell textbook (both referenced by the article). So, this is not just Morgan's opinion but the published opinion of at least 7 math professors (plus reviewers). -- Rick Block (talk) 18:34, 27 April 2009 (UTC)

Plus me!Nijdam (talk) 21:36, 27 April 2009 (UTC)

In what way does being a math professor qualify one to be the final arbiter of how to interpret what is universally agreed to be an ambiguous question? It seems presumptuous to me for ANYONE to state with 100% certainty what Whitaker DID NOT mean. And you've agreed above that this is what Morgan does. Without this 'presumption of what Whitaker is NOT asking', there is nothing wrong with the Combining Doors solution. And many math professors, subsequent to Morgan, along with millions of Parade Magazine readers, do NOT share his presumptive certainty. Glkanter (talk) 09:49, 28 April 2009 (UTC)

Although there are plenty of ambiguities in the Parade version of the problem statement, that the question is about conditional probability is not universally agreed to be one of them (in fact, basically the opposite). The fact that people have a very hard time correctly solving conditional probability problems is very well known - conditional probability confuses people. -- Rick Block (talk) 00:46, 29 April 2009 (UTC)

How do you know what it was that Whitaker actually wanted to know? It is quite likely that he actually wanted to know if it was generally better to swap than stick (the unconditional problem) and by what margin. Martin Hogbin (talk) 21:43, 29 April 2009 (UTC)

The question is not how do I know what it was that Whitaker actually wanted to know. I don't claim to know - what I claim is that the sources that know what they're talking about say that the question clearly asks about the conditional probability. The question is why do you think it is likely that he actually wanted to know if it was generally better to swap than stick? Are there any references that discuss these two interpretations that support this interpretation or is it simply your personal opinion based on the number of sources that present an unconditional solution (without saying anything about what they think the question means)? Let me turn this around - if you wanted to unambiguously ask the conditional question how would you phrase it differently? -- Rick Block (talk) 04:23, 30 April 2009 (UTC)

The reliable sources that you are talking about are statistics experts they are not experts in understanding the intended meaning of a question posed by a member of the public in a popular magazine. Morgan et al changed the question in two stages, first by misquoting it then by restating it so that it was clearly conditional. They then proceeded to answer it on that basis. The first to question whether that approach was justified was Seymann in his comment at the end of the paper.

Krauss and Wang do not take such a strong line over conditional probability. Although they eventually do refer to Morgan's result most of their paper ignores the issue of conditional probability. Although they do not expressly say so, K&W give the strong impression, to me at least, that the conditional issue is unimportant.

Regarding the original question, most likely Whitaker did not consider whether he meant the question to be taken conditionally or not. As we know, if the host chooses randomly, it makes no difference. If I thought that it mattered which door the host opened, I would phrase the question along these lines, '...the host has opened door 3...what difference would it make to my chances of winning by switching if the host had opened door 2 rather than door 3?'. How would you answer that last question?

Finally, Glkanter and I are not claiming to know what Whitaker meant, we are just pointing out, as Seymann did, that nobody else can claim to know for certain either, no matter what their academic credentials. Martin Hogbin (talk) 08:27, 30 April 2009 (UTC)

First, Whitaker did not invent the MHP - his was a slight rephrasing of a problem originally posed by Steve Selvin, or more likely (based on how similar it is), a rephrasing of Nalebuff's version of Selvin's problem. The problem didn't originate from a "member of the public" but from academia (specifically, from a statistics professor!). I think I've mentioned this before, but your focus on what Whitaker meant is misplaced.

Second, I'm not just referring to Morgan et al., but in addition Gillman, and Grinstead and Snell. If this is not enough, there are numerous others - Falk is another (you might find her book illuminating, see for example [4]).

You admit Krauss and Wang do not say the conditional issue is unimportant. Their version of the problem is constructed so that the conditional and unconditional answers are the same (i.e. they explicitly include the constraint that the host pick randomly if the player has initially selected the car). This by no means implies the problem is not asking the conditional question. You still have not a single source that says anything remotely like "the standard MHP is intended to refer to the general probability of winning by switching regardless of what door the host opens as opposed to the conditional probability". Until you find a source that says something like this your claim that this is the case is simply WP:OR.

That the question is conditional does not imply there's necessarily a difference between the answers given that the host opens door 2 or door 3. I don't know, but it seems you're still confused about this. If the answer is 2/3 given any door the player initially picks and any door the host opens, this doesn't mean the question is asking about the unconditional probability. Do you get this? The question is clearly conditional. The answer may not vary based on the specific door the host opens (and certainly most people would expect this to be the case), but assuming this is true and proceeding to "solve" the problem unconditionally is entirely different from showing this is true based on the problem statement. -- Rick Block (talk) 14:35, 30 April 2009 (UTC)

You keep referring to the question. If you do not take Whitaker's question to be the definitive statement of the problem that only makes it even less well defined. The problem clearly can be stated in a way that makes it unconditional.

I am not the least bit confused about conditional probability. Even if the problem is stated in a way that makes it conditional, if the host chooses randomly it makes no difference. I have explained before that maths can be made as hard as you like by making things more rigorous but it is never taught this way because it would put people off before they had made any progress. It is exactly the same with this article, we should start by treating the problem unconditionally then, after that has been full explained and discussed, we can talk about the conditional treatment. Martin Hogbin (talk) 22:26, 30 April 2009 (UTC)

By the way, does anyone know where I can find a copy of Selvin's problem statement? Martin Hogbin (talk) 22:27, 30 April 2009 (UTC)

Selvin's problem statement is referenced in the article. Nalebuff's, too. We don't agree and aren't going to, but as long as you stick to references (I will as well) I think we'll be fine. -- Rick Block (talk) 03:13, 1 May 2009 (UTC)

That is fine with me, I have always made it clear that I wish the article's quality to be maintained. On the other hand a certain degree of boldness may be required.

I have followed the links in the article but they do not seem to lead to Selvin's problem statement. Martin Hogbin (talk) 16:59, 1 May 2009 (UTC)

They are references, not links. Selvin's problem statement is in the Feb 1975 edition of American Statistician (volume 29, number 1, on page 67). Nalebuff's is in the Autumn 1987 edition of Journal of Economic Perspectives (volume 1, number 2, page 157).

Boldness is fine as well - but you should really get consensus for structural changes beforehand. -- Rick Block (talk) 18:34, 1 May 2009 (UTC)

Of course, consensus (but, as per WP, no vote counting)! Just look out for Rick Block's veto hammer! Glkanter (talk) 17:12, 2 May 2009 (UTC)

Frequentist approach

It is easiest to see the limitations of Morgan's solution using the frequentist approach. WP says, 'Frequentists consider probability to be the relative frequency "in the long run" of outcomes'.

So what we need to do is repeat the show and take the average for all players who switch (which we all know gives the uncondional solution of 2/3). No, say Morgan, we are only to consider the case where the player has chosen a specific door (say door 1) and the host opens a specific door (say door 3).

So we repeat the show but only consider cases where the player chooses 1 and the host opens door 3. Does that give Morgan's solution. Not necessarily. When we repeat the show what do we do? Do we put the car in the same place every time? Of course not that would be silly.

Maybe we could consider the actual repeats of the show where the producers and hosts act as they normally would. Unfortunately that does not necessarily give Morgan's answer. The producer might not place the car behind door 1 1/3 of the time. We have to place the car randomly for every show.

So we need to repeat the show withe the car randomly placed each time but with the same host. Unfortunately the host might change his policy for choosing which door to open from time to time. We need to keep a host who always has exactly the same door opening probability.

So, Morgan's solution applies only to the case where we repeat the show considering only cases with the same doors being chosen by the player and the host, the cars are initially placed randomly every time but the host has a fixed policy for choosing which door to open. Does anyone seriously believe that this was what Whitaker wanted to know? Martin Hogbin (talk) 20:19, 26 April 2009 (UTC)

I'm completely speechless. What do you suggest? Let us simulate the show, and then properly. For simplicity: "abc" means chosen a, car behind b, opened c.

112 1000x chosen door 1
113 1000x chosen door 1, opened 3, this is one possibility: conditional 1/3
123 2000x chosen door 1, opened 3, this is another possibility: conditional 2/3
132 2000x chosen door 1
221 3000x
223 3000x
231 6000x
213 6000x
331 2000x
332 2000x
321 4000x
312 4000x 

Comment? Nijdam (talk) 20:34, 26 April 2009 (UTC)

I do not understand what you are saying. Martin Hogbin (talk) 23:13, 26 April 2009 (UTC)

I think Nijdam is suggesting the numbers would be what we would find (but don't obviously correspond to any particular simulation?). For example, in 6000 times where the player picked door 1, the car was behind door 1 and the host opened door 2 1000 times, the car was behind door 1 and the host opened door 3 1000 times, the car was behind door 2 and the host opened door 3 2000 times, the car was behind door 3 and the host opened door 2 2000 times. Comparing the 2 cases where the player picked door 1 and the host opened door 3 shows a 1/3 probability of winning by not switching and a 2/3 probability of winning by switching.

What Morgan would say is that yes indeed for each trial you need to record the player's pick of door, the door the host opens, and the outcome. This is how you would discover whether the host has a preference for a particular door or not (or, alternatively, if the host is supposed to be picking randomly if the player has initially picked the car this is how you would verify the host is actually doing this). If you collected all this data you would see a 2/3 overall probability of winning by switching (which I think we all agree with), BUT you might see that (for example) if the player initially picks door 2 the host tends to open door 1 and if the player picks door 1 or door 3 the host tends to open door 2 (assuming the host and player are standing on the stage to the left of door 1 and the doors are laid out 1-3 left to right this seems like an entirely reasonable possibility - why walk across the stage to open door 3 rather than open the first door the host reaches?). If all you measure is the overall wins by switching you won't (can't) determine this. Asking about the conditional probability doesn't mean we're restricting the outcomes to those where the player picked door 1 and the host opened door 3, but that we're looking at each of the six possibilities of player pick and door the host opens in isolation. If the host picks randomly when the player initially picks the car these will all be the same, and they'll all show a 2/3 chance of winning by switching. If the host doesn't pick randomly (and, in reality, even if he's supposed to he might not), then these can vary anywhere between 1/2 and 1. -- Rick Block (talk) 00:49, 27 April 2009 (UTC)

Rick, you say,' Asking about the conditional probability doesn't mean we're restricting the outcomes to those where the player picked door 1 and the host opened door 3'. Of course it does, that is what conditional probability means. To solve a conditional probability question we only consider those events where the condition is met.

If our condition is 'player picks door 1, host opens door 3' then we only consider events where that condition applies. We could have chosen to apply a different condition, for example, 'player picks door 2, host opens door 1' but to get a conditional probability we must pick the chosen condition and apply it. Martin Hogbin (talk) 18:38, 27 April 2009 (UTC)

Martin - we analyze a specific outcome (like, say, player picks door 1 and host opens door 3) as a representative sample. The result applies to any specific choice of player door and any door the host opens. If the host is not constrained to pick randomly when the player picks the car, the chance of winning by switching if the player picks door 1 and the host opens door 3 is between 1/2 and 1, but it is the same for any door the player picks and any door the host opens. Do you really not understand this? -- Rick Block (talk) 18:50, 27 April 2009 (UTC)

Yes I do understand that. The specific example that Morgan pick (based on Whitaker's question) is that the player picks door 1 and the host opens door 3. They make this perfectly clear in their paper. I am well aware that they could have chosen the case where the player picks door 2 and the host opens door 1, or indeed any other legal combination. In all cases the probability of winning by switching is between 1/2 and 1, but this probability it is not necessarily the same for each combination.

Morgan actually consider the one specific case that the player picks door 1 and the host opens door 3 and they calculate the conditional probability that the player wins by switching, given these specific doors, as 1/(1+q) where q is the probability that the host will open door 3 given that the car is behind door 1. Had the specific case that Morgan chose to consider been that the player picks door 1 and the host opens door 2, the the probability of winning by switching would have been 1/(1+p) where p=1-q. In general p is not equal to q.

So what does it mean to a frequentist to say that the probability of winning by switching is 1/(1+q)? It means that if you repeat the experiment many times, applying the given condition, the player will on average win 100/(1+q) percent of the time if they switch. Martin Hogbin (talk) 20:08, 27 April 2009 (UTC)

Do I take it that everyone now agrees with this? Martin Hogbin (talk) 08:26, 29 April 2009 (UTC)

Do you? And, if you do, how can you not see that Whitaker's question asks about a player who finds herself in one (and only one) of the six possible conditional situations - mentioning the (presumably randomly selected) door 1/door 3 combination in order to make this absolutely clear? -- Rick Block (talk) 13:45, 29 April 2009 (UTC)

This is the question that Morgan answer, I have never disputed that. Whether it is what Whitaker actually wanted to know is another matter.

Do you agree that, from a frequentist perspective, to get a probability other than 2/3 we need to repeat the experiment considering only cases with the same doors being chosen by the player and the host, the cars being initially placed randomly every time but the host having a fixed policy for choosing which door to open? Martin Hogbin (talk) 17:01, 29 April 2009 (UTC)

Do I take it that you all agree with the above statement or does anybody think that it is incorrect? Martin Hogbin (talk) 16:59, 18 May 2009 (UTC)

It is not completely clear what you mean. If in the experiment the car is placed randomly, the player's choice is independent of the position of the car and the policy of the host is the "random policy" then when the experiment is repeated, in 2 out of 3 cases where a specific door is chosen by the player and also a specific door is opened by the host, the car will be behind the "other" door. To get another answer, at least one of the condition has to be different, but changing a condition doesn't logically lead necessarily to another answer. So where are you heading? Nijdam (talk) 17:37, 18 May 2009 (UTC)

Yes, I say above that the host has a fixed policy for choosing which door to open. For example he might always open door 2 if possible or more generally he might choose door 3 with a fixed probability of q. My aim is to show the very limited circumstances, from a frequentist perspective, in which the Morgan solution applies. To get Morgan's answer you must have random car placement, random player initial choice of door but the same host door choice policy every time, and the player and host must always choose the same doors. Do you agree with this? Martin Hogbin (talk) 19:18, 18 May 2009 (UTC)

I just noticed that I have made two contradictory statements above. It should be that either the player chooses any door but we only consider cases where she has chosen a specific door or the player is constrained to always pick a specific door. Martin Hogbin (talk) 19:27, 18 May 2009 (UTC)

From a frequentist perspective, what is required is keeping track of which door the player picks and which door the host opens. So, we don't just count how many switchers win and lose and how many stickers win and lose, but are able to look at each of the 6 combinations of initial player pick and host door independently. Depending on the host strategy, the answer to the conditional question might depend on the specific combination of player pick and host door or it might not. For a frequentist, the host's strategy doesn't have to necessarily be constant (and it certainly doesn't have to be predefined or conscious) - just observable over a large number of trials. This is entirely consistent with Morgan et al.'s interpretation - their q could be derived from observation. And, yes, the observed chance of a player winning by switching who picks door 1 and sees the host open door 3 might be different from the chance given any other combination. However, if we derived the door1/door3 q from observation we would also see the chance of a player winning by switching who picks door 1 and sees the host open door 2 would be 1(1+(1-q)) as shown by Morgan et al. -- Rick Block (talk) 01:00, 19 May 2009 (UTC)

My point was about what Morgan's solution of Pr(Ws|D3)=1/(1+q) means from a frequentist point of view. To get this answer you need to consider only the case that Morgan do, namely that the player has chosen door 1 and the host has opened door 3. Obviously other combinations could be considered and these would give different answers, as you say, if the player chooses door 1 then Pr(Ws|D2)=1/(2-q).

The other point to note is that the car must be placed randomly each time but the host must always have the same probability of opening door 3 when the car is behind door 1. Do you agree with this? Martin Hogbin (talk) 17:21, 19 May 2009 (UTC)

The only part of this I agree with is the assumption that the car is placed randomly (which is as much a background rule as the "assumption" that the host must open a door and make the offer to switch - your contention that this is a deficiency of the solution presented by Morgan et al. is simply ludicrous). Morgan et al.'s solution applies for any initial pick of door and any door the host opens (through renumbering of doors). A frequentist could measure any specific player pick (say door 2 rather than door 1) and any specific door the host opens (say door 1 rather than door 3), but would in all likelihood measure all iterations of the game but keep track by door the player picks and door the host opens (all 6 combinations). Any specific combination is addressed by Morgan et al.'s result - as well as the 3 pairs of combinations each involving a specific initial player pick. A frequentist wouldn't care whether the host's strategy is constant, only what it averages out to over a large number of trials. Whatever this average is would provide the expectation for a player in the same situation as the previous players (the host's preference is revealed by what the host has historically done). -- Rick Block (talk) 18:46, 19 May 2009 (UTC)

I am well aware that Morgan might have considered any legal combination of player and host door pick with similar results but I am asking the question as to what the specific example that Morgan deal with means to a relativist. Morgan are quite clear as to the situation that they are considering, they say, 'To avoid any confusion, here is the situation: The player has chosen door 1, the host has then revealed a goat behind door 3, and the player is now offered the option to switch'. Morgan then go on to calculate the probability of winning by switching for this specific case as Pr(Ws|D3)=1/(1+q) where q is the probability that the host will pick door 3 if the car is behind door 1. For the moment, I am just asking what this solution to this specific question means to a frequentist. Perhaps it would be easier if you tell me what experiment you think would need to be repeated to get this specific result (that is to say just considering the case where the player picks door 1 and the host opens door 3). In other words, what exactly does the statement Pr(Ws|D3)=1/(1+q) (where q is the probability that the host will pick door 3 if the car is behind door 1) mean to a frequentist?Martin Hogbin (talk) 22:14, 19 May 2009 (UTC)

A frequentist reading Morgan et al. would realize q (or it's inverse p) applies per initial player pick, and to be able to answer any player's question about what that player's chances of winning are at the point of deciding whether to switch would endeavor to collect the information I've described above (about all combinations of player pick and door the host opens). For players who have picked door 1 the trials of critical interest are the ones where it turned out the car was behind door 1; the proportion of times the host opened door 3 vs. door 2 in this case is the basis for estimating q (and also its inverse p). There are 3 pairs of p and q values (one pair for each initial player pick). If a player who has picked door 1 and has seen the host open door 3 asks the frequentist what are her chances of winning by switching, the frequentist would look at the relevant data (all prior cases where the player has picked door 1 and car turned out to be behind door 1), would figure out q, and would say "your chances of winning are 1/(1+q)" (this would be the same answer that would be given by an incurious frequentist who might reply with the sum of the number of previous switchers who won plus the number of previous stickers who lost divided by the total number of players but counting only players who initially picked door 1 and were shown door 3 by the host). -- Rick Block (talk) 03:00, 20 May 2009 (UTC)

You seem to be answering a more general question than the one that I am asking. For example if I have a loaded (unfair) six-sided die and I say to a frequentist that the probability of throwing a 6 is 0.5 (Pr(6)=0.5) they would take it to mean that 'in the long run', that is to say if the experiment of throwing the specific die in question were to be repeated a large number of times, 1/2 of these times would result in a throw of 6.

Morgan say that, given that the player has chosen door 1 and the host has opened door 3, Pr(Ws|D3)=1/(1+q) where q is the probability that the host will open door 3 if the car is behind door 1. They go on to discuss specific cases where q=0, 1, and 1/2. Let us consider just the specific case that the player has chosen door 1, the host has opened door 3, and q=1, where Morgan show that Pr(Ws|D3)=1/2. My question is simply, 'Exactly what experiment do we need to repeat for the player to win 1/2 of the time?'. Martin Hogbin (talk) 08:32, 20 May 2009 (UTC)

Why do you ask something for which the answer is obvious? This is the much discussed "rightmost door preference host" (see, for example, the results of the simulation I posted some time ago [5]). Tell the host to open the rightmost door whenever possible. Play. Players who pick door 1 and see the host open door 3 (or, equivalently, players who pick door 2 and see the host open door 3 or players who pick door 3 and see the host open door 2) will win about 1/2 the time. In this case what would you say is the chance of winning by switching for a player who has picked door 1 and has seen the host open door 3? -- Rick Block (talk) 12:38, 20 May 2009 (UTC)

There is no dispute as to the the chance of winning by switching for a player who has picked door 1 and has seen the host open door 3 when q=1. I agree with you and Morgan that the answer is 1/2. The question was, 'Exactly what experiment you need to repeat to get this result?'. You state the obvious conditions, that we only consider games where the player picks door 1 and the host opens door 3. But we also need to state that, when the experiment is repeated, the car must be placed randomly each time, not be replaced where it was before. In other words it is no use placing the car randomly once then repeating the experiment with the car in that same position. No doubt you will say that this is obvious, but it needs to be made clear. On the other hand, to get an answer (Pr(Ws|D3)) of 1/2 we must not have a host with a different policy each time that we repeat the experiment, we must always have a host who has the same probability of choosing door 3 when the car is behind door 1. All these conditions must be made clear in a frequentist interpretation.

As a separate issue, you are wrong when you say '...(or, equivalently, players who pick door 2 and see the host open door 3 or players who pick door 3 and see the host open door 2) will win about 1/2 the time. I have only specified q (probability of choosing door 3 when the car is behind door 1) to be 1. This does not give us any information about the host's probability of choosing door 3 when the car is behind door 2 or probability of choosing door 2 when the car is behind door 3; these can still have any value from 0 to 1. Martin Hogbin (talk) 13:42, 20 May 2009 (UTC)

I said to tell the host to pick the rightmost door. This makes q=1 for door 3 in the case the player picks door 1 and also the other cases I mention as well (so they are equivalent). And for the umpteenth time, a frequentist doesn't care if the host's policy is constant, only that it have an average over a large sample (which it will whether it's constant or not). If we observe over 1000 trials that q=.75 it doesn't make any difference (to a frequentist) if the host alternates between q=.5 and q=1 or uses q=1 3 out of 4 times and q=0 the 4th time. Is there any particular point you're trying to make here? -- Rick Block (talk) 21:55, 20 May 2009 (UTC)

Of course, if you change the question you get a different answer. It is quite possible to consider any form of host door preference, for example the host may prefer odd numbered doors, or maybe the nearest door to where he happens to be standing at the time. Having decided on some kind of door policy for the host it is then possible to assign probabilities of various doors being opened in specific cases.

For some reason you seem unwilling to address the question that I asked, which is the only question that Morgan actually deal with in their paper. In my question, 'The player has chosen door 1, the host has then revealed a goat behind door 3, and the player is now offered the option to switch'. Given this question only, can you explain exactly what experiment would need to be repeated to get the solution Pr(Ws|D3)=1/(1+q)? Martin Hogbin (talk) 08:26, 21 May 2009 (UTC)

How have I not addressed this already? Tell the host to pick the rightmost door. Play. If you don't care to use the equivalence given this setup between the door 1/door3 combination and door2/door3 and door3/door2 combinations then play enough times to have a meaningful sample of the door1/door3 combination to look at. Another way is to simply play (without telling the host anything). When there is a meaningful sample of the door1/door3 combination figure out q based on the cases where the player picked door 1. Verify (for the subset of these cases where the host opened door 3) that Pr(Ws|D3)=1/(1+q). I'm not trying to be difficult here - I really don't have any idea what point you're driving at. If you don't like this answer then please drop the Socratic approach and just say what you're trying to say. -- Rick Block (talk) 14:46, 21 May 2009 (UTC)

Fine, your second answer at least addresses the question asked, but you have still missed out some essential details. How is the car placed behind the doors each time? Do we use a method of generating a random number from 1 to 3? Do we leave it to the whim of the producer or stage hand? Do we always replace it behind the same door?

What about the host? I presume we keep the same host throughout, who has the same policy on door choice. Do we ask the host to toss a coin? Are we allowed to change the host for one with a different policy? Martin Hogbin (talk) 19:34, 21 May 2009 (UTC)

A frequentist who wants to know the probability of X does X over and over again until there's a meaningful sample size. So, a frequentist who wants to know Pr(Ws|D3) has to count only occurrences when the player picks door 1 and the host opens door 3. Use whatever rules you want, but for the probability to be meaningful you need to be reasonably consistent. Morgan et al. use the same rules vos Savant did (e.g. random initial car placement), which pretty much match how you'd really run a game show. For the probability to be meaningful you clearly want to use the same host (letting the host "naturally" decide in the case where the player initially selects the car) or assign the host a controlled policy (like secretly flipping a coin to decide which door to open if the player has initially picked the car) - if you assign a policy the specific host wouldn't matter.

But, you know all this. So, what's your point? -- Rick Block (talk) 01:45, 22 May 2009 (UTC)

At last we agree. My point is that Morgans solution only applies in a rather contrived case. The producer must place the car randomly. For some unexplained reason we do not take it that the producer chooses to place the car behind behind door 1 with probability c1 etc we assume that this is done randomly so that C1=C2=C3=1/3. Please tell me on what basis this assumption is made.

The host, on the other hand, is taken to open a door (in any given set of circumstance) with a fixed probability. Please tell me why we not assume that the host, like the producer, chooses randomly.

The frequentist view, in my opinion, makes this inconsistency particularly obvious. Martin Hogbin (talk) 09:03, 22 May 2009 (UTC)

Ah, so this becomes the same point you're driving at below. Let's discuss it in one place (below). -- Rick Block (talk) 13:24, 22 May 2009 (UTC)

Error in Morgan et al?

Firstly, before anyone else tells me, this is OR. However this is not the article, it is not even the article talk page, it is the arguments page and it is certainly relevant to the MHP.

At the bottom of the first column of page 286, Morgan calculate the conditional probability of winning by switching given that the the player has chosen door 1 and the host has opened door 3 (Pr(Ws|D3)) on the basis of a uniform prior distribution of the host door opening strategy parameter q. In other words they consider the case that that the host initially has a equal probability of having any value of q from 0 to 1.

Morgan calculate Pr(Ws|D3) by integrating the probability of switching (1 / (1+ q)) over the interval 0 to 1.

Nijdam shows Morgan's calculation thus :

$\int _{0}^{1}{\tfrac {1}{1+q'}}f_{q}(q')dq'=\int _{0}^{1}{\tfrac {1}{1+q'}}dq'=ln(2).$

The problem that I see is that after door 3 has in fact been opened, which is the condition under which we are trying to calculate the probability, the distribution of q will no longer be uniform. Because we know that door 3 has been opened it is more likely that the host will have a value of q closer to 1 than otherwise. The prior distribution of q is uniform but we must use the posterior distribution, which is not.

Using some dubious mathematical reasoning, which I am happy to explain to anyone interested, I believe that the correct calculation of Pr(Ws|D3) is:

$1/(1+\int _{0}^{1}\ {f_{q}(q)}dq)=1/(1+\int _{0}^{1}\ {q}dq)=2/3$

So what is the relevance of this? It certainly weakens the case for Morgan to be considered the definitive article on the MHP but it also resolves a paradox in Morgan's result that had puzzled me.

If we take a uniform prior distribution of q this means that we also have a uniform distribution of p (= 1 - q). Morgan calculate Pr(Ws|D3)to be 0.693. By the same token we could calculate Pr(Ws|D2) to be 0.693. There is the paradox. Given some particular situation (that the prior distribution of q is uniform) the probability of winning by switching is 0.693 if the host opens door 3 or if the host opens door 2, yet we know that the probability of winning by switching given that the host opened either door 2 or door 3 is always 2/3 regardless of the hosts strategy. This is easy to verify given any single value for q. I know that probability can be surprising but I find this hard to believe. Morgan must be wrong.

Perhaps someone who know about these things would like to comment. Do you see a paradox in Morgan's result? Are Morgan et al wrong? Is my result correct? Martin Hogbin (talk) 14:07, 4 May 2009 (UTC)

I'm not an expert in Bayesian reasoning, but I think that after the host has opened door 3, information is revealed about q, hence the posterior density of q is 2/3·(2-q). It also reveals info about the position of the car, hence the conditional prob. of the car behind door 2 is: 1/(2-q). We can ask for the expectation of this prob. with respect to q (being log(2)), but also for the conditional expectation given door 3 opened leading to the answer 2/3 (numerically equal to your answer, but I don't understand your derivation). Further I don't follow your criticism, don't even know what you mean. Nijdam (talk) 22:21, 4 May 2009 (UTC)

We agree that information about q is revealed by the opening of door 3. Do you also agree that that to calculate Pr(Ws|D3) we need to use the posterior distribution of q? Are you saying that you get a numerical answer of 2/3 for Pr(Ws|D3) given a uniform prior distribution of q? If so, perhaps you could show be how you derive this.

Well, assuming P(D3|C1)=q, we know that P(C2|D3) = 1/(1+q) and f(q|D3)=2/3·(1+q). (I prefer to speak of C(ar) etc. rather than of Ws, which is not properly defined.) Hence its conditional expected value, given D3 is simply $\int {\tfrac {1}{1+q}}f(q|D3)dq={\tfrac {2}{3}}.$ Of course derived from the apriori uniform disribution of q. Nijdam (talk) 21:47, 5 May 2009 (UTC)

While I take a while to consider your derivation perhaps you could comment on the fact that you get a different numerical result from Morgan. Who is right? Martin Hogbin (talk) 08:32, 6 May 2009 (UTC)

As you will no doubt have concluded, I have no formal training in statistics. Could you please explain to me what f(q|D3) is. I presume your integral is to be evaluated from 0 to 1. Martin Hogbin (talk) 09:05, 6 May 2009 (UTC)

Yes, please explain this. I assume f(q) = 1/(1+q), but then what is the meaning of f(q|D3)? With regard to probability computations q is an unknown constant. -- Rick Block (talk) 14:12, 6 May 2009 (UTC)

Well, the prior density of P(D3|C1)=q is f(q)=1 on [0,1], after door 3 being opened (D3), the postrior density is f(q|D3)=2/3·(1+q). Could be written down more extensively of course.Nijdam (talk) 08:38, 7 May 2009 (UTC)

Regarding my criticism, let me try to explain what I mean in stages. We are considering the situation where there is a uniform prior distribution of q. Do you agree that in this case the prior distribution of p (=1-q) would also be uniform? Martin Hogbin (talk) 16:34, 5 May 2009 (UTC)

Of course. Nijdam (talk) 21:47, 5 May 2009 (UTC)

OK. Give the uniform prior distribution of q, what do Morgan calculate the probability of winning by switching to be given that the host has opened door 3?

Given exactly the same distribution of q, using Morgan's method, what would be the probability of winning by switching if the host has opened door 2? Martin Hogbin (talk) 08:16, 6 May 2009 (UTC)

ln(2) in both cases. -- Rick Block (talk) 14:12, 6 May 2009 (UTC)

Given the same distribution again, what would be the probability of winning by switching given that the host has opened one (unspecified) of the unchosen doors? Martin Hogbin (talk) 21:49, 6 May 2009 (UTC)

What information is revealed about q by opening door 3? The fact that the conditional probability, given a uniform prior distribution of p or q, is the same whether the host opens door 2 or door 3 is hardly surprising (since the only difference between doors 2 and 3 is their arbitrary door numbers). That it is not 2/3 reflects the nature of uniform distributions. That it is not the same as "the probability of winning by switching given the host opened either door 2 or door 3" is because it is a completely different question. Rather than "Morgan must be wrong" the correct conclusion would be that Martin does not understand. -- Rick Block (talk) 03:06, 5 May 2009 (UTC)

The fact that the host has opened door 3 means that it is more likely that the host has a preference for door 3, in other words the posterior distribution is biased towards 1. Martin Hogbin (talk) 16:34, 5 May 2009 (UTC)

[ Outindented ] Martin, careful, you are making a common beginner's mistake: confusing a parameter that a proposition depends on with the proposition itself. I used to recommend my students to always separate the two things using a different notation, e.g. uppercase for propositions and lowercase for parameters. I don't have the Morgan paper, but it's easy to reconstruct their argument from the info provided in the MHP article and here. I will use in the following the same notation style as in the "Bayesian analysis" section of the WP article.

Begin by adding the following assumptions to the background $I\,$:

(a) The host "prefers" to open one of the two doors that are available after the player chooses the door with a car behind.

(b) His preference is independent of the car's location.

(c) His preference is established before the game is played.

Now add to the model one additional proposition:

$Q:\,$ "The value of the host's preference for opening the door most to the right of the one chosen by the player is q."

where q is a parameter in $[0,1]\,$.

Then, using the notation of the "Bayesian analysis" section in the article, the likelihood $P(H_{ij}|C_{k},\,Q,\,I)$ for opening a door is written as:

$P(H_{ij}|C_{k},\,Q,\,I)\,\,=\,{\begin{cases}\,\\\,\\\,\\\,\end{cases}}$	$\,0\,$		if i = j, (the host cannot open the door picked by the player)
	$\,0\,$		if j = k, (the host cannot open a door with a car behind it)
	$\,q\,$		if i = k and j > i, (of the two doors with no car, the right one has preference q)
	$\,1-q\,$		if i = k and j < i, (of the two doors with no car, the left one has preference 1 - q)
	$\,1\,$		if i $\neq$k and j $\neq$ k, (there is only one door available to open)

Then, using the same argument as in the "Bayesian analysis" section, and noting that it is $P(C_{i}|Q,\,I)=P(C_{i}|I)\,$ because of assumption (b) above, we find that:

$P(C_{2}|H_{13},\,Q,\,I)={\frac {1}{1+q}}={\frac {P(C_{2},\,Q|H_{13},\,I)}{P(Q|H_{13},\,I)}}\,$,

where the last equality stems from the definition of conditional probability.

However, we are interested in $P(C_{2}|H_{13},\,I)$, the probability that the player wins by switching regardless of the host's preference. We can compute it by marginalizing the previous result with respect to all possible values of the nuisance parameter q:

$P(C_{2}|H_{13},\,I)=\int _{0}^{1}P(C_{2},\,Q|H_{13},\,I)\,{dq}=\int _{0}^{1}{\frac {1}{1+q}}\,P(Q|H_{13},\,I)\,{dq}\,.$

So here we come to Martin's issue: is $P(Q|H_{13},\,I)\,$ the same as $P(Q|I)\,$, the latter being the prior distribution for the parameter q? The answer is yes because of assumption (c): The host decides on the value q of his preference before the game starts, therefore $Q\,$ must be independent of $H_{13}\,$. Morgan's result of log(2) for the value of the marginal then follows by assuming a uniform prior $P(Q|I)=f(q)=1\,$. glopk (talk) 05:51, 7 May 2009 (UTC)

??????In your notation (I drop the superfluous I) I calculate the posterior density of Q given H13 as:

$P(Q|H_{13})={\frac {P(H_{13}|Q)P(Q)}{P(H_{13})}}=$

${\frac {P(H_{13}|C_{1},Q)P(C_{1}|Q)+P(H_{13}|C_{2},Q)P(C_{2}|Q)}{P(H_{13})}}=const\cdot (1+Q)={\tfrac {2}{3}}(1+Q)$

Nijdam (talk) 08:50, 7 May 2009 (UTC)

Nijdam - Isn't your denominator $\,P(H_{13})=P(H_{13}|C_{1})P(C_{1})+P(H_{13}|C_{2})P(C_{2})$, in which case $\,P(Q|H_{13})=1$? In English, $\,P(Q)$ is not q, but the probability that the host's preference is q - which is 1. -- Rick Block (talk) 14:43, 7 May 2009 (UTC)

No, Nijdam is right, and I confused causation with logical independence (and should really stop doing math past 11 pm). $P(Q|I)\,$ is the prior distribution for the parameter q, which is uniform (i.e., 1) in our case. More precisely, expanding $P(H_{13}|Q,I)\,$ as Nijdam does:

${\begin{aligned}P(H_{13}|Q,I)&=\sum _{i=1}^{3}P(H_{13},C_{i}|Q,I)=\sum _{i=1}^{3}P(H_{13}|C_{i},Q,I)\,P(C_{i}|Q,I)\\&=\sum _{i=1}^{3}P(H_{13}|C_{i},Q,I)\,P(C_{i}|I)=q\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}={\frac {q+1}{3}}\end{aligned}}$

we have:

$P(Q|H_{13},I)={\frac {P(H_{13}|Q,I)P(Q|I)}{\int _{0}^{1}P(H_{13}|Q,I)P(Q|I)dq}}={\frac {(q+1)\times 1}{\int _{0}^{1}(q+1)\times 1\,dq}}={\tfrac {2}{3}}(q+1)$

and plugging this back into my last expression for $P(C_{2}|H_{13},\,I)\,$ above:

$P(C_{2}|H_{13},\,I)=\int _{0}^{1}{\frac {1}{1+q}}\,P(Q|H_{13},\,I)\,{dq}\,.=\int _{0}^{1}{\frac {1}{1+q}}\times {\tfrac {2}{3}}(q+1)\,dq={\tfrac {2}{3}}$.

So, kudos to Nijdam, ashes for Morgan and glopk (talk) 16:22, 7 May 2009 (UTC)

So do we all agree now that Morgan have got the answer wrong? Martin Hogbin (talk) 19:33, 7 May 2009 (UTC)

No. I'm good with $\,P(H_{13}|Q,I)={\frac {q+1}{3}}$ but given the Bayes expansion of $\,P(Q|H_{13},I)={\frac {P(H_{13}|Q,I)P(Q|I)}{P(H_{13}|I)}}$ I think the denominator also works out to ${\frac {q+1}{3}}$ since $\,P(H_{13}|I)=P(H_{13}|C_{1},I)P(C_{1}|I)+P(H_{13}|C_{2},I)P(C_{2}|I)$. Again, $\,P(Q|I)$ (which I read as the probability that the host's preference is q) is an invariant constant equal to 1 (by assumptions b and c). On the other hand, I've cut down on my coffee lately and this isn't precisely my field. What am I missing here? -- Rick Block (talk) 19:39, 7 May 2009 (UTC)

Rick, the equations in my paragraph above answer the question "What is the player's chance of winning by switching, assuming ignorance of host's preference for one door versus another?". Seen in this light, the answer must be 2/3, since we fall back onto the classic interpretation of the MHP from the player's POV. Treating q as a nuisance parameter requires the marginalization (i.e. sum) of $P(H_{13}|I)\,$ over all possible values of q, each value weighted with $P(Q|I)\,$ (== 1 with our uniform prior). Try choosing a different prior for Q and see what happens to convince yourself. I don't have the Morgan paper, so I won't comment further on what exactly is computed there, and whether it is wrong or not (no ashes on Morgan yet). glopk (talk) 00:24, 8 May 2009 (UTC)

Morgan et al. show the probability of winning by switching, treating the host's preference for the door that has been opened as an unknown constant q (in the case the player has initially picked the car), is 1/(1+q). The paragraph in question is this:

“ The above solution engenders disappointment in some students, who feel that there should be a single answer for Pr(W_s|D3) not depending on unknown probabilities. This provides an excellent opportunity to bring in the Bayesian perspective. For instance, the noninformative prior in the vos Savant scenario [the normal MHP rules, but minus the restriction that the host pick randomly with equal probability in the case the player initially picks the car] makes this probability

$\int _{0}^{1}{\frac {1}{1+q}}dq=ln(2)\approx .693$

”

Perhaps this is what you mean, but we're assuming the host might have a preference for one door over another without knowing exactly what value this preference has (i.e. it is equally likely to be any value between 0 and 1). Q seems to be the nuisance here - why not just put q as part of the background information (it's a constant, not changed by the player pick or the door the host opens)? And, even using Q as a "proposition" (it's simply a true statement, so treating it as a proposition seems somewhat curious) why isn't using the Bayes expansion of the denominator valid? -- Rick Block (talk) 01:43, 8 May 2009 (UTC)

I will leave the maths experts to argue out the maths for the moment but I do not understand why you insist that the opening of a door by the host does not change the distribution of q. Of course this distribution is initially fixed, just like the players choice of door is, but in the conditional case, that is to say only considering cases where door 3 has in fact been opened, we have taken a non-representative sample from the original set, so the conditional distribution of q changes in just the same way that the probability that the player has chosen the car (sometimes) changes when the host opens a door. Martin Hogbin (talk) 22:22, 8 May 2009 (UTC)

In fact with a uniform prior distribution of q, if door 3 is opened, a value of q=1 then becomes twice as likely as q=0. If we assume the distribution of q remains linear and is normalized, we get Nijdam's posterior distribution of 2/3(1+p), which I now agree with. Martin Hogbin (talk) 23:26, 8 May 2009 (UTC)

We're talking about the conditional case where the host has opened door 3. In this case, the host has a preference q. I assume you're thinking that if q is 0 the host opens door 3 only if the car is behind door 2 and if q is 1 the host opens door 3 if the car is behind door 2 or door 1, so if the host has opened door 3 q is twice as likely to be 1 as 0. I guess the question is are we assuming a uniform distribution of q given the host has opened door 3, or a uniform distribution of q before the host opens door 3. I think this again gets back to the nature of the problem. At the point of the player's decision, it's given that the host has opened door 3. If at this point we assume we know nothing about q, then I'm not sure if we should take its distribution as uniform before or after the host has opened the door. If we take the distribution to be uniform after the host opens the door then I presume you agree with Morgan et al.'s result. If we assume q is uniformly distributed at the start of the game (before the host opens a door), then q is influencing which door the host opens so we can then use the fact that the host has opened door 3 to adjust our assumption about the distribution. I'm not sure which interpretation is more reasonable, although my inclination is to trust a peer reviewed paper. -- Rick Block (talk) 02:01, 9 May 2009 (UTC)

What you are saying is exactly my point, however there is absolutely no doubt that Morgan are considering a prior distribution of q. They say, 'for the noninformative prior...'. Any attempt by anyone, including the authors themselves, to claim any other meaning is farcical and would only server to discredit the claimant even further.

Regarding the issue of peer reviewed sources, I have always supported the use of reliable sources over OR in Wikipedia and for that reason I have kept my unsourced opinions off the article page. Authoritative opinion is not set in stone and can be changed. If the experts here can agree that there is an error in the Morgan paper I would like to ask someone of sufficient standing in the statistics community to support me in writing to the American Statistical Association with a view to getting a letter pointing out the error published. This may be a first for Wikipedia but I believe that it is fully in accordance with its policies and ethos. On refelection, I accept that my calculation may well be wrong and I understand Nijdam's calculation of the posterior distribution. The first thing that must be decided is whether we agree that there is an error in Morgan's paper. So far Nijdam and Glopkg seem to think that there is. Martin Hogbin (talk) 10:43, 9 May 2009 (UTC)

(outindented) I'm back, see you folks have been diligent with math. Let me put nit straight: I think Morgan didn't calculate the right probability. I gave you my calcuations in short, and in the more or less clumsy notation of glopk, in which happens to figure Q and q without some reasonable system. I write down the calculation in full. C is the door with the car, H the door opened by the host (I write for short C2 for C=2 etc.) and Q is the parameter such that:

$P(H3|C1,Q=q)=q$

Prior density of Q (uniform):

$f_{Q}(q)=1,q\in (0,1)$

Posterior density of Q, given H=3 (door 3 opened):

$f_{Q}(q|H3)={\frac {P(H3|Q=q)f_{Q}(q)}{P(H3)}}=$

$=const\cdot (P(H3|C1,Q=q)+P(H3|C2,Q=q)=$

$=const\cdot (q+1)={\tfrac {2}{3}}(1+q)$

Don't bother about the other terms, they're all in the (normalizing) constant. Then:

$\!\,P(C2|H3,Q=q)=$

$={\frac {P(H3|C2,Q=q)P(C2|Q=q)}{P(H3|C2,Q=q)P(C2|Q=q)+P(H3|C1,Q=q)P(C1|Q=q)}}=$

$={\tfrac {1}{1+q}}$

Hence, using the law of total probability:

$\!\,P(C2|H3)=\int P(C2|H3,Q=q)f_{Q}(q|H3)dq={\tfrac {2}{3}}.$

There is no logical or mathematical argument for integrating here with respect to the prior density of Q. It's an easy made error. On the other hand we may start with a different prior distribution and end with any number we like. That's the power of Bayesian analysis. Nijdam (talk) 23:46, 10 May 2009 (UTC)

Do you think that we should write to the American Statistician with this result? Martin Hogbin (talk) 08:20, 11 May 2009 (UTC)

Thanks for confirming the result. On the "clumsy" notation, I will just redirect you to to a list notable users of it[6] (and if you happen to be in that list and use a different notation, well, congratulations). On the difference between Q and q, as you are (by your own admission) unfamiliar with bayesian terminology, I will point to you that in frequentist (and Kolmogorov's measure-theory) P.T., Q would be called a continuous random variable, and q would be called one value of that random variable. Is the difference clear enough now, or do I have to repeat it very very slowly? glopk (talk) 16:12, 11 May 2009 (UTC)

Well. I wouldn't call it confirmation. And on the clumsyness, it mainly concerns the difference between Q and q. As you know so well the difference between Q and q, then why do you mix them up? Nijdam (talk) 16:32, 11 May 2009 (UTC)

If you say A, then I say A (admitting a mistake, and graciously confirming that A is correct), and then you repeat A, that is a confirmation too. But if you want to call it a potato, bon appetit. On the "mixing", can you please point to a specific equation I've written where the two symbols appear together improperly? Note that any symbol $P(\cdot )\,$ that contains $Q\,$ in the $(\cdot )\,$ is a function of q by the definition of the proposition (or random variable) Q. Is your $f_{Q}(q|H3)\,$ a case of bad "mixing" too? glopk (talk) 19:14, 11 May 2009 (UTC)

Morgan is still on the faculty of Virginia Tech (see [7]). I emailed him a while ago on a different topic and he did respond. BTW - you do realize this paper was published nearly 20 years ago? Before claiming this is a new result I would suggest checking to see whether it's already been published looking first at papers (and books) that cite Morgan et al. I haven't personally chased down every published reference to this paper. References (even just academic references) to the MHP are so voluminous that verifying this is a new result would actually take some time. -- Rick Block (talk) 13:57, 11 May 2009 (UTC)

Write Morgan and ask him if the error has already been reported and published.Nijdam (talk) 16:32, 11 May 2009 (UTC)

Am back too. Yes, given Rick's summary of that paper's paragraph, I do think it is in error, i.e. that it is wrong to marginalize the result using the wrong using the prior for $Q$. The simplest explanation is that if the player has to assume complete ignorance of the host's preference for one door when two are available to open, then he is in the same position as in the classic interpretation of the MHP. glopk (talk) 16:12, 11 May 2009 (UTC)

While you are at it, can anyone confirm that the same result applies to any prior distribution of q that is symmetrical about 1/2. My simple symmetry argument, which is what got me started on this, suggests that it should. Martin Hogbin (talk) 17:41, 11 May 2009 (UTC)

You're right, with symmetric prior we get for the posterior density of Q, given H=3 (door 3 opened):

$f_{Q}(q|H3)=const\cdot (q+1)f_{Q}(q)={\tfrac {2}{3}}(q+1)f_{Q}(q),$

because EQ=1/2, and hence const=2/3. Then, as:

$\!\,P(C2|H3,Q=q)={\tfrac {1}{1+q}}$

it follows:

$\!\,P(C2|H3)=\int P(C2|H3,Q=q)f_{Q}(q|H3)dq=\int {\tfrac {2}{3}}f_{Q}(q)={\tfrac {2}{3}}.$

Nijdam (talk) 20:38, 11 May 2009 (UTC) I forgot to mention that the only thing needed is the expectation of Q being 1/2.Nijdam (talk) 20:41, 11 May 2009 (UTC)

Or more complete, in general:

$f_{Q}(q|H3)=const\cdot (q+1)f_{Q}(q)={\frac {1+q}{1+EQ}}f_{Q}(q),$

and:

$\!\,P(C2|H3)={\frac {1}{1+EQ}}\left(={\frac {1}{1+\int qf_{Q}(q)dq}}\right).$

So the general form of your dubiously (your words) obtained formula. Nijdam (talk) 20:54, 11 May 2009 (UTC)

Well done and thank you. Martin Hogbin (talk) 21:41, 11 May 2009 (UTC)

Nijdam, I think we should write to the American Statistician with this result. Martin Hogbin (talk) 08:05, 12 May 2009 (UTC)

Isn't assuming a uniform prior before the host has opened a door effectively transforming this into the unconditional problem? I.e this makes the Bayesian computation the chance of winning regardless of which door the host opens (which everyone knows is 2/3) as opposed to the chance of winning given the host has opened door 3. BTW - when I contacted Professor Morgan some time ago he said he'd already said all he has to say about this problem. If you do want to pursue this I'd suggest you run it by some more folks from Wikipedia:WikiProject Mathematics. I am not an academic, but since this is a peer reviewed academic paper my understanding of the proper procedure to follow is to contact the journal. If the editor agrees the point has any merit whatsoever then presumably Morgan and his co-authors would be given an opportunity to formally reply. In an academic sense, agreeing here (which I don't) that this is an error means nothing at all. Furthermore, by Wikipedia policy (Wikipedia:No original research), even if this were in the article (which it's not) agreeing here that it's an error without a supporting published source would not mean the article should be changed. -- Rick Block (talk) 16:18, 16 May 2009 (UTC)

I have not read anything but the post by Rick Block at 16:18, 16 May 2009. I cannot imagine there is much benefit to writing the editors of a journal about this sort of thing. If there were general agreement on the talk page that there is an error in the source, we could simply choose not to use it. — Carl (CBM · talk) 17:14, 16 May 2009 (UTC)

@Rick No, I do not believe this is the same as the unconditional problem. We are considering the case where the host (or a population of hosts) has uniform prior distribution of q. That is to say the host has an equal probability of having any value of q from 0 to 1. We the only consider the conditional case that the host has in fact opened door 3. It turns out that the answer is the same as for the unconditional case, as you might expect from the symmetry of the situation.

I am happy to discuss this further on Wikipedia:WikiProject Mathematics. What is the best way to do this? I have no intention of changing the article based on my (and nijdam's) OR. We intend to write to the journal but even its letters policy is very strict. We may need some support from a more prominent figure in the statistics community to get a letter published. Martin Hogbin (talk) 21:15, 18 May 2009 (UTC)

I've already solicited comments at Wikipedia talk:WikiProject Mathematics#Allegation of error in a peer reviewed source (CBM's response above is likely in response to this). -- Rick Block (talk) 01:07, 19 May 2009 (UTC)

New section

@Martin: Ok.

BTW: The whole calculation comes down to (using Bayes and the rules of the game):

$P(C2|H3)={\frac {P(H3|C2)P(C2)}{P(H3|C2)P(C2)+P(H3|C1)P(C1)}}={\frac {1}{1+P(H3|C1)}}={\frac {1}{1+EQ}}$

Nijdam (talk) 15:20, 12 May 2009 (UTC)

I gather you are academic. I therefore suggest that you write something and email it to me at monty@hogbin.org Martin Hogbin (talk) 16:48, 12 May 2009 (UTC)

The above result is quite obvious when you think about it in the right way. It simply says that there is no difference between a single host with a particular value of q and several hosts with the same average value of q. Martin Hogbin (talk) 08:43, 22 May 2009 (UTC)

Well, here you have to be carefull, as Morgan's error shows. If hosts has some result w(q) depending on their parameter q, distributed as f(q), the average would be $\int w(q)f(q)dq$ and not w(Eq). Nijdam (talk) 16:20, 22 May 2009 (UTC)

Surely you have got that the wrong way round. Your formula above is w(Eq), if I understand your notation. Martin Hogbin (talk) 20:41, 22 May 2009 (UTC)

Sure is, that why your 'obvious' isn't that obvious. Nijdam (talk) 21:26, 22 May 2009 (UTC)

It is only obvious when you think of it he right way. I had this thought after after a conversation with Rick in another section. Say the car is behind door 1, this is the only time host door preference matters. Imagine one host with q=1/2. He will pick door 2 half the time and door 3 the other half. Now replace him with two hosts, one with q=1 (who will always pick door 1) and one with q=0 (who will never pick door 1) from which one is chosen randomly for each game. Door 3 will still be chosen 1/2 the time. If you could not see the hosts, there would be no way to tell that there were two hosts instead of one. In fact any combination of hosts that opened door 3 half the time would be indistinguishable from one host with q=1/2. Martin Hogbin (talk) 22:33, 22 May 2009 (UTC)

Why Morgan's error matters

It may seem that the error in the Morgan paper is only relevant to the special case of a prior uniform distribution of q but I believe that it has more important consequences for the validity of the paper.

Morgan's main error, in my opinion, is to inconsistently apply the principle of indifference. In the question that Morgan choose to consider, that asked by Whitaker, the car is originally placed behind one of the three doors by someone I will call the producer, a door is chosen by the player, and a door is opened by the host. In no case is the choice specified to be random and we are given no other information as to how the choice was made.

As we (and the player) have no information as to how the producer initially places the car, Morgan, quite reasonably, apply the principle of indifference and take it that the producer places the car randomly.

As the player has no information as to what is behind any of the doors, Morgan, quite reasonably, apply the principle of indifference and take it that the player chooses a door randomly.

As we (and the player) have no information as to how the host chooses which door to open, the principle of indifference should also be applied and we should assume the host chooses a door randomly (within the rules of the game).

With the above consistent assumptions, although it might still be argued that the probability is still conditional, there are many arguments to show that it does not matter which door the host chooses. It is a degenerate case, by which I mean it is a condition that makes no difference to the answer (probability of winning by switching), which is still 2/3.

What Morgan do is fail to apply the principle of indifference to the host choice and instead give it a parameter q. They then show that this leads to an answer other than 2/3. In order to add weight to this unexpected result they then show (wrongly) that even with a plausible prior distribution of host door choice parameter q the answer is still not 2/3. Martin Hogbin (talk) 15:04, 20 May 2009 (UTC)

Just a short comment. Whatever arguments you may use, the problem will always ask for a conditional probability to be calculated. As I far, far above remarked: as soon as an event has happened, the probabilities are conditioned. And the reliability of Morgan's paper is not jeopardised by this (minor) miscalculation. Nijdam (talk) 21:50, 20 May 2009 (UTC)

Perhaps we could drop this conditional/unconditional argument for the moment, that is not the basis of by complaint. The point is that, if you apply the principle of indifference to the host's door choice policy, the answer (probability of winning by switching) to the MHP becomes 2/3. If it makes you happy to insist that the problem is still conditional that is fine with me. I agree that there is a condition if a specific identified door is opened but it is a condition that makes no difference to the answer if the host chooses randomly . The point that I am making is that all this nonsense about q, the host door choice parameter disappears if the principle of indifference is applied to the host, conditional or not. Martin Hogbin (talk) 22:36, 20 May 2009 (UTC)

Nijdam, what is you explanation or justification for the fact that Morgan apply the principle of indifference to the action of the producer but not the host? Martin Hogbin (talk) 09:08, 22 May 2009 (UTC)

I wouldn't give much thought about this. To see the MHP as a probabilistic problem, it has to be modeled. In the original wording not all parameters are explicitly given. Yet it lies at hand to model the placing of the car as random. It doesn't bring much extra to introduce preference parameters. As for the strategy of the host, his preferences tell much about the conditional probability of winning by switching. It also makes clear the flaw in the simple reasoning. I would have done the same analyzing this problem.Nijdam (talk) 16:08, 22 May 2009 (UTC)

I have no problem with using the MHP as an example of conditional probability but if you are intending to use it in that way you should point out exactly the formulation you are using. There is nothing wrong with stating the problem with the usual rules and then adding, 'assume that the car is initially randomly placed but that he host's action is not necessarily random'. The MHP then becomes a good example of a the unexpected effect of a seemingly unimportant condition. This formulation, however, is not a natural assumption or a reasonable application of the principle of indifference and to not make the exact problem formulation that you wish to be answered clear at the start is unfair misdirection. To do what Morgan do and claim in a published paper that everyone else has got the problem wrong because they did not happen to agree with your particular unstated assumptions is not right either. Martin Hogbin (talk) 12:54, 23 May 2009 (UTC)

(continuing from the thread apparently about the same point, above) Martin - Morgan et al. analyze the same problem vos Savant did (including her clarifications in later columns, note in particular the experiment she suggested). She made it absolutely clear:

1) the car is initially randomly placed

2) the player randomly selects a door (before the host opens a door)

3) the host knows what is behind each door

4) the host always opens a door revealing a goat

5) the host always makes the offer to switch

Do you see anything in any of the Parade columns about the host's protocol for which door to open in the case the player initially selects the car? I don't. The entire set of columns is reproduced here [8]. Although they don't say it quite this directly, Morgan et al. are clearly using vos Savant's interpretation of the question; they are specifically not applying the principle of indifference to the action of the producer, but rather using vos Savant's assumptions. They acknowledge it would be possible to analyze the case where the car is not placed with equal probability behind each of the 3 doors. Why are you so insistent that this is an "error"?

The only mistake that vos Savant (who was writing for a popular general interest magazine) made was not to make clear in her initial explanation that she took the host to choose randomly. She later did make this assumption clear.

Morgan et al were writing a paper for a peer reviewed journal. They have no excuse whatever for not making their initial interpretation and assumptions perfectly clear. Nowhere in their paper do Morgan actually state that they take the car to be placed randomly, we are left to divine this for ourselves from a vague remark at the end of page 285 and a comment at the end of the paper stating that that other possibilities might be considered. To blame vos savant for Morgan's shortcomings is absurd.

Even if we conveniently assume that Morgan have incorporated vos Savant's initial stated assumptions in their problem formulation they chooses not to criticize them. They accept without comment the assumption that the car is randomly placed but then choose to take it that the host acts non-randomly, even though neither we nor the player can have any idea of the host's door opening policy.

In fact Morgan do make quite clear the question that they are attempting to answer; they (mis)quote Whitaker's original question at the start of their paper. They then restate the question later in the paper 'To avoid any confusion' . To not make clear their initial assumptions with a rationale for each is inexcusable in a scientific paper that claims to be the last word on the subject. On the other hand it is the only way to get an 'elegant solution'. Martin Hogbin (talk) 16:31, 22 May 2009 (UTC)

If this were a real game show, with the rules as above but not specifying the host's protocol in the case the player initially selects the car, I would actually be surprised if the host chose randomly in this case. The probability of winning the car after picking door 1 and seeing the host open door 3 is 1/(1+q₁) and it's 1/(1+p₁) if the host opens door 2, with q₁ + p₁ = 1 (with similar pairs of p₂/q₂ and p₃/q₃). Before going on the show I'd study the past episodes and try to determine if the host has a preference. If I had done this would you agree my chances of winning would be 1/(1+q₁)? Even if they hadn't done this, wouldn't the chances of winning for anyone who picked door 1 and sees the host open door 3 be 1/(1+q₁) (whether they realize it or not)? -- Rick Block (talk) 14:09, 22 May 2009 (UTC)

Real world advice for a prospective contestant on the show is another matter. Studying previous showings to see where the car was most often placed and what the host usually did would be a good idea. My guess would be that both the producer's and the host's door choices would be approximately random. The producer probably told a stage hand to 'place the car behind one of the doors' and the host probably made a quick arbitrary decision at the time, as he would have nothing to gain from doing anything else. The point that I am making is that both decisions were made by people in a manner we will probably never know. We have two sensible options: apply the principle of indifference to both choices and therefore take both to be random; take a more strict line that we have no information about either so we have to assume arbitrary probabilities for both. Unfortunately neither of these options makes a very interesting paper. Martin Hogbin (talk) 16:31, 22 May 2009 (UTC)

You continue to miss the entire point of the paper. It was not published in a vacuum, but in direct response to the nationwide controversy in the U.S. following the publication of vos Savant's columns. The authors were obviously using the same interpretation of the question as vos Savant - even referring to it as the "vos Savant scenario". In her third column she suggested an experiment using the words "Set up a probability trial exactly as outlined below" - if this is not clear I don't know what is. The point of the paper is that the question is conditional and that the probability of winning is therefore a function of the host's preference in the case the player initially selects the car (which is completely ignored by unconditional solutions). And, BTW, where exactly did vos Savant make it clear that she was assuming the host picked randomly (perhaps you're thinking of the letter she wrote in response to this paper). -- Rick Block (talk) 02:15, 23 May 2009 (UTC)

For some reason, you continue to defend the indefensible. Even if vos Savant had clearly stated that she took the producer to place the car randomly but the host to choose a door non-randomly and Morgan et al had stated that they were starting with these assumptions because they were those that vos Savant had used, Morgan's first obligation would have been to comment on the inconsistent assumptions and application of the principle of indifference made by vos Savant. This was a scientific paper intended for publication in a peer-reviewed journal. To attempt to blame its deficiencies on a popular media furore is bizarre.

Regarding vos Savant's statement that she took the host to choose randomly, I am referring to her statement that she took the host to be acting as the agent of chance. But vos Savant's assumptions are quite irrelevant, it is Morgan et al who professed to be writing the definitive paper on the subject; it is Morgan et al who should have made their initial assumptions clear.

In any case, regardless of how they got there, Morgan do in fact base their argument on an inconsistent application of the principle of indifference. If this principle is applied consistently their main argument disappears. Martin Hogbin (talk) 10:03, 23 May 2009 (UTC)

And, for some reason, you continue to miss the entire point of the paper which is that since the problem asks about a conditional probability an unconditional solution is not the correct approach (this point is supported by the Gillman, Grinstead and Snell, and Falk references in the article). The assumptions vos Savant made are exactly relevant because the paper is written as a criticism of her solution, which obligates them to solve the same problem she was solving. Her statement that she took the host to be acting as an agent of chance is in her response letter to the Morgan et al. article. Since we're simply repeating ourselves is there anything in particular you're trying to accomplish with this thread? If not, I'd suggest we agree to disagree. -- Rick Block (talk) 14:20, 23 May 2009 (UTC)

I really am staggered that you believe that it is alright for a peer-reviewed paper to be inconsistent in its approach to a probability problem just because it is responding to a magazine column, but if that really is your view then, yes, we will have to agree to disagree.

Once you treat the problem consistently the conditional solution gives the same answer as the unconditional one and, although the point could be made that the problem was still strictly speaking one of conditional probability, this would be a very uninteresting point and certainly one not worthy of a paper. Martin Hogbin (talk) 21:43, 23 May 2009 (UTC)

There is not such as a conditional solution and an unconditional one. It's merely a way of speaking. That's one of the big problems in the MHP: if you give the analysis in words, you easily get confused. There is the solution, and it is necessarily in terms of conditional probabilities. No principle of indifference of whatever can change this. And the paper of Morgan, for one, is to emphasize this. Nijdam (talk) 20:25, 24 May 2009 (UTC)

Let me make clear: Morgan doesn't (explicitly) mention the distribution of the placement of the car. It may be considered an omission, but in the "context" of the discussions that time, one could accept it as an implicit assumption. Nevertheless it would have been correct if Morgan had mentioned it. On the other hand it doesn't affect the issue of the paper: the conditional nature of the solution. Nijdam (talk) 20:08, 23 May 2009 (UTC)

My main criticism of Morgan is not that they do not state that they have taken the original car placement to be random, it is that they have taken the producer's actions to be random but the host's actions not to be. There is no justification for doing this, regardless of what vos Savant or anyone else did. Martin Hogbin (talk) 21:43, 23 May 2009 (UTC)

Starting from Morgan et al., it's a trivial extension to show that if the car is placed with probabilities c1, c2, c3 the chance of winning by switching if the player picks door 1 and the host opens door 3 is c2/(c2 + q*c1). The note at the end of the paper suggests this is of less interest. They could of course have expressed it in this form, although then your criticism would no doubt be that they solved a different problem than vos Savant. What is your reason for studiously ignoring Gillman who also says it's a conditional problem and the probability is 1/(1+q)? And Falk, who also says it's a conditional problem? And Grinstead and Snell, who also say it's a conditional problem? -- Rick Block (talk) 01:43, 24 May 2009 (UTC)

You seem to be trying to avoid accepting a simple error. In their published paper, Morgan et al apply the principle of indifference inconsistently. Nothing vos Savant did before the event and nothing you do after the event can change this. Morgan made a mistake. Martin Hogbin (talk) 08:45, 24 May 2009 (UTC)

(outindented) They made a mistake, in calculating the wrong posterior probabilities. They were sloppy in not mentioning the distribution of the car. But they certainly were not mistaken in varying the hosts strategy. In fact it is one of the merits of the article. Nijdam (talk) 20:32, 24 May 2009 (UTC)

Suppose that we start with Whitaker's question. Do you agree that neither the original distribution of the car nor the host's strategy is given in the question?

Assuming that you agree with the above, do you the accept that to answer the question it is therefore necessary to have some kind of rationale for dealing with the missing information? Martin Hogbin (talk) 22:35, 24 May 2009 (UTC)

Why start with Whitaker's question and not vos Savant's explicit experiment [9]:

One student plays the contestant, and another, the host. Label three paper cups #1, #2, and #3. While the contestant looks away, the host randomly hides a penny under a cup by throwing a die until a 1, 2, or 3 comes up. Next, the contestant randomly points to a cup by throwing a die the same way. Then the host purposely lifts up a losing cup from the two unchosen. Lastly, the contestant "stays" and lifts up his original cup to see if it covers the penny. Play "not switching" two hundred times and keep track of how often the contestant wins.

Clearly, the initial placement is random (as is the initial player choice). The host selection in the case the player initially picks the cup with the penny underneath is not mentioned (so could be anything). This version matches a game show where the contestant is told a car is randomly placed behind one of three doors but is not told anything about the host's strategy in the case the player initially selects the car (only that the host will always open a door and reveal a goat and make the offer to switch). The experiment as suggested measures the unconditional chance of winning, not the conditional chance given the player initially selected #1 and the host revealed #3. Unless you're arguing Whitaker's question asks the unconditional question (which I think you aren't) then I think you're agreeing vos Savant is not answering the appropriate question (which is the main point of the Morgan et al. paper). -- Rick Block (talk) 20:25, 25 May 2009 (UTC)

To answer your question. When you answer a question you start with the question, not someone else's attempt to answer it, especially if you think that attempt is flawed.

You seem determined to defend the Morgan paper at all costs. The primary purpose of the Morgan paper was not to attack vos Savant (or any of the others mentioned) but to give a correct solution to the problem that Whitaker asked, as Morgan themselves put it, 'at least partially put the problem to rest'. Surely you must agree with this?

So let me ask again, regardless of any other attempts including Morgan to answer Whitaker's question, do you the accept that it is necessary necessary to have some kind of rationale for dealing with the missing information? Martin Hogbin (talk) 21:32, 25 May 2009 (UTC)

The question is Whitaker's question, as interpreted by vos Savant - as extensively discussed following the publication of her columns. The rationale for dealing with the missing information is match the assumptions made by the popular discussions at the time (i.e. random initial car placement, host must make the offer to switch, host must reveal a goat). These are the assumptions nearly everyone made (and still makes). Morgan et al.'s contribution is the insight that the host's preference matters. Fully explicit versions (e.g. Krauss and Wang) now include the constraint that the host pick randomly if the player initially selects the car. Before Morgan et al. (and Gillman who you're still ignoring for some reason) pretty much no one specified this detail of the problem setup (although Martin Gardner's version of the Three Prisoners problem includes such a constraint). This is not a detail to be left to the principle of indifference. It is either a constraint on the host or it's not. In either case, the problem is a conditional problem and Morgan et al.'s solution is correct. -- Rick Block (talk) 22:36, 25 May 2009 (UTC)

If you can tell me where I can get a copy of the Gillman paper I would be happy to comment on it. I note, however, that it was published after Morgan and have assumed therefore that it just propagates Morgan's original error.

Are you saying that Morgan have intentionally chosen to answer the formulation where the car is placed randomly, the player chooses randomly but the host chooses non-randomly? There is nothing in the paper to indicate this. If this is indeed the case then perhaps somebody should publish a paper that actually does 'solve the problem at hand', which most people take to be Whitaker's actual question. Martin Hogbin (talk) 11:30, 26 May 2009 (UTC)

As I've mentioned before the Gillman paper is available (hard copy) in the 1992 American Mathematical Monthly volume 99, pp 3–7. First page preview is at jstor [10].

I do not have easy access to such things. I will try my local library to see if they can get a copy.

Yes, Morgan et al. intentionally chose to answer the formulation (widely discussed at the time) where the car is placed uniformly randomly, the player chooses randomly, the host's action in the case the player has initially chosen the car is not specified, and the specific probability of interest is the (conditional) probability of winning given the player has chosen door 1 and the host has opened door 3. What indicates this is the very first sentence of the Morgan et al. paper "In a trio of recent columns ..." which establishes the context as the problem vos Savant analyzed. Most people understand Whitaker's actual question is underspecified (for example doesn't say the host must make the offer to switch or that the host necessarily won't reveal the car) but make the reasonable assumptions. What most people don't understand is that because the question is about a conditional probability, the host's choice in the case the player has selected the car is critically important to the answer and using an unconditional solution technique masks this factor. -- Rick Block (talk) 13:51, 26 May 2009 (UTC)

Then perhaps somebody should attempt to answer the question that Whitaker actually asked. You say that the question is about a conditional probability, but that is only because Morgan set it up that way. If you take the host door choice to be random and you rephrase the original question as, '...the host, who knows what's behind the doors, opens another door, which has a goat', the essentially conditional nature of the problem disappears. Martin Hogbin (talk) 16:24, 26 May 2009 (UTC)

Martin, you took the words right out of my mouth.

Take the door #s out of Whitaker's question, which imho are there for example (clarity?) only. Even with the door #s, I would, and do, argue that Whitaker did not mean to ask specifically only about doors 1 and 3.

Recognize that the 'host behaviour' by definition of a game show, and by US law, cannot demonstrate any bias observable to the contestant.

Acknowledge the above statement regarding 'host behaviour' need not be stated as a premise, precisely for the 2 reasons given.

And there you have it, Morgan's paper holds no water. His statement that the unconditional solutions 'solve the wrong problem' is factually incorrect. Glkanter (talk) 19:14, 26 May 2009 (UTC)

Given the way Whitaker (and, originally, Selvin) phrased it, there is no doubt that the entire point of the question is to make the reader imagine the player is standing in front of two closed doors and one open door in a circumstance in which the location of the car is still unknown but the probability of the two doors at this point is not equal. Specific door numbers are used to make it absolutely clear that this is the problem (not to limit the result to these specific two doors, but using them as representative of any two specific doors). Martin's rephrasing is equivalent to the player knowing the rules of the game but picking whether to switch or stay before the host opens a door, which misses the entire point of the problem as does any unconditional so-called "solution". The problem is a conditional probability problem where the probability of interest is the probability given that the host has already opened a door, as opposed to the probability knowing that the host will open a door. That neither of you can apparently see that these are completely different questions does not alter that fact that they are indeed different questions. -- Rick Block (talk) 01:12, 27 May 2009 (UTC)

Rick, do you really believe that I do not know the difference between the conditional and unconditional formulations of this problem?

The point is that Whitaker's actual words are ambiguous. Morgan slightly changed them to make the question clearly conditional, I slightly changed them to make the problem unconditional. Why is their interpretation better than mine?

Of course, we can use our real world knowledge to imagine the likely scenario to which the question refers, but that is not what Morgan do, they refer to the 'stated conditional problem' and 'information not given in the problem'. It is quite clear that Morgan claim to be answering Whitaker's stated question. Once we move into the real world, we need to ask ourselves what the state of knowledge of the various characters is. As Glkanter has repeatedly stated, the player has no knowledge of the host's strategy and we must therefore take it that they are equally likely to open any legal door. Martin Hogbin (talk) 09:02, 27 May 2009 (UTC)

There is too much talking about conditional and unconditional. Rick has made it completely clear, but if anyone wants to know: unconditional is only the placement of the car: P(C1)=P(C2)=P(C3)=1/3 (or any other distribution if you like), the choice of the player P(H1), P(H2) and P(H3) (uniform??), independent (?) of the position of the car. What is here of interest? Nijdam (talk) 10:08, 27 May 2009 (UTC)

What is of interest is the way that Whitaker's original question is converted into a precise problem formulation. Apart from the game rules, which we all agree on, there are several other matters that must be decided before the question can be answered. The final problem formulation, and therefore the correct method of solution, depends on how we decide to deal with the information that is not provided in the original question. I would refer you again to the wise Prof Seymann's comments at the end of the Morgan paper.

If these matters are not of interest to you that is fine, perhaps you are only interested in the mathematics of the problem after it has been unambiguously formulated, but I am personally interested in the complete process from initial problem statement to final mathematical solution. Martin Hogbin (talk) 15:47, 27 May 2009 (UTC)

And, do you not agree that the paradox is that you're standing in front of two closed doors, don't exactly know where the car is, but the probability of the two doors you're looking at aren't equal? What Nijdam is saying is if you don't make it conditional, you're just putting a whole lot of extra words around "what is your chance of picking a car from behind three doors" - which is hardly paradoxical at all. -- Rick Block (talk) 01:22, 28 May 2009 (UTC)

I think you have completely misunderstood the MHP. Even when unambiguously stated as an unconditional problem, most people get it wrong. But this is a very interesting question that is very relevant to the article. What exactly is the Monty Hall problem. I think it is worth a new section just to at least understand how different people see it.

Okay, Martin, you made your position clear. But ... why al this criticism of Morgan. Morgan is, like me, mainly interested in the mathematical well-formed problem. And actually this well-formed problem is mostly addressed as the MHP. So if you want to follow the evolution of the problem, fine, but let's skip the mathematical discussion for the time being. Nijdam (talk) 11:38, 28 May 2009 (UTC)

I criticize Morgan because they set themselves up as the last word on the subject. That is to say, in my opinion, they claim to be providing the definitive solution to Whitaker's question, both the formulation and the maths. In my opinion they make serious errors in both. Martin Hogbin (talk) 18:34, 29 May 2009 (UTC)

Winning by switching

May be this is a good moment to comment on the "event" 'winning by switching'. Not everyone who uses this term, uses it in the right way. The following reasoning, that some people use, is wrong: initially the player has probability 1/3 to get the car. Hence if she switches the probability increases to 2/3. In this way it is equivalent to the "simple solution". Swichting can only be judged after a door has been opened by the host, leading to the known calculation of the conditional probabilities. Nijdam (talk) 11:54, 28 May 2009 (UTC)

Please explain why your opinion is that this is wrong. Perhaps you could critique the 'Combining Doors' solution, which assigns the value of the selected door at the outset as 1/3, then shows the value is still 1/3 after a door is opened. It would be even more valuable if you could provide a reference in addition to your OR. Glkanter (talk) 12:15, 28 May 2009 (UTC)

I've explained this to you before, many, many times so I won't try again. Perhaps you'll believe Nijdam, who (like Boris Tsirelson and C S) claims to be a professor of mathematics (Nijdam: care to be more specific about your real life identity? Glkanter at least claims to understand that math professors know more about math than he does). BTW - Morgan et al. is the reference you're asking for. As an unconditional solution, "combining doors" is in the general family of solutions they criticize. It is equivalent to their F1: If, regardless of the host's action, the player's strategy is to never switch, she will obviously win the car 1/3 of the time. Hence the probability that she wins if does does not switch is 2/3. -- Rick Block (talk) 13:44, 28 May 2009 (UTC)

Nijdam appears to have many skills. Ventriloquism among them. Glkanter (talk) 15:27, 28 May 2009 (UTC)

That's the least. Nijdam (talk) 18:50, 28 May 2009 (UTC)

Ah, yes, not to forget C S. Here's a complete paragraph he posted:

"Actually, I don't believe this material is over any reasonably intelligent adult's head. So let's give it a go. Let's address why people might think Marilyn's statement is conditional. By the way, I hope I haven't given the impression that I think the problem must be understood as conditional. As for Marilyn's exact statement, I think it's ambiguous enough on several fronts, and there's no reason to state with absolute confidence it should be conditional."

Of course, one is encouraged to read his entire response, and all his others, in archive 9 of the talk page. Glkanter (talk) 17:35, 28 May 2009 (UTC)

Here is another quote of C S. from your talk page: Since you like to misrepresent Boris' quotes, perhaps we could press him on whether the turbulence is caused by your refusal/inability to understand basic probability. --C S (talk) 05:12, 15 February 2009 (UTC)Nijdam (talk) 11:56, 29 May 2009 (UTC)

Yeah, that's been out there. Please provide where I actually mis-quoted Boris. Because it didn't happen. Maybe you could focus instead on the questions at hand. Glkanter (talk) 12:07, 29 May 2009 (UTC)

Morganian Fallacy

The argument for the continued emphasis on Morgan's paper (and those written by 2 or 3 others) has been primarily that they are published, reliable sources that are (allegedly) peer reviewed. Any logical counter arguments are therefore wasted effort.

This is a dangerous conceit, and should be rejected.

I could provide links to official US Government documents written by highly educated men and women that have been reviewed by the highest members of government. These documents would indicate that waterboarding suspected terrorists is not torture, and is therefore legal. There is great dispute about these papers and their conclusions. Fortunately, thinking, rational people are engaged in debating the merits of this proposition. Simply being written does not make it so.

I could provide a link to the US Constitution, in which slavery in certain of the 13 original states is allowed. Again, highly intelligent people wrote and reviewed this document. This same document denies women the right to vote.

Martin has pointed out an error in the Morgan paper. Once corrected, if I understand correctly, the formula determines that the likelihood of winning by switching is 2/3, rather than some other result. This is the same outcome as the unconditional solution, which is no coincidence.

I'd like to point out a logical fallacy in the Morgan paper. That is that the host may provide information as to the car's location based on which goat door he reveals. This contradicts the very notion of a game show. And in the US, is illegal. The producers of the show are obligated not to reveal any information. So, the probability of this happening is 0.

Morgan's paper was written by human beings, and as such, is subject to error. To blindly accept these errors as gospel only compounds these errors. Many papers have been written subsequent to Morgan's that solve the MHP using an unconditional solution. Devlin is listed in the article's references as one, there are countless more.

But, the Morganians discount these sources for various specious reasons. Essentially arguing of their sources, "Mine is bigger."

I'd still like to know a logical argument explaining how the original 1/3 likelihood of selecting a car is dependent on which goat door Monty opens. The strategy is to always switch. When you (the contestant) select a goat, this works 100% of the time. When you select the car, this works 0% of the time, regardless of which goat Monty reveals. As you will select a goat 2/3 of the time, the strategy works 2/3 of the time. Glkanter (talk) 08:11, 26 May 2009 (UTC)

As I have explained before, it is a conjuring trick. Just like the rabbit that the magician pulls out of a hat is a real rabbit, Morgan's solution is a real solution. Just like it is no use checking out the rabbit to see how the magic trick is done, it is no use arguing about Morgan's solution to find out the problem. Like many good magic tricks, it is accomplished before the main action starts.

The main bit of trickery is what I have been discussing above. In Whitaker's statement there is no information given about the way that the producer initially places the car, the way the player chooses a door, or the way that the host chooses a door. To formulate the problem we need decide how we will deal with this lack of information. What Morgan do, possibly unwittingly, is to take it that the car is placed randomly, the player chooses randomly, but the host chooses non-randomly. There is no justification whatsoever in Whitaker's original question for this assumption although it is possible that others had previously made the first two assumptions and Morgan merely followed suit. Once this decision has been made, all that is needed is a little sleight of hand and, hey presto, you have an 'elegant solution'.

I would be happy to explain why Morgan's solution is correct once you have set the trick up, as understanding this properly helps you see how limited their solution really is. Martin Hogbin (talk) 11:09, 26 May 2009 (UTC)

Oh, no, the canard occurs when Morgan says the unconditional solution is false. First, they offer a 'definitive' interpretation of what Whitaker is asking. As if somehow, they alone can divine this. Then they dismiss solutions like (but not exactly) the 'Combining Doors' solution by saying something like 'it answers the wrong question'. What hogwash! The MHP itself is a trick, as someone else posted a couple of months ago. By opening a door, the reader is lured into the 50/50 trap. Nothing has changed, (it is still 1/3 for the door the contestant selected, right?), but it 'appears' substantially different. That's why it's the world's most famous paradox.

That was the sleight of hand that I was referring to. Morgan do what I say above but also change the question asked in two stages to one that, given the non-random host door choice they have assumed, must be treated conditionally. Martin Hogbin (talk) 15:58, 26 May 2009 (UTC)

Good luck with your Morgan error effort. I'm doubtful that you'll get any 'acceptable' experts to help you out. My guess is that such experts studiously avoid the MHP (the Wikipedia Math Braintrust sure does!), and specifically Morgan. What is there to gain for someone with 'standing' to get involved in this quagmire? Heck, we all know we're wasting our time trying to fix this insignificant Wikipedia article. But, I've been at it for over 7 months now, and you, even longer. Of course, we're all essentially 'anonymous', and have no real 'standing' to lose. No, I can see no benefit to any Professor in either taking on Morgan, or proving his rantings invalid. To me, that would be like challenging some 'expert' who claims the sun rises in the west. Why even bother? Glkanter (talk) 13:34, 26 May 2009 (UTC)

Well at least no one seems to be arguing that Nijdam and I are wrong. Martin Hogbin (talk)

While we're out fishing, why don't we look at the MHP FAQs and the Bayes' Theorem article.

The FAQs are pretty much one person's personal POV.

The Bayes' Theorem page is interesting, in that it has a MHP example. http://en.wikipedia.org/wiki/Bayes%27_theorem#Example_3:_The_Monty_Hall_problem

I'm no expert, so could someone tell me what this means:

In the situation where the prize is behind the red door, the presenter is free to pick between the green or the blue door at random. Thus, P(B | Ar) = 1 / 2

Somehow, the doors are colored, rather than numbered. But what's that 1 / 2 stuff all about? I thought we've been told that it's not 1 / 2? And certainly not random!

Here's the conclusion:

Note how this depends on the value of P(B). Another way of looking at the apparent inconsistency is that, when you chose the first door, you had a 1/3 chance of being right. When the second door was removed from the list of possibilities, the conditional probability for the chosen door to be the winning door is also 1/3 and hence this left the last door with a 2/3 chance of being right.

And here's how it was before Nijdam's edit:

Note how this depends on the value of P(B). Another way of looking at the apparent inconsistency is that, when you chose the first door, you had a 1/3 chance of being right. When the second door was removed from the list of possibilities, this left the last door with a 2/3 chance of being right.

I think the word 'still' was intended rather than 'also' in Nijdam's edit. Is this correct?

[Nijdam]Definitely not. I mean 'also', because 'still' suggests it is the same probability, and it is not. Nijdam (talk) 18:16, 29 May 2009 (UTC)

At what point did it change from 1/3, and what did it change to? Then, when did it change back? Please provide a reference. Thank you. Glkanter (talk) 18:30, 29 May 2009 (UTC)

Apparently at a point far beyond your reach. I have explained this a dozen or more times, and won't repeat it any more. Nijdam (talk) 10:56, 31 May 2009 (UTC)

Still unwilling to provide a thoughtful response, Nijdam? Glkanter (talk) 09:31, 4 June 2009 (UTC)

Help a guy out. Can you provide a link to one of those diffs? Thanks. Glkanter (talk) 11:11, 31 May 2009 (UTC)

So, is the 'Combining Doors' (with a meaningless, superfluous 'conditional probability' thrown in there) solution good enough for the Bayes' Theorem article, but not the MHP article? Glkanter (talk) 12:35, 29 May 2009 (UTC)

The MHP article is a featured article, the Bayes' Theorem article is not. The line you quote

In the situation where the prize is behind the red door, the presenter is free to pick between the green or the blue door at random. Thus, P(B | Ar) = 1 / 2

is an assumption not present in this version of the problem statement (or the one presented in Parade) that many people think is reasonable but if it's not given in the problem statement it shouldn't be used in the computation of the conditional probability. Assuming instead that this is an unknown probability q leads to the 1/(1+q) solution. -- Rick Block (talk) 14:06, 29 May 2009 (UTC)

You mean just like the assumption that the car was initially randomly placed or that the player picks randomly? Above all we should be consistent in our assumptions. Martin Hogbin (talk) 17:23, 29 May 2009 (UTC)

How about consistency between Wikipedia articles? It seems downright unfair that the MHP gets all the Morganian's attention, but the Bayes' Theorem readers remain unenlightened. But then, the Bayes' Theorem article wasn't personally 'shepherded' through the FAR process.

So, I guess it's valid to throw a 'condition probability' phrase in there, but not bother addressing any of the other obvious issues a Morganian would have with it. Glkanter (talk) 18:07, 29 May 2009 (UTC)

I can't exactly put my finger on it, but the Morganian's resistance to answering simple and relevant questions brings to mind this link, plus the 4 days that follow it. http://www.doonesbury.com/strip/dailydose/index.html?uc_full_date=20090427 Glkanter (talk) 23:58, 29 May 2009 (UTC)

What exactly is the real Monty Hall problem?

Before we start arguing about who is right, which I am sure that we will, can we just make sure what people actually think is the real MHP. Let me give some options. Please agree or disagree or add your own options to clarify your opinion. Normal game rules assumed.

1 The unambiguously worded problem treated unconditionally. In other words, what is the best player strategy, decided in advance. Although this should be a very simple probability problem, most people still get it wrong.

2 The unambiguously worded problem where the host chooses randomly, thus the issue of conditional probability is ignored or considered irrelevant. The player decides after a door has been opened but it is known that the door chosen by the host makes no difference to the result. Although this should be a very simple probability problem, most people still get it wrong.

2a The problem where the host door policy is not stated but the issue of conditional probability is ignored. The player decides after a door has been opened but it is assumed that the door chosen by the host makes no difference to the result. Although this should be a very simple probability problem, most people still get it wrong.

3 The problem described by Morgan et al, where the host is assumed to act non-randomly, thus the door opened by the host could affect the probability of winning by switching. The confusion arises because people do not realize that the host door opening policy is important. Were it not for this fact most people would get the answer correct. Martin Hogbin (talk) 19:27, 29 May 2009 (UTC)

4 Same problem as your #2, but entirely different "thus": The unambiguously worded problem where the host chooses randomly (if the player initially picks the car), thus the conditional solution is forced to be the same as the unconditional solution. The unconditional solution produces the correct answer, but it needs an argument for why it is correct involving the constraint that the host pick randomly. Any solution that does not mention this constraint is incomplete, since the solution then applies to versions of the problem without this constraint (and is then incorrect). Although this is a very simple probability problem, nearly all people (including most of those who think they understand it because they know one of the unconditional solutions) get it wrong. -- Rick Block (talk) 00:33, 30 May 2009 (UTC)

Thanks, I understand your position now. Martin Hogbin (talk) 10:26, 30 May 2009 (UTC)

Due to the popularity of the Parade version, there's a reasonable argument in favor of #3 as well. Most people get this wrong, and most explanations are incomplete (per above). -- Rick Block (talk) 00:38, 30 May 2009 (UTC)

As you know, I see no justification for the inconsistent application of the principle of indifference to the Parade problem. Martin Hogbin (talk) 10:26, 30 May 2009 (UTC)

Martin, regarding #3, how do you square it with the fact that Monty is prohibited from 'telegraphing' any bias to the contestant? Remember: "Suppose you're on a game show". #3 can't exist. Nor can any so-called 'variants', for the same reason. Glkanter (talk) 00:53, 30 May 2009 (UTC)

As I said above, I am trying initially just out find out what people think rather than to criticize or argue. I will answer your question in another section below. Martin Hogbin (talk) 10:26, 30 May 2009 (UTC)

I would go with #1 or either of the #2s. As there can be no new knowledge gained when Monty opens a door, there is no statistical (logical?) difference among them. You didn't list it, but I reject that Whitaker (or anyone who has ever been associated with the MHP) was interested in doors #1 and #3 to the exclusion of any other combination. This may be a nitpick, but I don't agree/understand the "most people" portion of choice #3. "Most people", including 1,000 PHDs got it wrong, hung up on the 50/50 trap. "Most people", to paraphrase Freud, think a goat is just a goat. They have never heard of Morgan, or the "equal goat door constraint", and there lives are no less rich for this gift. Glkanter (talk) 14:55, 30 May 2009 (UTC)

Some philosophical issues

Glkanter, this is to show how answer 3 in the section above can be right.

There are several definitions of probability. Below are some edited extracts from relevant WP articles to show the principles of each. (Note for Rick: yes, this is the same old host chooses randomly argument).

Classical definition: Initially the probability of an event to occur was defined as number of cases favorable for the event, over the number of total outcomes possible in an equiprobable sample space.

1. Frequentists talk about probabilities only when dealing with experiments that are random and well-defined. The probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment. Frequentists consider probability to be the relative frequency "in the long run" of outcomes.

2. Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, given the evidence.

Modern definition: The modern definition starts with a set called the sample space, which relates to the set of all possible outcomes in classical sense... It is then assumed that for each element [in the sample space] an intrinsic "probability" value, is attached...

I do not believe that any of the statements above are contentions. It is certainly not my intention that that should be the case.

In all cases there are two philosophical issues that must be addressed before we can come to any conclusion. The first is how to formulate the problem and the second is how to interpret the answer once we have done our calculations. The different definitions can treat these issues differently.

Consider this (badly posed) example:

A bag contains 20 red balls and 80 white ones. You, who know only that the bag contains a mixture of red and white balls, pick a ball at random from the bag, what is the probability that the ball picked is red?

a) Using the modern definition we define our sample space, based on the information given in the question, and arrive at the answer 1/5 as the ball is defined to be taken at random from this sample.

b) On the other hand, you do not know the ratio of red to white balls in the bag. From your state of knowledge (Bayesian perspective) the probability of picking a red ball is 1/2 as you have no more reason to believe that you will pick a red ball than a white ball. (Of course, from our perspective, the answer is the same as above)

So we have two answers to the same question. This paradox is resolved when we consider what the two answers mean. The first answer is to the question, 'given 20 red and 80 white balls ... what is the probability...'. The second answer is to the question, 'from your perspective, as you have no idea how many of each colour there is... what is the probability...'. Not that hard to understand, so what is the point of all this? The point is that it is no use thumping the table and saying answer a) or b) must be right; there is no right answer to the question until you have decided what it means. Martin Hogbin (talk) 16:49, 30 May 2009 (UTC)

Martin, I appreciate your efforts to educate me. As you may have already inferred, I am a little slow on the absorption of this new information. Help me narrow down your alternatives as it applies to the MvS recitation of the MHP, specifically, "Suppose you're on a game show". So, this excludes the knowledge of the producer, at one extreme, but does recognize that the contestant knows the 'history' of the two remaining doors.

Is this a narrow view point? Absolutely. I understand that the unconditional solution does not generalize to other permutations of the puzzle. Why would I expect it to? For purposes of improving the article, I'm only interested in the most efficient solution to the world's most famous paradox, not a comprehensive understanding of probability theory. Which I would be happy to engage in at the appropriate place and time. Glkanter (talk) 17:21, 30 May 2009 (UTC)

Don't forget, I agree with most of what you say. "Suppose you're on a game show" is an important part of the MH question and it does indeed imply that the question should be answered from the point of view (state of knowledge) of the contestant. But I am trying to show that there is another way of looking at it. What would be your answer to my question above? Martin Hogbin (talk) 17:45, 30 May 2009 (UTC)

As far as the balls go, I guess I would have to say 'Hell, I don't know'. Or, 'something between 0% and 100% inclusive.' But the guy who placed the balls in the bag would know the precise answer. But the MHP starts: "Suppose you are on a game show'. And then I'm (the contestant) provided the 5 premises that you, me, Rick and others have previously agreed upon.

But, there is an assumption that you're making. And that is that those are the only techniques available to derive an answer. I'll be consistent and refer to the 'Combining Doors' solution. I think it's more "Logic' than 'Probability'. (Otherwise, please help me understand which definition the 'Combining Doors' falls under.) Equally valid approach. And it renders moot the entire discussion of host behaviour, etc. Maybe the first sentence of the article: "The Monty Hall problem is a probability puzzle..." itself points us in the wrong direction.

I don't agree with your comment that while this is a game show problem, there is another way of looking at it. For purposes of this puzzle, I don't think these ambiguities exist. And I'm still waiting for someone, anyone, to demonstrate any failings in the 'Combining Doors' solution, other than 'Morgan calls it false'. Glkanter (talk) 18:42, 30 May 2009 (UTC)

The essential point is whether the probability is the average probability across all initial player picks and doors the host happens to open (this is the "unconditional" probability) or the probability that applies when we know which door the player has picked and which door the host has opened (the "conditional" probability). From the Parade wording of the problem statement, the latter is what we should be talking about. The use of door 1 and door 3 does NOT mean we're only interested in the case where the player picked door 1 and the host opened door 3, but that we're interested in a SPECIFIC case where we know what door the player has picked and what door the host has opened (focusing on door 1 and door 3 should end up with the same answer as any other door the player picked and door the host opened).

From Martin's description of a frequentist, the difference is whether we watch the show (of course, there never was a show quite like this) and count only 4 things (player switches and wins, player switches and loses, player stays and wins, player stays and loses) or we watch the show and keep track of these four things for each initial player pick and door the host opens (i.e. we keep track of 24 things). If we keep track of all 24 things, then if asked by a player who has picked door 1 (or any other door) and has seen the host open door 3 (or any other door) what are her chances of winning by switching, we can look at our data samples and produce an answer specific to this player's condition (the "conditional" probability).

An unconditional solution (e.g. combining doors) answers the first question (average across all players and any door the host opens), not the question specific to any individual player's case. If the car is initially randomly located, and if the host chooses randomly when the player initially selects the car, the answer must be the same - but exactly why these conditions force this to be true is not exactly obvious.

Since the question seems to be asking about a specific case (door 1 and door 3), responding with the overall average doesn't really answer the question. It's as if someone asks what a baseball player's batting average is when playing against the Yankees and the answer provided is the player's overall batting average. These might be the same, but they might not.

Back to MHP, what you're saying is because it's a game show the conditional probability must (well, at least should) be the same as the unconditional. I'm not going to disagree with this, but if you only solve the problem unconditionally you really can't tell whether the show is "honest" or not. Placing the car randomly at the beginning (and having the host know where the car is, open a door showing a goat, and always make the offer to switch) is not sufficient to guarantee every player has the same odds. You also have to make sure the host picks randomly if the player initially selects the car. The unconditional solution doesn't make this clear, because it averages out the host's preference across all initial player picks and door the host opens. If you worked for the FCC and wanted to make sure the show was honest, you'd have to record the results for each pair of player pick and door the host opens. What it means for the show to be honest is that each of these will be the same (and they'll each be 2/3).

What Morgan et al. show is that each of the conditional probabilities (assuming the car is randomly placed) is of the form 1/(1+q) where q is the host's preference when the player initially selects the car. If you don't solve the problem as a conditional probability problem, you don't discover this. If you're working for the FCC and you want to make sure the show is honest, you need to check 6 things: is the car randomly placed, is the initial player pick random (if the car is randomly placed, this doesn't actually matter), does the host always open a door showing a goat (and, thus, knows where the car is), does the host always make the offer to switch, does the player get to decide to switch after seeing which door the host opens, AND does the host open a random door when the player initially picks the car. If this last point is not satisfied, different players can have different odds even though the overall odds will still be 2/3. -- Rick Block (talk) 20:12, 30 May 2009 (UTC)

You just wrote that if the host doesn't pick randomly when the contestant picks the car, he would be in violation of the FCC. Hence, it need not be stated outright as a premise, but it is, indeed a premise of the MHP. Glkanter (talk) 21:15, 30 May 2009 (UTC)

You both seem to be determined to stick to your standard lines on the MHP, I am trying to establish some philosophical points that might help to reach some sort of agreement.

Glkanter, you keep saying that the player does not know about the host's policy. I know that. In my question, you say that there is no answer, but the question asks for 'the probability that the ball picked is red' and we know that 1/5 of the balls are red. Surely you must see the argument for saying that the answer to the question is 1/5.

Suppose the question told us that the balls are all white, could there be any answer other than 0?

Rick, can we please forget conditional/unconditional for the moment, I will come to that later. Do you agree that, from Bayesian perspective of someone who does not know the proportions of the balls, the probability is 1/2?

Martin Hogbin (talk) 21:46, 30 May 2009 (UTC)

Martin - In a strict sense I agree, but I don't see what your question has to do with the problem at hand. We're NOT asked what does the player think her probability is, but what IS her probability given the problem statement (in the context of your problem the answer would be 1/5 if we're given the distribution of balls). If in your problem we're told there are red and white balls in the bag but not how many of each, 1/2 is only a weakly correct answer. A better and much more useful answer is R/(R+W). -- Rick Block (talk) 22:07, 30 May 2009 (UTC)

That R(R+W), where R and W are both completely unknown, is a better answer than 1/2 is debatable. In fact it can be argued that it is not an answer at all, having no defined numerical value. It is no better than saying 'I have no idea'. Supposing that you wer pressed to give your best estimate numerical value, it could only be 1/2.

If you wanted to do things the hard way, you could take your formula and using a noninformative Bayesian prior calculate the probability to be 1/2. Martin Hogbin (talk) 08:53, 31 May 2009 (UTC)

Can we leave the MHP for the moment, it is all a matter of how you interpret the question. which I will move on to later.

Martin, I answered 'I don't know' on the understanding that I hadn't been told the distribution of the red and black ball in the bag. (Re-read your paragraph, it's tricky.) I'll stand by that. I'm a 'premises' guy. If you told me that it was 20/80, then I could confidently answer 1/5. So I focus on the premises of the MHP contestant only. I, for one, think the ambiguity of the 'random goat door constraint' has been eliminated. That the host picks randomly is a premise. But I'm more interested in your response to the questions I posed above. Where does the 'Combining Doors' solution fit in your narrative? Thanks. Glkanter (talk) 23:39, 30 May 2009 (UTC)

In my (deliberately ambiguous) question you are clearly told that there are 20 red and 80 white balls. Why is the actual probability of picking a red (rather then your estimate from a position of ignorance) not 1/5? Martin Hogbin (talk) 08:53, 31 May 2009 (UTC)

Glkanter, do you see my point, the answer depends on who is being asked the question. Martin Hogbin (talk) 13:47, 1 June 2009 (UTC)

Yes, I do. I got confused by this portion of your question:

"You, who know only that the bag contains a mixture of red and white balls, pick a ball at random from the bag..."

So I answered, "I don't know". But my point is, in the MHP, it's the contestant being asked the question. Of this, there can be no doubt. As I said, ask the producer, and you'll know with 100% certainty where the car is.

By the way, I don't think I would ever answer 50%. This is just my opinion, but I can't see 'indifference' making that assumption for a bag with an unknown quantity of balls. (Maybe, 50% of the time I would guess each color, but that's not the question you asked). For contrast, with 100 doors and one car, I can see 1% from 'indifference'.

Please leave the MHP out of it for the moment. Both you and Rick seem to keep going back to it. At the moment I am trying to make some general points and will then go back to the MHP when we have agreed them. Let me rephrase my question. A person picks a ball from a bag containing 20 red and 80 white balls. What is the probability that they will pick a red ball? Martin Hogbin (talk) 18:04, 1 June 2009 (UTC)

As I have been informed of the distribution of the balls in the bag, I am comfortable saying there is a 1 in 5 chance that the first ball chosen randomly from that bag will be red. Glkanter (talk) 18:30, 1 June 2009 (UTC)

What if the person picking the ball does not know the distribution of balls in the bag? Martin Hogbin (talk) 19:49, 1 June 2009 (UTC)

As I don't believe the picking person's knowledge state affects his random selection of balls from a bag, I'll stick with 1/5. But you're scaring me Martin. I can't tell yet where this is all going. Glkanter (talk) 20:31, 1 June 2009 (UTC)

Will you answer my question? Where does the 'Combining Doors' solution fit in your Probability narrative? Glkanter (talk) 14:28, 1 June 2009 (UTC)

I will answer that in a new section below. Martin Hogbin (talk) 18:04, 1 June 2009 (UTC)

About Martin's problem. If I'm just shown a bag with 100 balls, red and white, without any further info, and I'm asked to tell the probability a random drawn ball is red, I really wouldn't know. Also a Bayesian will not say 1/2. He would model the situation with a parameter R, the number of red balls. However instead of merely saying: "I don't know the value of R", he would assume a prior distribution of R, to be used to improve his knowledge of R by calculating the posterior distribution after one or more drawings have been done. Of course he may compute the prior expected value of R if you want to know. That's just calculus. Nijdam (talk) 18:32, 31 May 2009 (UTC)

Nijdam, are you saying that you would never estimate the probability as 1/2 in such circumstances? Martin Hogbin (talk) 13:47, 1 June 2009 (UTC)

Well, I wouldn't speak of estimating. No info from a sample is available. You are merely asking about a good guess. I put some euro coins in a bag. How many did I put in? Nijdam (talk) 17:42, 1 June 2009 (UTC)

As I am sure you must know that is not the same type of question as there is no upper limit on the number. A bag contains only red and white balls, you pick one, what is the probability that it will be red? Do you assert that the only answer that can be given to this question is 'unknown' (or maybe a formula such as R/(R+W) where R and W are both unknown)? Martin Hogbin (talk) 18:04, 1 June 2009 (UTC)

You have gone quiet, Nijdam, perhaps you can see where I am going. Martin Hogbin (talk) 08:20, 2 June 2009 (UTC)

For the relevant issue it is the same type. The answer 'unknown' or R/100 or R/(R+W), would be in my question: 'unknown' or N (=the number I put in). In the end all answers mean: 'unknown'. Nijdam (talk) 08:52, 2 June 2009 (UTC)

[Outdent] How about this question then. A car is placed behind one of three doors numbered 1 to 3. What is the probability that it is behind door 1? Martin Hogbin (talk) 11:37, 2 June 2009 (UTC)

Strictly speaking, and in line with the above: I don't know the probability. But similar as when you say: I throw a die, instead of: I throw a fair die, often randomness is implicitly assumed. Nijdam (talk) 17:15, 2 June 2009 (UTC)

Nijdam, is randomness of the host's behaviour implicitly assumed when you are talking about a game show problem, from the contestant's point of view, in which it would be illegal and illogical for the host to provide any information to the contestant as to the location of a car when faced with a choice of 2 doors? If no, why not? Glkanter (talk) 23:52, 2 June 2009 (UTC)

It was hard to make you assume randomness when I was talking about the balls. The one thing that is needed is consistency of approach. Martin Hogbin (talk) 22:22, 2 June 2009 (UTC)

There is quite a difference between the two situations. To compare them: with the car the experiment may be considered as the choice of 1 out of 3 doors. With the balls, the experiment is the choice of 1 out of 100 balls, but with unknown distribution of the colors. Even if you wouldn't explicitly say "draw random", mostly this is assumed, but the big difference is the unknown number of red balls. Nijdam (talk) 13:38, 4 June 2009 (UTC)

Nijdam, do you see my point now? There are three door choices made in the MHP, the producer's choice of door to place the car behind, the player's choice of door, and the host's choice of door. None of these is defined in the question to be random. There are only two consistent ways to treat the problem, treat them all as random or treat them all as non-random. What is your justification for treating the host choice differently from the others? Martin Hogbin (talk) 09:48, 3 June 2009 (UTC)

I wouldn't speak of "consistent way" to treat the problem. The placement of the car to be random seems quite natural. The distribution of the initial choice of the player is unimportant for the problem and is not specifically considered to be random. The strategy of the host is considered to be random in the MHP and doesn't take away the necessity of the conditional probability to be considered. Only for the sake of showing the dependency of the host's strategy it is varied, as any mathematician would have done. So, I'm sorry, I do not see your point. Nijdam (talk) 13:38, 4 June 2009 (UTC)

Saying, 'The placement of the car seems natural' is no way to deal with a probability problem. We must answer the question based on the information given in the question. What may seem natural to you may not seem natural to anyone else. If we are to apply the principle of indifference to the producer's door choice we must apply it to the host's door choice. I agree that, as it turns out, the player's choice is unimportant if the car is placed randomly (but not otherwise). Why would a mathematician not show how the producer's choice is important also? Only because this is obvious and does not lead to an 'elegant solution'.

I really do not understand you Nijdam, you seem like someone who likes to do things properly yet you are content to approach this problem in a thoroughly sloppy manner. This is a highly contentions probability problem and we therfore must approach its formulation in a consistent manner. Saying 'this seems natural' but 'that does not' is simply not good enough, especially as it has no basis in the real world. Martin Hogbin (talk) 19:35, 5 June 2009 (UTC)

Why the combining doors solution fails if the host chooses non-randomly

This section is to answer a question put by Glkanter in the section above.

Note that, in my opinion at least, the host must choose non-randomly for this argument to be valid. In other words when the host has a choice of which door to open, for example when the player chooses door 1 and the car is behind door 1, the host has a preference for opening one particular unchosen door. An extreme case would be that the host always choses door 2 when possible.

The questions of whether in the real MHP the host chooses randomly or not, and whether we are assumed to know that, and what difference that might make, are being addressed in the section above. This section is strictly concerned with the case where the host does in fact have a preference for one door.

Before I go any further, Glkanter, do you accept what I have said so far? Martin Hogbin (talk) 19:46, 1 June 2009 (UTC)

Thanks for asking, as I do not agree with the implied conclusion, to wit: By showing other puzzles have different results (actually, I'll even add in 'the same results'), the 'Combining Doors' solution is disproved. I believe the only appropriate method to disprove the 'Combining Doors' solution would be to show that:

with the 5 previously agreed premises

+ what I have demonstrated is an unstated, but existing premise: any host behaviour cannot impart information to the contestant as to the location of the car

+ the Combining Doors solution

there can be results other than: always switching results in winning 2/3 of the time. Glkanter (talk) 20:25, 1 June 2009 (UTC)

As discussed in the section above, depending on how you read the question, it may not matter what the contestant knows. Martin Hogbin (talk) 22:30, 1 June 2009 (UTC)

By the way, the question was 'Where does the 'Combining Door' solution fit into the Probability narrative?' It's multiple choice! Glkanter (talk) 20:35, 1 June 2009 (UTC)

That is what I am trying to answer. But as I have said, the combining doors solution only fails if you accept that the host acts non-randomly. Whether that is a reasonable assumption is another matter. Martin Hogbin (talk) 22:30, 1 June 2009 (UTC)

If you're asked what a player's batting average is against the Yankees and you say "divide the player's total hits by total at bats" is this a legitimate solution? Or does it answer a different question? Does it answer the question if you then say "well, only in cases where it's right"? -- Rick Block (talk) 23:56, 1 June 2009 (UTC)

Rick, I understand what you are saying but I am trying to address one issue at a time. For the moment let me retract the word 'only' from my statement above. In this section I am now making only this assertion:

If the host chooses non-randomly, the combining doors solution fails.

I am not, in this section, making either of the following assertions:

That the host does choose non-randomly.

The combining doors solution does not fail if the host chooses randomly.

Of course, both of the above statements do need to be addressed, but can we agree one thing at a time. Does everyone agree with my first statement? Martin Hogbin (talk) 08:18, 2 June 2009 (UTC)

Martin Hogbin (talk) 08:18, 2 June 2009 (UTC)

The 'combining doors idea' always fails (for the given problem). When the host acts 'randomly' it produces the number 2/3, but so does the division 4/6 or the addition 1/3+1/3. The flaw, as Rick and I have stipulated over and over, lies in the arguments. Nijdam (talk) 08:57, 2 June 2009 (UTC)

With the host acting 'randomly', will it ever produce a result other than 2/3? Glkanter (talk) 09:08, 2 June 2009 (UTC)

No. Martin Hogbin (talk) 17:45, 2 June 2009 (UTC)

Glkanter, as you are not prepared to even consider the possibility that the question might be interpreted in a way that the host acts non-randomly, I will not continue with this argument. Martin Hogbin (talk) 18:29, 2 June 2009 (UTC)

G - the problem is "combining doors" says 2/3 whether the host acts "randomly" or not. You're asking if something that always says 2/3 is a correct solution if the problem is constrained to have the answer 2/3. It is producing the correct number, but so does anything else that results in the number 2/3. Consider "Rick's stupid approach" - with your first pick your probability is 1/3 and if you switch it's your second chance so the numerator changes from 1 to 2 (i.e. your chance of winning by switching is your guess number divided by 3). This produces the correct numeric answer, so it must be correct reasoning, right? Does it ever produce a result other than 2/3? Is this not the correct numeric answer? What's confusing you about "combining doors" is that there is a question that it correctly answers and the answer to this question is 2/3 - but it's not the question that we're asking. The way to see it is not the question we're asking is to change the assumptions slightly so the answer to the question we're asking is different from the answer to the question "combining doors" solves. This is exactly like using an overall batting average when asked about a team specific one. If the player's overall average happens to be the same as his average against a specific team the error in how we're computing the average is not readily apparent. We can see the error by looking at players for whom these two averages are different. -- Rick Block (talk) 18:43, 2 June 2009 (UTC)

Martin, I think that's the best course. I thought I read a few days ago where a Morganian agreed that the FCC would require Monty to act randomly. Because it's in the definition of a game show. And it's US law. How is it still a debatable point?

Rick, how come when I disagree with a professor, I'm described as ignorant and arrogant, but when you disagree with Devlin and dozens of other published authors, you claim to have a valid point? As always, you did not address a single facet of the 'Combining Doors' solution in order to refute it. You just go off on that gibberish, whether it's changing the premises, or cards, or now, batting averages.

Martin, could you indulge me? Into which category does the 'Combining Doors' solution belong in your Probability narrative? Glkanter (talk) 20:04, 2 June 2009 (UTC)

I do not understand what you mean. It is an unconditional solution, thus it has all the problems of unconditional solutions, if you believe that there are any. Martin Hogbin (talk) 22:16, 2 June 2009 (UTC)

Martin, which category does the 'Combining Doors' solution belong to:

Classical definition - Frequentists

Classical definition - Bayesian

Modern definition

Other - 'Logical Notation'

Glkanter (talk) 00:33, 3 June 2009 (UTC)

I'd say

Other - Informal explanations meant to provide an intuitive understanding for why the solution is not 1/2.

Although I sincerely believe such explanations are offered with the best of intentions, I think they have interfered with the general public developing an actual understanding of this problem. Approaching it as a conditional probability problem works if the host is constrained (or assumed) to pick randomly if the player initially picks the car, or if the host picks non-randomly in this case, or if the host forgets where the car is and happens to open a door showing a goat, or many other versions. Teaching people an unconditional approach gives them a "solution" that works in only one of these cases. -- Rick Block (talk) 01:29, 3 June 2009 (UTC)

Which case does the 'Combining Doors' solution work for? Glkanter (talk) 02:04, 3 June 2009 (UTC)

I assume this is a rhetorical question, but I'll answer it anyway. It "works" in any case where the answer to a) "what is the chance of winning by switching for a player deciding to switch when standing in front of two closed doors and one open door showing a goat" (given the other assumptions we all agree on) is the same as the answer to b) "what is the chance of winning by switching for a player who must decide whether to switch before the host opens a door". In my opinion, this means it doesn't ever actually "work" if you want the answer to a) because it's always telling you the answer to b). If you constrain the problem so a) and b) have the same answer it seems to work. But this is like saying your stopped clock that says "2:15" works as long you only ask what time it is at 2:15. -- Rick Block (talk) 02:50, 3 June 2009 (UTC)

Not rhetorical at all. You wrote:

"Approaching it as a conditional probability problem works if the host is constrained (or assumed) to pick randomly if the player initially picks the car, or if the host picks non-randomly in this case, or if the host forgets where the car is and happens to open a door showing a goat, or many other versions. Teaching people an unconditional approach gives them a "solution" that works in only one of these cases."

So, which of the above cases does the 'Combining Doors' solution work for? It's the first one, isn't it? Glkanter (talk) 03:00, 3 June 2009 (UTC)

Note the quotes around "solution". What about my explanation above don't you understand? It never answers a) since it actually answers b). It turns out they have the same numeric answer if the host picks randomly in the case the player initially selects the car, but unless you can give a reason for this you're not answering the question that's asked - which is a) not b). -- Rick Block (talk) 03:28, 3 June 2009 (UTC)

You're not arguing with me. You're arguing that Devlin and many other published sources are not as reliable as Morgan. How can you be so certain that they're all wrong? Glkanter (talk) 03:43, 3 June 2009 (UTC)

You also wrote this:

"Although I sincerely believe such explanations are offered with the best of intentions, I think they have interfered with the general public developing an actual understanding of this problem."

The way I read that is you feel that while they may provide a valid solution, your personal opinion is that they're not properly addressing general probability theory. This strikes me as a POV contrary to those published sources. I don't believe that published sources are to be discounted based on an editor's personal POV. Glkanter (talk) 04:01, 3 June 2009 (UTC)

I've emailed Devlin and asked him to comment. Maybe he will, maybe he won't. I've read and understand the Morgan et al. paper (and Gillman, and Grinstead and Snell, and Falk). What they all say is that the MHP is a conditional probability problem. The Morgan et al. paper is specifically about this exact issue - and it's in a peer reviewed math journal which means it is more reliable than Devlin's column (and BTW, I was the one who found Devlin's column and added it as a reference for the "combining doors" section of the article). Searching for references for the article I've read probably 50 papers on this problem. I really do understand it. I understand POV as well and have never argued that reliable published sources should be stricken from the article. -- Rick Block (talk) 04:40, 3 June 2009 (UTC)

I believe your personal POV as described above, plus your strong affinity for the Morgan paper, have been the primary barrier to the unqualified recognition of the validity of the various published unconditional solutions. Such unqualified recognition would, of course, diminish Morgan's claim as the only 'true' solution. Noteworthy, perhaps, but not due the emphasis it currently enjoys. There are numerous citations in the article of the unconditional solutions being false, etc., and these would be removed. Extensive discussion of 'variants' would also be less important, as 'host behaviour' would no longer be a necessary component in understanding the solution. I believe these changes would result in a more concise article, leading the reader to a better comprehension of the Monty Hall problem.Glkanter (talk) 05:57, 3 June 2009 (UTC)

Mutually Exclusive Published Reliable Sources

Are you trying to understand the problem, or suggesting a change to the article, or what? Suggestions for changes should be at talk:Monty Hall problem. If you're seeking to understand, this is the appropriate place. -- Rick Block (talk) 12:23, 3 June 2009 (UTC)

I created a new section with the same name on the regular talk page. Glkanter (talk) 12:35, 3 June 2009 (UTC)

Analysis

I model the problem as follows:

C₁ is the event that the car is behind door 1. Similar C₂ and C₃. We have:

$P(C_{1})=P(C_{2})=P(C_{3})={\tfrac {1}{3}}.$

K₁ is the event that the player chooses door 1. Similar K₂ and K₃.

The choice of the player is independent of the position of the car, hence:

$P(C_{i}\ and\ K_{j})=P(C_{i})P(K_{j})={\tfrac {1}{3}}P(K_{j}).$

M₁ is the event that the host opens door 1. Similar M₂ and M₃.

According to the rules:

$P(M_{i}|C_{i}\ and\ K_{j})=0.$ (The door with the car is never opened)

$P(M_{j}|C_{i}\ and\ K_{j})=0.$ (The chosen door is never opened)

Thus:

$P(M_{3}|C_{2}\ and\ K_{1})=P(M_{2}|C_{3}\ and\ K_{1})=1.$

$P(M_{3}|C_{1}\ and\ K_{2})=P(M_{1}|C_{3}\ and\ K_{2})=1.$

$P(M_{2}|C_{1}\ and\ K_{3})=P(M_{1}|C_{2}\ and\ K_{3})=1.$

According to the rules we also have:

$P(M_{2}|C_{1}\ and\ K_{1})=P(M_{3}|C_{1}\ and\ K_{1})={\tfrac {1}{2}}.$

$P(M_{1}|C_{2}\ and\ K_{2})=P(M_{3}|C_{2}\ and\ K_{2})={\tfrac {1}{2}}.$

$P(M_{1}|C_{3}\ and\ K_{3})=P(M_{2}|C_{3}\ and\ K_{3})={\tfrac {1}{2}}.$

The probability the car is behind door 3 is:

$P(C_{3})={\tfrac {1}{3}}.$

What happens when the host opens door 3? What is the probability the car is behind door 3? As above:

$P(C_{3})={\tfrac {1}{3}}.$

But there is a goat behind door 3? There is no chance there is a car? Right, but the car is placed randomly, hence in 1/3 of the cases the car is behind door 3. Also is the probability the car is behind door 3, 1/3. A new situation is originated, but that doesn't change the probabilities. In the new situation we have also new probabilities. And the new probability the car is behind door 3 is clearly 0! To distinguish them from the old ones, the new probabilities are called 'conditional' probabilities. They differ from the original, unconditional probabilities. They refer to the new situation. The conditional probability the car is behind door 3 is 0.

$\!\,P(C_{3}|M_{3})=0.$

After the choice of door 1 by the player all probabilities are conditional probabilities , given the event K₁. We have i.e.:

$PC_{1}|K_{1})=P(C_{2}|K_{1})=P(C_{3}|K_{1})={\tfrac {1}{3}}.$

Then the host opens door 3. All probabilities are from now on conditional probabilities , given the events K₁ and M₃. What will be the probability the car is behind door 3? As always:

$P(C_{3})={\tfrac {1}{3}}.$

But!! the conditional probability the car is behind door 3, given K₁ and M₃ is:

$P(C_{3}|K_{1}\ and\ M_{3})=0.$

And what will be the probability the car is behind door 1? No different:

$P(C_{1})={\tfrac {1}{3}}.$

And what will be the probability the car is behind door 2, the unopened door? Well, also:

$P(C_{2})={\tfrac {1}{3}}.$

We have:

$\!\,P(C_{1})+P(C_{2})+P(C_{3})=1.$

But in the new situation the probability the car is behind door 2, i.e. the conditional probability the car is behind door 2, is:

$P(C_{2}|K_{1}\ and\ M_{3}).$

We don't know it's value, we have to calculate it.

For the the conditional probabilities also holds:

$P(C_{1}|K_{1}\ and\ M_{3})+P(C_{2}|K_{1}\ and\ M_{3})+P(C_{3}|K_{1}\ and\ M_{3})=1.$

And we already know:

$P(C_{3}|K_{1}\ and\ M_{3})=0.$

hence:

$P(C_{1}|K_{1}\ and\ M_{3})+P(C_{2}|K_{1}\ and\ M_{3})=1.$

It is thus sufficient to calculate one of these probabilities.

With the use of Bayes' law:

$P(C_{2}|K_{1}\ and\ M_{3})=$

$={\frac {P(M_{3}|C_{2}\ and\ K_{1})P(C_{2}|K_{1})}{P(M_{3}|C_{1}\ and\ K_{1})P(C_{1}|K_{1})+P(M_{3}|C_{2}\ and\ K_{1})P(C_{2}|K_{1})}}$

$={\frac {1\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {2}{3}}.$

As the choice by the player is independent of the position of the car and door 1 is chosen, we may only consider the situation K₁ and leave the mentioning of K₁ as a condition.

Another possibility is the use of the symmetrie in the problem to proof that the conditional probability the car is behind door 1, given K₁ and M₃ is also 1/3, just like the unconditional.

We have:

$P(C_{1}|K_{1})=P(C_{1}|K_{1}\ and\ M_{3})P(M_{3}|K_{1})+P(C_{1}|K_{1}\ and\ M_{2})P(M_{2}|K_{1})$

Because of the symmetry:

$\!\,P(C_{1}|K_{1}\ and\ M_{3})=P(C_{1}|K_{1}\ and\ M_{2})$

Hence:

${\tfrac {1}{3}}=P(C_{1}|K_{1})=$

$=P(C_{1}|K_{1}\ and\ M_{3})P(M_{3}|K_{1})+P(C_{1}|K_{1}\ and\ M_{2})P(M_{2}|K_{1})=$

$=P(C_{1}|K_{1}\ and\ M_{3})\left[P(M_{3}|K_{1})+P(M_{2}|K_{1})\right]=$

$=P(C_{1}|K_{1}\ and\ M_{3}).$

If someone thinks any solution may be formulated without conditional probabilities, please use the above notation to proof it.

Nijdam (talk) 23:35, 3 June 2009 (UTC)

[Please do not put comment in the text of the analysis]

Yes indeed, symmetry is a very good, simple and obvious option. Why not use it. Martin Hogbin (talk) 17:23, 4 June 2009 (UTC)

Okay, but so is Bayes' law. Nijdam (talk) 21:50, 4 June 2009 (UTC)

The main problem with your analysis is with the problem formulation. What is the question that you are answering? Is it Whitaker's question in Parade? How are you interpreting the problem? Are you giving an answer to what you believe Whitaker wanted to know, in which case you should make clear what that is, or are you treating the question as a formal probability problem, to be interpreted from what is actually written, in which case you must make clear your treatment of all points that are not made clear in the question and your rational behind each decision.

You cannot have a definitive answer without a definitive question. Martin Hogbin (talk) 09:39, 5 June 2009 (UTC)

Quit right of course. My question is clear. What question do you want to consider?Nijdam (talk) 07:52, 7 June 2009 (UTC)

Suppose we take the question to ask whether is it better to switch or stick and by how much, for a player the show Whitaker describes. This is, in my opinion, what Whitaker actually wanted to know. The door numbers etc were just to make clear what happened in the show. Martin Hogbin (talk) 16:57, 7 June 2009 (UTC)

When?

Consistent with "Suppose you're on a game show", these premises seem warranted:

The puzzle is to be solved from the contestant's POV/knowledge-state

The host acts randomly when faced with 2 goat doors

Could someone please explain to me at what point the contestant's selected door does not have a probability of 1/3 of being the car, and what is that probability?

It seems to me the only value it could change to is 1/2. I don't think anyone would argue that it goes to 0, 2/3, or 1.

And the only point at which it could change is when Monty reveals a goat. Otherwise, nothing changes.

I look forward to your responses. Glkanter (talk) 13:39, 5 June 2009 (UTC)

As Nijdam and I have said, more times than I care to count:

Initially, the chances are 1/3 + 1/3 + 1/3 = 1

After the host opens (say door 3), the chances are a + b + 0 = c

Since the host doesn't always open door 3, c != 1 (c is the probability the host opens door 3).

Since the host always opens door 3 if the car is behind door 2, b=1/3.

Since the host doesn't always open door 3 if the car is behind door 1, a != 1/3.

If the host acts randomly when faced with 2 goats, this happens 1/3 of the time (the times when the car is behind door 1), in which case c=1/3 + (1/3)*(1/2), which totals 1/2, and a=(1/3)*(1/2) = 1/6. So, under your assumptions, the chances after the host opens door 3 are:

1/6 + 1/3 + 0 = 1/2

Under the "normal" rules, the host opening door 3 doesn't affect the chances that the car is behind door 2, but halves the chances that the car is behind door 1. So switching doubles the player's chances of getting the car. -- Rick Block (talk) 15:28, 5 June 2009 (UTC)

Thank you, Rick, for the thorough response. As you know, I am not an expert in probability. I'm only familiar with the branch of probability theory where the sum of the possible outcomes must = 1. Would I be following proper form in adjusting the "1/6 + 1/3 + 0 = 1/2" by multiplying it by 2 on both sides of the '=' sign? Glkanter (talk) 16:35, 5 June 2009 (UTC)

When you do this (multiply by 2) you're changing what you're talking about from the original (unconditional) probabilities to the conditional probabilities that are in effect in the case where the host opens door 3. This makes the statement "the host opening door 3 doesn't change the player's initial 1/3 chance of picking the car" sort of true, but entirely misleading since it is simultaneously referring to the initial (unconditional) 1/3 probability and the resultant (conditional) 1/3 probability. Looking at it this way, the host opening door 3 certainly does change the player's initial 1/3 probability - it changes into a conditional probability that is also 1/3. These are both 1/3, but they are completely different probabilities.

A less confusing way to look at it is to keep the unconditional perspective throughout. The other 1/2 is the case where the host opens door 2, so the whole (unconditional) equation is:

Initially 1/3 + 1/3 + 1/3 = 1

We know the host will open door 2 or door 3

(1/6 + 1/3 + 0)_{host opens door 3} + (1/6 + 0 + 1/3)_{host opens door 2} = 1

By keeping the unconditional perspective we avoid the switch to conditional probabilities and the host opening either door doesn't change any of the initial probabilities (player's door is 1/6+1/6, and the other doors are 1/3+0) and you can see switching doubles your chances of winning the car (from 1/6 to 1/3). -- Rick Block (talk) 03:11, 6 June 2009 (UTC)

Clear as mud. Martin Hogbin (talk) 09:52, 6 June 2009 (UTC)

What about this do you find hard to follow? -- Rick Block (talk) 15:34, 6 June 2009 (UTC)

The question asked is "Could someone please explain to me at what point the contestant's selected door does not have a probability of 1/3 of being the car, and what is that probability?"

You wrote:

"1/6 + 1/3 + 0 = 1/2", which, despite your response, is equivalent to 1/3 + 2/3 + 0 = 1

and

"(1/6 + 1/3 + 0)_{host opens door 3} + (1/6 + 0 + 1/3)_{host opens door 2} = 1", which is equivalent to 1/3 + 1/3 + 1/3, which is where we obviously started

and

"These are both 1/3, but they are completely different probabilities."

So, you've agreed that it is 1/3. But you haven't shown where it ever was not 1/3, much less what that value might have been. Which, of course, were the only questions asked. Glkanter (talk) 16:40, 6 June 2009 (UTC)

I think I've been perfectly clear that there are two different probabilities that are both 1/3. You don't seem to think there's a difference between 1/3 of something and half of 1/3 (which is 1/3 of something else). Please answer the following questions (with numbers, not fractions):

1) If 300 players all pick door 1, about how many of these would you expect to have picked the car?

2) Of the 300 players in #1, about how many would you expect to see the host open door 3?

3) Of the players who see the host open door 3 (your answer to #2), about how many have picked the car?

Are your answers to #1 and #3 the same? If you say "the host opening door 3 doesn't change the player's probability of having selected the car" how many players does this statement refer to, or does "probability" in this statement mean two different things? I think a better way to say this is "the player's initial probability of having selected the car is the same as the conditional probability in the case where the host opens door 3". These are different probabilities (100 out of 300 vs. 50 out of 150), and (yes) they are both 1/3. The constraints on the host force these both to be 1/3, but they are distinctly not the same thing. -- Rick Block (talk) 17:41, 6 June 2009 (UTC)

You know, Rick, absent the canard that Monty may 'tip off' the contestant in some way by having a bias, your arguments just doesn't have much heft. Like a cigar, sometimes 1/3 is just 1/3. Glkanter (talk) 04:02, 7 June 2009 (UTC)

I would answer: the contestant's door (always) has probability 1/3 to hold the car. But it is not obvous that the conditional probability to hold the car is 1/3. So better to formulate your question and ideas in the terminology I proposed. And yes: 1/3 is always 1/3, but not everything with the value 1/3 is the same as something else with that value. I.e. my wife is called Petra, just like one of her friends. I better be careful not to think they are the same. Nijdam (talk) 08:00, 7 June 2009 (UTC)

Well, I understand (and agree with) this:

"I would answer: the contestant's door (always) has probability 1/3 to hold the car. "

Which differs from this, just a week ago:

"At what point did it change from 1/3, and what did it change to? Then, when did it change back? Please provide a reference. Thank you. Glkanter (talk) 18:30, 29 May 2009 (UTC)"

"Apparently at a point far beyond your reach. I have explained this a dozen or more times, and won't repeat it any more. Nijdam (talk) 10:56, 31 May 2009 (UTC)"

It appears you changed your mind on this question. But I would not want to put words in your mouth. Can you explain this apparent dichotomy?

As you probably expected, I don't understand the relevance of rest of your response. In what situation would 'the 1/3 probability of the contestant's door holding the car' not exactly equal 'the 1/3 probability of the contestant's door holding the car'?

If I didn't know better, I would think you're just playing word games with me. But a professor of Mathematics wouldn't do that, would he? Glkanter (talk) 10:36, 7 June 2009 (UTC)

Glkanter - are you unable to answer my questions above? -- Rick Block (talk) 17:35, 7 June 2009 (UTC)

Rick, given that Nijdam just 2 hours ago proposed an entirely new "Probabilistic solution" section which is entirely devoid of any 'host behaviour', or references to the unconditional solutions being 'false' or 'incomplete', I see no need to continue these topics on this 'arguments' page. I hope we can focus our attention on attaining a consensus that will propagate these long over due enhancements throughout the entire article and the FAQs. Glkanter (talk) 18:15, 7 June 2009 (UTC)

You've missed the point of Nijdam's suggestion - he's suggesting starting the Solution section with this, not replacing the "Probabilistic solution" section. I'd still like your answers to my questions above. -- Rick Block (talk) 20:00, 7 June 2009 (UTC)

Not much interested. And just so no random reader thinks I'm insane, here's what you wrote 3 hours ago on the talk page:

"Just to be clear, Nijdam is suggesting this new text come before (or perhaps even instead of) the current "Popular solution" section. Are you saying you'd be OK with this? -- Rick Block (talk) 17:57, 7 June 2009 (UTC)"

Why not just drop it, OK? Glkanter (talk) 21:01, 7 June 2009 (UTC)

Why?

Why is it that in the 18 years since Morgan and at least 3 others published their papers, other reliable sources continue to publish solutions to the MHP using the unconditional solutions? And, as Rick points out, no reliable source has ever published a paper directly challenging Morgan and the others? Although, "The American Statistician" felt it appropriate to include a 'comment' from Professor Richard G. Seymann immediately following Morgan's paper. I presume they had Morgan's approval to include this.

Since Morgan's POV is that the unconditional solution is 'false', they can't both be right, can they? Glkanter (talk) 14:05, 5 June 2009 (UTC)

I suggest that you write a paper yourself if so inclined, or blog about it, or take it privately with the editors of "The American Statistician". However, I fail to see how the above is a "mathematical argument concerning the MHP", which is the subject of this page (or any argument at all, for this matter).glopk (talk) 22:14, 5 June 2009 (UTC)

Thank you for the thoughtful suggestions. I will certainly take them under advisement.

In the context that on these pages Morgan is often trotted out as the definitive word on the MHP because no subsequent publisher has challenged their analysis directly, and that being peer-reviewed makes it the 'uber'-reference, I saw fit to comment that even in the absence of direct conflict, the fact that many reliable sources continue to use unconditional solutions indicates that many professionals do not accept Morgan's argument. I think it's worth re-examining many previous arguments on these pages that have occurred over 4 years and 9 archives, in light of two recent developments: A mathematical error has been found in Morgan's work, and that the premise that the host must act randomly when faced with two goats may indeed be part of MvS's problem statement.

I didn't edit the article. I didn't even put this on the talk page. I think this is the appropriate venue for this discussion. Glkanter (talk) 02:31, 6 June 2009 (UTC)

Do you live anywhere near a major city that might have a university library that has copies of The American Statistician? If you've purchased the online copy of the Morgan et al. paper, it has a truncated version of Seymann's response and is missing Morgan's rejoinder. If it's at all possible, please go to a university library and look up these references. While you're there you might look up papers that reference Morgan's paper (the librarian will be happy to help you). Contrary to what you apparently think, Morgan et al. is not some perverse oddity that mathematicians dismiss as the ravings of some lunatic. It is instead the first published rigorous treatment of the MHP (deserving a prominent place in Wikipedia's article on this problem). Do some subsequent sources not just ignore but conflict with this paper? Absolutely. Shame on them. Perhaps they might read the Wikipedia article and learn something. -- Rick Block (talk) 05:03, 6 June 2009 (UTC)

Do you continue to claim you have a NPOV on this article? Glkanter (talk) 06:30, 6 June 2009 (UTC)

WP:NPOV applies to articles, not to editors. Of course I have a POV. You do, too. Everyone does. The point is for articles to fairly represent what is published about the topic. -- Rick Block (talk) 15:31, 6 June 2009 (UTC)

Rick, your claim that Morgan's paper is it all rigorous does not stand up to scrutiny. It has a misquotation and several inconsistencies, it does not make clear the premises on which it reaches its conclusions, and most amazingly it contains a error in the application of conditional probability! These are matters of fact that I will be pleased to point out to you. Martin Hogbin (talk) 10:23, 6 June 2009 (UTC)

Continuing to harp on the alleged error in this forum is relatively pointless. Morgan and his co-authors have (as far as I know) never commented on this article. As I've said, I've previously contacted Morgan by email and he has politely refused to comment here. If you want to pursue this, please contact the journal. Regardless of what you think of the paper, it is the first peer reviewed paper published in a math journal about the MHP. As such it should be a primary reference for the article here. -- Rick Block (talk) 15:31, 6 June 2009 (UTC)

I will be happy to stop pointing out the deficiencies of the Morgan paper if you stop promoting at as the last word on the subject. I accept that it is a reliable source but there are others, such as Krauss and Wang, that are much more balanced in their approach and more relevant to the article in that they discuss the basic problem and why people get it wrong. I also see no reason for us to duplicate the unscholarly tone of the Morgan paper in the article, with descriptions of the good work of others as 'false' and insistence on 'one correct solution'. We have an equally reliable source, which no one has challenged, in the form of Seymann's comment that, 'Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem'. Morgan's paper does indeed present a correct solution to their particular formulation of the problem but there are clearly other ways to interpret the problem and these require different solutions.

Regarding two of the errors in the paper, the misquotation of Whitaker's question is there for all to see and the error in the application of conditional probability is being pursued via the journal. Martin Hogbin (talk) 12:54, 7 June 2009 (UTC)

What was Whitaker's question?

I think it is universally agreed that to answer a question we must understand what the questioner is actually asking. In the case of Whitaker's question I see two ways of doing this. We can either take an open view of the question and ask ourselves, regardless of what he wrote, what do we think Whitaker actually wanted to know. Although this, in my opinion, is the better way to answer the problem it has the difficulty that the intended question becomes purely a matter of opinion that can never be settled. If we are to take this approach the least that we can do is make clear the exact formulation that we are providing a solution to.

The other way is to answer the problem as written. This has the advantage that we have a clear problem statement to start with although, personally, I doubt that this is the right approach, bearing in mind that the question was a letter from a reader of a popular general interest magazine, rather than one in a statistics exam. If we do take this approach, where we have a clear problem statement, there are many issues to be decided before we have a precise enough formulation of the problem to get a single correct solution. Here is Whitaker's question:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Firstly the game rules. Certain aspects of these are not clear from the problem statement but there is general agreement that the standard game rules should state that the host always offers the swap and always opens an unchosen door to reveal a goat, even though this may not have always been the case on the assumed show.

Starting with the stated question, first we have, 'Suppose you're on a game show...'. What do we take this to mean? To many it might be taken to mean that we should answer the question from the perspective or state of knowledge of a contestant on a game show, that is to say someone who has no idea of the host door choice strategy. Many might argue that this is the only way to take it. For some reason Morgan ignore this part of the question.

Next we come to, You pick a door, say No. 1. Here we could have two contradictory possible meanings depending on which part of the phrase we give priority to. 'You pick a door', means that you could pick any door but, say No 1 strongly suggests that we do, in fact, pick door 1. To most English speakers this phrase means that we pick any door, an example of such a door being door 1. In other words Whitaker does not want to ask specifically what the best action is only if you happen to open door 1 but he is asking a general question.

Then we come to, the host, who knows what's behind the doors, opens another door, say No. 3. Once again we could take this to mean either that the host opens another (unspecified) door or that the host does, in fact, open door 3. Which of these two meanings we give to this phrase is of vital importance for it determines whether the questioner considers the door opened by the host to have any significance and therefore be identified in the question. Curiously, Morgan misquote this section as the host, who knows what's behind the doors, opens No. 3, giving the strong impression that Whitaker actually intended the door opened to be identified.

It may remain debatable which of these interpretation we should assume but we should at least be consistent in our choice. Either we take it that Whitaker meant only for the specific case that the player chooses door 1 and the host opens door 3 to be considered, which seems unlikely to me, or we must take it that the question means that the player chooses any one of the three doors and the host then opens either one of the two unchosen doors.

To sum up, there are two issues to be decided before we can provide a solution. Should we answer the question from the state of knowledge of a player on the game show? Does the questioner intend the door opened by the host to be identified? Martin Hogbin (talk) 22:18, 8 June 2009 (UTC)

Why are you so focused on Whitaker's version of the problem statement? As I've mentioned before (the article says this) the Monty Hall problem predates Whitaker's question to Parade (by 15 years!), and even Selvin's original version is clearly a reformulation of the Three Prisoners problem, which itself did not even originate in Gardner's 1959 Scientific American column. The problem is a classic problem in elementary conditional probability. It has gained popular notoriety precisely because people are really bad at correctly reasoning in the realm of conditional probability. -- Rick Block (talk) 04:19, 9 June 2009 (UTC)

I have focussed on Whitaker's question because it is the question that Morgan quote in their paper and it is by far the most well known problem statement. What can Morgan be referring to when they talk of the 'problem at hand'? They insist that the problem is one of conditional probability, yet it is easy to formulate the problem such that this is not the case. What Morgan do is start with Whitaker's problem, reformulate it so that it is a conditional question, add in bogus 'information' that makes the condition important, them criticize everyone else for not treating the problem the way that they do. Martin Hogbin (talk) 08:51, 9 June 2009 (UTC)

I'm sorry Martin, but the MHP is always a question for a conditional probability. Already the choice of the player is in principle a condition, although one may omit it, but the opening of the door certainly is. That's why they're critical, and so am I. Any effort to answer the question without referring to a conditional probability is wrong.Nijdam (talk) 18:01, 9 June 2009 (UTC)

Even if the numbers of the chosen and opened doors were not mentioned in the problem statement, the solution would be the conditional probability as a function of all the possible combination of doors. That's why it is made more specific and one of the combination - doors 1 and 3 - are taken as an example.Nijdam (talk) 18:06, 9 June 2009 (UTC)

So this is yet another attack on the Morgan et al. paper? We all know you don't like this paper, why not just drop it? -- Rick Block (talk) 12:57, 9 June 2009 (UTC)

It is not specifically an attack on the Morgan paper, it is an attack on the concepts, that Morgan started, that the question identifies the door opened by the host and that the question should not be answered from the state of knowledge of the player. How about you address the content of what I have written, which is that the question in general, and Whitaker's statement in particular, is open to other interpretations.Martin Hogbin (talk) 16:24, 9 June 2009 (UTC)

I guess it's my turn. For purposes of editing the article, it doesn't matter which is right. They both go in. The question is how much credence/emphasis do we, as thinking, analytical editors, give to each solution? Specifically, since the conditional solution does not rely on them, should 'host behaviour' or 'unconditional is false' be part of the conditional solution? Or where do they best belong, if at all? Nijdam has presented a conditional solution which does not include either of them. I heartily support that approach. Glkanter (talk) 18:20, 9 June 2009 (UTC)

@Martin - we've been through your questions about the interpretation of the problem innumerable times. Is Whitaker's question open to interpretation? Of course it is. Do you realize yet that Whitaker was paraphrasing an already well-known probability puzzle he'd no doubt heard elsewhere? Regarding your questions:

Should we answer the question from the state of knowledge of a player on the game show? If the player knows everything given in the problem statement, yes. We should answer the question from the state of knowledge of someone reading the question (which may or may not match the player's knowledge).

Here you have neatly sidestepped answering the my question, which was whether we should answer the question from the state of knowledge of the player. You seem to have given two answers.Martin Hogbin (talk) 20:29, 10 June 2009 (UTC)

No, I've given one answer. What is not clear about it? If you think the player knows everything stated in the problem statement, then the answer is yes. Otherwise, the answer is no. If you insist on a one word answer, then it would have to be no. -- Rick Block (talk) 02:09, 11 June 2009 (UTC)

Does the questioner intend the door opened by the host to be identified? In the sense you mean, yes, absolutely. There is simply no other reasonable way to read the question. And, if you think you can read it some other way Whitaker's paraphrasing was lousy. The classical question is constructed to contrast the initial situation in which the player clearly has a 1/3 chance of having selected the car, with a conditional case in which the player is standing in front of two closed doors and an open door, not knowing which door the car is behind, and deciding whether to switch. If you're going to take the player's viewpoint the player can clearly see which door the host opened (whether you say which one it is in the problem statement or not). Anyone reading the problem statement can infer the player can see which door the host has opened (again, whether or not it's specifically mentioned in the problem statement). Your continued insistence that not identifying the door the host opens means it is not a conditional probability problem is simply bizarre. Perhaps this is not so clear to you since probability is apparently not your field - but this would be like someone arguing a problem that starts with "in a situation where your velocity is close to the speed of light ..." is not a relativity problem. Of course it is. -- Rick Block (talk) 04:17, 10 June 2009 (UTC)

What classical question is, 'constructed to contrast the initial situation ... with a conditional case...'? How do you know this? Not identifying the door that the host opens does make the problem unconditional. Nijdam has already agreed this. If the question merely states that the host opens one of the other two doors to reveal a goat, what is the condition? Martin Hogbin (talk) 20:29, 10 June 2009 (UTC)

Let's be specific. What I agreed on in not identifying doors, is that another type of problem arises. One in which the player doesn't know which door is opened. The MHP definitely is about identified doors. And as I wrote somwhere above: the MHP always concerns conditional prob's. Nijdam (talk) 10:35, 11 June 2009 (UTC)

On what basis do you make the statement, 'the MHP always concerns conditional prob's'? You seem to be resorting to a circular argument that the doors must be identified because that makes the problem conditional and the problem is conditional because the doors are identified.

This is a little unfair: I never said the doors must be identified to make the problem conditional.Nijdam (talk) 14:02, 11 June 2009 (UTC)

If you ask yourself what Whitaker (or any one else) actually wants to know. It could quite easily be that all he wants answered is the question, 'You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?'. Martin Hogbin (talk) 13:06, 11 June 2009 (UTC)

Well we discussed this before and I repeat: the decision would be based on the conditional probabilities as a function of the numbers of the chosen and the opened door. Nijdam (talk) 14:02, 11 June 2009 (UTC)

I cannot understand why you are so fixated on conditional probability. The formulation I have given above is that given by Morgan as a statement of the unconditional problem. Martin Hogbin (talk) 20:21, 12 June 2009 (UTC)

@Martin: let us be practical, the problem we want to address is the formulation of Krauss & Wang. This is generally considered the MHP. At least it has all the aspects needed. May be we will never be sure what Whitaker really wanted to ask, although I'm pretty sure it was what K&W formulated. May be even Whitaker himself didn't exactly know what he wanted. The article already accounts for the ambiguities in his formulation. Happily this gives the opportunity to discuss the alternative host strategy. Please what exactly do you want to discuss in the article other than these aspects. Nijdam (talk) 11:32, 10 June 2009 (UTC)

I agree that K & W do give a formulation that they consider is the way that most people see the problem. From their formulation I see two directions to go in, both leading to valid solutions. Firstly, the academic direction in which we consider that the host door choice may not be random (although car placement remains so). This is what Morgan do and it has a place in the article.

Alternatively we can notice that in the K & W formulation there are several very simple arguments to tell us that the 'condition' that the host opens a specific door is irrelevant and thus this factor can be completely ignored in our solution. In other words the problem can be treated unconditionally. Martin Hogbin (talk) 13:06, 11 June 2009 (UTC)

No, the arguments that tell us that the 'condition' that the host opens a specific door is irrelevant, lead to the conclusion that the conditional probability, that needs to be calculated, equals the unconditional, and hence is also 1/3. That's the only simplification in what may be referred to as "unconditional solution". There is no way the decision is based on the unconditional probability. Nijdam (talk) 13:37, 11 June 2009 (UTC)

I think you are just arguing about words here. The simple solution is correct in principle and practice. Martin Hogbin (talk) 20:21, 12 June 2009 (UTC)

Host acts randomly?

Whitaker begins, "Assume you are on a game show..."

Rick, given that Whitaker's question is a story problem about an American game show, please expand on your response on the talk page that the host doesn't necessarily have to act randomly when faced with two goats. Thank you. Glkanter (talk) 02:14, 11 June 2009 (UTC)

This either is simply setting some informal context for what we should take as a word problem, or is suggesting the answer should consider a real life scenario. The predominant interpretation is by far the first. If instead we're to treat this as a real life scenario, US laws about game shows could be considered to be in effect (which would mean the car should be placed randomly and the host should pick randomly if faced with two goats). On the other hand, in a real life scenario the motives of the host distinctly come into play, and since we're not told the host makes this offer to every player we should consider the possibility that the host is only (or more often) making this offer when the player has initially selected the car. I mean, what kind of show is going to create a situation where with optimal play 2 out of 3 players can win a car? If we allow the host to make the offer more often (or exclusively) when the player has initially picked the car, the player's chances of winning by switching are anything from 0 to 2/3.

So, clearly, it's a word problem. As many, many sources point out, as a word problem it's underspecified. Any solution that clarifies it's assumptions and solves the problem according to those assumptions is therefore valid. Whitaker's version doesn't say anything about the host preference when faced with two goats, so assuming the host might have a preference is at least as valid as assuming the host will pick randomly in this case. But, whatever you assume, the problem is a conditional problem.

It seems to me you're wanting to have it both ways. The answer is 2/3 only if the host always makes the offer to switch (which wouldn't be the case on a real game show) AND the host chooses randomly if picking between two goats (which might be considered to be required on a real game show). -- Rick Block (talk) 04:02, 11 June 2009 (UTC)

Relative to the question I asked, I haven't assumed anything. It's in the dna of a game show. I'm talking dictionary, not judicial statutes.

Always offers the switch? I've never commented on that. From what I've read, MvS agrees she should have stated it, believes her answer implies it, and everyone agrees its a premise to the MHP. Just more stalling. Glkanter (talk) 04:23, 11 June 2009 (UTC)

Of course it is. Surely we can take it as read now that the host always opens and unchosen door to reveal a goat and always offers the switch. Martin Hogbin (talk) 12:46, 11 June 2009 (UTC)

You both seem to be missing the point. Because we're assuming the host always makes the offer to switch we're obviously not considering a real-life game show scenario, but rather a word problem. The argument that the host must choose randomly when faced with two goats is based on the premise that a real-life game show would have to work that way in the US - but since it's not a real-life scenario, this constraint (or any other based on how real-life game shows operate) doesn't necessarily apply. -- Rick Block (talk) 14:11, 11 June 2009 (UTC)

"Assume you're on a game show, well, not exactly a game show, I'll specifically use the words 'game show', but anyways, don't feel compelled to even understand what a game show is, just make up any exceptions that don't fit your narrow paradigm..." Glkanter (talk) 14:53, 11 June 2009 (UTC)

Actually Rick, you are confusing the definition of a game show with the rules of the two goats and one car game on Lets Make A Deal. Whitaker says, 'Assume you are on a game show..', then spells out the unique rules of his Lets Make A Deal contest, specifically, that the host will always offer the switch. This does nothing to negate the obvious understanding (the guys who didn't understand this went to jail) that the host will not actually tell the contestant where the car is. Glkanter (talk) 15:02, 11 June 2009 (UTC)

Whitaker spells out that the host will always offer the switch? Here's the quote from Parade:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Where does this say the host always makes the offer to switch? In the typical analysis we assume this to be true even though if you were on a game show I think we both know the host wouldn't always be giving you a 2/3 chance to win a car. In any event, you asked me to expand on my response from the talk page. I have. It's clear you think "Suppose you're on a game show" means the host must pick randomly in the case in question. I've said why I disagree. Can we drop this now? -- Rick Block (talk) 18:46, 11 June 2009 (UTC)

You're right, as I posted earlier, MvS agrees she should have stated "the host always offers the switch", believes her answer infers it, and it is universally accepted as a premise of the MHP. Of course, you ignore the point, which is by the definition of a game show, the host will not tell the contestant where the car is. "Offering the switch" was brought up by you in an attempt to negate that point.

No, we can't drop it. Morgan says it. NPOV says we have to include it. It has to be addressed. It's not needed for the conditional solution, so my point is it shouldn't be in the solutions section. It is there now. And Nijdam, seemingly, proposed a conditional solution without it. And without 'host behaviour', Morgan's only remaining point is that the conditional solution provides a solution to the MHP, as does the unconditional solution. So, the question will ultimately be, 'how extensively should the 'dispute' be covered in the article'. Since we won't try to prove one side or the other wrong or right in the article, my proposed statement "'Morgan says the unconditional solution is false. This viewpoint is not shared universally in the professional community." may be all that is required in the article. Outside of this Wikipedia article, where does the so-called 'controversy' exist, anyway? The professional community certainly don't address it. How does the reader benefit, when both solutions are deemed valid by reliable published sources? Glkanter (talk) 19:17, 11 June 2009 (UTC)

Self-analysis tool

The following is a method designed to help people who are uncomfortable with the puzzle's resolution to pin down which specific features lead their intuition astray, without requiring them to delve into any but the most superficial aspects of probability maths.

After reading some of the preceding discussion, I should clarify right away that it's based on the only version of the puzzle I was familiar with before reading the article - namely one in which one considers the three doors to be a priori indistinguishable. This is what justifies using an urn problem analogy by removing most of the ambiguities that otherwise need to be addressed.

Incidentally, the reason for my not having come across the version the article focusses on before may have to do with my being German. While the name "Monty Hall problem" is in use here, the names "Let's Make a Deal", "Parade magazine" and "Marilyn vos Savant" were completely unknown to me. If there is a German standard phrasing of the problem at all, it may be based on an earlier version than the Whitaker one, or on a very free translation that didn't preserve all the details.

No matter, the aspect that this is meant to address is one that comes up in any version.

Thought experiment

We can construct an equivalent problem by using two urns and three balls, one white and two blacks. The host puts one (random) ball into the first urn and the other two into the second. The player knows the number of balls in either urn, but not the colour distribution. The game consists in choosing one urn or the other, and the aim is to end up with the white ball.

• In version A, the player simply picks an urn.

• In version B, the host first looks into the second urn, takes out a black ball, shows it to the player, and drops it back in. Then, the player picks an urn.

For version A, it is intuitively obvious that it is favourable to pick the second urn, and that the odds are 2:1.

However, it is also intuitively obvious that the difference between the versions is immaterial - all the host does is confirm that there is at least one black in the second urn, which was already a certainty.

What the experimental subject has to decide is what their intuition tells them about the favourable choice and the odds for version B. If it is the same as for version A, one now transforms version B step by step into the MHP (with indistinguishable doors), which is trivial, and marks the point at which their intuition flips. This may, for example, be to discard the black ball instead of putting it back, or the replacement of a single urn by two separate doors, or ...

If it is different from version A without any transformation, the subject is confronted with a simple and objectively contradictory situation consisting of the two versions and their equivalency, which can only be resolved by recognizing and localizing the flaw in their intutitive understanding.

Unfortunately, none of the regulars here will make good test subjects themselves, I imagine. :)

62.180.36.2 (talk) 04:23, 28 June 2009 (UTC)

On the German Wikipedia a user, Wilbert, proposed the same, in his opinion, similar experiment. I already answered there that this experiment is not similar to the MHP, because of the unindistinguishability of the black balls. Nijdam (talk) 11:43, 28 June 2009 (UTC)

I thought I addressed that above, sorry if I wasn't being clear enough - this is meant to mirror a version of the MHP in which the doors (and, if you like, the goats, though unless I'm mistaken this is not strictly required) are indeed indistinguishable. Whether such a version is too different from the standard one to still be called "the Monty Hall Problem" is a matter for debate, I suppose.

Either way, the purpose of the thought experiment isn't to solve the problem, it's to help anyone who's at the "I have to accept the numerical solution but I still don't really believe it" stage to figure out where the disbelief stems from. As such, I find it helpful even with regards to the standard version, as the reluctance it evokes is qualitatively the same, I should think.

(dynIP) 62.180.36.10 (talk) 12:26, 28 June 2009 (UTC)

Thanks for pointing me to the German discussion page, I found the passage you were referring to. However, my version is quite a bit more powerful as a tool for understanding, IMO, as it reduces the complexity by several more steps. This makes the intuitively contradictory situation much tighter and thus easier to resolve - the equivalency between versions A and B has to be overwhelmingly obvious.

62.180.36.10 (talk) 12:53, 28 June 2009 (UTC)

To be equivalent, I'd suggest using balls of three different colors, perhaps white, red, and green, with one ball in one urn and two in the second urn. Then, in version B, the host withdraws a non-white ball, say the red one, and doesn't drop it back in but discards it leaving both urns with only one ball. Note that if this is your own idea it can't be incorporated into the article as it would be original research, which is prohibited (all material added must be from reliable sources). This is actually fairly close to one of the experiments reported by Fox and Levav (referenced in the English version of the article). In their experiment they dealt 5 cards including only one ace face down, 2 to the subject and 3 to the "host" and then the host either looked at the host's cards and put them back face down, or looked and turned a non-ace card face up, or looked and turned two non-ace cards face up, and then asked the subject to estimate the probability that the "host" had the ace and whether the subject wanted to trade their two cards for the host's remaining face down cards (3, 2, or 1). In about 35 trials each way, the mean estimate of the probability that one of the host's cards was the ace was .6, .5, and .33 depending on whether the host was left with 3, 2, or 1 face down cards (with a total of 5, 4, or 3 face down cards between the host and the subject) and the percentage of subjects willing to trade their two cards for the host's remaining face down cards was .58, .23, and .15 respectively. Assuming the host was randomly selecting non-ace cards to turn face up, the actual probability was always .6 and to maximize their chances of winning (winners were actually paid a small sum of money) all subjects should have switched. Like it says in the article here, people apparently strongly believe that if there is one winner among N unknowns, the probability of each is 1/N whether this is true or not. The Fox and Levav experiment would suggest in your urn version that it's important to actually remove and discard one of the balls rather than put it back in the urn.

BTW - in German the problem is apparently often called "Das Ziegenproblem". There's a book about it by van Randow called Das Ziegenproblem: Denken in Wahrscheinlichkeiten [11] (I don't read German so haven't read this book, but it is referenced in the German Wikipedia article about the problem). -- Rick Block (talk) 16:22, 28 June 2009 (UTC)

It's entirely OR, which is why I posted it here (looking for feedback) rather than on the main talk page.

I agree that changes along the lines you suggest would make the thought experiment more compatible with the standard version, but any added complexity runs contrary to the experiment's purpose, which relies on statements being what I call "intuitively obvious".

Basically, I'm asserting that the reluctance of many people to accept the puzzle's solution is based on features other than the distinguishability of the doors and/or goats. Anyone unhappy with the idea that there is an objective advantage in switching in the standard MHP will be just as unhappy with the idea that there is an objective advantage in switching in a related urn experiment (one urn, three balls, player picks one without looking, host removes a non-white ball, stay or switch), and vice versa. If this is so, any insight gained about the intuitive thought processes leading to the disbelief should transfer from one to the other.

I'm fairly certain that the explanation that "people apparently strongly believe that if there is one winner among N unknowns, the probability of each is 1/N" is at best incomplete. People are familiar with non-equiprobable distributions (you seem to be using the term "non-random" on these talk pages, btw, which seems imprecise to me), and would be perfectly happy with accepting and using reliable information along the lines of e.g. "most of the time, the stage hands put the car behind the middle door". My impressions is that it has more to do with the notion that information revealed about one door should not affect the probabilities assigned to another door.

Your impression that many people will see a crucial difference between the host's replacing or discarding a black ball after showing it is likely correct, and is precisely part of the reason for the design of the experiment as it stands. These people should agree that the odds in version B are still 2:1, and will disagree with the step along the subsequent back-transformation to an MHP-like form which changes this aspect. All that remains is to convince them (or wait for them to convince themselves) that the transformation step is valid, without being encumbered by any other aspect of the puzzle.

ps: As a sidenode, I'd just like to say that after reading a significant portion of the talk page archives, I'm very impressed with the amount of patience you have shown with even the most stubborn and quarrelsome parties.

62.180.36.29 (talk) 17:38, 28 June 2009 (UTC)

Why is the chance 50/50 when the host doesn't know?

Can someone explain why it doesn't matter if you stay or switch when the host doesn't remember which door has the goat and which the car? Assuming he still (randomly, by accident) chooses a door with a goat, wouldn't it still be a 2/3 chance of winning if you switch, simply because you had a 2/3 chance of initially choosing a goat? --JohnJSal (talk) 15:46, 29 June 2009 (UTC)

This confusion is exactly the problem with the so-called "unconditional" solutions. At the point there are two doors and a goat, however you get there, you're in a different situation than you were originally. To figure out the probabilities in this (new) situation you have to use concepts from conditional probability. In the "standard" version, the host opens Door 3 whenever the car is behind Door 2 (which is 1/3 of the time) and half the time the car is behind Door 1 (i.e. 1/2 * 1/3, which is 1/6), so the car is behind Door 2 twice as often as it is behind Door 1. The overall probability of being in this case (where the host has opened Door 3) is 1/3 + 1/6 = 1/2, and the conditional probability of the car being behind Door 1 is (1/6)/(1/2) = 1/3, while the conditional probability of the car being behind Door 2 is (1/3)(1/2) = 2/3. In the "host forgets" version, the host opens Door 3 half the time regardless of where the car is. So the chances for each door are 1/6, 1/6, and 1/6. But in addition, we know the host didn't reveal the car so the 1/6 case where the car was behind Door 3 simply hasn't happened. We're left with a 1/6 chance of the car being behind Door 1 and a 1/6 chance the car is behind Door 2, so switch or not doesn't matter. Note that these only add up to 1/3, which means if the host opens Door 2 or Door 3 completely randomly, the case where the host opens Door 3 and happens to not reveal the car occurs only 1/3 of the time. In the "host forgets" version this is the only case we're interested in, and so the conditional probability the car is behind Door 1 (or Door 2) is the same, i.e. (1/6)/(1/3) = 1/2. -- Rick Block (talk) 16:33, 29 June 2009 (UTC)

Yikes! I appreciate the in-depth response...just a little beyond my league I think. :) But it definitely helps me to see the overall shape of the problem, at least. --38.108.205.180 (talk) 17:45, 29 June 2009 (UTC)

Conceptually, you can think about it like this: If player initially picked car, there is no chance of host revealing car. If player initially picked goat, there is a chance of host revealing car. Hence, the observed fact that host didn't reveal car makes it more likely that player initially picked car.

62.180.36.26 (talk) 12:04, 1 July 2009 (UTC)

The basics of conditional probability are not that difficult. The probability of this given that is written P(this|that) and is the joint probability of this and that together divided by the probability of that, i.e. P(this and that)/P(that).

In the normal version of the problem, assuming we're asking about the probabilities when the player has picked Door 1 and the host has opened Door 3, the probabilities of interest are the conditional probability that Door 1 hides the car given the host has opened Door 3 and the conditional probability that Door 2 hides the car given the host has opened Door 3. P(Door 2|host opens Door 3) is slightly easier to figure out, and is P(Door 2 and host opens Door 3) / P(host opens Door 3). P(Door 2 and host opens Door 3) is the probability the car is behind Door 2 and the host opens Door 3. We're assuming the player picked Door 1, so since the host always opens Door 3 if the car is behind Door 2 and the player picked Door 1, this is the same as P(Door 2), which is 1/3. P(host opens Door 3) is the probability the host opens Door 3. We might assume the problem is symmetric, meaning we're assuming the answer is the same whether the host opens Door 2 or Door 3 - another way to approach this is to specifically figure out the probability the host opens Door 3 as opposed to Door 2 (perhaps assuming the host picks randomly if the car is behind Door 1). Either of these approaches results in the probability of the host opening Door 3 being 1/2. Putting these together, we get P(Door 2|host opens Door 3) = (1/3) / (1/2) = 2/3. Using similar reasoning P(Door 1|host opens Door 3) works out to 1/3. The above is the "show your work" step missing from all of the "unconditional" solutions that say something like "when the host opens Door 3 the initial 1/3 chance of the car being behind Door 1 doesn't change" (or "your initial chance of picking a goat is 2/3, so ...").

In the host forgets version, the probabilities of interest are a little bit different and are the conditional probability that Door 1 (or Door 2) hides the car given the host has randomly opened Door 3 and has not revealed the car. Lets figure out P(Door 1|host opens Door 3 and car is not behind Door 3). The joint probability, P(the car is behind Door 1 and the host opens Door 3 and the car is not behind Door 3), is the same as P(the car is behind Door 1 and the host opens Door 3) because if the car is behind Door 1 it is not behind Door 3 and this is (1/3)*(1/2) = 1/6. The probability the host opens Door 3 (by itself, this is 1/2) and the car is not behind Door 3 (i.e. the probability Door 3 hides a goat) is (1/2) * (2/3). So, the conditional probability Door 1 hides the car given the host has randomly opened Door 3 showing a goat is (1/6) / (1/3) = 1/2.

If you approach this problem (or any of its variants) as a conditional probability problem you shouldn't get confused. The approach works for other "paradoxes" as well, like the Boy or Girl paradox (I know a family has two children, I see one and the one I see is a boy, what is the probability both are boys?). The instinctive answer is 1/2, but the question here is actually P(both are boys|at least one is a boy) which if you work it out you'll find is not 1/2. -- Rick Block (talk) 14:26, 1 July 2009 (UTC)

"I see one and the one I see is a boy" is usually not taken to be equivalent to "at least one is a boy", but to "one which was chosen at random is a boy", which makes all the difference, no?

62.180.36.6 (talk) 00:22, 2 July 2009 (UTC)

Ah yes, that is the problematic wording. How about "You run into the mother of a family you know has two children and since you're involved with the local Boy Scouts organization ask her if at least one is a boy. She says yes. What is the probability both are boys?" (the intent is "at least one is a boy"). -- Rick Block (talk) 12:23, 2 July 2009 (UTC)

The common feature is the method in which partial information (location of goat, minimum number of children of given sex) is obtained: Deliberately, by a knowledgable agent (Monty-with-good-memory/mother), as opposed to random sampling (Monty-with-bad-memory/observer). The solutions are different because the methods affect the sample space differently.

62.180.36.7 (talk) 14:36, 2 July 2009 (UTC)

This is Marilyn Vos Savant's answer:

"Here’s one way to look at it. A third of the time, the clueless host will choose the door with the prize, and the game will be over immediately. In our puzzle, that didn’t occur. So we’re considering the two-thirds of the time when either: 1) You have chosen the door with the prize; or 2) The prize is behind the unopened door. Each of these two events will occur one-third of the time, so you don’t gain by switching."

Please note there are 3 different premises to this puzzle from the MHP:

Host does not always show a car

Host does not always offer the opportunity to switch

Contestant doesn't always makes a decision

Since it's random (as per vos Savant), anyone could pick the 2nd door. Even the Contestant. And now you have Deal Or No Deal with just 3 suitcases. Glkanter (talk) 19:13, 29 June 2009 (UTC)

(from talk) Knowing the host will open a door makes the probability 1/2

Moved from Talk page, 28 July 2009 per Martin's suggestion.

Given the statement of the problem, that the host knows what is behind all 3 doors, and he is going to reveal a door with a goat (or whatever) behind it, the original choice made by the contestant had a 1 in 2 chance of winning from the beginning. Not after the reveal, but before it. Because the host was ALWAYS going to reveal a booby prize behind one of the unchosen doors. Although the contestant might not know he has a 1 in 2 chance from the beginning, that's what he has. And that probability remains the same whether he switches or not. I think this is a misunderstanding in the statement of the problem. I agree with the logic leading to the 2 in 3 chance of winning if the a priori condition is left out. However, that's a different problem, not the one that was in Parade or discussed in this article. PatPM (talk) 04:24, 27 July 2009 (UTC)

If you want a response to this, I suggest that we move to the /Arguments page. Martin Hogbin (talk) 08:21, 27 July 2009 (UTC)

PatPM (talk) 00:47, 29 July 2009 (UTC)

So, is it your claim that all the references about the problem including probability textbooks are incorrect and all the simulations showing a 2/3 chance of winning by switching are flawed? Are you actually interested in understanding why 2/3 is the correct answer even though you don't apparently believe it, or have you completely made up your mind that you're right and the rest of the world is wrong? -- Rick Block (talk) 01:40, 29 July 2009 (UTC)

Rick, this is another example of where the current article fails to do its job. By obfuscating the central problem with talk of conditional probability we fail to address the real problem of why people get the answer wrong. Martin Hogbin (talk) 13:34, 29 July 2009 (UTC)

I am of the exactly opposite opinion on this. By highlighting that the problem is a conditional probability problem we address the precise reason people get the answer wrong, which is that they fail to realize the problem is about conditional probabilities. The structure of the problem puts the reader in a conditional case (thinking about the player in front of 2 closed doors and one open door). The "simple" approach says to just ignore that and think instead about all possible outcomes rather than the specific case the problem asks about (with the implicit assumption either that the answer in the specific case matches the answer considering all outcomes or, more dubiously, that the probability of interest is actually the one considering all outcomes). My claim is that most people read the problem as asking about the conditional case but then fail to correctly analyze the conditional probability, and the article should therefore show them how to do this correctly. The "simple" approach (ignore what's actually asked, and answer something else that is easier to figure out) doesn't generalize to any other conditional probability problem and doesn't even work for all variants of the MHP. It is by obfuscating the conditional nature of the problem that we fail to address the real problem of why people get the answer wrong. -- Rick Block (talk) 02:39, 30 July 2009 (UTC)

So would you claim that, if the question were asked in a completely unconditional format (as in the example given in Morgan) most people would get it right? Martin Hogbin (talk) 15:44, 30 July 2009 (UTC)

The completely unconditional format is, for example, "the host will open a door showing a goat, what is your chance of winning if your strategy is to switch" (i.e. deciding whether to switch before the host opens a door). If you don't switch, this problem is logically equivalent to "what is your chance of correctly choosing the car given it is behind one of 3 closed doors", which is a completely uninteresting problem - one that nearly all people would get right. On the other hand I suspect most people would probably get the "unconditional" form of the MHP wrong, because they'll actually treat it like the conditional problem (not realizing it is simply the completely uninteresting problem with a bunch of extra words around it) - meaning they'll imagine the situation after the host has opened one of the two unpicked doors and (mis)judge that the probabilities are 50/50 because there are only 2 doors left. I know you don't agree with this, but IMO the unconditional form completely misses the entire paradoxical nature of the MHP which is that you're deciding whether to switch when you're looking at two doors (not three) and don't know where the car is, but the chances are not 50/50. Understanding that picking one out of 3 doors has a 1/3 chance is trivial. Understanding how two unknowns can end up with different probabilities is not so trivial, but not so hard that it's beyond the grasp of, say, a reasonably bright high school student. -- Rick Block (talk) 00:03, 31 July 2009 (UTC)

Pat, is what you are saying essentially this? If the host opens a door to reveal a goat right at the start, the contestant clearly then has a 1/2 chance of initially choosing a car and the same chance of getting the car if he later switches. Because we know that the host is going to open a door later on, the odds are exactly the same in the Monty Hall problem. Martin Hogbin (talk) 13:34, 29 July 2009 (UTC)

What I interpret Pat to be saying is that the knowledge of the game rules, in particular that the host will with certainty open a door revealing a goat, makes the odds of picking a car or a goat the same (each 1/2) before the host opens a door. You know you will be left with two alternatives, so your chance is 50/50 of initially picking either one of them. Then, when the host follows through and does open the door, the odds remain 1/2.

Pat - the reason this is not what happens is because there are three different initial conditions (assuming the player picks door 1), each equally likely, as follows:

CASE1) Car is behind door 1. In this case the host can open either door 2 or door 3.

CASE2) Car is behind door 2. In this case the host must open door 3.

CASE3) Car is behind door 3. In this case the host must open door 2.

If we see the host open door 3 we must be in CASE1 or CASE2 and cannot be in CASE3. If we're in CASE1, and if we assume the host picks which door to open randomly in this case, and we see the host open door 3, we're in this case 1/2 of 1/3 of the time (i.e. 1/6 of the time). We're in CASE2 1/3 of the time. These are the only possibilities if we've seen the host open door 3, so we have a 1/6 chance of the car being behind door 1 and a 1/3 chance of the car being behind door 2 when combined with the fact that we've seen the host open door 3. The crux of the problem is that if we pick door 1 the host always opens door 3 if the car is behind door 2 but only opens door 3 1/2 the time if the car is behind the door 1 (the other 1/2 the time the car is behind door 1 the host opens door 2, but since we've seen the host open door 3 this didn't happen). So if we pick door 1 and see the host open door 3, the car is twice as likely to be behind door 2 as behind door 1.

The same argument applies if we pick door 1 and see the host open door 2, in which case the car is twice as likely to be behind door 3 as behind door 1.

Make sense? -- Rick Block (talk) 02:39, 30 July 2009 (UTC)

New book as mentioned on Talk Page: search for: Morgan

http://www.amazon.com/Monty-Hall-Problem-Remarkable-Contentious/dp/0195367898/ref=sr_1_1?ie=UTF8&qid=1249163746&sr=8-1#reader —Preceding unsigned comment added by Glkanter (talk • contribs) 22:10, 1 August 2009 (UTC)

Is there some point here you're wanting to discuss? This book was already mentioned and it's account of the exchange between Morgan et al. and vos Savant was discussed on the talk page at talk:Monty Hall problem#Refereed paper agreeing with Morgan et al.. -- Rick Block (talk) 14:54, 2 August 2009 (UTC)

JeffJor's comments to your redirection here from the MHP discussion page.

Martin, you are wrong about Morgan's so-called "surprising error." The integral on the bottom of p286 expresses an a posteriori probability in terms of the related a priori probabilities according to the definition of conditional probability. Correctly. You erroneously translated it to a formula based on a posteriori probabilities: your q' is literally "the probability the host opens Door #3 given that the host opens Door #3." And whatever your "symmetry" argument is, it seems to be incorrect. That same argument should apply to any value of q, and different values of q can (with a catch) lead to answers other than 2/3.

The "catch" I referred to is subtle, and often confuses people about probability. In fact, it is ultimately the cause of the MHP's controversy, even among PhDs. I'll explain by using an example: Suppose I have a bag with 101 coins. Each coin has an integer N printed on it, ranging from 0 to 100, and has the property that it lands on heads N% of the time. I draw a coin out at random, and flip it. What is the probability that it lands heads? There are actually two answers, based on whether I tell you what N is. P(heads|know N) = N/100, and P(heads|don't know N)=1/2. The first probability can be thought of as representing the fraction of the time I would get heads if I flip that same coin over and over; while the second probability is when I flip different coins, replacing each and drawing another one at random each time.

The subtle point is that probability is a property of the process, not of the coin, per se. The problem statement I gave you only described a single example of the process I intended, and I left whether the process required replacement ambiguous. So the two kinds of knowledge represent different processes. In Morgan, if q=0 is a known property of the process that is repeated every time the show airs, then the probability P(Ws|D3)=1; and if q=1 the same way, then P(Ws|D3)=1/2. But it is 2/3 without that knowledge, since you can't attribute any specific value to it. The ln(2) answer represents an average where the host picks a q at random each show, and tells it to the contestant; the 2/3 answer represents results over many shows where we do not know q. (NOTE: This seeming paradox is not Bayesian Inference, as some people will claim. It is a result of including parameters that are not specified in the problem as though they are both included and known.)

It is an odd way of putting it, to say, 'the host picks a q at random each show, and tells it to the contestant', this never actually happens, but I understand what you mean so let us consider that to be that case. At the start of the show let us say that the host randomly chooses a value of q, this the a priori distribution of q is uniform (over the interval 0 - 1). The quesion that Morgan are answering is 'what is the average probability of winning by switching, given that the host has opened door 3 ?' Although the distribution of q was uniform at the start of the game, we are only considering cases where door 3 has been opened. this is not a random selection from the original sample set thus the posterior (meaning in this case after the host has opened a door) distribution of q is no longer uniform.

Rick, you either didn't read, or didn't understand, what I wrote. (1) The concept of the uniform q is not consistent with the problem statement, so yes I am saying it is absurd, and (2) the integral equation on p286 is a correct expression for the a posteriori probability expressed in terms of the a priori probabilities. There was no error there; and if you think there was, you need to revisit your elementary course in probability. For a similar example, the conditional probability that a playing card is a spade, given that it is black, is 1/2. P(spade|black)= [a priori probability that a card is both black and a spade]/[a priori probability that a card is black]=(1/4)/(1/2)=1/2. Changing q to q', as was done to correct this non-error, is equivalent to changing that card formula to P(spade|black)= [a priori probability that a card is both black and a spade]/[a posteriori probability that a card is black]=(1/4)/(1)=1/4. That "correction" is wrong. JeffJor (talk) 21:42, 28 October 2009 (UTC)

It was Martin who wrote the above comment, not Rick. Let us go through this more slowly. You say that a uiniform distribution of q is inconsistent with the problem statement. Why do you say this. The problem statement (Whitaker's) says nothing whatever about the host's strategy thus the prior distribution of q could be anything. Martin Hogbin (talk) 17:12, 29 October 2009 (UTC)

However, Morgan is still a mixture of inconsistent pedantry that is both correct, and incorrect, at points. Ignoring the misquote of the problem (which invalidates much of Morgan's argument, but does not invalidate the correct results if you adjust them properly), it treats "Player chooses Door #1" as a fixed part of the probability experiment (i.e., a priori information) while treating "Host opens Door #3" as the condition which separates a priori from a posteriori. Since most of Morgan's argument is based on "Solution FX is wrong because it did not prove why an a posteriori probability is equivalent to an a priori one," that was an inconsistency - Morgan ignored the same proof w.r.t. Door #1.

I agree completely that Morgan is a mixture of inconsistent pedantry. This is the main error in their paper. Above, you say that 'the host picks a q at random each show, and tells it to the contestant'. Of course we know that the host does not, in fact, tell the contestant of his door choice policy, this the parameter q should be fixed at 1/2, just as Morgan tacitly fix the initial distribution of the car at 1/3 per door.

I also notice the result at the top of page 286 is wrong: P(Ws|D3)=(p23+p33)/(p13+p23+p33). This allows for the possibility that p33>0, which is the probability the host opens Door #3 when the car is there. It also assumes the contestant will always pick the car when she can deduce where it is, something Morgan's pedantry at this point does not allow us to assume. But p33>0 is what I mean by Morgan's pedantry, since including p33 as a parameter requries us to allow that p33>0. Morgan also ignored the possibility the host would not open a door, which is more justifiable (Monty Hall did not always open a door) than p33>0.

Assuming the contestant does not know the host's strategy (remember my catch?), Morgan's approach can be treated as correct if you make one simple adjustment: The player ignores any numbers printed on the doors, and mentally labels the door she choose Door #1, the unopeneed door #2, and the opened door #3. This can be done with no loss of generality, since each door is clearly specified in this system and the contestant does not know how they relate to the host's choices in the process I am assuming. But we can also deduce p=q=1/2 by the principle of indifference. Morgan's formula on the top of page 286 is correct, and strictly speaking was derived while many of the so-called incorrect student answers were not. But the rest of the discussion, including that integral, really have no bearing on the problem as asked, because the problem does not say we know what q is. We can't assume it is a constant from show to show. The only possible answer is 2/3, by the formula on the top of page 286. JeffJor (talk) 14:58, 28 October 2009 (UTC)

I agree that the only possible answer is 2/3. The article gives gives the K&W statement as the problem to be answered, where the host chooses randomly. If this is the question then the answer is 2/3 plain and simple and Morgan is an irrelevance. Martin Hogbin (talk) 18:07, 28 October 2009 (UTC)

I've had this discussion with Martin before, but I'll note that the context given for the problem ("Suppose you're on a game show") means this problem is not an abstract urn problem with indistinguishable doors.

Yes, it is an abstract "urn problem." Even if the host distinguishes the doors, the player does not now how he is doing that. Whatever the host does differently to distinct doors is not revealed to the player. In her context, they are indistinguishable. So the rest of your comment, which I indented for consistency, is irrelevant. If you need to re-think this, look at what I said about the player using her own numbers for the doors, not the numbers the host uses. JeffJor (talk) 21:54, 28 October 2009 (UTC)

I guess we'll have to agree to disagree. The doors on a game show are clearly distinguishable since they have (as I say below) distinct, persistent, physical locations and, since we're later told (in the Parade version) to imagine the player picking "say, No. 1" the doors in addition are almost certainly labeled (consistent with the photo of the Let's Make a Deal stage). The Parade version doesn't say the host chooses randomly (if given the choice) - analyzing the effect of a preference (even if the contestant is unaware of it) seems perfectly reasonable. -- Rick Block (talk) 23:16, 28 October 2009 (UTC)

(A) Whether or not a game show typically labels doors, this problem didn't say they were labeled. Stop projecting your experience into the problem. (B) The mention of labels in the problem we are dealing with were examples, not meant to be taken literally in the answer. Morgan MISQUOTED the problem, (C) Even if they are labeled, and the problem were about Doors #1 and #3, the probabilities you calculate based on the labeling only apply if the player knows how the host uses the labels. Without that knowledge, the probability is 2/3 for switching. (D) No, it is not a contradiction that two people can attribute different probabilities to the exact same action in the exact same game. Say I draw a card, and note to myself that it is the King of Spades. I then tell Alice it is black, Bob it is a spade, Cindy it is a face card, and Doug I tell nothing. I ask them the probability it is the King of Spades. To me, it is 100%. To Alice, it is 1/26. Bob says 1/13, Cindy says 1/20, and Doug says 1/52. All are different, all are correct, and there is no contradiction. (E) It is an abstract urn problem to the player, based on the information the player has according to the problem statement. JeffJor (talk) 11:02, 29 October 2009 (UTC)

Oh, and finally: (F) The Principle of Indifference (used in Baysian Perspective) does not mean "assume a uniform distribution for anything you need a distribution for." It means you can assume that the same distribution values can be assumed for elements that are indistinguishable. It is only if all elements are indistinguishable that you can assume a full uniform distribution. Applied to this problem, it would mean you assume p=q=1/2, not that q is uniformly distributed between 0 and one. Because each q is definitiely distinguishable from another. The ln(2) calcualtion is correct when you do assume that distribution, but there is no justification for assuming it. JeffJor (talk) 16:47, 29 October 2009 (UTC)

This context provides ample justification for assuming:

1) the doors are distinguishable (physical objects on a stage with at least persistent physical identity if not big numbers 1, 2, 3 on them like the photo at http://www.letsmakeadeal.com/), and because of this we can mentally label the player's door Door #1, the unopened door Door #2, and the opened door Door #3

2) the player has the opportunity (by having viewed past airings of the show) to observe p relating to her specific initially chosen door - if players initially pick randomly, roughly 1/3 will have picked the same door our player has picked. Of these, roughly 1/3 will have initially picked the car. By observing how many of these players win by switching, p can be estimated.

Perhaps this is debatable, but I think the problem is obviously intended to be a straight conditional probability problem where the probability of interest is exactly what Morgan et al. analyze, i.e. the player has picked a specific, identified, door and the host has in response opened a different, also identified, door. Removing these aspects of the problem essentially turns it into a very wordy version of "what's the probability of correctly picking a car from behind 3 doors". Perhaps the wordiness obfuscates the problem sufficiently that people get it wrong as well. However, the conditional probability problem where your initial 1/3 pick is now one of only two possible choices is (IMO) FAR more paradoxical. -- Rick Block (talk) 20:15, 28 October 2009 (UTC)

But note, Rick, that the article now gives the K & W unambiguous description as a statement of the problem. In this the host is stated to choose randomly. The problem that Morgan answer thus clearly becomes not just an obfuscation but an extension of the MH problem in which the host is allowed to chose non-randomly. Martin Hogbin (talk) 21:22, 28 October 2009 (UTC)

But note, Martin, that even the K&W version is clearly a conditional probability problem. Morgan et al. address both the "usual" case where the host must choose randomly as well as variants (like the Parade version) where this aspect of the host behavior is not specified. -- Rick Block (talk) 23:16, 28 October 2009 (UTC)

This is pretty much where we started. To a degree it is a matter of opinion. Let me state some things that I think we agree on in the unambiguous K & W case.

The doors are distinguishable.

The sample set is reduced when the host opens a door and the conditioned set is dependent on the door opened.

On the other hand I think that you must agree (and other have noted) that there is an obvious symmetry between the two conditioned sample sets. In other words the answer clearly must be the same whichever door the host opens.

As well as the symmetry argument I am sure that you agree that it is generally accepted that no information is revealed by a random choice thus the probability that the player has originally chosen a car must be unchanged by the host's action.

I accept that there is no reliable source that treats the problem this way. (If there were I would be busy editing the article).

My opinion is that, in the light of the above, a conditional treatment of the problem is an unnecessary complication and an obfuscation of the real problem. You think that the problem must still be treated conditionally. On that point we must agree to differ. Martin Hogbin (talk) 17:00, 29 October 2009 (UTC)

(1) The doors are indistinguishable, and the problem is not supposed to be about Doors #1 and #2. This is clear from reading K&W, since thier solutions actually use other arrangements. THE DOOR NUMBERS ARE USED ONLY AS EXAMPLES.

(2) Even if they are distinguished, it does not matter UNLESS THE PLAYER KNOWS HOW THE HOST USES THE METHOD OF DISTINGUISHING THEM. This is how "the answer clearly must be the same whichever door the host opens" is achieved.

(3) There is a more convienent way to distingusih the doors: Chosen, Unopened, Opened. Once again, these names are not associated to any bias the Host has, so they can (and must, by the principle of indifference) be treated as having equal a priori probabilities for anything that needs a probability. The (a priori) chances are 1/3 that the car is behind each, and 1/2 that the two unchosen doors would be opened by the host. By the principle of indifference.

(4) No problem "needs" to be treated conditional, it just helps express the answer sometimes. Bayes' Theorem works to solve this particular problem. P(Unopened has car|Host can find a door with a goat to open)=P(Unopened has car AND Host can find a door with a goat to open)/P(Host can find a door with a goat to open)=(2/3)/(1)=2/3. The a priori probability that "Unopened has car AND Host can find a door with a goat to open" is 2/3 because the Host can always find such a door, and in 2/3 of the them the door he declined to open has the car. No, this isn't a strict proof of the answer, but a strict proof can be made of it. 23:55, 29 October 2009 (UTC) —Preceding unsigned comment added by JeffJor (talk • contribs)

The so-called “conditional solution” is invalid.

(1) The problem statement Morgan uses is a misquote. The source Morgan attributes actually says “You pick a door, say #1, and the host … opens another door, say #3.” Not “You pick Door #1, and the host … opens Door #3.” The door numbers are meant to be examples, not conditions of the problem, so that the solutions are easier to describe. It is just shorter to say “Door #1” than “The door that the contestant originally chose,” “Door #2” rather than “The door the contestant didn’t choose and the host didn’t open,” and “Door #3” rather than “the door the host opened.”

(2) While this problem sounds similar to many games used on Let’s Make a Deal, no game like it was ever played there. Specifically, Monty Hall never offered a player the chance to swap to a prize that player had already turned down (see teh letter at http://www.letsmakeadeal.com/problem.htm). He did offer players the chance to trade in a prize for new options, but never old ones. And the only game that used door numbers was the “Big Deal of the Day,” and it was played with two contestants, and all prizes had significant monetary value. The doors were revealed in the reverse order of value, regardless of whether they were chosen. So don't refer me to pictures of the show's set. Since the problem sttement didn't say the doors had numbers printed on them, or say they were distinguishable in any way, THEY AREN'T. That doesn't mean that a contestant can't create names for them, like #1, #2, and #3; it just means that they can't be related a priori to anybody else's names.

(3) The 1/N discrete probability distribution is required by the Principle of Indifference when the random elements in a discrete random process are indistinguishable, except for any names given to them. Door numbers are names. So in the prior distribution, this principle applies even when doors are numbered. They need to be treated identically.

(4) The Principle of Indifference is very hard to apply to a continuous random element, since what makes the different values “indistinguishable” is all but impossible to define. For an example, see Bertrand’s Paradox. The three answers there try to use the same technique that Morgan used with q.

(5) If we are going to assign random probabilities to strategies, based on door numbers, we also need to consider the possibility that the car’s location is determined with a similar biased strategy. (Math removed - I shouldn't try to do math and figure out how to display it, all while hurried, as I did yesterday JeffJor (talk) 14:39, 6 November 2009 (UTC).) I'm not sure what that would do to the "conditional solution," but I doubt it leaves it unaffected. The point is that the Principle of Indifference tells us what probabilities to use for the strategies, not how to vary them. JeffJor (talk) 20:47, 5 November 2009 (UTC)

[Quoted forom the Monty Hall discussion page, as stated by Martin Hogbin on 11:32, 17 November 2009.]

To start with, is there anyone who can tell me whether Morgan are answering the problem from:

1. The point of view (state of knowledge) of the player.
2. The point of view (state of knowledge) of the audience.
3. Only on the information given in Whitaker's question.
4. Only on the information given in Morgan's restatement of the question.
5. Some other basis - please specify.

This is not a minor detail but something that is universally accepted as an essential requirement in statistics - that the question being asked is clear.

Exactly right, Martin. But I prefer to use this list, of States of Knowledge (SoK), for the general problem. I'll try to divorce them from anybody's problem statement for now, and define them a little better than you did:

1. The contestant's state of knowledge, which does not include any way to assign a value to p.
2. The audience's state of knowledge, which we can assume represents a weighted average of the many different values p could take on based on any number of factors. But those factors are not seen by the audience, which is why they see a weighted average.
3. The producer's state of knowledge. She does know the value of p Monty Hall will use today, but not where the car is nor the final choice Monty makes based upon that value of p.
4. The mysterious Morgan Solver, who doesn't know p, but suspects it may not be 1/2 and assigns a uniform distribution to it without giving a valid justification.
5. Monty Hall's state of knowledge. Yes, the Host's himself. He knows where the car is, and so knows that P(win|switch) is either 0% or 100%.

To answer your question, nobody explicitly says a contestant has an SoK other than #1. This includes those who mis-quote, or misinterpret, the problem statement as "the conditional problem." Now, some solve for #2 or #3 after stating it is an alteration of the actual problem. Even Morgan starts that way. They show that the solution still has the property that, at worst, switching does not reduce the probability. Which is true. They don't always make it as clear as they should that such a solution is to a DIFFERENT PROBLEM (and I don't mean "the conditional problem," I mean "the problem from a different SoK") than what is stated, but it usually is presented along the lines of "If the SoK included a p, then the probability would be 1/(1+p)". Where they lack accuracy is that, even if the question is the conditional problem, the SoK being used has to be associated with every answer. Because from SoK #1, even the conditional probability is 2/3. And I think some people, like Rick Block, feel that is a contradiction. Is that true, Rick? It really isn't a contradiction, you know.

The way to see that it isn't a contradiction is to consider SoK #5. It is just as valid an SoK to evaluate the probability within, as any of the others except #4. But it allows only two values for P(win|switch), either 0% or 100%, and is independent of p. The contestant, stuck in SoK #1, just can't know which of those values applies to him. Just like the contestant has no way to know what p is. Another way to say that, is that it is just as valid to say "When Monty Hall knows the car is actually behind Door #1, then P(win|switch)=0 and the contestant should not switch" as it is to say "when the producer knows what p is, P(win|switch)=1/(1+p)>=1/2 and the contestant should switch. Both statements are true. They give different answers for "the probability." Neither is directly useful to a contestant in SoK #1. There is not such thing as the one-and-only value of the probability. The latter answer does bound it, and is helpful, so it makes an interesting problem extension. What it doesn't do is tell the contestant what "the probability" is.

However, SoK #4 is wrong. Morgan tries to calcualte "the probability" from any SoK. There is no law, theorem, or proposition in the field of Probability that allows Morgan, et al, to say "the noninformative prior in the vos Savant scenario makes..." That portion of their analysis is wrong.

Now Rick, I agree that the article has to be based on citable references. But that does not mean that everybody is interpreting those references correctly, or even that all are correct. Not every reference has to be used, especailly if the valid point they are trying to make is contained in opthers. And as Martin points out, whenever you say "the probability is..." you have to make it clear what SoK it applies to, and that is where references are being misinterpreted (or if SoK cannot be made clesr from them, they can't be used at all because interpretation is impossible).

We can apply some educated intelligence to clearing that mess up. In particular, any statement like "The probability is 1/(1+p) has to be qualified: "The the probability from an SoK where p is known is 1/(1+p). If p is not known in the SoK, the probability is 2/3." This is said, with varying degrees of clarity, in most of your references. But the article fails completely to make that distinction in the "Probabilistic solution" section. In particular, it is not true in general that "The conditional probability may differ from the overall probability depending on the exact formulation of the problem" as the statement is intended. What is true is "The conditional probability from an SoK that includes p may differ from the overall probability depending on the exact formulation of the problem." And it needs to be clear that no version of the problem allows for the contestant to have that SoK. We can say that an extension problem using SoK #3 can be helpful to the contestant because they all say "switch" regardless of p. JeffJor (talk) 20:36, 17 November 2009 (UTC)

[Rick Block's reply begins here, indented once. My counter-comments are interspersed, indented twice and ID'ed individually] JeffJor (talk) 14:49, 19 November 2009 (UTC)

1) Yes, the door numbers are examples. I don't think anyone (but perhaps Martin) is reading Morgan et al. to mean anything different. The problem is seemingly about the conditional probability a player faces after having first choosing a door and then seeing which door the host opens. The entire point of the problem is the difference between the situation before the host has opened a door and after the host has opened a door. Talking about this in the general case, i.e. the overall success of a pre-selected "switch" strategy vs. a pre-selected "stay" strategy destroys the essence of the problem. The analysis of a specific case, using door 1 as the player pick and door 3 as the door the host opens is representative (by renumbering the doors) of all cases. This doesn't mean the analysis is of an unconditional problem, merely that the same analysis of the door 1/door 3 combination applies to any of the 6 possible combinations of player pick and door the host opens. Similarly, an analysis of a problem about roulette using the number 17 as an example pertains to any specific number - not just the number 17.

If you allow the doors to be renumbered, it can only be about the general (unconditional) problem. That is what renumbering them accomplishes, after all. If you don't allow them to be renumbered, but the problem does not state what q is, or that a bias is even possible, you have to treat them as unbiased. I am not saying that the conditional problem, even using a specific value for q, is not useful. It shows that the host can't be trying to trick the contestant into making a choice that reduces her expected value. It just has no other relationship to the problem that was asked. JeffJor (talk) 14:49, 19 November 2009 (UTC)

Also, both "versions" of this problem represent a conditional probability. The difference between the "unconditional" and "conditional" versions, is whether a specific door or a gerneral door is opened and/or chosen. JeffJor (talk) 14:55, 19 November 2009 (UTC)

Later editing

Yes, I have tried to explain this before. The problem is that some people seem to treat the Morgan paper as a religious text and give it meaning to suit their own POV. Morgan do actually state the question that they are answering, quite clearly, they say (with my emphasis), " To avoid any confusion here is the situation: The player has chosen door 1, the host has then revealed a goat behind door 3, and the player is now offered the option to switch" The problem is that it is hard to imagine that Morgan were only considering this specific case but that is what they actually say. Despite this, some people like to try to interpret Morgan's clear statement to mean that the door numbers are examples only. The problem with this is that, as Jeff says, it blows the conditional answer out of the water. Again sometimes the Morgan faithful try to explain that Morgan really meant to say that that the player chooses any door and the host then opens a numbered and identified door and then offers the swap. It really is a great pity that Morgan did not say exactly what they meant. Martin Hogbin (talk) 18:01, 23 December 2009 (UTC)

2) Yes, no game show was ever played by these rules. However, the context of "game show" certainly implies the doors are real physical objects on the set of a TV studio. If nothing else they have positional identities (left, middle, and right). I don't know what universe you live in, but in the one I live in any two doors that are not the same door ARE DISTINGUISHABLE (and, if two doors are not distinguishable they are the same door). This means even if they aren't already numbered they can be referred to by their role in the problem as the door the player initially picks, the door the host opens, and the other door. Calling them #1 and #3 is a notational convenience.

And I don't know what mathematical universe you live in, but "able to be distinguished from each other" is not what allows the solver to treat them differently. I keep linking you to the Principle of Indifference, but you refuse to look at it. "The principle of indifference states that if the n possibilities are indistinguishable except for their names, then each possibility should be assigned a probability equal to 1/n." It is only when the problem statement allows you to distinguish how the random factors apply to the named doors, that you can assign a probability other than 1/n. "They could apply differently" is not enough, you need to know how. The problem statement does not do this. Having numbers, or having defined physical placements that allow you to create names but not to divine how random factors are applied, do not alter the problem for a simialr one where teh doors are completely indistinguishable.

It is the exact same reason EVERYBODY IN THE KNOWN UNIVERSE assumes that the car has a 1/3 probability of being behind any of the distinguishable-only-by-name doors at the start of the game, even though the problem doesn't say that, either. If we assume a bias can exist, by door number, for the host's chocie of a door to open, then we also must assume a bias is possible, by door number, for the stagehands when they drove the car into the studio. If the probability that livestock can be placed behind door #2 (the furthest from the stage hands who might have to deal with accidents) is zero, q is irrelevant and the probability the contestant wins by switching is 100%. JeffJor (talk) 14:49, 19 November 2009 (UTC)

3) I don't think anyone (again, excepts perhaps Martin) is challenging the assumption of the prior 1/3:1/3:1/3 distribution of the car. Actually, Morgan et al. note that this assumption could be relaxed as well, although they suggest this is likely of less interest.

It is included essentially as an afterthought. What they say is that possibility is "unlikely to correspond to a real playing of this particular game show situation." The same argument applies to q. Sponsers are very particular about being able to estimate the expected payouts, and demonstrating a bias that can be observed doesn't allow that. But the point is, that unlikely possibility has just as much relevance to the problem statement whether or not it is interesting. If you ignore it for lack of relevance, you are admitting that using q is also not relevant to the problem as asked. JeffJor (talk) 14:49, 19 November 2009 (UTC)

4) I think Morgan et al. would agree that applying the principle of indifference to a continuous random variable is difficult - SO THEY DON'T. I don't think anyone disagrees that the formula for the conditional probability of winning by switching (i.e. the probability in the specific example case of player picks door 1 and host opens door 3) is

P23 / (P13 + P23)

where Pij is the probability the host opens door j given the car is behind door i. By the standard rules, P23 is 1 (host must open door 3 if the car is behind door 2). The Parade description (and vos Savant's clarifications) never address P13, SO THEY LEAVE IT AS A VARIABLE. This makes the conditional probability 1/(1+p) where p is the probability the host opens door 3 given the car is behind door 1. If the host is constrained to act equally randomly in this case, p is 1/2 and the resulting probability is definitely 2/3 - but if the host is not so constrained Morgan et al. don't insist on picking a specific value. They do suggest one way to arrive at a definite value is to use a non-informative prior for the distribution of p resulting in the seemingly nonsensical answer of ln(2). I think this is more like the equivalent of picking one of the Bertrand Paradox methods - but Morgan et al. clearly are not expressing any sort of preference for this answer.

Yes, they do apply it. They don't say so, but few people do when they use it. They just give no reason for assuming "the noninformative prior" means that q is uniformaly distributed, but that principle is the only possible justification, especially when it is just presented without an attempt at justification. They assume such a distribution when they decide P(W_S|D3)=ln(2). Do you want references for why "noninformative prior" does not mean a uniform distribution?

And the point is not that they use it, but that they think there is such a prior (other than q=1/2) at all. There isn't - a basis for such a prior has to be described in the problem statement. Assuming any one prior means that you have to give equal weight to the exact opposite prior (because you can only distinguish the doors by name, not intent), and "equal weight" is what makes it difficult to apply distributions to priors. That's why the only possible prior to assume is q=1/2, since it is the only prior that needs no opposite. JeffJor (talk) 14:49, 19 November 2009 (UTC)

5) Morgan et al. suggest analyzing the effect of non-random initial placement of the car is possible, but likely of less interest. WP:OR: I think if the probabilities are C1, C2, C3, and the player has selected door 1 and the host opens door 3, the probability of winning by switching is

C2P23 / (C1P13 + C2P23)

which, because P23 is 1, simplifies to C2 / (C1P13 + C2). In this version, one might further conjecture the host and player both know the non-random C1, C2, and C3 values and figure out the optimal "play" on both the player's and host's part. I'm sure I've seen this published somewhere.

Again, the point is that any formulation of a solution with a q in it must (yes, MUST) also include C1, C2, and C3, and is incomplete without it. Isn't Morgan's entire thesis based on others' solutions to the problem are incomplete? JeffJor (talk) 14:49, 19 November 2009 (UTC)

Regarding the points about the probability being a function of the SoK, I'm fine with different SoK's resulting in different probabilities. I don't think there's any controversy about this and that everyone basically agrees the relevant SoK when solving a problem is that of someone who knows what is given in the problem statement. In the case of the MHP this can be taken to be that of the player, or the audience. However, (per my comments just above) the structure of the problem seems to be asking about the probability of winning by switching in a specific example case, leading to a straightforward conditional probability treatment. Looking at it this way makes the dependency of the answer on P13 (or, more generally, the probability that the host opens the door he's opened in the case the player has initially selected the door hiding the car) quite obvious. Unless you know this probability you can't compute an exact answer for any example case. The question is literally "should you switch". The answer, even not knowing P13 (or Pij for whatever door you've picked and door the host has opened) is YES - your specific probability of winning if you switch is between 1/2 and 1 so you're no worse off switching and you might be far better off if you do. Will players in your specific situation who have initially picked door i and have seen the host open door j win 2/3 of the time? If you say yes, you're asserting Pij (for your case) is 1/2. -- Rick Block (talk) 06:49, 19 November 2009 (UTC)

And here the point is that there is one, and only one, SoK that can be applied to the question "Should the contestant switch?" The contestant's, AS CONTAINED IN THE PROBLEM STATEMENT. And it is impossible for "the structure of the problem" to ask "about the probability of winning by switching in a specific example case" without stating what the random factors that affect that alternate case are. Branching out beyond the problem statement this way is "unlikely to correspond to a real playing of this particular game show situation."

Morgan, et al, address only the SoK of someone who knows what q is, calling their solution "the probability." They do it parametrically, but this is not the contestant's SoK, and the result is not "the probability" for anything to do with the problem as stated. Essentially, it represents an SoK from after the host chooses a q to use today, but before he decides between #2 and #3 based on that q. And it includes knowledge of what q is, which may not be observable from past history. To answer Martin, and contradict what you claim in that other discussion page about Morgan's question, it is the answer to "If you were watching the show with this fabricated SoK, how would you assess the contestant's options to switch?" It is only one possible answer among many possibilities once you start introducing unspecificed SoKs. Relying on it is no more valid than those who insist on using the SoK of Marilyn vos Savant's little green woman. It is indeed far more interesting, and has application, but not directly to the problem as asked. JeffJor (talk) 14:49, 19 November 2009 (UTC)

Oh, I just noticed I didn't address your last point, about "you're asserting [that] Pij (for your case) is 1/2." I guess I am, sorta. Because that value represents the typical host as represented in the problem statement. And yes, if the host is atypical, the calculated probability will not match experimental evidence over many games. THIS IS NOT A CONTRADICTION, because the actual game does not match the game implied by the problem statement. 2/3 is the correct probability for the game represented, but the actual game was misrepresented. Try it a different way: say the car never gets put behind door #2 on Tuesdays, but this Tuesday's contestant does not know that and is not told. The host tells her that he picks randomly if able, and opens #3 when she picks #1. To the best of that contestant's knowledge, should she switch? Yes. (If you say "no," what basis do you have for that decision, how does the contestant know when it applies, and why does it not apply to Morgan's problem?) To the best of her knowledge, her probability after switching is 2/3. The supposed "contradiction" here is that the game was not correctly represented to her, not that she miscalculated anything. JeffJor (talk) 19:25, 19 November 2009 (UTC)

What exactly makes a probability problem conditional.

The WP article on the subject states that, 'Conditional probability is the probability of some event A, given the occurrence of some other event B'. This means that we must make some kind of judgement as to whether event B can possible affect the probability of event A, otherwise we would have to take every event into account for every probability problem. Martin Hogbin (talk) 21:44, 20 December 2009 (UTC)

Your searching in vain for an argument to avoid conditional probabilities. The event "opening door 3" is part ot the "experiment" and hence we have to condition on it. Nijdam (talk) 13:42, 21 December 2009 (UTC)

Exactly why is this so and not me clapping my hands, for example. Martin Hogbin (talk) 21:34, 21 December 2009 (UTC)

As I said: clapping your hands is not an event in the problem (read the problem statement). Nijdam (talk) 16:30, 22 December 2009 (UTC)

So you re saying that any event mentioned in the problem statement is a condition? Martin Hogbin (talk) 17:14, 22 December 2009 (UTC)

So far I have.

1) Conditional probability is the probability of some event A, given the occurrence of some other event B.

2) The event must reduce the sample space.

3) The event must occur before the question is asked.

4) The event must be 'part of the experiment'.

5) The event must be mentioned in the problem statement.

For an event to be a condition must it be one of these, some of these, all of these?

Martin Hogbin (talk) 20:31, 22 December 2009 (UTC)

Do not make to much fuss about it. If you perform some experiment, the possible events are known to you. If you know one of these events has occurred, you just have to account for the outcomes that are still possible. In theory this is achieved by reducing the sample space to the given (occurred) event. Draw a card from a complete deck. If you see the card is a spade card, the outcome can only be a card of spade. That's all. Nijdam (talk) 10:03, 23 December 2009 (UTC)

But I am not the one making the fuss. All I want is a simple explanation of a simple problem but people keep telling me that the problem must be treated conditionally, yet no one can tell me precisely why. Martin Hogbin (talk) 10:29, 23 December 2009 (UTC)

You're the one asking about what conditional prob. means. And you have been told more then once precisely how things are. Please study them. Nijdam (talk) 14:20, 23 December 2009 (UTC)

I only ask because you and others insist it is mentioned in every solution to the MHP, if you stop insisting that, I am happy to drop the subject. I have studied what you have said, plus the WP article on the subject but unfortunately no answer is provided. Why not just answer my question? What are the exact criteria for a given event to be considered a condition in a probability problem? Martin Hogbin (talk) 15:10, 23 December 2009 (UTC)

I explained some lines above completely what cond. prob. is about. And: the only MHP that will be of interest to, I guess, all interested readers, is the one where a door has been opened and then the player is offered the choice to switch. If you have another version in mind, please tell me. Nijdam (talk) 13:09, 24 December 2009 (UTC)

BTW: I laid my cards completely open to anyone. But I have no idea what your opinion is of what version or versions to consider the MHP. You mentioned somewhere: what sources say. So please tell me what sources say. And formulate a solution. This is not to try you, I will be glad in assisting you doing so. Nijdam (talk) 13:14, 24 December 2009 (UTC)

What Exactly Makes Probability A Requirement To Solving This Puzzle?

I'm no expert on these things, but do the various published simple solutions rely on Probability? Glkanter —Preceding unsigned comment added by Glkanter (talk • contribs) 23:22, 20 December 2009 (UTC)

Yes. Martin Hogbin (talk) 00:06, 21 December 2009 (UTC)

Maybe for the original 1/3, I can see that. And for the total of all outcomes equaling 1. Am I allowed to mix disciplines when solving a problem logically? Glkanter (talk) 01:03, 21 December 2009 (UTC)

The question, in pretty well any form, asks the question, 'What is the probability...'. That makes it a probability question in my book. 86.132.187.240 (talk) 15:28, 21 December 2009 (UTC)

The question is 'Should the contestant switch?' Glkanter (talk) 18:15, 23 December 2009 (UTC)

Yes you are right, but if this question is not to be answered on the basis of the probability ow winning by switching, on what other basis should it be answered? Martin Hogbin (talk) 17:14, 14 January 2010 (UTC)

I was only pointing out to 86.132.187.240 that the actual question does not include the word 'probability'. He needs a different argument. Oh, no. That's not you, is it? Besides, is 'What is the probability the sun will rise in the east tomorrow?' a probability question?

Yes. (86.132.187.240 was me, by the way, I forgot to sign) Martin Hogbin (talk) 17:40, 14 January 2010 (UTC)

Maybe to a Mathematician. And I suppose he can solve it. Would an Astronomer call it a probability question? Could the question reasonably go on an Astronomy test? Would it be an Astronomy problem or a Probability problem? Maybe a little of both? Glkanter (talk) 18:00, 14 January 2010 (UTC)

The real response to your question is 'Logic'. Here's part of it. There's 2 goats, and the contestant can only choose 1 of them at most. Hence, there is always a goat for Monty to show. That's not probability. It's logic, and a very big part of the simple solutions. The first sentence of the Article is wrong. The MHP is not strictly a probability puzzle. In fact, it's a story problem that begins, "Suppose you are on a game show...". Maybe that's where certain editors get confused. It's not a test question on a university exam. Glkanter (talk) 17:31, 14 January 2010 (UTC)

Because the result of what the player will get if they choose to switch is uncertain, it depends on the random position of the car, the problem must be about probability.

On the other hand, I completely agree that, "It's not a test question on a university exam", I have made this same point myself many times. Martin Hogbin (talk) 17:40, 14 January 2010 (UTC)

Is the problem to be answered from the player's point of view?

We are to see the problem through the eyes of the player. The host offers her the opportunity to change her choice. And, as far as I know, almost everyone (except you and ...) put themselves in the position of the player and consider the door opened. Actually we also have the same information: door 1 (or another) chosen and door 3 (or another) opened. Nijdam (talk) 13:56, 21 December 2009 (UTC)

This depends on the basis on which we are addressing the problem. As I understand it there are three possible bases on which probability can be dealt with: frequentism, Bayesian probability, and the more formal modern probability theory.

The Morgan paper seems to be promoting the modern probability theory, in which the problem must be answered strictly from information given in the question, they make this point several times in their article. What we assume that a person involved may or may not know is unimportant, we must solve the problem only on the information given in the question, which is used to assign intrinsic probabilities to events in the sample space. Information not given cannot be used to do this. If the problem statement does not say which door the host has opened then this data cannot be used in answering the question. Rick and others have made the point that according to Morgan it does not matter whether the player knows about the hosts door opening policy. This only applies if they are using modern probability theory.

If on the other hands we are Bayesians we can decide on whose state of knowledge we are to answer the question from. I agree the players POV is the most obvious to take. We might also choose to make some reasonable assumptions about reality. How a host might act, what a player might see etc. If that is to be the case I would have to rephrase my question to make clear that the player does not see which door was opened. Or do you insist that, even in this case the problem is conditional? Martin Hogbin (talk) 21:54, 21 December 2009 (UTC)

In its simple form it may be addresed in a frequentistic or formal way, both essentially equivalent. And both asking for conditional probs. The Bayesian approach is a theoretical extension, meant for experts, in my opinion not to be mentioned in the article. But it also asks for conditional probs. Where are you heading at? Nijdam (talk) 16:38, 22 December 2009 (UTC)

In my problem statement I did not say which door the host opens, thus, according to your propose rule above, the door opened cannot be taken as a condition. My problem statement is exactly equivalent to Rick's first problem, the door opened is not identified in the problem statement just as the we have no information on which ball the host reveals in Rick's problem. Of course we are free to speculate on whether the host might actually be able to distinguish between the two red balls just as we may speculate on what the player might see but neither is part of the problem as stated. Martin Hogbin (talk) 17:21, 22 December 2009 (UTC)

What urn problem is the appropriate model

There are basically two ways to turn the MHP into an urn problem.

1) The host puts three marbles, indistinguishable except two are black (representing goats) and one is white (representing the car), in a bag. The player withdraws one without looking at it. The host looks in the bag and withdraws a black marble and shows it to the player and the audience. The player can now switch the marble she's withdrawn for the remaining marble in the bag. The choice is between two marbles, one that is white and one that is black. Should the player switch?

2) The host puts three marbles, indistinguishable except one is red (representing a goat), one is black (representing a goat), and one is white (representing the car), in a bag. The player withdraws one without looking at it. The host looks in the bag and withdraws a losing marble, and shows it to the player and the audience. The player can now switch the marble she's withdrawn for the remaining marble in the bag. Let's say the host withdraws the red one meaning the choice is between two marbles, one that is white and one that is black. Should the player switch?

I think those who are favoring the unconditional solution are thinking the first version is the most appropriate analogy. For this problem there are two scenarios

1a) Player withdraws a white marble, leaving two black marbles in the bag. Host now withdraws a black marble leaving a black marble in the bag. (probability 1/3)

1b) Player withdraws a black marble, leaving a white and black marble in the bag. Host now withdraws the black marble, leaving the white one. (probability 2/3)

The second version is more complicated and seems to have the same definitional issues as the MHP. Is the question specifically about the case where the host withdraws a red marble or not? Whatever the question is, it's clear that there are two actions (initial player pick and then host withdrawal) whose probabilities combine in ways that are not immediately obvious. Enumerating the possible cases as above is not that helpful

2a) Player withdraws the white marble, host withdraws the black one.

2b) Player withdraws the white marble, host withdraws the red one.

2c) Player withdraws the black marble, host withdraws the red one.

2d) Player withdraws the red marble, host withdraws the black one.

It's perhaps easy to see that the first two of these are subcases of a case with probability 1/3 where the player withdraws the white marble and the other two cases each have probability 1/3. However, if the question really is asking about the probability given the host withdraws the red marble we clearly have a conditional probability problem.

The sources presenting unconditional solutions to the MHP are treating it as if it is equivalent to the first of these urn problems. The sources that say it should be treated as a conditional probability problem are saying it's equivalent to the second of these urn problems. This is a POV issue. Neither one is "more right" than the other. It's simply different POVs. -- Rick Block (talk) 01:04, 22 December 2009 (UTC)

Only Bosons are truly indistinguishable so I suspect that Nijdam might insist that even this case [the first version] is strictly one of conditional probability as one or other of the black marbles must in fact have been shown.

There really is no fundamental difference between these two cases, in both cases there are two goats represented by two different marbles. My point is that there is no absolute clear distinction between a conditional problem and an unconditional one, your problems above demonstrate this clearly. This it is, to some degree a matter of choice as to how we treat a given problem. The question that we must ask is whether the potential condition could affect the probability that we are trying to calculate. Martin Hogbin (talk) 12:55, 22 December 2009 (UTC)

Some sources say that the MHP should be treated as a conditional probability problem, some do not, so you are quite right it is a POV issue but there are clearly two valid POVs, both of which should be represented in the article. All I have ever asked for is that a convincing explanation the simple problem, or what I will call the non-conditional problem (meaning that the problem is either stated in an unconditional form, or that it is treated unconditionally by virtue of its inherent symmetry) is shown at the start of the article, together with a discussion of reasons for confusion for the non-conditional case. After that the issue of conditional probability can be discussed for those interested. Martin Hogbin (talk) 12:43, 22 December 2009 (UTC)

To me the first representation is not equivalent to the MHP. I know some sources say it is. May be there even are "sources" advocating the 50/50 "solution". What ever. Anyone (almost) accepts a door has been opened. Only because one wants the simple solution to work, one tries to argue the opened door is not identifiable. Nijdam (talk) 17:05, 22 December 2009 (UTC)

You are quite right, I do want to use a simple model to start with, but as you can see from the two problems above there really is no sharp division between a conditional and an unconditional problem. If you disagree you must tell me exactly what makes the distinction. Martin Hogbin (talk)

Martin - Nijdam has answered this repeatedly. There is a sharp difference, and it is whether the condition reduces the size of the event space. For example in the second problem above, withdrawing the red marble means we're only talking about cases 2b and 2c. The only way the host opening a door in the MHP does not reduce the size of the event space is if the doors are indistinguishable. You can certainly change the problem description to make this the case (the contestant is blind and the host says he's opened a door revealing a goat but doesn't say which door, or the two unselected doors are hidden behind a single curtain and a goat is produced from behind the curtain, or the player must decide before the host opens a door, or ...). Do you agree any of these are changes?

Rick - Reduction of the sample space is not an answer because it then depends on how you set up your sample space. It is also not supported by Conditional probability which says just, 'Conditional probability is the probability of some event A, given the occurrence of some other event B'. In my example in which I clap my hands before choosing a door I can set up my sample space to include the options of clapping and not clapping. This makes it a condition. Martin Hogbin (talk) 19:36, 22 December 2009 (UTC)

Nijdam - Do you agree the first problem is NOT "conditional"?

Assuming we get to mediation, we'll rehash all of this there so there's really not much point in continuing this conversation here. I offered these two urn problems as a way to illustrate the difference between the approaches. As Boris says The coexistence of the conditional and the unconditional can be more peaceful. Ultimately, these are two sides of the same coin. The coin is not fully described unless both sides are considered. NPOV says we must present both, and can't favor either one. -- Rick Block (talk) 18:52, 22 December 2009 (UTC)

Rick, would you describe the current article as adhering closely to "we must present both, and can't favor either one."? Glkanter (talk) 19:06, 22 December 2009 (UTC)

You're talking about this version, and I wouldn't claim it's perfect but I think it is fairly close - I would strongly prefer a single solution section. Note that "neutral" here means that it says what the sources say fairly, in a disinterested tone. -- Rick Block (talk) 19:51, 22 December 2009 (UTC)

Rick, what about the text and graphic volume of the content? Or all the time spent on so-called 'variants'? Or the talk page FAQs, which are barely sourced, more like riffed? These are all part of the 'Article', not just the Solutions section. If they were allegedly NPOV when you insisted Morgan was dominant, how can they still be NPOV when you now argue both sides should be equal? Glkanter (talk) 20:19, 22 December 2009 (UTC)

But that must be all the sources, without giving one source the right to veto the others. Martin Hogbin (talk) 19:56, 22 December 2009 (UTC)

I agree that, 'The coexistence of the conditional and the unconditional can be more peaceful', this is what I have always wanted. One solution might be to treat the sources in rough chronological order, with earlier solutions first. As far as I know no sources before Morgan state the problem must be conditional. Martin Hogbin (talk) 19:59, 22 December 2009 (UTC)

MHP?

1. The player is informed about the rules, but still in the dressing room, before being on stage, she is asked whether she will switch. Simple solution, no need of conditional probs.

I consider this version as the standard MHP: (Insert your name here)

Nijdam - Let us assume that we know that the host will always choose the highest available door, within the rules, to open. Can you show me how you would calculate the probability of winning by switching in this case using correct notation. Martin Hogbin (talk) 19:42, 22 December 2009 (UTC)

2. The player is informed about the rules, and after she made her first choice, the host tells her he will shortly open one of the remaining doors. Before doing so, he asks her whether she likes to switch to the door he will leave closed. Simple solution possible, although one formally has to condition on the player's choice.

I consider this version as the standard MHP: (Insert your name here)

3. The player is informed about the rules, and after she made her first choice, the host opens one of the remaining doors. He then asks her whether she likes to switch. The solution needs to consider the conditional probability. (And I add: the player knows which door she has chosen and sees which door has been opened, hence the solution is a function of these door numbers.Nijdam (talk) 10:06, 23 December 2009 (UTC))

I consider this version as the standard MHP: (Insert your name here)

Nijdam

4. Any other scenario.

I consider another version as the standard MHP: (Insert your name here)

For me only the 3rd form is of interest. Nijdam (talk) 17:39, 22 December 2009 (UTC)

Just out of curiosity and to know where everyone stands, I invite all the participants in the discussion to just mention there name.Nijdam (talk) 10:17, 23 December 2009 (UTC)

Nijdam, I do not think that you will get a good response to your request. As I am sure Rick will agree, it is not up to us to determine what the 'standard MHP' is, or even if there is such a thing, we must rely on sources. On the other hand I do not know of any source that says 'This is the standard MHP', we therefore have some freedom to decide what the problem is and how it should be answered. My suggestion is that we should use this freedom to start with a formulation that allows us to provide a simple, non-conditional solution. We can then discuss other formulations and their solutions later. So, to continue with a question you asked elsewhere: yes, I do want to change the question to suit the answer but I believe we can do this and still maintain the quality of the article. Martin Hogbin (talk) 10:43, 23 December 2009 (UTC)

I know it is not up to us to determine, yet I'm interested in were you and others stand. Nijdam (talk) 14:12, 23 December 2009 (UTC)

I cannot answer the question as asked as there clearly is no 'standard' formulation. I would say we should start with any non-conditional formulation (certainly this is the notable case) and then proceed to a conditional one. This is in line with the sources, some of which treat it one way and some the other. Martin Hogbin (talk) 17:44, 23 December 2009 (UTC)

Nijdam - You now seem to be using time to determine what is conditional and what is not, can you show any reference that states that this is what makes an event a condition. In the first two examples we know the host will have to open a door and that it might matter which door he opens. Why can we ignore this just because it will happen in the future? Martin Hogbin (talk) 19:42, 22 December 2009 (UTC)

Here's An Opportunity:

Convince me that Huckleberry could have done better if someone had explained the 'equal goat door constraint must equal 1/2' to him. Or, that 'maybe it didn't equal 1/2'.

I'll become open to compromise, and I'll stop believing that some editors are being obstructionist.

On the other hand, if no one comes forward with a convincing argument, my views will remain unchanged. Glkanter (talk) 13:15, 23 December 2009 (UTC)

Okay then, an oppoprtunity for you: how do you know Huckleberry wins 2/3 of the times?Nijdam (talk) 14:16, 23 December 2009 (UTC)

Various reliably published sources say so, some using the Combining Doors solution, like Devlin. Glkanter (talk) 14:46, 23 December 2009 (UTC)

Nijdam, please don't treat this as you giving me a test I must pass. I think prominent Wikipedia editors have answered all your questions and challenges about symmetry, and whatever else, and I'm not aware that the 2/3 is even being debated. I don't think it's necessary for each editor to 'prove' to your satisfaction his or her master of Probability. This is the MHP paradox article, only. Please respond only about the Huckleberry section as I originally started this section, thank you. Glkanter (talk) 15:02, 23 December 2009 (UTC)

Lets say Huckleberry is not a contestant but works for the FCC. His job is to make sure the show is fair. He doesn't know much about probability so looks up the problem on Simplepedia. The article has the "combining doors" solution and various other "simple" explanations that convincingly say that 1/3 of the time you should win by staying and 2/3 of the time you should win by switching. He watches the show. He keeps track of how many times players who switch win and how many times players who stay with the first choice win. He sees 3000 shows in which 2700 players stay with their original choice and only 300 switch (people think it doesn't matter so most stay with their original choice). Of the 2700 "stickers" 900 win, and of the 300 switchers 200 win. 1/3 of the stickers win and 2/3 of switchers win. Perfect. He sleeps soundly, convinced everything is OK.

But, it's a game show, so it's run by humans. It turns out what's really going on is they perfectly randomize the initial car placement but the host is told only "open one of the other doors if the player initially picks the car". The host figures it doesn't matter which one he picks in this case, so doesn't pay any particular attention to it.

Huckleberry's boss also doesn't know much about probability and looks up the problem on Wikipedia which has a reference to the Morgan et al. paper and says that how the host picks between two goats matters. The boss thinks about it and decides to see if Huckleberry is really doing his job. The boss looks at tapes of the show and keeps track of the 6 individual pairs of player pick and door the host opens separately, i.e. player picks door 1 host opens door 2, player picks door 1 host opens door 3, etc. The boss determines that even though 2/3 of the players who pick door 1 win by switching (and 1/3 win by staying) players who pick door 1 and see the host open door 2 would win by switching with probability about .625 (rather than .66) and players who pick door 1 and see the host open door 3 would win by switching with probability about .714 (rather than .66). Using the Morgan et al. formula he found on Wikipedia, the boss works backwards, and solves .625=1/(1+p) and .714=1/(1+q) and sees that instead of p=q=1/2 the host is exhibiting a p=.6 preference for the leftmost door and a q=.4 preference for the rightmost door. That's odd, he thinks. Exactly 2/3 of the players who switch win and exactly 1/3 of the players who stay win, but yet the host is exhibiting a slight preference for one goat door over another.

Looking into it some more, the boss notices that the host seems to open door 2 more often on Mondays, Wednesdays, and Fridays, and less often on Tuesdays and Thursdays. The boss talks to the host and the host says to make it "random" he always picks the leftmost door (if he has a choice) on Mondays, Wednesdays and Fridays, and the rightmost door on Tuesdays and Thursdays.

Huckleberry is fired.

All because Simplepedia never said anything about how much it matters how the host chooses between two goats when he gets the chance. -- Rick Block (talk) 20:21, 23 December 2009 (UTC)

Let's limit this to only those variants consistent with 'Suppose you're on a game show', and answers that are intended for the typical Wikipedia reader. The FCC has plenty of books they can rely on. Thanks. Glkanter (talk) 20:51, 23 December 2009 (UTC)

In your opinion Rick, are the explanation and the risk you described above your reasons for advocating for 'equal treatment' in the article of the simple solution and Morgan's solution? Glkanter (talk) 23:31, 23 December 2009 (UTC)

No. My reason for advocating for 'equal treatment' is because the simple solution and the "conditional" solution are both significant POVs published in reliable sources. How about if we just leave this discussion for mediation? -- Rick Block (talk) 05:24, 24 December 2009 (UTC)

Rick forgot to mention the show where the car is placed behind door 1 on Mondays, door 2 on Tuesdays and door 3 on Wednesdays, the only days that the show is on. Huckleberry, who knows all about the effect of host preference from reading the paper by Morgan is still puzzled to discover that the chances of winning by switching are not what he expects. Finally he wakes up from his dream to watch the real show where the contestant has no idea of either the host preference or the car placement so he just chooses randomly and wins 2/3 of the time by switching. Martin Hogbin (talk) 00:03, 24 December 2009 (UTC)

More detailed response to Rick's contrived scenario

This is an interesting but somewhat contrived scenario that indicates the errors of thinking by those who support the Morgan approach that deserves a detailed response.

Lets say Huckleberry is not a contestant but works for the FCC.

This is a bad start. I and others have explained many times that any probability problem must be answered from a defined state of knowledge (or, if you are using modern probability theory, defined distributions based on the information given). I do not think this statement is in serious doubt. It has always been assumed by everyone that we treat the problem from the SoK of a contestant on the show. Whitaker's questions starts, 'Suppose you are a contestant...'. But let is see how it looks from anothe angle.

His job is to make sure the show is fair. He doesn't know much about probability so looks up the problem on Simplepedia. The article has the "combining doors" solution and various other "simple" explanations that convincingly say that 1/3 of the time you should win by staying and 2/3 of the time you should win by switching. He watches the show. He keeps track of how many times players who switch win and how many times players who stay with the first choice win. He sees 3000 shows in which 2700 players stay with their original choice and only 300 switch (people think it doesn't matter so most stay with their original choice). Of the 2700 "stickers" 900 win, and of the 300 switchers 200 win. 1/3 of the stickers win and 2/3 of switchers win. Perfect. He sleeps soundly, convinced everything is OK.

And so he should, the show is indeed perfectly fair to the contestant in that the overall odds come out as would be expected. The players who lose might complain that it was unfair to then but that is just bad luck. Overall the show is fair.

But, it's a game show, so it's run by humans. It turns out what's really going on is they perfectly randomize the initial car placement but the host is told only "open one of the other doors if the player initially picks the car". The host figures it doesn't matter which one he picks in this case, so doesn't pay any particular attention to it.

This is bizarre in the extreme, they randomise the car placement but implement some peculiar rule to decide which door to open rule. I take your statement to mean that the host is told to implement some unspecified rule to decide which door to open when there is a choice. Still, I suppose it could happen.

Huckleberry's boss also doesn't know much about probability and looks up the problem on Wikipedia which has a reference to the Morgan et al. paper and says that how the host picks between two goats matters. The boss thinks about it and decides to see if Huckleberry is really doing his job. The boss looks at tapes of the show and keeps track of the 6 individual pairs of player pick and door the host opens separately, i.e. player picks door 1 host opens door 2, player picks door 1 host opens door 3, etc. The boss determines that even though 2/3 of the players who pick door 1 win by switching (and 1/3 win by staying) players who pick door 1 and see the host open door 2 would win by switching with probability about .625 (rather than .66) and players who pick door 1 and see the host open door 3 would win by switching with probability about .714 (rather than .66). Using the Morgan et al. formula he found on Wikipedia, the boss works backwards, and solves .625=1/(1+p) and .714=1/(1+q) and sees that instead of p=q=1/2 the host is exhibiting a p=.6 preference for the leftmost door and a q=.4 preference for the rightmost door. That's odd, he thinks. Exactly 2/3 of the players who switch win and exactly 1/3 of the players who stay win, but yet the host is exhibiting a slight preference for one goat door over another.

Huckleberry's boss then thinks a bit harder and realises that the host action is not important from the contestant's perspective unless the contestant is aware of what the host's policy is. He decides to consider the to possibilities separately. Firstly he imagines that the contestant has not watched tapes of the show and thus will have no idea of how the host decides which door to open. He concludes that, as the contestant has no knowledge of the host policy, and, as the overall chances of winning by switching are fair, there is no problem. He briefly worries about how this relates to an individual contestant but soon realizes that although from the Sok of the producer the odds are either 1 or 0 depending where the car was placed the odds from the SoK of the player are 2/3.

The boss then wonders about the player who may have studied the show and found out about the host's strange door policy. He soon realises that the player could use this information to gain an advantage over the TV station, although it looks as though no one has done this so far. Anyway that is the station's problem, if they want to give the player an advantage by using an non-random policy that is their problem.

Looking into it some more, the boss notices that the host seems to open door 2 more often on Mondays, Wednesdays, and Fridays, and less often on Tuesdays and Thursdays. The boss talks to the host and the host says to make it "random" he always picks the leftmost door (if he has a choice) on Mondays, Wednesdays and Fridays, and the rightmost door on Tuesdays and Thursdays.

The boss wonders why they choose to give the player an advantage in that way and buys Huckleberry a drink.

Martin Hogbin (talk) 11:38, 24 December 2009 (UTC)

Sorry lads, for all your theoretical considerations, but the shows only took place on Fridays (except on Good Friday of course). Nijdam (talk) 13:02, 24 December 2009 (UTC)

Of course, the shows were completely different from the MHP as it is generally known. Monty in fact never offered the swap on the real show. I was responding to Rick's rather contrived situation and erroneous conclusions. When are you going to tell me exactly what makes an event an condition in a probability problem. Yes, you have told me before but four different answers, none of which agrees with the answer given in the WP article. Martin Hogbin (talk) 13:36, 24 December 2009 (UTC)

Yes it's a contrived scenario. The point is only that the chance of winning for an individual player on the show or even of potentially identifiable subgroups of players on the show does not have to be the same as the overall average across all players. All players get the same odds (2/3 chance of winning by switching) if and only if all of the following conditions are met

You are quite right that if the host has a particular door opening policy (say, always opens rightmost door permitted by the rules) then the subgroup of players who themselves happen to have a (probably unwitting) appropriate 'policy' of deciding whether to switch or swap will do better than those who have the wrong policy (it might turn out that the player's initial door choice is important too). But, the players cannot know any of this as they do not know the host's door opening policy. A player, or a member of the the audience, or us (who only have the information given in the question) cannot know which subgroup a player is in thus from that particular state of knowledge they all have the same chance.

So you are right insofar that subgroups of players who are identifiable to someone who knows the host door choice policy may not all have the same chance of winning by switching, but who cares? Certainly not the player (or the audience or the FCC). For any given player it may turn out that they are in the 'lucky' subgroup or it may turn out that they are in the 'unlucky' group but they still have a 2/3 chance of winning by switching. To explain this again, overall the players who switch have a 2/3 chance of winning therefore from the player's state of knowledge she has a 2/3 probability of winning, that is what probability means. Sure, someone with greater knowledge, say the producer who is watching the show or even someone who knows only the host door opening policy, may have a better idea of which players will win by switching, but neither the player, nor members of the audience, nor those who only have the information presented in the question have any basis on which to assign any probability other than 2/3 to this event. Martin Hogbin (talk) 12:57, 30 December 2009 (UTC)

1) the car is uniformly placed

Yes.

2) the player's choice is independent of the initial car placement (it doesn't do any good to randomize the car placement if a stagehand tells the contestant where the car is)

Sure.

3) the host must open a door with knowledge of where the car is and must reveal a goat

Never disputed.

4) the host must choose which of two "goat doors" to open equally randomly

See my reply above.

5) the host must make the offer to switch

Of course. In fact he must be known to always do this.

6) We should add that the player has not studied the history of the show. Otherwise this becomes more of a game theory problem (in which the host can do no better to act randomly). Martin Hogbin (talk) 12:54, 30 December 2009 (UTC)

If conditions 1-5 are in effect, it doesn't matter if the player has studied the history of the show - the probability for each and every player on the show will be exactly the same as the probability for any other player. This is my exact point. And this is exactly what the FCC cares about, i.e. whether it is possible for a player to gain an advantage over other players. -- Rick Block (talk) 17:44, 30 December 2009 (UTC)

You are right in that 'if conditions 1-5 are in effect, it doesn't matter if the player has studied the history of the show ' but as conditions 1-5 may not always be the applied. I think it is wise to make clear whether we consider the possibility that the player might study form. I have to admit that I do not know exactly what the FCC cares about, I am not sure that they would worry about a player being able to gain some advantage over the expected odds by studying form, especially as this is does not reduce the odds for other players, but maybe they would object to this. If I were them I would not care if TV stations were silly enough to let viewers gain an unexpected advantage over them. Nobody would mind if a casino gave a 40:1 payout for a single number in roulette, in fact there is no legal objection to players of blackjack card counting to gain an advantage over the house. It is up to the casino to take whatever steps they feel necessary to avoid losing money. Martin Hogbin (talk) 12:59, 1 January 2010 (UTC)

The "unconditional" solutions ignore #3 through #5 and say the answer of winning by switching is 2/3 regardless of whether these are in effect. Another way to put it is that these solutions assume these conditions. Yet another way to put it is that these solutions are valid only if these conditions hold. All of them, except #4, were explicitly addressed by vos Savant in her followup columns, in particular the explicit experiment she suggested in her second followup. In my opinion, the solution is not complete unless each and every one of these conditions (including #4) are mentioned. -- Rick Block (talk) 17:10, 24 December 2009 (UTC)

That is exactly why I mention my point six. We can drop point 4 and, provided the player has no information on the hosts door opening policy, either from studying history of the show or in some other way, the odds of winning by switching from the players SoK are still always exactly 2/3, individually, overall, any way you like.

I am not sure why you continue to complicate this d1scussion with points 1,2,3,5. We have long since agree these standard game rules. The only issue being considered is whether the host door choice policy matters. To those who do not know what it is it does not. Martin Hogbin (talk) 12:54, 30 December 2009 (UTC)

See example in new section below. 86.132.187.240 (talk) 15:18, 30 December 2009 (UTC)

Note: If X (choice) and C (car) are independent: P(C=X)=SUM(c) P(X=c|C=c)P(C=c)=SUM(c) P(X=c)P(C=c). As a consequence if either X or C is uniformly distributed P(C=X)=1/3. Hence instead of #1 the choice may be uniformly made.Nijdam (talk) 13:15, 28 December 2009 (UTC)

I think not. If the host has a choice for a particular numbered door for example, the initial car placement and the player's choice should both be random. This is what I was trying to show in my analysis page. I would be interested in completing this page with you, using whatever notation you prefer, as I think it shows some interesting results. Martin Hogbin (talk) 13:07, 30 December 2009 (UTC)

In formulas

2/3=P(Winning by switching)="average over the possible scenario's"=

=SUM P(winning by switching|scenario)P(scenario)

If we know P(winning by switching|scenario) is the same for every scenario, we may conclude that for every scenario:

2/3=P(winning by switching|scenario)

And note: we decide on the basis of the conditional probability.Nijdam (talk) 18:08, 26 December 2009 (UTC)

For better understanding

Much of the confusion lies in the formulation. So, for a better understanding, I give you the following consideration:

What is the probability the car is behind the chosen door? Most people will not hesitate to answer: 1/3, assuming of course the car is placed randomly. And they are right, numerically, but do they understand the situation? Let us put the car half the times behind door 1, and 1/4 of the times behind each of the other doors. Again: what is the probability the car is behind the chosen door? Answer? Nijdam (talk) 10:34, 27 December 2009 (UTC)

If the choice is random without knowledge of which door is which, then from the player's state of knowledge 1/3. This is (of course!) a conditional probability problem. The player's random choice has a 1/3 chance of being each door, so the player's chance is (1/3)(1/2) + (1/3)(1/4) + (1/3)(1/4) = (1/3)(1/2 + 1/4 + 1/4) = 1/3. Uniform distribution means the choice might as well be random. Random choice means the distribution might as well be uniform. -- Rick Block (talk) 18:19, 27 December 2009 (UTC)

I never had any doubt you wouldn't know the answer. The point is of course, we don't know the distribution of the choice. Nijdam (talk) 23:22, 27 December 2009 (UTC)

Two urns

There are two identical urns. One contains 10 white balls and 20 black balls, the other contains 20 white balls and 10 black balls. A person picks a ball from one of the urns. Using only the information given, what is the probability that it is a black ball? Martin Hogbin (talk) 15:22, 30 December 2009 (UTC)

This seems the same as Nijdam's question just above. How about if there are 20 white balls (and no black balls) in one of them and 40 black balls (no white) in the other? Or 20 white balls (no black) in one and 20 each white/black in the other? I'm not sure where you're going with this, but my point is that these are all conditional probability questions. -- Rick Block (talk) 17:32, 30 December 2009 (UTC)

Perhaps you would care to answer my question. What is the probability that a black ball is chosen? Then I will answer yours. Martin Hogbin (talk) 17:45, 30 December 2009 (UTC)

I thought I did answer it by referring to the section above. More explicitly, there's a 2/3 chance of picking a black ball from one of them and a 1/3 chance from the other. Assuming the person is picking randomly between the urns, the overall probability of ending up with a black ball is therefore (1/2)(2/3 + 1/3) = 1/2.

And, ah yes, where you're clearly going is that this means that the host's preference between two goats doesn't matter since the player doesn't know it. In your urn problem what you're asking is the overall chance. No one is arguing that the overall chance of winning by switching in the MHP is anything other than 2/3. We all agree on average, across all players, 2/3 who switch will win regardless of any host preference between two goats. However, just like your urn problem where an individual player has a 1/3 or 2/3 chance depending on which urn they pick from which averages out to a 1/2 chance, in the MHP an individual's chance depends on the host preference and only averages out to a 2/3 chance.

That is what probability means. A single player on the show has a 2/3 probability of winning, that is it. There is no separate probability and overall probability, except in the unrealistic case that the host's door opening policy is known.

To make your urn problem more similar to the MHP rephrase it like this. There are two urns, A and B, containing only white and black balls. Randomly choosing between urns, a person's probability of picking a black ball is 1/2. If a person picks a ball from urn A what is this person's probability of picking a black ball?

I suspect your answer would be 1/2, and there is certainly a sense in which that is the right answer. But if we aren't given that the fraction of black balls is the same in both urns, another answer is "anything at all" since all we really know is that if the fraction of black/total balls in urn A is X it must be (1-X) in urn B. -- Rick Block (talk) 19:07, 30 December 2009 (UTC)

Your problem above is not a good model of the MHP unless the question is changed to, 'If a person picks a ball from one of the urns what is this person's probability of picking a black ball', or in the MHP the player is constrained to always initially choose door 1. Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

The point is that whatever host policy you decide upon, (host always pick door 1, picks leftmost door, picks centre door, in all cases, where allowed by the rules) the player's initial door choice can change the effect of the host's choice. The result is that, for an initial random choice by the player and a fixed host door opening policy, the players chance of winning by switching is 2/3. Martin Hogbin (talk) 20:41, 30 December 2009 (UTC)

The point Nijdam is making above is that there are two different events. The first is where the player chooses a door. This event can hardly be construed as anything other than a random choice by the player, even if the initial distribution is non-random. The second event is where the host opens a door. We all agree this event is done with knowledge of where the car is (or else how does the host always show a goat). The host is therefore not acting randomly. You're saying in the one case where the host picks between goats, the player must assume the host IS acting randomly or, perhaps more accurately, that we must average all possible host preferences. [Rick]

That is not quite what I meant but, if the player has no knowledge of the host policy, this would not be unreasonable. It is what Morgan attempt to do but get wrong, the correct answer in this case is 2/3 again. However even if we adopt a single fixed host policy, say host always opens the highest numbered door (within the agreed rules), if we average over all possible games you will see that the probability of winning by switching is still 2/3, try it for yourself. So even for a fixed host policy, the chances of a player winning by switching are 2/3 (unless of course the player knows the host policy, in which case she might be able to improve her odds of winning, but this is a different question and part of game theory). By the way, I made a mistake above which I have struck out. Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

You haven't responded to the point I made above about studying the history. I think we all agree that if the host chooses randomly between two goats each and every player has a 2/3 chance of winning by switching and studying the history cannot possibly reveal anything else. Do you disagree that if the host isn't constrained to choose randomly between two goats there is a possibility of obtaining relevant information from studying the history? [Rick]

Yes, of course. It then becomes a matter of game theory but has already been pointed out, if the host has a policy and the player finds this out then the player can do better than 2/3 overall so, in fact the best policy for the host is to act randomly (when there is a choice). Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

And, yes, it might be fairly complicated (e.g. if player picks door 1 host picks door 2 if possible, if player picks door 2 host picks door 1 if possible, if player picks door 3 host chooses randomly if the car is behind door 3). All I'm saying is that there's a difference between saying the player's chance is 2/3 on average and saying the player's chance is 2/3 in all cases. You seem to be denying that these are different, or perhaps saying the former implies the latter. -- Rick Block (talk) 22:18, 30 December 2009 (UTC)

Except in the case where the player knows, either by studying the history of the game or by some other means, the host door opening policy the odds of winning (from the player's SoK) by switching are 2/3 plain and simple. The overall odds are the odds, since there is nothing known to the player to distinguish different circumstances. Martin Hogbin (talk) 12:16, 31 December 2009 (UTC)

I cannot see your point about studying the history, perhaps you could remind me please. My opinion is that it is assumed by nearly everyone that the player has not studied the history of the game and that this possibility forms no part of the standard MHP. I would also note that anyone who studied the history of the game would not only attempt to gain information about the host's door opening policy but also about the initial car placement, which also might turn out to be non-random.

Studying history, along with trying to guess the show's strategy all belong to a variant section where game theory aspects are discussed. They only serve to confuse the standard MHP even further. Martin Hogbin (talk) 13:09, 31 December 2009 (UTC)

My point about studying the history is that there's a difference between saying the player's chance is 2/3 on average and saying the player's chance is 2/3 in all cases (or in some particular case, such as if the player picks Door 1 and host opens Door 3).

Sorry but you are just plain wrong. The probability of the player winning by switching is always 2/3 from her SoK (does not know the host door opening strategy). I would like to convince you of this, regardless of how the article progresses. Martin Hogbin (talk) 02:47, 1 January 2010 (UTC)

And I would like to convince you that the probability given a specific initial player pick and specific door the host has opened (from the perspective of the puzzle solver, regardless of what we consider the SoK of a contestant) does not have to be 2/3 and is indeed an undetermined value between 1/2 and 1 with an average of 2/3 unless the host is constrained to pick between two goats randomly. But, alas, I think this is not going to happen.

I take you now agree that from the SoK of the contestant the odds are simply 2/3 since you no longer want to treat the problem from this perspective. That is fine, let us treat the problem from the point of view of the puzzle solver. We have to solve the problem on the information given in the problem statement, this is what Morgan purport to do. Whitaker's problem statement gives us no information as to how the car was originally placed. What do we do? We could assign probabilities to each door and conclude rather uninterestingly that the player's odds of winning by switching are from 0 to 1. The only alternative is to apply the principle of indifference and take it that it is equally likely to initially be placed be behind any door. This may not, in reality, be true, perhaps it is more often placed behind door 2, perhaps it is always placed behind door 2, the problem statement does not tell us. We thus have two options, apply a standard principle of probability from which we hope to answer the question or give what is essentially a non-answer. Morgan do as I would do and chose the former.

Whitaker's problem statement gives us no information as to how the host chooses which door to open if the player has originally chosen the car. What do we do? We could assign probabilities to each door and conclude that the player's odds of winning by switching are from 1/2 to 1. The only alternative is to apply the principle of indifference and take it that it is equally likely that the host will open either door in the above case. This may not, in reality, be true, perhaps the host always opens the highest numbered door allowed by the rules. We thus have two options, apply a standard principle of probability from which we hope to answer the question or give what is essentially a non-answer. I would chose the former and thus take it that there is an equal chance that the host will choose either door.

However, it is not even necessary for us to apply the principle of indifference to the host's action. We could take a non informative (uniform) prior distribution of host door choice parameter (Morgan's q) and integrate this to get the answer of exactly 2/3 again.

In fact, it is not even necessary to assume anything at all about the hosts door opening policy, he could have a fixed and policy of always opening the highest numbered door possible for example and still the odds of an individual player winning by switching would be exactly 2/3, unless we restrict ourselves only to the specific case that the player initially chooses door 1 and the host reveals a goat behind door 3. This is shown on my analysis page (this is currently being rewritten to use standard notation, any help welcome). Alternatively you could just consider these two cases: the player initially chooses door 1 and the host reveals a goat behind door 3 and the player initially chooses door 1 and the host reveals a goat behind door 2. Both are always possible for any host strategy. If we do not restrict our problem to the specific case that the host opens door 3, the host door choice parameter is unimportant.

A sudden rush of blood to the head caused me to be a bit overenthusiastic above. Martin Hogbin (talk) 12:42, 2 January 2010 (UTC)

This is an interesting discussion which I would like to continue but it need not stop us reaching agreement on how to best write the article. Martin Hogbin (talk) 13:56, 1 January 2010 (UTC)

Perhaps one last try. If you worked for the FCC, would it be OK with you if the host said he always opens the rightmost door (if he can) on Mondays and Wednesdays, leftmost door on Tuesdays and Thursdays, and literally flips a coin to decide which of two goat doors to open on Fridays? Average chance of winning by switching is 2/3. [Rick]

That would be absolutely fine by me. I might wonder why the TV station would give away more cars than they need to, but that is their problem. Martin Hogbin (talk) 14:08, 1 January 2010 (UTC)

Average chance of winning by switching given the player picks door 1 and the host opens door 2 (or door 3) is 2/3. Average chance of winning for players who initially pick door 1 on Mondays and Wednesdays is 2/3. But on Mondays and Wednesdays if the player picks door 1 and the host opens door 3 the chance of winning (whether the player knows it or not) is 1/2, while on these same days players who pick door 1 and see the host open door 2 win with probability 1 (the latter case happens less often than the former case to make the average 2/3). Are you really saying in your view with this host the chance of winning by switching for a player who picks door 1 and sees the host open door 3 on a Monday is 2/3 unless the player knows the host's policy? And, if a different player does know the host's policy this player has a probability of 1/2?

Players who do not know the hosts policy have exactly a 2/3 chance of winning by switching, from their SoK, from your SoK, who somehow knows the host strategy the odds are 1/2 or 1, you will know which and thus your overall odds will be better than 2/3. Players who discover the host's strategy from studying previous shows will also have this advantage. You are forgetting a basic principle that probability is a state of knowledge. Martin Hogbin (talk) 14:08, 1 January 2010 (UTC)

I'm fine with deferring any discussion of anything other than the symmetric case to a "Variants" section, but I would really like to see a conditional probability analysis in the initial "Solution" section - again, much like talk:Monty Hall problem#Proposed unified solution section. -- Rick Block (talk) 06:11, 1 January 2010 (UTC)

You think it's confusing to approach the problem using conditional probability. I think the wording of the problem deliberately leads people to approach it conditionally, and they do - thinking about the specific case where the player has picked Door 1 and the host has opened Door 3. Furthermore, they typically do this incorrectly thinking 2 unknowns must have equal probability, and this is exactly why people arrive at the 1/2 answer and exactly why they so insistently believe 1/2 is correct.

I suggest we stop trying to convince each other and have the article present both approaches (to the "symmetric" problem). My goal here is not to knock the unconditional approach, but to also show how to correctly address the problem conditionally (since this is how I think people screw it up). I think this means explaining how the unconditional and conditional approaches differ. Can we find a way to do this that you don't find offensive? -- Rick Block (talk) 00:00, 1 January 2010 (UTC)

I am happy to present more than one approach but I think that initially the solution should be in what I will call a non-conditional form. This does not mean necessarily mean unconditional just that we do not mention the issue at the start. Many sources do not mention this issue. As you will see from my statement to the mediator I am happy to include Nijdam's proposed statement about the probability of having the car not changing when the host opens a door, or perhaps we could use footnotes. What I do not want to do at the start is mention that the host might have a choice of doors to open. I do not think that we are too far apart. Martin Hogbin (talk) 02:47, 1 January 2010 (UTC)

I think we have to mention the "issue" of why the probability in the given case (player picks door 1 and host opens door 3) is the same as the average probability across all cases. The probability in this case is a conditional probability. Asserting it is the same or simply ignoring it does not explain it, and (IMO) fails to address the core of the problem. -- Rick Block (talk) 06:11, 1 January 2010 (UTC)

Rick, let us make clear that it is not my and I guess not your intention to give a first solution in complete technical terms using the word conditional probability. I've got the idea, that's what some are afraid of, whether they understand the conditional nature or not. Like you, I want to make sure the "simple solution", not taking the conditional nature into account, is not taken for granted. I think you also will support a first formulation without mentioning the term conditional, but in some way indicating the incompleteness. Nijdam (talk) 11:36, 1 January 2010 (UTC)

I am not going to argue about your statement 'The probability in this case is a conditional probability'. although, I think the issue is not as clear cut as you and Nijdam think. I accept that there is an event, the host opens one of the doors, that might possibly affect the probability that we are trying to calculate and thus we might be wise to treat this as a condition, just to be on the safe side. To make this clearer, I am prepared to accept that the K&W formulation is, strictly speaking, conditional.

Therefore we need, as you say, to give some rationale as to why the host opening a does does not affect the probability that the player has chosen the car. There are two that I can see, the first is symmetry, there is nothing to suggest that opening of one door is different from opening the other when the player has originally chosen the car. The other argument is that, the host chooses chooses randomly so no information can be conveyed by this action, the player's state of knowledge is therefore unchanged.

The problem, as you have pointed out before, is that neither of these arguments is specifically mention in any reliable source, as far as I know. Maybe a footnote of some kind might help.

Alternatively, perhaps we could start with Whitaker's question, give the u/c solution then state that some sources consider this solution to be the solution of a particular interpretation (a bit of licence here) of Whitaker's statement and give Morgan's unconditional formulation, and that a better formulation (here we give the Morgan restatement of the problem) requires a conditional solution. This discussion should probably return to the main talk page now. Martin Hogbin (talk) 11:45, 1 January 2010 (UTC)

Issues about the Morgan paper

1. Method of numbering doors

In the Morgan paper, the assumption is made that the doors are numbered statically, i.e. the doors have a static identity through repeated experiment. An alternative assumption would be that the door initially picked by the contestant is always called No. 1, while the door opened by the host is always called No. 3. Based on this, I postulate:

a. The assumption should be made explicit.
b. "The elegant solution that assumes no additional information" is incorrect, in that it needs additional information.
Heptalogos (talk) 20:57, 31 December 2009 (UTC)

There is plenty wrong with the Morgan paper. Here is my comprehensive criticism of it. Martin Hogbin (talk) 02:50, 1 January 2010 (UTC)

Little is wrong with the paper, following science rules. Most of your critics is about the application of it, i.e. mismatch with reality. Some of it I agree with, some not. Anyway, I found an error in the paper which is an error following their own rules. You mention another relevant aspect not explained in the paper, which is the implicit law (?) that the problem must be treated as one of conditional probability. Heptalogos (talk) 11:49, 1 January 2010 (UTC)

No one has yet given me a clear description of what exactly makes it necessary to treat a given event as a condition in a probability question. Martin Hogbin (talk) 12:31, 1 January 2010 (UTC)

No one??Nijdam (talk) 13:42, 1 January 2010 (UTC)

2. The strategy of Morgan et al.

Morgan chooses not to write an article about the Monty Hall problem, but about several solutions offered to a specific question asked in Parade. They reject all solutions and then present the only possible elegant solution, which is very complex and of no obvious use. What is their strategy? Do they choose a problem of public interest every year? Are they requested by society? Is there any scientific relevance? Do they help anyone by providing their solution? Are they constantly spying for an opportunity to show their own quality, even if they have to misinterpret a public issue, from the public point of view? The relevance of their action is in their strategy, ironically similar to the subject of their solution. Heptalogos (talk) 16:29, 3 January 2010 (UTC)

3. The need for conditional solution

Morgan stated the problem to be conditional and start solving it using conditional formula. Also, they disapprove other unconditional solutions given, which they say would only be right as a solution to the unconditional problem. Thus, the assumption seems to be made implicitly that all conditional problems should be solved using conditional method, for which no scientific proof is given. Actually, it is scientifically valid that if two events A and B are statistically independent, P(A|B) = P(A), which is exactly the case in the 'conditional' problem they are solving. What we need next, is a scientific source which states that there is no need for conditional methodology when conditions are independent on the calculated probability. Simple as that. Heptalogos (talk) 16:39, 3 January 2010 (UTC)

4. The position of a single event

I was hesitating to add this fourth issue, because it's a very formal point of criticism to some probability statements in general. But then I reminded myself of the extreme formality of the arguments of Morgan to find a condition in a number. Moreover, I find the following formality much more inspiring. Morgan calculated some probabilities of winning by switching, considering the entire sample space of probabilities. However, they had to deal with one unique event within that space. The third (!) assumption implicitly made by Morgan is that the event of interest (suppose you're in...) is placed randomly within the sample space. Amen. Heptalogos (talk) 22:28, 3 January 2010 (UTC)

To put it differently: no assumption of this kind would have to be made if the question had been something like: "what would the switching chance of contestants be if they..." instead of describing a single situation. Heptalogos (talk) 22:36, 3 January 2010 (UTC)

5. Inconsistant formality

Morgan is being as formal as stating that some unconditional answers given are not formally answering the exact question in which a door number is given as a condition, even if the detail of the number has no statistical dependeny on the probability of winning by not switching. Next they present a solution that needs no assumptions. But at the end of the paper is mentioned that prior information of the player as to the location of the car, or the car not being placed randomly initially, "are unlikely to correspond to a real playing of this particular game show situation". So, we have aspects which, in relation to the requested probability, are irrelevant and those that are unlikely, of which the first are welcomed and the last ignored. Heptalogos (talk) 22:33, 5 January 2010 (UTC)

A conceptually equivalent urn problem

I have always maintained that there were two issues in Morgan's treatment of the MHP, the basic problem and Morgan's conditional treatment, and that each issue obfuscates the other. This is simple urn problem which shows the meaning of Morgan's conditional treatment. It is intentionally not mathematically equivalent to the Monty Hall problem. However the concepts of knowledge and probability involved are the same.

A host has ten white balls and ten black balls. He must put ten balls in to each of two urns, numbered 1 and 2. He may do this any way he chooses so long as there are ten balls in each urn. For example he may put all the black balls in one and all the white balls in the other, or he may put five of each colour in each urn, or any other combination that he chooses, so long as there are ten balls in each urn.

The player then picks one of the urns and takes a ball from it.

We assume that no one has studied the history of the game.

The unconditional formulation

This might be stated thus. The host will place the balls in the urns as described above and the player will then choose an urn and pick a ball from it. What is the probability that the ball will be black?

The answer is 1/2 because there is nothing which makes it more likely that the player will pick a black ball than a white ball.

The main point is: what is the mechanism of chance, that determines the probability? Only if we know this, we can answer the question. People easily use the term probability, without specifying its meaning. There are two mechanisms, the policy of the host and the policy of the player. If both are random, the answer is 1/2.Nijdam (talk) 12:43, 2 January 2010 (UTC)

What you say is, of course, true. For all we know, the host might always place all the black balls in urn 1 and the player might always pick urn 1. If you take that position the problem is clearly insoluble.

Rather than ask about mechanisms I think it is better to ask about initial probability distributions for the sample set. What is the probability that the host puts N black balls in urn 1 and what is the probability that the player picks urn 1? How should we answer such questions? As you imply, we could say that these distributions are unknown and thus the problem cannot be solved. This is exactly the same with the MHP, for all we know the producer might always place the car behind door 1 and the player might always pick door 1. In a real-world situation these questions are vitally important and might make the actual answer very different from that expected from assumed random action. For example ask one person to think of a digit from 1 to 9 then ask another to guess what it is. The probability of a correct guess is, in reality, likely to be better than 1/9.

The question we have to answer here and for the MHP is what initial distributions do we assume? We might start by admonishing the questioner for not defining the question with sufficient precision. Fine, I accept my admonishment, I cannot speak for Whitaker. So now what? We have, as I see it, only two alternatives. Insist that the problem cannot be solved, or make some reasonable assumptions, state what they are, and answer the question on that basis.

Before we make our decision we should perhaps consider the origin and purpose of the question. Is it a real-world question, in which case we maybe should think very hard about the likely starting distributions, based on our general knowledge, or is it a mathematical puzzle, in which case we might suppose that we should take unknown distributions to be uniform, especially as we cannot answer the question any other way. There are two things that I would add to this, firstly that we should be consistent in or decisions, and secondly that we should state the assumptions that we have made. I think that in this case application of the principle of indifference is justified for all unstated distributions. Martin Hogbin (talk) 13:50, 2 January 2010 (UTC)

Conditional formulation

The host has place the balls in the urns as described above and the player has chosen urn number 1 and taken a ball from it. What is the probability that the ball is black?

Some points to note

1. The host, by their choice of ball distribution, cannot change the overall odds of a player winning, just as the MHP host cannot, by means of their door opening policy, change the player's overall odds of winning by switching.
2. The host, by their choice of ball distribution, cannot change the odds of an individual player winning, because the host cannot know which urn the player will chose. In the MHP, the host cannot, by means of their door opening policy, change an individual player's odds of winning by switching.
3. The urn chosen is clearly identified, just as in the MHP the door opened by the host is identified.
4. The host knows what is in each urn, just as the host knows his door opening policy in the MHP.
5. The player does not know what is in each urn, just as the player does not know the host door opening policy in the MHP.
6. The problem statement does not tell us what is what is in each urn, just as it does not tell us the host's door opening policy in the MHP.

The conditional answer

So what is the probability the the player has drawn a black ball.

From the SoK of the host it is N/10 where N is the number of black balls in urn 1.

From the SoK of the player it is 1/2 as the player does not know what is in the urns.

From the SoK of the problem solver it is 1/2 as no information about the distribution of the ball in the urns is given in the problem statement.

Does anyone disagree with these answers? Let us leave for later discussion the best way to get them. Martin Hogbin (talk) 10:32, 2 January 2010 (UTC)

The strategy of science

Indeed, the distinction between conditional and unconditional is in the number of the urn. What you are demonstrating, is that this number (the event of the specific host choice) has no statistical dependency on the player's choice. When no statistical dependency exists between events A and B, P(A|B) = P(A). But the problem is still conditional simply because we ask P(A) given event B. So far most of us agree.

What we don't agree about is the need for using conditional formula. In math there appears to be no common rule or law which tells us exactly when to use conditional formula. There even seems to be no reliable definition of "conditional". Some say "any given event", others say "any event reducing the sample space". The use of such definitions is unclear. Both actually don't have any value regarding the method of solution. Heptalogos (talk) 14:06, 2 January 2010 (UTC)

I agree. I have seen five definitions of what is conditional as, as you say, even for conditional problems there may be a clear rationale for using an unconditional solution. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

For the future, I foretell that the common definition of conditional probability will include the statistical dependency between probability and condition. However, the paradox will still not be solved! Because, how do we prove independency? Should conditional formula be used? There seems to be a logical loop. It's all in the question. What should we ask? Should we use all given information in the question, or does some of it have no relevance? How can we be sure?

The paradox is that of learning through mistakes, of evolution, iteration, efficiency. Should we try as much as possible, use our energy to think out our best chances, or don't take any chance at all? Good knowledge needs good questions, but good questions need good knowledge. When we finally reach the perfect question, we already have the answer!

Conditional methodology doesn't take any chances, but consequent use of it would kill us all, because of it's unefficiency. Most of the times there's no use or sense in it. So we have to consider when to use it, which means taking a chance! Heptalogos (talk) 14:06, 2 January 2010 (UTC)

I agree again. We cannot possible take every event into account for every problem. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

The strategy of Morgan et al. has considered it quietly: they ripped the information which simply couldn't be used scientifically, but they used everything that could be calculated. However, if the question would hold multiple conditions which had no statistical relevance, but could all be calculated through, they would probably call it a useless question and don't even start at it. So the randomness of their science is in deciding which questions to aks and answer. Which is exactly what makes science an overall subjective process, namely the strategy of it. Heptalogos (talk) 14:06, 2 January 2010 (UTC)

I agree the distinction between conditional and unconditional is not absolute. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

The conclusion is that we are all right. We cannot speak about the 'need' for conditional calculation; it is a matter of probability itself. In most cases we quietly agree on the use of it, but here we found an example in the twilight zone of effectiveness and efficiency. Even Nijdam is right (;), taking absolutely no risk at all. Using this strategy he could e.g. simply refuse to solve any problems for which the solution takes more energy than it produces. At least from his own perspective. Heptalogos (talk) 14:24, 2 January 2010 (UTC)

That is why I would like to start with a simple and convincing solution that does not mention conditions then, after that has been properly explained, we can move on the discuss the issue of conditional probability. Martin Hogbin (talk) 14:55, 2 January 2010 (UTC)

Where is it mandated that 'probability' is the only discipline that may be used to solve this problem? The article says it's a 'probability problem'. Some of it, yes: the original 1/3 distribution and the total = 100%. Beyond that, it's logic, as per the simple solutions. That's why, imho, it does not require conditionality. A mental check that I am indifferent to which door the car is behind does not make the conditional formulas a requirement. Nobody, after all this time, has answered my Huckleberry question. You all just go on with all this OR, and classroom stuff. ITS A GREAT PARADOX BECAUSE LAY PEOPLE CAN START OUT CONFUSED AND THEN GET INFORMED IN A SIMPLE MANNER. AND THAT HASN'T CHANGED SINCE 1975. Glkanter (talk) 15:44, 2 January 2010 (UTC)

You may be both right if you state that this very old paradox has been taken hostage by fighting groups because of a silly (only in an absolute way) question being asked in Parade. The old unconditional dilemma is wounded by someone accidentally conditioning a door number. On the other hand, the old dilemma is honoured to be the center of a new paradox, which is about the use and sense of conditional probability methodology. The first and original paradox however is now totally snowed under by the new paradox.

What I like to suggest is to divide this article into two parts. The first one describing the old paradox, unconditional. It should indeed not mention conditional probability and logically explain why chances are not 50-50. And the second part to actually present another paradox, which started almost immediately after the initial commotion that vos Savant started. It is about the exact (un)conditional questions and their solutions, mainly the unconditional (popular) and conditional approach.

Finally something could be said about the risk of asking less usefull questions. Because the conditional solution is correctly taking no risk in judging the relevance of any condition, but the unconditional approach is wisely using the common sense of understanding what not to ask. Heptalogos (talk) 21:49, 2 January 2010 (UTC)

There's no 'new paradox'. Whatever Morgan may have come up with is not even 'a' paradox, and certainly did not supplant Selvin's as 'the MHP paradox'. And Rick has said Selvin came up with 'it' first, way back in 1975. Check this out: The Meta Paradox. Glkanter (talk) 22:37, 2 January 2010 (UTC)

It's not at all something that Morgan comes up with; it is treating any given condition implicitely as a 'relevant' condition, just because it's presented as a condition, when the question is taken literally. What else should science do? Whatever the answer is, it will be in the outskirts of science, or abroad. I think your problem is in fact that you overestimate science. Try to keep it within your framework, instead of being the twist in it's framework. You are in another dimension. Heptalogos (talk) 23:22, 2 January 2010 (UTC)

You may be right that paradox is not the right term. It's about judging the methods of judging, about finding the answers to ask the right questions, about the usage of logic, which may or may not be determined logically. I used the term iteration to give the idea of not starting with a bird or an egg specifically, but to start with 'shady shapes' that gradually become eggs and birds. It means that even 'pure logic' doesn't start from scratch, but always uses assumptions and common sense at a certain level. We only have to be clear about that and not pretend that our solutions are perfect. The shortcoming of science is it's choice of conditional questions to answer, which may not be science at all. Heptalogos (talk) 23:48, 2 January 2010 (UTC)

There is one and only one paradox, and it's paradoxical precisely because it's on the edge between conditional and unconditional. Because of the way the problem is phrased, lay people create a mental model of a (conditional) situation where there are two closed doors and one open door (the player has picked door 1 and the host has opened door 3). In this case, there are two unknowns and people naturally jump to the assumption the probability must be equal. These people are willing to argue that the simple solution is wrong not because they don't understand it but because it does not match their mental model of the problem. In particular, it requires them to ignore the specific case they're thinking about - or, more sophisticated, see that the specific case and the general case must be the same. Most people have a very hard time switching mental models after one has been created, and an even harder time thinking about two models at the same time and seeing that they're equivalent, and therefore find the "simple" solution unconvincing.

Rather than force them to switch their mental model or bludgeon them into agreeing that the question is actually talking about the overall probability rather than the probability in the specific case they're thinking about, another approach is to show them exactly how even in a specific concrete case (such as the one where the player has picked door 1 and the host has opened door 3) the probability of the two closed doors is not the same. This is what the conditional probability approach does. -- Rick Block (talk) 00:28, 3 January 2010 (UTC)

Thank you for that outstanding piece of OR, Rick. The only facts are: Selvin came out with this great paradox in 1975. Morgan and others wrote papers critical of it, perhaps unaware of the meaning of 'Suppose you're on a game show...' Other papers are subsequently written using Selvin's original solutions. It remains a great paradox. Glkanter (talk) 08:59, 3 January 2010 (UTC)

I agree with Glkanter here. Rick's statement, '...it's paradoxical precisely because it's on the edge between conditional and unconditional' is not supported by any source that I am aware of. Although Rick is fully entitled to express his personal opinion here it should not be allowed to control the article. Martin Hogbin (talk) 11:24, 3 January 2010 (UTC)

The one and only paradox, the old one, has little to do with the specific question asked in Parade, which mentions door numbers. Starting the article almost immediately with this specific question, it already introduces a complexity which kept us all busy for years, while the simple paradox has not even been explained yet.

The complexity in solving any specific question asked, divides people mainly into two groups: those who insist on the need of conditional methodology and those who prefer the unconditional approach when this seems effective and efficient. It was only yesterday that I found that the discrepancy is not really contradictory because people argue at different levels.

The controversy over the conditional approach should actually not be based upon taking the question literally or not, but upon taking the question seriously, i.e. regarding it as usefull. It is the usually unquestioned scientific strategy that may be failing. Science is always searching for better questions to ask, using it's autonomy to improve knowledge. Sometimes it is asked by society to answer specific questions. And sometimes it is playing around, choosing a specific question to solve, for let's say, demonstrating it's superiority. It is such random strategy that to my opinion is felt by people as arrogant and useless. And it is interesting to see that when science loses it's legitimate drive, it risks the blame of being corrupted. Heptalogos (talk) 12:42, 3 January 2010 (UTC)

To rephrase: the paradox of the three door problem has in principle nothing to do with conditional probability. There also doesn't have to be a shady edge between conditional and unconditional. The shadow is actually on the benefits of conditions in questions, which is intrinsically the same as the choice of answering such a question.

Glkantor, btw, Morgan did not write about the great paradox, but about the specific question asked in Parade, which is significantly different. Heptalogos (talk) 13:07, 3 January 2010 (UTC)

You may be right. I've read it numerous times. I still don't know what their point is. But the Wikipedia article is about The Monty Hall problem, and the Paradox Selvin describes in 1975. Really, these 4 great scholars pick apart a general interest Sunday newspaper supplement brain teaser, and find it insufficiently rigorous for a Probability text in its problem statement? I'm shocked. Shocked! And then they feel this analysis is worthy of being published in a peer-reviewed professional journal? No wonder Seymann wrote his companion piece. And still, Morgan's paper becomes the focal point of this whole article? Unimaginable, really. Glkanter (talk) 13:44, 3 January 2010 (UTC)

I agree. They kidnapped the old dilemma, introducing a new problem to it that they created themselves, shining a light on their own superior solution to that. That's maybe why they chose to answer a question that is quite silly taking it literally. It is silly because there is no meaning to the door number, unless in providing the scientists an opportunity to present their ultimate elegant solution that needs no assumptions. In which they failed for that matter, because they assumed static door numbering. Anyway, I am very thankful for the event to happen, because I got inspired by it, and I know you all experience the same. Heptalogos (talk) 14:09, 3 January 2010 (UTC)

An interesting result

My analysis page now confirms a fact that have always suspected which is this: Within the standard rules (which we all know), for the answer (chances of winning by switching) to depend on the host's door opening strategy and be anything other than exactly 2/3, all the following conditions must be true. Specific doors must be mentioned in the problem statement, the host choice of door must be non-uniform (in other words he must be assumed to choose non-randomly) but the initial placement of the car must be uniform (random).

It is this last condition that I bring to your attention. It has sometimes been claimed that, provided the player chooses randomly, the answer may still not be exactly 2/3. My (with Nijdam) complete analysis of the problem shows that this is not the case. As is often the case, this is obvious when thought about in the right way.

Suppose the initial car placement is non-uniform, let us take an extreme example, let us say the producer always puts the car behind door 1. If we ask what the answer is if a player chooses a door and the host opens another door to reveal a goat, the answer is exactly 2/3. If, on the other hand, we restrict ourselves to the case that specific doors are given in the question, say the player chooses door 1 and the host opens door 3, the answer clearly must depend on the probability that the car was initially placed behind door the door chosen by the player, in this example 1. In this case the player is certain to lose if she switches. If you are not convinced by this simple argument have a look here. [[12]]

The point I am making is this. We cannot use a random player choice to 'randomise' the initial car placement. For the answer to depend on the host door choice strategy the producer must initially place the car randomly but the host must choose (within the standard rules) non-randomly. This assumption cannot be justified on any logical grounds. Martin Hogbin (talk) 10:26, 3 January 2010 (UTC)

"...the host must choose (within the standard rules) non-randomly." AND convey this is some way to the contestant. But he can't. Because it's a problem about a game show. And game show hosts don't tell contestants where the car is. Glkanter (talk) 16:55, 3 January 2010 (UTC)

They don't? But keep in mind: In telling her the rule they tell her where a GOAT is. They tell her that "1/3+1/3" never is conclusive. Never. As written above: She cannot choose BOTH goats, and there is only one car. She made her choice, and she definitely separated the group of three doors in two groups: Into one single door and into a PAIR of two doors. So the refused PAIR of two doors never can contain two cars. This pair of two doors always is to contain ONE GOAT, at least: Just ONE unavoidable goat!

Exactly ONE unavoidable goat. No more and no less. No matter what door she has chosen. And no matter "which one" of the two refused doors contains a goat. Remember: Because "one goat" will be shown there anyway, within this PAIR of two doors, still before the questeion "which one?" even can be put! In any case: Exactly ONE unavoidable goat with a chance of zero. Only one! Since the time she made her first choice. Doesn't matter if (in 1/3) BOTH goats will be there. Remember: Exactly ONE unavoidable goat. And when "one goat" with a chance of zero is shown there: A second goat never is "mandatory", for only ONE is "given" there, with a chance of zero. She knows that, and she pays attention to this fact. Hope you will finally pay attention to this fact, too.

One mandatory goat with a chance of definitely zero. Exactly ONE mandatory goat. Maybe there will be even two goats. Maybe. But remember: There is only ONE mandatory goat. Exactly one. She knows this, the rule says so. If one goat is shown there, the other door can impossibly be a "mandatory" door, also. Because there's ONLY ONE mandatory goat. No matter which one of the two doors is concerned. No matter, for one goat will always be shown there. Regardless behind which door! Only one mandatory goat. The "other door" however always is free to contain whatever it contains. In 1/3 the second door will also contain a goat, but never a mandatory one. Please pay attention to this fact: Never a mandatory one. And guess what in 2/3 the second door will contain? The combined chance of the PAIR of two doors is 2/3. And only ONE of them is a mandatory goat with a chance of zero. Read the rule. So it is always given that the 2/3-chance of this PAIR is always and in any case solely consolidated by "the other" door, even if it contains (in 1/3 of events) the second goat. And she knows which one is the other door, "the privileged door", because that "one goat" has already been shown there.

The chance of the "other" door offered, of that "privileged door", always is 2/3 (not 1, but 2/3, FROM THE BEGINNING, when she made her choice). Hence: Those two doors never are equal "1/3+1/3", but the rule says: In each and every case, those two doors have to be "0+2/3", from the beginning, since she made her decision. She, at least, pays attention to the fact that the refused PAIR of two doors is to contain exactly ONE mandatory goat. The position of the "mandatory goat" never is relevant, it will be known when "one" goat is shown there. Because there's only ONE mandatory goat. If you ignore this fact (0+2/3) you unnecessarily will be operating with "conditional" chances, regrettably. If you still ignore the fact that she has definitely been told where a GOAT is. She knows it since she made her decision. You don't? If you ignore the fact of "one unavoidable goat" within this pair of two doors you unnecessarily will continue to claim "conditional chances" a necessity. And you are wrong if you argue that, from the time of her decision on - not "0+2/3" but "conclusively 1/3+1/3" - is all you know. Obsolete. Sorry for that.   Regards,   Gerhardvalentin (talk) 20:27, 20 January 2010 (UTC)

There Are No Previous Game Plays to Study

"Suppose you're on a game show..." does not imply that the show existed before, or will exist after you play it. There's no history to possibly study. Just this one playing to decide upon. Glkanter (talk) 16:59, 3 January 2010 (UTC)

That's why assumptions are made about the host's strategy. It's all in the strategy of that moment, even without the past or future. Heptalogos (talk) 18:23, 3 January 2010 (UTC)

So, can Huckleberry do better than he already did using a simple solution, if someone tells him about the equal goat door constraint equaling 1/2? And then explains what that means? I don't see how. Glkanter (talk) 18:49, 3 January 2010 (UTC)

No, his common sense creates the same result. But Morgan state that the door number is presented as a condition, so it should be used as a condition, even if there is no use or sense in it otherwise. It's a formality that either gives them a chance to validate themselves, or is the result of some imaginary rules of science. Heptalogos (talk) 20:10, 3 January 2010 (UTC)

Quite. Martin Hogbin (talk) 09:54, 4 January 2010 (UTC)

Explaining the paradox

Let us return to the original paradox. I believe that many intelligent people get fooled because of the specific question asked. You see, not all necessary information is given right away. This is accepted by most people, because information can be completed by making reasonable assumptions. However, these assumptions are just not reasonable enough to be made right away. It's a strategy of concealing information. What I am saying is that a significant part of the paradox is actually an illusionist's show.

In the graphics below can be seen that all necessary information forms a rectangle (1). Taking out some information randomly still shows the rectangle (2). We fill in the missing outlines intuitively. In our mind, we probably don't even notice it missing. However, taking out certain information creates an imaginable square for many people (3). It is only after realizing that we are missing information, that we think all given information through and are able to make the reasonable assumption, thus showing the big picture (4).

In the Monty Hall problem, the significantly missing information is obviously the fact that the host always opens another door than the one chosen by the contestant. He is not just opening another door, another door is always opened. It's only a reasonable assumption if you get the notion to make an assumption anyway. For some people it will still not make any difference to have an explicit idea of the exact host strategy, but I'm very curious what a difference it does make. Let's try to describe the dilemma without the necessity of any additional assumption:

---

Imagine a game show presenting three doors to a contestant, who may open one door, receiving whatever is behind it. One car and two goats are placed randomly, each behind one door. These are the game rules: first the contestant chooses a door randomly, which keeps closed. Then the show host always opens another door with a goat behind it, choosing randomly if possible. Finally the contestant is offered to open one of the remaining closed doors. To win the car most of the time, should contestants stick to their choice or open the other door?

---

Heptalogos (talk) 10:51, 8 January 2010 (UTC)

I agree that this is the Monty Hall problem as it should be stated. This means that the problem and solution are uncomplicated by the host's door preference, making it a simple mathematical puzzle, which most people still get wrong. The problem for us as editors of Wikipedia is that what we say should be based on reliable sources. Are there any sources which state the problem the way you do? It would be very good if you can find one. Martin Hogbin (talk) 09:53, 4 January 2010 (UTC)

I just added ", choosing randomly if possible", which I forgot. Anyway, I think it's fine that the question (at first) is quoted as it is asked most of the times, historically. It would only be nice to know what happens if we ask the question above, in order to understand the paradox better. I haven't found any example. Maybe we could arrange an experiment, in which different questions are asked to people. If the outcome would be (very) significant, media would like to write about it. Voila. Heptalogos (talk) 20:21, 4 January 2010 (UTC)

What experimental conditions are required for the 2/3 answer

I have a question for all you anti-conditionalists out there. Assuming the door numbers are persistent, e.g. Door #1 is always the leftmost door on the stage, Door #2 is always the middle door, and Door #3 is always the rightmost door (if you'd like to argue about whether this is a reasonable assumption can we please do it in a different thread?), if I want to run an experiment where the observed frequencies of winning by switching or staying are the same (in the limit, as the number of trials approaches ∞) for each of the 6 possible combinations of initial player pick and door the host opens, then what experimental conditions must I ensure are true? There are 12 frequencies I'm talking about, i.e.:

1-2-stay: (number of players who pick Door 1 and see the host open Door 2 and stay and win) divided by the total number of (players who pick Door 1 and see the host open Door 2 and stay)

1-2-switch: (number of players who pick Door 1 and see the host open Door 2 and switch and win) divided by the total number of (players who pick Door 1 and see the host open Door 2 and switch)

1-3-stay: (number of players who pick Door 1 and see the host open Door 3 and stay and win) divided by the total number of (players who pick Door 1 and see the host open Door 3 and stay)

1-3-switch: (number of players who pick Door 1 and see the host open Door 3 and switch and win) divided by the total number of (players who pick Door 1 and see the host open Door 3 and switch)

etc.

I'll start the list.

a) the car is initially uniformly distributed

b) the host, who knows what's behind the doors, always opens a door showing a goat

c) the host always makes the offer to switch

My desired experimental outcome (as n approaches is ∞) is

1-2-stay = 1-3-stay = 2-1-stay = 2-3-stay = 3-1-stay = 3-2-stay (= 1/3)

1-2-switch = 1-3-switch = 2-1-switch = 2-3-switch = 3-1-switch = 3-2-switch (= 2/3)

Aside from the list I started, is there anything else I need to ensure is true to get my desired experimental outcome? I think we all know that the answer is yes (the host must also choose between two goats randomly), but I just want to check and make sure there's no disagreement about this. -- Rick Block (talk) 20:30, 6 January 2010 (UTC)

Is the contestant aware of the situation (bias) if the host doesn't choose between the goats randomly? Is this consistent with 'Suppose you're on a game show'? Hosts and producers don't tell contestants where the car is located. The puzzle says nothing of any prior plays, from which to learn a history or pattern. So, I disagree. But I'm not a professor. And this is OR, anyways. Glkanter (talk) 20:53, 6 January 2010 (UTC) Glkanter (talk) 20:55, 6 January 2010 (UTC)

I, the experimenter, am aware of what door the contestant picked and what door the host opened. I'm measuring the success of switching or staying in the situations I've enumerated above. The question is what must be true for this to be 1/3 win by staying and 2/3 win by switching in each case I've enumerated. Yes, it is OR (of a sort), but I posted this on the Arguments page. -- Rick Block (talk) 01:13, 7 January 2010 (UTC)

Yes, if specific doors are specified in the problem statement (as you have done above) then I think that you are right, the host must choose a legal door randomly, in addition to the conditions a) to c) you quote, for the 1/3 vs 2/3 result to be true.

If the problem statement were to be that the player chooses any door and the host opens any one of the other doors to reveal a goat then the host is not required to choose randomly and your condition a) can also be dropped, the car could always be behind door 1 for example and the host could always open the leftmost legal door, provided that the distribution of the players original choice of door is uniform. The result in this case is still swap 2/3 win, stick 1/3 win. Martin Hogbin (talk) 22:39, 6 January 2010 (UTC)

Sounds like "yes" to me. I'm using specific doors in my experiment. I haven't connected this is a statement of the MHP yet, just talking about an experiment that could be run in the real world (well, n won't get particularly close to ∞ - but close enough to show a meaningful trend). -- Rick Block (talk) 01:13, 7 January 2010 (UTC)

Why do you want to run an experiment? Martin Hogbin (talk) 16:59, 7 January 2010 (UTC)

Any other takers? -- Rick Block (talk) 01:13, 7 January 2010 (UTC)

What is your realistic experiment? Is it a computer simulation, or did you collect humans randomly, or what? If it's a simulation, what does 'knows what's behind the doors' mean? The switching choice should be random. I miss a description of all actions, so I cannot tell if all requirements are met, if that's what you mean by 'condition'.

If it's a real show experiment, requirements would be about the knowledge and participation of the contestants, for instance. I miss a full description of the experiment.

Why should the host choose randomly, or should the car be uniformly distributed? Anyway, I am willing to accept the last requirement if that's simply the case. Heptalogos (talk) 22:12, 7 January 2010 (UTC)

The question is if I were to run such an experiment, what experimental setup would be required. It's the same point as the "Huckleberry works for the FCC" scenario above, in the forward direction rather than reverse. Each n-m-stay frequency measured by the experiment is, of course, the frequency probability of winning by staying given you've initially picked door n and have seen the host open door m while each n-m-switch frequency is the frequency probability of winning by switching in the same cases. It is a way to objectively measure probability, independent of "state of knowledge". Based on this experiment I could say, for example, what the (objectively measurable) probability is of winning by switching given the player picks door 1 and the host opens door 3.

Part of the problem here may be that you all are arguing that "state of knowledge" is the ONLY way to determine probability and you are specifically excluding knowledge of which particular door the player picked and which particular door the host opened from the player's state of knowledge. This makes "the probability" 1/3:2/3 (stay:switch) regardless of the host's preference (I'm perfectly willing to agree with this). However, if the doors have persistent numbers the actual probability (actual in the sense that we can measure it) in the 6 individual cases must be 1/3:2/3 (stay:switch) if and only if the host chooses between two goats randomly (in addition to the other conditions) regardless of the player's "state of knowledge". I'm not exactly sure how to reconcile these two statements, both of which seem to be true. One approach is to say the frequency probability is from a different state of knowledge. But I think another might be to say that if the doors are persistently numbered and the contestant knows which door she initially selected and which door the host opened, then the host's bias is within the contestant's state of knowledge. This information definitely makes the host's bias knowable to the contestant.

As the article on probability makes clear there are two ways of defining probability, Bayesian (state of knowledge) and frequentism. I tried to discuss this issue long ago.

In the Bayesian approach we need to consider a particular well-defined state of knowledge. This could be the SoK of a character within the problem scenario or the SoK of us, the problem solvers. The important point is that we are consistent and stick with the same clearly defined state of knowledge. If we agree that we are going to consider the problem from the state of knowledge of the player (which many here think is the only valid option, bearing in mind Whitaker's question starts, 'Suppose you're on a game show') then exactly what the player knows or does not know is of critical importance. As I have said before we have an option here, we can either consider a real-world situation, which is very complicated, or take the problem to be mathematical puzzle (again many here think this is the essence of the problem), where certain reasonable assumptions are made, such as that the cars are initially randomly placed and thus we have no knowledge about where they might be, and that we have no knowledge of the host's door opening policy and we must therefore take this to be random.

An alternative, and more formal, approach is to use only information given in the problem statement. This is the approach used by Morgan et al. However, once we adopt this approach we must continue with it consistently. What we think the player might know is not important, thus using this approach, it does not matter whether the player knows the host preference or not, but it also does not matter whether the player sees specific doors opened or not. All that matters is what the problem statement tells us.

A frequentist approach does not avoid any of the issues discussed above. The answer still depends on how you set up your experiment. We need to decide what changes and what stays the same between trials. Specifically you need to consider which of, the initial car placement, the initial player door choice, the host door choice policy, changes between trials. Obviously if they are all fixed the result is always the same. The point is, as Seymann says, we must decide on exactly what the question is before we can provide an answer. Once the question is properly settled there is no need for simulations (except perhaps to convince sceptical members of the general public) the mathematics is well understood. Martin Hogbin (talk) 13:03, 8 January 2010 (UTC)

"we must decide on exactly what the question is before we can provide an answer." This is the main source of confusion. How do we strip (communicated) phenomenon from irrelevant complexities, resulting in elementary code? Is Seymann a reliable source? Did he state that the decision mentioned is a main source of discussion? Heptalogos (talk) 20:50, 8 January 2010 (UTC)

The point I have been making all along is that, whichever of the approaches to probability you use, provided that you are consistent, there is no justification for taking the host to choose a legal goat door other than randomly. Martin Hogbin (talk) 13:03, 8 January 2010 (UTC)

In any event, the main point is and continues to be that the conditional probability in any of the 6 specific cases is mathematically different from the unconditional probability. These can have the same numeric value, or they can be different. An unconditional solution tells you one but not the other. Since many sources present both solutions, the article should as well. -- Rick Block (talk) 01:34, 8 January 2010 (UTC)

I think the essence of your message is that not all information is used in the unconditional approach. A consequence is that conditional solutions "can have the same numeric value, or they can be different." This is generally true. But certainly not in the Monty Hall case, because the door number has no statistical dependency on the requested probability. In plain logic: there is no relevance in the number. As there is no relevance in the knowledge of the host, for which reason Morgan et al. chose to ignore this information also.

The dispute is not about maths, but about it's practice. Three basic elements are involved: phenomenon, theory and communication. Maths can be communicated (written) fully objective; phenomenon cannot. Language is interpretated to strip the phenomenon from all irrelevant information (game, show, host, door, number) into objective maths. The math calculation is fully valid, but the risk is in the interpretation. Real experiment (show, host, game, contestant) has less risk of interpretation, but has the risk of bias.

Knowledge is the only way to determine probability. By knowledge you mean knowledge that influences the phenomenon or knowledge to determine the probability with. Actually the difference in 'door knowledge' is not essentially in the number of the door; it might also be e.g. the knowledge of the door closest to the chosen door. Another issue: regardless of the contestant's knowledge, it is not enough to require random host behaviour, because in that case the contestant might simply know where the car is. Or she may know when the host is behaving randomly, concluding that both unchosen doors have goats.

Is it true that bias means systematic distinct from random? Knowledge of the host's choice in a single event does not give information about the existence of such bias, does it? Heptalogos (talk) 11:16, 8 January 2010 (UTC)

Making the specific general, explicitly justified.

The probability of winning by switching is not at all effectively influenced by the knowledge of the particular door that is opened; it doesn't matter a bit whether door 2 or door 3 is opened. If this (extensive) logic is included in the correct unconditional solving method, it is fully proof. One could even state that this kind of solving is conditional eventually, because it does use the specific information, which is then identified as part of a bigger, more general, pool of events with uniform effect on the requested probability. Uniform meaning that each specific event in the pool is as likely to happen; all types of specific events are equally distributed. By accurately replacing the specific event with the more general pool of events, all possible consequences are fully accounted within the entire scope of the problem at hand.

Do you agree? Heptalogos (talk) 14:44, 10 January 2010 (UTC)

Have a look at this page, which I am writing with Nijdam (although he seems not to be around at the present). This does what I think you are talking about, it starts with the general case and considers various specific cases. The most important thing is to know which of: the producer's initial car placement, the player's initial door choice, and the host's legal door choice are to be considered random. Your comments, on the associated talk page, are welcome.Martin Hogbin (talk) 15:23, 10 January 2010 (UTC)

OK, I'll check it later anyway, although it doesn't seem familiar at this point. Heptalogos (talk) 20:19, 10 January 2010 (UTC)

I think Heptalogos's point is that if

$P({\text{car behind door i}}|{\text{player picks door i and host opens door j}})\,$

is the same for all i,j in {1,2,3} (with i != j), and these are non-overlapping events that exhaust all possibilities, then these all must be the same as

$P({\text{car behind door i}})\,$

And, if an "unconditional" solution is accompanied by a logical argument showing the relevant statements hold (all conditional probabilities must be equal and the conditions exhaust all possibilities) then this solution is "complete". This is, of course, true. The issue some sources have with the "unconditional solutions" is that rather than make such an argument they implicitly assume BOTH of these statements, sometimes using wording implying they are always true. Falk (in particular) goes into much more detail about this. For example, many solutions say something like "since you already know one of the two unchosen doors must be a goat, by opening one of them the host has not given you any additional information about the probability that your chosen door hides the car" - as if it is a mathematical axiom that P(A|B) = P(A) if B "only reveals information you already know". There is no such axiom. The reasoning is WRONG. The host picking the leftmost door if possible variant is the common counterexample showing how this is false. What is true is that P(A|B) = P(A) if A and B are independent.

"P(A|B) = P(A) if A and B are independent" is fine and I do not disagree. The problem is that it is somewhat tautologous. To say that event A is independent of event B is to say that the occurrence of event B does not affect the probability of event A (and vice versa). So this gets us nowhere. We have to use out judgment to decide whether events are independent and thus we have to use our judgment to decide whether a given event must be treated as a condition of the problem. There is no simple rule that answers this question. The only basis on which we can objectively address this question is a Bayesian one. That is to say does the occurrence of event B give us any more information about the probability of event A? There is no other formal basis. Martin Hogbin (talk) 16:51, 10 January 2010 (UTC)

This is exactly the point! You can't just "use your judgment" because people's intuitions about whether events are independent or not are often completely wrong. If the problem gives you a condition, you shouldn't ignore it but rather you should verify whether it affects the solution or not. -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

The only formal basis: yes. Intuition not adequate: yes. Plain logic not objective: wrong. Every formal mathematical axiom has naked logic as it's mother. The essence here is: seemingly solid intuition needs explicit arguments when asked for it. Heptalogos (talk) 21:48, 10 January 2010 (UTC)

Rick, if you are saying that it is a good precaution to treat any event which might affect the probability of interest as a condition just in case it does then I agree, as my statement at the end shows. 86.132.191.65 (talk) 22:46, 10 January 2010 (UTC)

If the host always picks the leftmost door if possible, his choice actually does reveal new information. So I don't see how this could prove the 'axiom' false. Furthermore, some true axiom's may still be missing in maths. Accurate logic should be enough. Heptalogos (talk) 20:29, 10 January 2010 (UTC)

Yes. I had not noticed Rick's claim to the contrary.

The common argument is that because you already know one of the doors hides a goat, the host showing you that a specific door hides a goat does not give you any additional information. This is the statement for which the leftmost host variant is the counterexample. If you precisely define "no new information" to be the same as independent, then you're OK - but the intuition and the mathematical precision often don't align. And, if you're using a "new" axiom, you really need to prove it is true first (this might be what you mean by using accurate logic). -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

The leftmost variant is not valid with the assumption of random door choice. Indeed, 'no new information' means 'no new relevant information' means 'no information of statistical dependence'. That's the difference between logic and maths; logic is much more tricky in it's many interpretations. So, I do hope that you consider non-maths logic as a valid solving method anyway? Heptalogos (talk) 21:26, 10 January 2010 (UTC)

Another example is many solutions say something like "your initial chance of picking the car is 1/3, if you never switch then after the host has opened a door your probability must also be 1/3", as if it is an axiom that P(A) = P(A|B) regardless of B! The "host forgets" variant is a common counterexample showing how this is false.

Except in that case the host action does give you information. Martin Hogbin (talk) 16:51, 10 January 2010 (UTC)

The assumption must be that the host always opens another door with a goat (randomly), so it doesn't really matter what the host forgets. Heptalogos (talk) 20:39, 10 January 2010 (UTC)

Heptalogos, the "host forgets" variant is a standard term here for the case where the host may reveal a car. Martin Hogbin (talk) 21:00, 10 January 2010 (UTC)

Yes, which is not possible with the reasonable assumptions made. Heptalogos (talk) 21:37, 10 January 2010 (UTC)

Perhaps your intuition is better than most people's - but I don't think it's at all obvious how the host's action gives you information in this case but not in the normal one. You know it MUST be true since the probability is different, but that gets back to your tautology issue. -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

If, instead of this sort of wrong reasoning, an "unconditional" solution were accompanied by an argument that all the conditional cases are equally probable and they exhaust all the possibilities then the solution is indeed addressing the conditional case. -- Rick Block (talk) 16:17, 10 January 2010 (UTC)

I would accept this statement. If there is any doubt as to whether a given event B might affect the probability of event A occurring then it might be wise to treat B as a condition and deal with the problem on that basis. If it turns out that B makes no difference that is the answer. Martin Hogbin (talk) 16:51, 10 January 2010 (UTC)

Nice to agree on that one! I think this opens possibilities. Heptalogos (talk) 20:41, 10 January 2010 (UTC)

I think a better way to say this is if the problem asks for P(A|B) and you're going say the answer is P(A) you HAVE to show B makes no difference. -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

Exactly, why didn't I say so. Heptalogos (talk) 21:39, 10 January 2010 (UTC)

Rick, in the case of a random legal door choice by the host you can do just this. If the host chooses a door, when there is a legal choice, randomly the door he opens can convey no information. This is a fundamental fact of information theory. If no information is conveyed then the probability cannot be affected. As I agree above, it might be a wise precaution to treat the problem conditionally but you cannot insist that the problem must be treated this way if it can be shown that the condition makes no difference. Martin Hogbin (talk) 22:50, 10 January 2010 (UTC)

Of course Rick can say, and has said for years, it must be conditional. So did a bunch of those Mathematics guys last spring, and Nijdam and Kmhkmh to this day. Because Morgan calls the simple solutions 'false' in a published source. Glkanter (talk) 23:26, 10 January 2010 (UTC)

Do you want us to agree anyway? Showing the condition not to be dependent can be regarded as a first phase conditional solution, because in that phase it changes the problem from conditional (how it is indentified) to unconditional. Heptalogos (talk) 12:00, 11 January 2010 (UTC)

Your argument isn't with me. Rick and I agree on nearly nothing at all. I'm trying to play within the rules of Wikipedia, and unfortunately, Morgan is reliably published. I'm trying to demonstrate that Morgan adds nothing, and should, therefore be minimized. I've advocated for simple solutions since October, 2008. Morgan calls them 'false'. We have a consensus. Let's get through Mediation, Arbitration, or whatever. Glkanter (talk) 15:43, 11 January 2010 (UTC)

Rick is very close to what I mean. Using his method, which is very nice, I come up with this:

$P({\text{car behind door 1}}|{\text{host opens door 2 or 3}})\,$ is the same as $P({\text{car behind door 1}})\,$

$P({\text{car behind door 1}}|{\text{host opens door 2}})\,$ is the same as $P({\text{car behind door 1}}|{\text{host opens door 3}})\,$

so $P({\text{car behind door 1}}|{\text{host opens door 3}})\,$ is the same as $P({\text{car behind door 1}})\,$ Heptalogos (talk) 20:16, 10 January 2010 (UTC)

That seems a good argument for not worrying about the 'condition' if the host chooses a legal door randomly. Martin Hogbin (talk) 21:04, 10 January 2010 (UTC)

Yes, but the question is why are

$P({\text{car behind door 1}}|{\text{host opens door 2}})\,$

and

$P({\text{car behind door 1}}|{\text{host opens door 3}})\,$

the same? I think you can't conclude this without mentioning something about the host's preference in the case where there are two goats (unless you're going the "I can't tell the difference between 2 and 3" route). -- Rick Block (talk) 21:07, 10 January 2010 (UTC)

Indeed, we're still using the most reasonable assumptions here. So, then you agree with this method? Heptalogos (talk) 21:30, 10 January 2010 (UTC)

Agree in what sense? It is logically correct, however (as far as I know) there's no source that actually treats the problem this way. The popular sources (no doubt following vos Savant's lead) gloss right over this. The more academic sources solve the conditional case directly (using Bayes' theorem or a probability tree or some other conditional case analysis mechanism). Without a source it becomes OR to put it in the article (even though it's true). From the standpoint of the article, I think we should do what Kmhkmh and several others have suggested which is "stick to summarizing all reputable literature in a representative and readable fashion". -- Rick Block (talk) 21:55, 10 January 2010 (UTC)

True, but let's not bother about the article for a moment; we're still on the arguments page. Now I'll try to write down the maths that we agree about in normal language, given the same reasonable assumptions:

The chance that the car is behind door 1 is still the same, given the goat revealed behind door 3, since the host would anyway open door 2 or 3 revealing a goat, for which the chances are both the same.

Would such an answer be correct?Heptalogos (talk) 22:12, 10 January 2010 (UTC)

"...for which the chances are both the same". Marilyn didn't say so. She used the example of the sea shells.

• MvS: "Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger."
• Morgan: "You may do this regardless of my choice, but you can still give a hint on the winning shell."
• MvS: "Good heavens, we made so many reasonable assumptions, isn't my random behavior simply another reasonable assumption?"

(Actually, the table below the sea shells example, with all 6 possibilities, shows the random behavior.)

Formally, Morgan are right in their criticism: one can never be sure. But then they make the same mistake of assuming things within their own solution. The lesson is that it's all in the interpretation, for every phenomenon described. Otherwise it would be written in mathematical code. Heptalogos (talk) 23:36, 10 January 2010 (UTC)

The problem is vos Savant's quote, "As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger.", is exactly one of the kinds of statements that assumes P(A|B) = P(A) if B is "known". She makes this claim without even hinting that how the host chooses between two goats might matter (it's not in her problem description or any of her followup columns). It's not enough to know that the host can and will show a goat. The host also has to choose between two goats randomly. This is not a minor point. What she should have said is something like "As I can (and will) do this regardless of what you've chosen and the specific shell I flip tells you nothing, we've learned nothing to allow us to revise the odds on the shell under your finger. Without some "and" here at least implicitly saying the host will choose randomly if he has a choice, this statement is demonstrably false (per the leftmost preference variant). -- Rick Block (talk) 17:00, 11 January 2010 (UTC)

So, out of all other assumptions, which seem reasonable to you, you find this one not reasonable: the host choosing randomly between goats. Why not? What else would be more reasonable? Heptalogos (talk) 19:48, 11 January 2010 (UTC)

Bear in mind that Whitaker's question was from a member of the public to a column in a popular general interest magazine. It can hardly be expected that he frame his question with mathematical precision. Vos Savant made an excellent job of interpreting it. Morgan, on the other hand wrote a paper for a peer-reviewed statistics journal. Martin Hogbin (talk) 23:47, 10 January 2010 (UTC)

Quite. Vos Savant is effectively trying to solve a realistic problem; Morgan are trying to explore, analyze and educate the domain of scientific statistics, using the opportunity of public attention. Heptalogos (talk) 12:13, 11 January 2010 (UTC)

Firstly, I would say that vos Savant was trying to solve a mathematical puzzle. It terms of mathematics and probability theory there is nothing at all new in the Morgan paper. All it does is conjure up an additional layer of pointless obfuscation for a simple mathematical puzzle. Martin Hogbin (talk) 14:03, 11 January 2010 (UTC)

It's not a maths problem basically. Maths may be used though. Which offers the opportunity to make it a maths problem. MvS would have liked to avoid maths, as she did initially. I'm not even sure if she used it anyway, I guess not. Heptalogos (talk) 21:04, 11 January 2010 (UTC)

Editing break

I completely disagree with this. What the Morgan et al. paper does is argue that the problem is conditional. Per Krauss and Wang most people (nearly all) initially think about the problem in terms of the player having picked door 1, door 2 remaining closed, and the host having opened door 3 - i.e. they think about the specific conditional case given as an example in the problem description. The unconditional solution is a completely different approach requiring a different mental model (per Krauss and Wang) that people have a hard time accessing after they've created their initial mental model (also per Krauss and Wang) and (IMO) this is why people argue that the unconditional solutions must be wrong. I also think that by ignoring the conditional case the popular solutions have contributed to probabilistic innumeracy reinforcing WRONG intuitions like "if the outcome of B is known, P(A|B) = P(A)". The fact that I know B is going to happen does not mean it is statistically independent of all other events. Many of the popular solutions confuse prior and posterior probabilities so much that it's hard to argue anything other than that they are simply wrong. -- Rick Block (talk) 17:00, 11 January 2010 (UTC)

Morgan show that in a specific case the legal door opened by the host could be important, in which case the problem must be solved conditionally. I agree, in the case they initially consider, that the host may choose a legal door non-randomly, and the doors are identified, the problem must be treated conditionally.

In the more realistic, and consistent, case that the host opens a legal door randomly, the Morgan paper shows us that we might be wise to treat this case conditionally too, just in case the door opened matters. I agree with this too.

But, Morgan or you are claiming that this is the only possible way to treat the symmetrical case then I disagree. We know from symmetry of the situation that the the result for either door opened must be the same. We also know that if a door is opened randomly, knowledge of the actual door opened cannot give us any information and thus cannot alter the probability the car is behind the initially chosen door (Rick, do you agree that this is true?). We therefore have a sound basis on which to say that the probability the car is behind the initially chosen door is independent of the opening of a door by the host and thus it need not be included as a condition in our calculation. Martin Hogbin (talk) 18:05, 11 January 2010 (UTC)

Martin has explained this very nicely. I would like to add towards Rick the following. The difference in mental models is of course completely true, but it makes no sense to state that only one party fails to understand one of them. Also, I see nobody here claiming the unconditional solution to be wrong. It's just not very efficient because it uses conditions which are not dependent. I'll start a section on the dependency. Heptalogos (talk) 20:33, 11 January 2010 (UTC)

Martin, in your last section you just formulate that the conditional probability to be calculated, because of the symmetry, equals the unconditional. But it is the conditional one the decision is based on, as we mentioned over and over!Nijdam (talk) 18:13, 11 January 2010 (UTC)

Even if door numbers are specified, I am not obliged to set up my sample space using them, as they are not significant. Suppose I set up a sample space with just two events: player chooses a goat(2/3), player chooses the car(1/3). So, I start with of the options one of the events occurs, the host opens a door, which I have shown is independent of the original choice, I now have to calculate the probability of winning by swapping but I know for sure that I will get the opposite of my original choice, which I still know is 1/3 chance of getting the car. Martin Hogbin (talk) 19:06, 11 January 2010 (UTC)

The important point to add to this, which was raised by Morgan, is that if the host might choose non-randomly the above method fails, and conditional probability using a sample set based on door numbers must be used. Martin Hogbin (talk) 10:25, 12 January 2010 (UTC)

If door numbers are specified, you are obliged to formulate a sample space - relating to the discussed problem - taking account of the door numbers. Nijdam (talk) 13:15, 12 January 2010 (UTC)

I disagree, it might be wise to use door numbers, it may have great instructional value to treat the problem conditionally using door numbers, but there is never an obligation to treat any mathematical problem is a specific way. For all I know there may be a geometric solution to the problem, but I cannot declare it invalid just because it is not my chosen solution. There is also no obligation to use redundant information in any probability problem.

Please tell me the flaw in my solution.

I set up a sample set containing two events and assign probabilities as shown. The player initially chooses a goat(2/3), the player initially chooses a car(1/3).

>>Okay: events G(oat) and its complement C(ar). Now describe to me in your probability space: "the host opens door 3 and reveals a goat" or "the host opens a door and reveals a goat".Nijdam (talk) 14:46, 12 January 2010 (UTC)

These events are not in my probability space any more that the event that the host sneezes is. Non of these events affects the probability that the car is behind the door originally chosen. They can be ignored in the probability calculation. They cannot be ignored in deducing that the player always get the opposite of her original choice if she swaps, but that has nothing to do with the probability calculation. Martin Hogbin (talk) 19:03, 12 January 2010 (UTC)

This set meets the requirements for a sample set, namely: '...the probability function f(x) lies between zero and one for every value of x in the sample space Ω, and the sum of f(x) over all values x in the sample space Ω is equal to 1', taken directly from the probability theory article. There are also no possibilities omitted from the set.

Now the host opens a door to reveal a goat, choosing randomly when he has a legal choice. At this point we are asked to calculate the probability of winning the car by switching doors. We note that the events in our sample set are independent of the event that the host has opened a door to reveal a goat on the following basis:

1) This event happens after the player's initial choice of door it cannot possibly affect the player's initial choice of door.

2) It is always possible for the host to reveal a goat, so the revealing of a goat cannot reveal any information that might change the probability that the player has originally chosen the car.

3) As the door number opened has been chosen randomly, the number of the door opened cannot reveal any information that might change the probability that the player has originally chosen the car. (Not that this condition fails if the host might choose non-randomly)

We now simply have to observe that if the player switches she gets the opposite of her original choice with probability 1.

What is the problem? Martin Hogbin (talk) 14:23, 12 January 2010 (UTC)

If you must have door numbers then change, 'The player initially chooses a goat(2/3)' to 'Door 1 has a goat(2/3)', and 'the host opens a door to reveal a goat' to 'the host opens door 3 to reveal a goat'. Martin Hogbin (talk) 14:36, 12 January 2010 (UTC)

Nijdam, you are still arguing that even if the host chooses a legal door evenly, the simple solution is wrong. I have just given a valid solution above. You might not like it and maybe you would not choose to do things that way but that is not reason enough to say that it is actually wrong. Martin Hogbin (talk) 17:09, 14 January 2010 (UTC)

Door numbering matters

Consider the following game. Roll a (fair) dice, call the outcome X. Put as many white balls as the outcome in an urn and complete with black balls till 6 balls all together. Then before you draw a ball from the urn, predict its colour. Then draw a ball; if your prediction was right you win a car, otherwise a goat. What will be your prediction? Nijdam (talk) 17:04, 11 January 2010 (UTC) [Copied from the main talk page]

If I am not aware of the dice throw then white. Otherwise, I would, of course, predict the colour which had the most balls in the urn. Martin Hogbin (talk) 17:47, 11 January 2010 (UTC)

How can you compose the urn without being aware of the dice throw? Would you say the outcome X matters?Nijdam (talk) 18:06, 11 January 2010 (UTC)

Yes, it matters if I know what it is. Martin Hogbin (talk) 18:42, 11 January 2010 (UTC)

As Martin says, the unconditional probability says the choose white. Why? Because by your problem, the possible ball combinations (each with 1/6 probability of occuring) is: (w/b) 1/5, 2/4, 3/3, 4/2, 5/1, 6/0. Since the probabilities are equal, there is 1+2+3+4+5+6 (21) probability of picking a while ball vs. 5+4+3+2+1+0 (15) probaility of black (21/15) = 1.4:1 odds of while. I'm not sure if this was your intent, Nijdam; if your question was "add balls until there are SEVEN (not six), there would be equal odds of black or while being picked.

The bottom line is that this has NOTHING to do with the Monty Hall Problem. Your problem is analoguous to: ALWAYS being six doors, but the number of cars vs. goats are unknown. What I'm not sure of is what you mean when you say "door numbers matter".

Are you suggesting that the number of total doors affects the probability of winning by switching (it does - 2/3 with 3 doors, 3/4 with 4 doors, 99/100 with 100 doors)? Or are you suggesting that it matters which door number is picked by the contenstant or host? (eg: matters if I pick door 2 and host opens door 1? As stated in the FAQ re: conditional vs. unconditional, this matters, but the answer only changes in non-standard MHP questions. Or are you actually saying that the number of goats vs. cars matters, which is what your analogy matched? (it certainly does affect the odds of winning by switching). TheHYPO (talk) 21:28, 11 January 2010 (UTC)

I think Nijdam has not finished yet. Heptalogos (talk) 21:47, 11 January 2010 (UTC)

Title reads: Door numbers matter, not: Door number matters. Nijdam (talk) 13:07, 12 January 2010 (UTC)

Nijdam, I have given my answer to your question. Is there more? Martin Hogbin (talk) 13:50, 12 January 2010 (UTC)

@Martin: No, my comment is meant for TheHYPO. As for you: do you see that the player on stage knows which door she has chosen and which door has been opened? Nijdam (talk) 14:39, 12 January 2010 (UTC)

I cannot tell what the player on stage knows for sure but I expect that they will probably know the door opened by the host, they are very unlikely to know the host's door opening policy. The problem statement tells us the door number opened by the host, but not the host's policy. So, yes, on strictly the information given in the problem statement, the door opened by the host is known. However, this information is not necessary in order to calculate the probability of winning by switching if you make consistent assumptions about unknown probability distributions. Martin Hogbin (talk) 16:01, 12 January 2010 (UTC)

You almost become a fan of Morgan! The player has chosen door 1 and the host has opened door 3. Should she switch? As you mention, she may not know the host's policy. Hence she models it using a parameter q. The rest you know. Nijdam (talk) 16:14, 12 January 2010 (UTC)

This so-called analogy is meaningless. First off, knowing what door numbers are chosen at any point does not, per se, affect the answer. It is only knowing how door numbers affect the random choices that can influence the results. Nijdam, you apparently wanted to include that information in the probability you asked for. But to do that, you had to include how the information was used in your problem statement. No such information is included in the MHP brainteaser, which is why this analogy is meaningless. Now, had you asked "A random number of white balls, 0<=W<=N, are placed in an urn by a game-show host, based on the roll of a die (which you see). He then places enough black balls into the urn to make N total. What is the probability of drawing a white ball from this urn?", the answer is 1/2. Note, please, that I did not say you know (1) What N is, (2) What W is, or (2) How the host might use the die to pick W. This is the form of your question that is analoguous to the MHP. The probability is 1/2 because you don't know these things, even if you see the die roll.

Anyway: if you were the one who rolled the die and composed the urn, would you know the outcome of the die roll? And in the other form, with 7 balls in total (which I deliberately didn't choose) what would you predict if the die showed 1? Nijdam (talk) 23:34, 12 January 2010 (UTC)

It crossed my mind we may actually play the game (I mean the one with the 7 balls, and hence a probability 1/2 on a white ball). I pay you 10 pound, and roll a die. After the composition of the urn I predict the outcome of the draw af a ball. When right, you pay me 20 pounds. Seems fair from your point of view. Are you willing to play this game for let's say a 100 times?Nijdam (talk) 12:18, 14 January 2010 (UTC)

Since the MHP (which, by the way, describes a game that has never appeared on an American TV game show, so it is purley hypothetical) does not include any information about bias, no such information can be assumed when answering the question. A real host might have a bias, but a different host with the opposite bias is just as likely. Since it is hypothetical, we don't have a real host, and we must represent all possible hypothetical hosts. As a result, we must assume that all such choices are made uniformly, which represents the average of all such biases. The car is placed with 1/3 probability behind each door; and when the host has N possible doors he could open, he opens each with probability 1/N. And no published source addressing the MHP has suggested that these probabilities are any different. Some have allowed them to be defferent by deriving a parametric solution, but any actual application of that solution to the MHP (as opposed to specific hypothetical setting with stated biases) has always set those parameters to the uniform values. And to repeat: This doesn't mean we are assuming that we know what the probabilities are, or that they actually are 1/2. It means we are assuming we don't - in fact, can't - know, and use 1/2 to represent that lack of knowledge.

Anyway: if you were the player on stage, would you know which door you'd chosen and which one is opened? And what would your calculations be? Nijdam (talk) 23:34, 12 January 2010 (UTC)

(1) Door numbering can not matter, unless we are explicitly told how the numbers affect probabilities. (2) If you assume you can use unknown probabilities, then the MHP is unanswerable because the car might not be placed uniformly. The fact that Morgan ignored this invalidates their analysis - their answer is wrong by the assumptions they claim to have (and explicitly claim to not have) made. (3) But we know (see Seymann, and MvS' response to Morgan) that without a statement of bias, the host must be treated as an agent of chance. That means we assume he acts without bias. (4) We know that MvS did not intend the door numbers to be part of the problem. We know this, because it is the only interpretation difference between MvS and Morgan, and she said they did misinterpret. And finally, (5) We also know that (see K&W), the door numbers mentioned are not sematically part of the problem statement. They are examples, that can be used without loss of generality if and only if no bias is allowed. Which is good, because no bias is allowed. JeffJor (talk) 17:32, 12 January 2010 (UTC)

If Morgan take it that the host legal door choice might not be random they must also take it that the initial car placement and the players initial door choice might not be random. This makes a solution impossible. The only logical and consistent assumptions are to do as Jeff says and take all the undefined distributions to be uniform. Martin Hogbin (talk) 18:49, 12 January 2010 (UTC)

He's just stringing you guys along. It is my understanding that Boris left this discussion only after he had demonstrated to Nijdam's satisfaction that the simple solutions are not inferior to the conditional solutions in any way. Do I have that right? Glkanter (talk) 17:41, 12 January 2010 (UTC)

Independence

In probability theory, to say that two events are independent intuitively means that the occurrence of one event makes it neither more nor less probable that the other occurs.

Event A = contestant wins
Event B = host opens door number 3

other data:

• three doors with fixed numbers (1,2,3)
• car and goats placed randomly, each behind one door
• contestant picks door number 1
• host always opens another door
• host always reveals a goat
• host always chooses randomly between goats
• host always offers to switch
• contestant switches after the host opens a door

Statement HL1: event B makes it neither more nor less probable that event A occurs. Heptalogos (talk) 20:33, 11 January 2010 (UTC)

In fact, as the door opened by the host is random, the number of the door opened is automatically independent of all other events. Martin Hogbin (talk) 20:44, 11 January 2010 (UTC)

Small chance that he opened the door randomly. Heptalogos (talk) 20:52, 11 January 2010 (UTC)

I do not understand what you mean. You say above, 'host always chooses randomly between goats'. Would you prefer me to say that the goat chosen by the host is random and thus independent of any other event. Martin Hogbin (talk) 22:33, 11 January 2010 (UTC)

I mean that the host most of the times has no other choice than to reveal the goat, which is no random choice. So only between goats he chooses randomly. But I now see that you may mean that overall, through repeated experiment, doors 2 or 3 are equally distributed in opening? Which you call random? Heptalogos (talk) 20:03, 12 January 2010 (UTC)

Yes, I usually say the host opens a legal goat door randomly with legal being a short way of saying that the host always opens an unchosen door to reveal a goat. Thus of the doors available to the host to choose from, according to the rules, he chooses one randomly. Sometimes he only has one to choose from. I should have said ' the host always chooses randomly between goats, when he has a choice'.

Also, for completeness, I might have mentioned the case where the host has no choice of door to open. In this case the door that he must open is decided by the random initial placement of the car, but nobody ever asserts that the door he opens could be significant in this case, as far as I know.Martin Hogbin (talk) 14:58, 13 January 2010 (UTC)

That's right. Whoever is responsible for the initial placement of the car would be equally likely to have a contrived bias as the host. Glkanter (talk) 16:16, 13 January 2010 (UTC)

Perhaps you could also confirm that the revealing of random information (note that is not the same as revealing of information randomly) cannot affect the probability of an earlier event. Martin Hogbin (talk) 22:44, 11 January 2010 (UTC)

Is "random information" a term generally used? If not, it's tricky without definition. I think I know what you mean by it, but why not call it independence? It's also not a valid general statement about "random information", since the host may randomly open a door with a car, which for sure affects the probability of the earlier event. Heptalogos (talk) 20:11, 12 January 2010 (UTC)

Yes, I am sure that there is better terminology that I should have used. The point that I am getting at is that, if the host opens a door to reveal a goat randomly, when he has the choice, revealing of the door number opened by the host cannot affect the probability that the player has initially chosen the car. Martin Hogbin (talk) 20:08, 13 January 2010 (UTC)

What do you mean precisely with: ...revealing of the door number opened by the host cannot affect the probability that the player has initially chosen the car? Because here lies the only point you seem not to understand. Nijdam (talk) 12:24, 14 January 2010 (UTC)

Independence doesn't play a big role in the MHP, except the independence of the placement of the car and the first choice of the player of course. Nijdam (talk) 13:10, 12 January 2010 (UTC)

And as explained above. Martin Hogbin (talk) 14:49, 12 January 2010 (UTC)

If A and B are independent, P(A|B) = P(A). I know your opinion is that, even when P(A|B) = P(A), the problem still needs to be solved as P(A|B). Maybe because they may coincidentally be the same in this case, but not consequently, or proven. But what if we can prove that P(A|B) = P(A) in the given case, even without calculating P(A|B)? I think I also know your answer to that, because you wrote that "one is obliged to take account of whatever is specified", which is probably a holy dogma of statistics. On that matter your opinion differs from Rick's, who maybe has/uses another discipline. The problem here may be that different persons use different methods, some of them implicitly judging that their discipline is the only true one. Heptalogos (talk) 20:28, 12 January 2010 (UTC)

Why would you prove that P(A|B)=P(A) without calculating P(A|B)? Indeed do different persons use different methods, even false methods.Nijdam (talk) 23:39, 12 January 2010 (UTC)

Science

Science at it's best is proven formula, nothing to argue about. Sense or logic is of no use, other than changing the formula or rules. Which makes sense. So within the wide scope of science, there may be consensus about the discipline used to solve a specific problem. In our case, this may be statistics. In that case, statisticians totally rule. Even if others argument reasonably that some rules of statistics may be changed to become more efficient, this is another discussion than solving the actual problem scientifically.

In that case I don't understand why Nijdam participates in the discussion at the point where we're stuck in the scientific dogma. Because there's nothing to explain really. It's simply: "one is obliged to take account of whatever is specified", because that's the rule. Which is a good rule in science! It's scientific, yes!

So, what if Einstein explained his theories, and most of us would really understand it. We would fully, logically understand, and be able to share our thoughts, write them down, over and over, backwards, diagonal, and it would all fit. Then the practicing scientists of that time would not agree, because it's no proven law (yet). End of discussion. Heptalogos (talk) 21:00, 12 January 2010 (UTC)

What is so very nice about the history of this problem is that science and logic are really in debate: best proven system (science, Morgan) against best proven process (intelligence, Vos Savant). So which one is leading? Intelligence is constantly improving (and therefore disproving) science, but science is constantly disproving (claims of) intelligence. Which one is mostly leading in which field? Then there is no science without intelligence. Science as nothing but proven formula is not real. Even mathematicicans use their intuition to decide what formula to use on what problem, and no rules for that even exist in many cases.

Instead of "popular solution" and "probabilistic solution" I think it's best to use "logical solution" and "scientific solution". Heptalogos (talk) 21:57, 14 January 2010 (UTC)

This is such a great story, because Vos Savant may not be the most intelligent person existing, but scientifically she has the highest proven intelligence! Then we have some scientists who actually agree with her logic. They may be very intelligent to share this logic, but they confused logic and science. Then we have some scientists who disagree on the method. They may be very intelligent to see that the logical arguments are not scientific, but they confused science with truth. They should not have criticized the logic of Vos Savant, but only the scientific proof. Which is about what Seymann was trying to explain. Heptalogos (talk) 09:19, 15 January 2010 (UTC)

Thoughts from user:Raccoon.zip

After spending hours of sorting out where my confusion was, I'd like to add the following which is not obvious on the wiki page. Granted, after the host revealed a goat door, at that moment, the probability whether the player's initial pick is a winning choice becomes 1/2.

No, that is not true. After a goat has been revealed the players chance of holding the car remains at 1/3. It would change to 1/2 if the host had opened a door at random and had just happened to reveal a goat by chance, but, under the usual rules, the host knows where the car is and always picks a goat, knowing where one is.

Racoon.zip (talk) 22:27, 14 January 2010 (UTC) I think this is a little misguided answer. Whether it makes a difference that the host had the inside knowledge, would depend on the question we want to answer. His inside knowlege would have had no impact on the outcoming, if we first agree that the question at hand is what the palyer's chance is *NOW* in this *particualr*, *single* run of game (vs. in an arbitrary game taken from the "statistically sufficiently large set" of game outcoming data space). Then it is 1/2 after the host reveals a goat gate in *this* particular run of the game. On the other hand, if we are talking about the player's chance in a statistical sense over the "statistically sufficiently large set" of game data space, then his *overall* (or statistic) chance is 1/3 if the host happened to have opened a goat gate with no inside knowledge. Note that in this latter case, the host could have picked a car door as well, in which case it obviously makes no sense to ask the player whether he wants to make a switch, and the player obviously ends up with a goat in 2/3 of the cases statistically.

No, the chance that the player has the car is 1/3 plain and simple. I am not sure of the distinction that you are trying to make here by talking about this run if the game. I presume you agree that the players chance of holding the car before the host opens a door is 1/3 for this run of the game?

After the host opens a door (I will assume that he opens it evenly when he has a choice of two goats, even though you may think this does not matter) the probability for this game is still 1/3. Why do you say it is 1/2? Martin Hogbin (talk) 23:35, 14 January 2010 (UTC)

There are two different cases I was referring to, and I did not explain well. The case where we differ is as follows. In the original game, let's assume the host has no inside knowledge, and we start the game for just one run. At step 2 the host happens to opena goat gate. Now, the question I'd like to pose is: what is the chance of the player having the prized car in such instances of games that satisfies this sequence of events? Note that I have changed the game outcoming space by eliminating 1/3 of space where the host happened to have picked up the car door because he had no prior knowledge. Surely, the player's winning chance was 1/3. But wait, what's the chance of the unchosen door having a car behind? It's also 1/3. So the player's chance vs. the unchosen door is 1/3:1/3, which means 50:50 chance for getting the car if the player chooses to make a switch. Or look at it another way, since in my newly defined game spaces I have removed 1/3 of the original full outcoming space of the game where the host happened to have picked up the car door because he had no prior kowledge, the player's chance to win is 1/2 as long as he sees the host open a goat door. Racoon.zip (talk) 01:34, 15 January 2010 (UTC)

And, whether the host opens a door with "inside knowledge" or not affects the probability. This point is very hard for many people to wrap their head around. If the host opens a door randomly and fortuitously reveals a goat, the chances are 50/50. If the host opens a door knowing what's behind the doors (picking randomly if he has a choice of two goats) the chances are 1/3:2/3. There's an argument that from the player's perspective if she doesn't know the host "forgot" where the car was and the host fortuitously (but randomly) revealed a goat the player's chances are 1/3:2/3, or conversely if the player thinks the host is opening a door randomly but wasn't, then from the player's perspective the chances are 50/50 - but this only comes up if we're trying to evaluate the situation based solely on what the player knows as opposed to what we know as the problem solvers. -- Rick Block (talk) 00:03, 15 January 2010 (UTC)

Your last case is somewhat similar to my variation of the game above, at least in how the 50/50 number would come about. Racoon.zip (talk) 01:34, 15 January 2010 (UTC)

Raccoon, the answer to the question in your first sentence above is 1/2. Te make clear what we are talking about let me restate the facts. The car is placed at random, the player picks a door at random, the host now opens either of the two remaining doors at random, which happens to reveal a goat. Given the above information, what is the probability that the the car is behind the door originally chosen by the player? Answer 1/2.

Now let me change the facts to represent the Monty Hall problem. The car is placed at random, the player picks a door at random, the host now opens either of the two remaining doors that he knows conceals a goat, if there are two such doors he chooses at random between them. Given the above information, what is the probability that the the car is behind the door originally chosen by the player? Answer 1/3.

As Rick says, whether or not the host is known to have inside knowledge of where the goat is does affect the probability that the car is behind the door that the player has initially chosen. Martin Hogbin (talk) 10:01, 15 January 2010 (UTC)

Maritin, I agree with both your answers. However, even your two cases can be read in different ways (and raise questions about your answer). As discussed in this thread, the answer can be 1/3 for your first case (if you add clearly that the host has no inside knowledge, which I believe is implied as it's the only difference between the two csaes you stated).

I think all the troubles with MHP is that there are multiple ways to misread it, some are more incorrect than others. I think it would be very helpful to list them in the main wiki page about common confusions and misinterpretations. For example, we may misread the MHP as to derive

(1) the player's winning chance in the sub-space of the game where the host has no inside knowledge but only happened to have picked the goat door (thus 1/3 of the space where the original pick was a goat gate needs to be excluded):- it's 1/2*0 + 1/2*1 = 1/2 (after removing 1/3 of game space, each of the original 1/3 becomes 1/2).

(2) the player's winning chance by assuming that the game is equivalent to that we re-started it after the host's pick by first asking the player to place his initial pick back into the unchosen door pool (thus 1/2:1/2 = 50/50)

(3) the ratio of winning probability of player's original pick vs. the unchosen door (1/3:1/3=50/50 if the host has no inside knowledge, or 1/3:2/3=1/2=50% if the host has inside knowledge)

(4) the player winning chance by double counting goat-1 and goat-2 end results as of the same weight as car door end result if the player makes a switch (thus 2-loses:2-wins = 50/50 chance) Racoon.zip (talk) 06:24, 17 January 2010 (UTC)

However, what his winning possibility becomes now is not the problem we are trying to solve. What we want to solve is what is the winning chance if he makes a switch. If his initial pick was a goat (remember, he had a chance of 2/3 to have picked the goat), he'd win by switching as we are certain the unchosen door has a car behind. This gives us the answer of 2/3 because if his initial pick was a car, he would lose 100% of the time by switching.

That is correct.

The answer is numerically correct, but not the argument. Before the opening of the door by the host the chance of hiding a car is 1/3 for each door. Afterwards however this has changed. It's obvious for the opened door: chance 0. It has to be, and may be, proven that for the originally picked door also afterwards the chance is 1/3 to hide the car. From this one may conclude the 2/3 chance for the remaining door. (Not this has also changed.)Nijdam (talk) 16:43, 14 January 2010 (UTC)

You have yet to show that if the host chooses evenly, the simple argument is wrong. I mention the non-symmetrical case below and refer to them the appropriate section of the article. Martin Hogbin (talk) 17:04, 14 January 2010 (UTC)

You have to understand the difference between before opening the door and after. I explained it a thousand times, and even somewhwere above I left you with the question how you would explain the difference. You didn't react. But anyway, here we go again: before opening: probability to hide the car: 1/3, 1/3, 1/3 for the 3 doors. And now for something completely different: after opening: in order of chosen, remaining and opened door: x,1-x, 0. That's all we know for the moment. x will turn out to be 1/3, showing the chances: 1/3, 2/3, 0. They must be different probabilities, as you will understand, because they differ from the original uniform 1/3. And they belong together!! They belong to the situation after opening. Symmetry may be helpfull in calculating x, but it doesn't help in avoiding the notion of x, which is in technical terms: a conditional probability. I find it extremely important that students who read the article understand this. And happily, or better, naturally, there are sources to rely on. A final remark about the simple solution. It does not address the above problem, but a much simpler version: the player knows what is going to happen, and before she even made her first choice she is forced to make a decision about switching later on. I'm actually convinced this is not the Parade version, and certainly not K&W and others. Unfortunately MvS gave the simple solution, and as I understood it, when attacked about it, tried to find a way out and invented the "unconditional" version.Nijdam (talk) 00:19, 15 January 2010 (UTC)

This is not the place for this discussion. I have referen the OP to Morgan. Martin Hogbin (talk) 11:56, 15 January 2010 (UTC)

Racoon.zip (talk) 22:27, 14 January 2010 (UTC) I agree that whether the host picks the two goat doors evenly affects the formula, but the end result is the same. It suffices for the original problem to assume the host does his pick evenly.

I happen to agree with your last statement above but there are plenty here who do not and there has been a long running argument about this. I suggest that you leave this subject until we have finished the bit about the chances of the player holding the car after the host has opened a door. Martin Hogbin (talk) 23:35, 14 January 2010 (UTC)

One may argue (as I had), that since we have two possible outcomings of goat 1, and goat 2, depending on what the host picked, for the case where the initial pick was car and he switched, we should count the two different outcomings as "two losing games" when compared to the two winning outcomings for the case where the initial pick was a goat and the player switched. The fallacy here is that the goat 1 and goat 2 cases together should only be counted in terms of the weight of the "car pick" chance (1/3). While they are indeed distinct outcomings, they are confined to the event space of "initial pick is a car", which occupies 1/3 of the outcoming space.

Yes, this is shown in the second large diagram.

One may argue (as I had), that for the case where the initial pick was car and a switch is made afterwords, since we have two possible outcomings of ending up with goat 1, or goat 2, depending on which goat the host had picked, we should count the two different outcomings as "two losing games" when doing tally, as done with the two winning outcomings for the case where the initial pick was a goat and the player switched. This would then yield a tally of (2-losses + 2 wins) which means 50% of winning if the player switched. The fallacy here is that the tally of cases of ending with goat 1 and goat 2 should only be done in terms of the weight of the "car pick" chance (1/3). While they are indeed distinct outcomings, they are confined to the event space of "initial pick is a car", which occupies 1/3 of the outcoming space, with the other 2/3 being the case of initial pick of a goat. In probability terms, the losing probability with initial pick of car and making a switch, is (1/3 for initial pick of car) x ((1/2 for goat 1 picked by host) x (100% for switching car with the remaining goat 2) + (1/2 for goat 2 picked by host) x (100% for switching car with goat 1 that is left)) = 1/3 x (1/2 x 100% + 1/2 x 100%) = 1/3 x (1/2 + 1/2) = 1/3. Since the player is guaranteed to win in the case of initial pick being a goat and making a switch after, i.e., his losing chance is 0 there. Combining the two, the player's total losing probability with a switching is 1/3 + 0 = 1/3, meaning a winning probability 2/3.

Yes. Note that if the host does not choose evenly between the goats the above calculation becomes more complicated. This is what is discussed later on in the 'Probabilistic solution' section. Martin Hogbin (talk) 16:17, 14 January 2010 (UTC)

[END of Racoon.zip (talk) 00:05, 14 January 2010 (UTC)]

Summary

I'm aware that much of what I summarize here is a repetition of what is said earlier. But it may be of help in organizing our thoughts in the mediation process. I also have to confess that from a private discussion with Gill I learned that even some experts consider a rather simplified version as the MHP. And much of the confusion and discussion stems from the diffent views of what is considered the MHP. It mainly comes down to:

MHP

We are the audience and on stage we see three doors. Also on stage is the player and the host. The rules are explained and

This is already not the MHP, which begins, 'Suppose you're on a game show...' 'We' are not the audience. 'We' are the contestant. Glkanter (talk) 12:58, 15 January 2010 (UTC)

There is no "the MHP". Why would it be the Parade statement and nothing else? Heptalogos (talk) 20:20, 15 January 2010 (UTC)

A: before anything happens on the stage we (with no additional information) are asked to answer the question: "should she switch (in the end)".

B: the host asks the player whether she will switch in the end.

B': the player makes her initial choice and is then asked the same question.

C: the host opens the door with the goat and then offers the player to switch.

Remarks

Some consider A to be the MHP (to my surprise). I know of course it is a situation one may like to analyze, but I never would call it the MHP. The version A is the (fully) unconditional situation.

This is certainly a valid view. Seymann's comment at the end of the Morgan paper reflects a standard view of statisticians. In order to answer any statistics question we must consider the intent of the questioner. It was quite likely that Whitaker actually wanted to know the answer to this question. This is addressed in detail on my Morgan criticism page. Martin Hogbin (talk) 11:12, 15 January 2010 (UTC)

Of course it is a valid view, whatever that is supposed to mean. My point is, most people have a different view of what the MHP is. Nijdam (talk) 17:14, 15 January 2010 (UTC)

Some may consider B the MHP. It is equivalent to B', as the player will know which door she will choose. It is what Boris (I'm too lazy to look back) called "semi-conditional", as it formally needs conditioning on the choice. As the choice and the placement of the car are to be considered independent, this may in a certain way also be treated unconditional.

My opinion is that C is to be considered the MHP. We see on stage the player pointig to a door and an opened door with a goat. There is now the choice between two still closed doors. This picture causes the confusion, and generates the 50/50 idea. C needs to be solved with conditional probabilities. It doesn't allow the simple solution, nor the combined doors solution, nor the many doors solution. I consider the Parade version as well as K&W as from the form C.

You continue to say, without proof, that C needs to be solved with conditional probabilities, but I have shown a proof on the arguments page starting with a sample set of two events, noting that both these events are independent of the opening of a legal goat door randomly by the host, and proceeding to a valid solution.

I just know. Nijdam (talk) 17:14, 15 January 2010 (UTC)

Part of the controversy stems from the different opinions on what to consider the MHP. Another confusion comes from people who think of C as the MHP, but solve it as being A.

Maybe some people are confused about this but I am not. View A is certainly a valid one. Martin Hogbin (talk) 11:25, 15 January 2010 (UTC)

I didn't suggest you are confused, hope you aren't. What is the content of your remark?Nijdam (talk) 17:14, 15 January 2010 (UTC)

As the only person in these discussions who read vos Savant's columns in real time, as they were published, I can assure everyone that the only question the readers had was, 'Why isn't it 50/50?'. Nobody gave two hoots about which door. Per my recollections, the door #s never came up in her columns, other than as examples. That's not what 10,000 people wrote letters about, or why 1,000 PhDs told her she was wrong. Then changed their minds. Glkanter (talk) 13:14, 15 January 2010 (UTC)

I notice Nijdam's examples steer clear of any 'host bias'. So, using symmetry, wouldn't A - C all be equivalent? I thought Boris proved this to Nijdam's satisfaction. I demonstrated with 'Huckleberry' that Monty may open either door in each playing of the game, and Huckleberry will win or lose based solely on his original choice, regardless of 'when' he makes his decision.

Nijdam keeps saying that the original 1/3 is not the same as the ending 1/3. But it is. It never stopped being 1/3, then became some other value, then returned to 1/3. If so, when does it change, what does it change to, and when does it change back? And why?

And without a contrived 'host bias', of what value is Morgan's paper? Selvin already provided a conditional solution 16 years earlier. Glkanter (talk) 13:35, 15 January 2010 (UTC)

The symmetrical problem can be solved without conditional probability. I have done it and no one, including Nijdam can state what is wrong with my solution. Martin Hogbin (talk) 15:06, 15 January 2010 (UTC)

The point here is again: do you mean A, B, C or even something else?Nijdam (talk) 17:14, 15 January 2010 (UTC)

Enlightening example

Now why do I think C is the MHP? That's why I gave above the following example: You are to roll a die. Put as many white balls as the outcome in an urn and complete with black balls until a total of six balls. You are to draw a ball from the urn, but before you do this you may predict the outcome. Of course you win a car if your prediction is right, otherwise you get a goat. How will you solve this problem? Do you say: the overall probability of a white ball being drawn is 21/36, hence I predict always white? Or do you say: if the die shows 1 or 2, I predict black. If it shows 3 it is indifferent. Does the die show 4, 5 or 6 then I predict white. I would know what to do!

I understand your troubles about the page to post this issue on. My reaction would first of all simply be: is it OK for us, editors, to present the MHP in a way that we find consensus on, or should we use resources? Problem with the consensus is the small timespace we're in, practically. Formally this may be OR. Heptalogos (talk) 11:00, 15 January 2010 (UTC)

Nijdam, your example may correctly represent what you understand the MHP to be but this does not explain why you believe that it must be option C. I agree that it may be option C, and many sources, including Morgan, clearly take this to be the case, but that may not be what the actual questioner wanted to know. You keep ignoring the fact that the question was not in a statistics exam it was a letter to a popular magazine. It is highly unlikely that the writer had considered events in such detail. There are other sources that, indisputably, give a correct answer to question A. We might assume that this is what they take the question to be. We have no right to assume that all those who give the correct answer to question A actually intended to answer question C but got it wrong.

All you are doing here is essentially restating your opinion. Martin Hogbin (talk) 11:38, 15 January 2010 (UTC)

This example is of course not enlightening at all, because your winning chance here is in fact dependent on the information in the number, which is not the case in the number of the door in MHP. Heptalogos (talk) 11:24, 15 January 2010 (UTC)

This example is completely invalid. In the example we are presumed to know how many ball are in each urn, in all versions of the MHP we are not told the hosts door policy, thus we cannot use that information. Martin Hogbin (talk) 11:41, 15 January 2010 (UTC)

I only ask you both: what would be your solution?Nijdam (talk) 17:18, 15 January 2010 (UTC)

(rearranged)Nijdam (talk) 14:46, 15 January 2010 (UTC)

If I knew the throw of the die, I would do as you do and choose according to whether there were more white or black balls in the urn. How about you answer my questions below. Martin Hogbin (talk) 17:24, 15 January 2010 (UTC)

Why not do the same in the MHP??Nijdam (talk) 17:39, 15 January 2010 (UTC)

Because, in the MHP, we do not know the hosts door preference. Martin Hogbin (talk) 19:34, 15 January 2010 (UTC)

Because, in the MHP, we assume random host behavior. Heptalogos (talk) 20:28, 15 January 2010 (UTC)

Quite contradictory. isn't it? Of course I take C to be the MHP. That is comparable to my example. Just like what you propose in my example, the player will use all the info she can get. It doesn't help her to know that on the average there is a 2/3 chance to win the car when switching. She wants to know what the chance is in her situation. With the random host behavior also in her situation (conditionally) the chance is 2/3. And now your reaction, Martin. When you don't know the hosts preference, you have to model it. Read Morgan. Nijdam (talk) 23:06, 15 January 2010 (UTC)

I have read Morgan thoroughly as you well know, I was the first to find an actual error in it, which no one has noticed in nearly 20 years. How is that going by the way?. I do not see why you do not understand my argument about unspecified distributions. There are three unspecified distributions in the MHP. The original car placement, the player choice (which is not that important here) and the host door choice. There are only two consistent ways to treat these unspecified distributions.

1) Model them all, as you call it. So we start with a parameter C1 representing the probability that the car is behind door 1, and then C2 plus parameters representing the host choice. In this case the problem is insoluble.

2)Take them all as uniform.

There are no other consistent ways to deal with the problem. There is no logical basis on which to assume one unspecified distribution is uniform but the other might not be and should therefore be parametrically modeled. Martin Hogbin (talk) 23:24, 15 January 2010 (UTC)

"With the random host behavior also in her situation (conditionally) the chance is 2/3." So you agree on that, but you don't agree on the assumption of random host behavior, is that it? Heptalogos (talk) 11:25, 16 January 2010 (UTC)

Whom is this comment addressed to, Heptalogos? I do not understand it. Martin Hogbin (talk) 11:38, 16 January 2010 (UTC)

To Nijdam, as can be seen by the 'tab' (an extra :), the same as you did. Heptalogos (talk) 11:40, 16 January 2010 (UTC)

Thanks for making that clear, Heptalogos, not everybody is as consistent with their indenting as you. You seem to be the only on that understands one of the principal deficiencies of Morgan's paper.

If we treat Whitaker's question (somewhat perversely in my opinion) as a formal probability question then it cannot possibly be answered without deciding how to treat the information missing from the question. Most people, take it that the host always offers the swap and the he never reveals the car (the statement does give is a strong clue here, that the host knows where the goats are). Even when these standard rules are agreed we still have missing information.

1. What is the distribution for the producer's initial car and goat placement?
2. What is the distribution for the player's initial choice of door?
3. What is the distribution for the host's choice of goat door when the player has originally chosen the car?

The problem statement tells us nothing about any of these these distributions. Strictly speaking, we should say that we cannot answer the problem, after all, for all we know they may all be non-uniform, the car might always be behind door 2, the player may always choose door 1, and the host may always open door 3.

The only way to get an answer is to apply the principle of indifference to all the above distributions on the basis that we have no information to prefer any one possibility over any other. This means we must take the host legal door choice as uniform.

Does anyone disagree with this? Martin Hogbin (talk) 12:41, 16 January 2010 (UTC)

I see these valid options:

1. Not make any assumption and solve it conditionally the way Morgan tried. (but they failed)

I have just shown above that this is completely impossible. If you make no assumptions and only use the information given in the question then the answer (probability of winning by switching) is from 0 to 1. Not a very useful result. Is that what you mean by (but they failed)?

They failed because they made assumptions. I meant the options as general options, but even in the MHP I don't think you proved that it's impossible to solve without making assumptions. I agree that it seems unlikely, but this may be another sort of discussion as about proving independence: it may seem so logical, but where's the solid proof? Consider that Morgan came up with a very creative way to get around some problems that you probably also couldn't imagine. For practical reasons however we may skip this option since nobody succeeded in this. Heptalogos (talk) 14:40, 16 January 2010 (UTC)

It is obviously impossible to solve the problem with no additional assumptions. The two cases below are both are consistent the problem statement:

The car is always be behind door 2, the player always chooses door 1, and the host always opens door 3. Chance of winning by switching 1.

The car is always be behind door 1, the player always chooses door 1, and the host always opens door 3. Chance of winning by switching 0.

Martin Hogbin (talk) 14:50, 16 January 2010 (UTC)

2. Make assumption, by these rules, in order:

a. Make reasonable assumptions. (reality approach)

b. Treat anything undefined as random. (mathematical approach)

c. Do a, but do b if there's argue about what's reasonable.

If there's no consensus on (within) 2, it cannot be solved. Heptalogos (talk) 13:37, 16 January 2010 (UTC)

Fine, we could base our answer on what we might expect in a realistic situation for a hypothetical game show. My assumptions would be:

1 The distribution of the car's initial placement car is approximately uniform. A stage hand is probably told to to place the car randomly but I doubt that they would use any true randomising process, like throwing dice.

2 The distribution of player initial door choice is probably approximately uniform. There is no reason for the player to prefer any particular door but people do not actually choose uniformly is such circumstances

3 The distribution of host's legal door choice is approximately uniform. The host has no reason to prefer any particular door when he has a choice and probably would just choose on the spur of the moment. The choice would be unlikely to be completely uniform.

So in the realistic case we cannot actually solve the problem at all. An approximate solution is 2/3.

We have only one choice. Treat everything undefined as random. Nothing else produces a solution. Martin Hogbin (talk) 14:21, 16 January 2010 (UTC)

OK, now you're being very, very realistic. One could question anyway how realistic a calculated probability is. For a single event, it doesn't make sense actually. We can only map the phenomenon to a logical system that we find consensus on as being the most reasonable system to use. This is a system which includes repeated experiment, within which we can state that event X is happening with a frequency of y as part of a total frequency of Y. Which we then call the probabiliy of X to happen, but actually it's not the same.

This logical system may assume random behavior, which is then perfect random behavior, as the most reasonable logical mapping of what the realistic behavior is supposed to be. So we don't need to mention 'approximately random' or alike. Heptalogos (talk) 14:56, 16 January 2010 (UTC)

An alternative example

How would you answer this question? There are 6 white balls and 6 black balls. Six of these are randomly chosen and placed in an urn. You then pick a ball from the urn. What is the probability that it is white?

A better example

This is a much more realistic example, already given above. You did not answer it. There are two identical urns. One contains 10 white balls and 20 black balls, the other contains 20 white balls and 10 black balls. A person picks a ball from one of the urns. Using only the information given, what is the probability that it is a black ball? Martin Hogbin (talk) 14:36, 15 January 2010 (UTC)

An unconditional solution to the symmetrical problem.

We start with two events: the player initially chooses a goat (G), the player initially chooses the car (C). These two events form the whole of our sample space space. The probabilities associated with these two events are 2/3 and 1/3 respectively and they are independent of all other events.

So the sequence of events is.

1) One of the two events in our probability space occurs with probabilities as shown.

2) The host opens a door to reveal a goat, when he has a choice he chooses randomly. We may be told that, in fact, he opens a door with a number 3 on it. The probabilities of the events in our sample space are independent of this event, thus we do not need to take any account of it or include it in our sample space, just as we do not include any other events of which the probabilities of C and G are independent.

3)The probability that the player still has a goat is 2/3.

4)The player now makes the decision to swap or not.

5)A player who has a goat will with certainty get the car if she swaps, and vice versa.

6)The probability that the player gets the car if she swaps is 2/3.

Martin Hogbin (talk) 15:28, 15 January 2010 (UTC)

I'm sorry Martin, I guess you are an expert on the speed of light, but not in probability theory. This sounds to me like the following will sound to you. Suppose you are on a train and you notice a source of light next to you. The train is increasing speed every time the driver blows his whistle. Connected to the wheels is a mirror, rotating with double frequency. I look into the mirror and notice that the speed of light reduces every time the mirror has made one turn. So it is obvious nothing can go faster than the speed of light. Nijdam (talk) 17:09, 15 January 2010 (UTC)

I make no claim to be an expert in probability theory but your reply might be more convincing of you told me what is wrong with my solution. Martin Hogbin (talk) 17:21, 15 January 2010 (UTC)

Okay, remove it from here and put it on User:Martin Hogbin/Monty Hall analysis. Nijdam (talk) 17:37, 15 January 2010 (UTC) ‎

You still have not told me what is wrong with my solution. Martin Hogbin (talk) 19:35, 15 January 2010 (UTC)

I more liked to discuss this on your analysis page, but okay here we go. You said: we start with two events G and C (=complement of G). What do you mean by "start with", are these the only events to consider, or will you introduce others later on? Nijdam (talk) 17:50, 18 January 2010 (UTC)

Martin, better not to talk about sample space and then exclude events that do happen from it because they're independent. That indeed doesn't make sense when using these mathematical terms that have a different meaning. However, you and others have been very clear in showing logically that there is not at all any relevance in the door number, not any influence on the probability of winning by switching. Nobody ever disproved that logic, and in 99% of the time nobody even tries. All they do say is that the number should be used as it is given as a condition. Furthermore they use examples in which it does indeed matter to use the given conditions, as if they don't see the difference. So, how should they see the difference using math rules? They can't, because those rules don't exist. You are trying to show something to a blind man, who is btw perfectly professionally blind because he shouldn't see all kinds of things which are 'apparently so logical' or something like that. Heptalogos (talk) 20:51, 15 January 2010 (UTC)

There are always events that we exclude from our sample space because they are independent. Trivial events like the host sneezes, for example.

I agree that it is good practice to include events that you are not sure about just in case it turns out that they are important, but I am not claiming that this is the best way to solve the problem, just that it is a valid way. Martin Hogbin (talk) 23:28, 15 January 2010 (UTC)

I don't think that any event mentioned and happening in the experiment can be ever excluded from the sample space. The host sneezing is not such. Heptalogos (talk) 11:04, 16 January 2010 (UTC)

Of course it can. We always have to use our judgment to decide which events to include in our sample space, based or our assessment of whether they might affect the probability of interest. For example in the MHP problem statement we have the event that the host says the word 'door'. It seems fairly obvious that just saying the word 'door' will not affect the probability of interest (this is not an entirely trivial example, words spoken by the host could easily affect this probability) so we leave this event out of our sample space. The sample space does not create itself, we create it using our judgment. Martin Hogbin (talk) 11:46, 16 January 2010 (UTC)

Let's rephrase: any effect mentioned, with a probability of happening, that influences the sample space should be specifically measured. In reliable sources, you won't find any exception. But this is indeed no explicit rule, apparently. It's very typical that I have to mention this, while Nijdam is replying with funny metaphors. I think he is not explicitly aware of this; it's rather something that grew in his system over time which is so familair and obvious to him that he cannot even externalize it. Heptalogos (talk) 12:32, 16 January 2010 (UTC)

I think we are in basic agreement? I would not necessarily recommend the above solution because it relies too much on intuition. One of the strengths of mathematics is that by systemising things we have to think less and therefore are less prone to errors. Some may think that a particular approach, which may be a good one, is the only approach. Martin Hogbin (talk) 12:57, 16 January 2010 (UTC)

If A and B are independent, P(A|B) = P(A). We already agreed on that. There is also no disagreement on specific independence in the MPH. Rick is even willing to accept unconditional solution when independence is proven or assumed. However, Rick and Nijdam are not willing to apply logic to prove the independence. They need some law, whether implicit or explicit. That's all I can make of it. Which is not wrong, but it would have helped if they told so earlier. Heptalogos (talk) 13:22, 16 January 2010 (UTC)

There is law. Hosts don't tell contestants where the car is. It's illegal in the US. Which is where Selvin's and vos Savant's and Morgan's papers were all published. Everybody in the US knows this, as we watch game shows, beginning at the age of 2, instead of reading books or playing outside. That the contestant gains no knowledge from which door Monty opens is a premise of the puzzle, as it starts, 'Suppose you're on a game show...' Glkanter (talk) 13:33, 16 January 2010 (UTC)

That's cool. Btw, the MHP is not about Monty Hall. Heptalogos (talk) 13:43, 16 January 2010 (UTC)

If A and B are independent, P(A|B) = P(A). We already agreed on that. There is also no disagreement on specific independence in the MPH. Rick is even willing to accept unconditional solution when independence is proven or assumed. However, Rick and Nijdam are not willing to apply logic to prove the independence. They need some law, whether implicit or explicit. Rick, is this correct? Is this the essence of argue? Heptalogos (talk) 23:18, 17 January 2010 (UTC)

Are we talking about the article here, or personal beliefs? For the article I need sources. I think what we have are 1) sources that explicitly say the unconditional solutions don't address the (conditional) problem and, 2) no sources using an unconditional solution explicitly saying it addresses the conditional problem. I've said this before, but personally I think the crux of the Monty Hall problem is that (no matter how it's exactly phrased) people are led into thinking about the conditional situation and that unconditional solutions require a different mental model that most people find difficult to switch to. Frankly, I don't think ANY unconditional solution (no matter how convincingly presented) will satisfy most people if they are initially presented with anything remotely like the "standard" version of the problem. As a matter of logic, if you're going to address the conditional situation with an "unconditional" solution you have to make some argument for why it applies which must implicitly or explicitly use the "equal goat" assumption. Again personally, I find a direct conditional solution a far simpler approach. -- Rick Block (talk) 03:28, 18 January 2010 (UTC)

The 'equal goat' assumption is a premise of the MHP. It begins, "Suppose you are on a game show..." This precludes you, the contestant, from knowing about any potential host bias. The only assumption you can make is that the host choice is uniform. That means symmetry. And with symmetry, we know that the simple solutions are equivalent to the conditional solutions. Glkanter (talk) 04:16, 18 January 2010 (UTC)

Rick, not talking about the article here, but trying to find the essence of argue. Your answer: "if you're going to address the conditional situation with an "unconditional" solution you have to make some argument for why it applies which must implicitly or explicitly use the "equal goat" assumption". I agree. We might simply explicitly assume the equal goat assumption. Then we are indeed stating that P(A|B) = P(A), when A and B are independent, is that true? I believe that we already agree, implying that the argument for the independence is possible, is that also true? Heptalogos (talk) 09:13, 18 January 2010 (UTC)

The equal goat assumption is exactly what makes A (car is behind door 1) and B (host opens door 3) independent, and A and B being independent implies the equal goat assumption, so if one of these is assumed then there are plenty of ways to show the other. Is that what you're asking? Maybe I'll start another section on this, but I think the mathematical crux of the MHP is that P(A) is obviously 1/3 (which requires initial random car placement) and we're asked to think about whether P(A|B) is the same or different. It seems to me if the problem is stated in such a fashion that we can mathematically model the answer as P(A) completely ignoring P(A|B) we've entirely missed the boat. I mean, the whole question is does P(A) = P(A|B), and if so why and if not why not? What I and I think Nijdam are objecting to about the "unconditional problem" is it turns the question into an obscurely worded way of asking whether P(A) = P(A). Well, duh. This makes it hardly paradoxical at all, but more akin to childish word problems like "a plane crashes exactly at the North Pole, where are the survivors buried?"

The player picks a door (say Door 1) and clearly has a 1/3 chance of having picked the car - P(A) is obviously 1/3.

The host now opens a door, say Door 3. Does this have any effect on the player's initial chance - is P(A|B) the same or different from P(A)?

versus

The player picks a door (say Door 1) and clearly has a 1/3 chance of having picked the car - P(A) is obviously 1/3.

The host now opens another door. Considering all possibilities (i.e. keeping the initial sample space the same as the final sample space) what is the chance the player has picked the car - what is P(A)?

-- Rick Block (talk) 15:58, 18 January 2010 (UTC)

Rick, you seem now to be saying that you think the answer to the unconditional problem is obvious. Well, I suppose it is. The question just boils down to, what is the probability that you initially chose a goat. But, with a little diversion, in the form of opening a door which makes no difference at all, most people get it wrong. That is the whole point of the MHP, it is really easy when looked at correctly, but for some reason most people do not do, and sometimes cannot do, this.

As to why P(A|B) = P(A) in the symmetrical case, I have given two reasons, based on sound established mathematical principles, why this must be so. Firstly symmetry, secondly no information is conveyed by a random event. Martin Hogbin (talk) 16:45, 18 January 2010 (UTC)

What I'm saying is that the "unconditional problem" boils down to a completely uninteresting mathematical tautology. You're saying people don't understand that the MHP is simply an obscurely worded way of asking whether P(A) is the same as P(A) and if you look at it "correctly" this is what you see, i.e. they have trouble arriving at the mathematical question that is asked. I'm saying people understand the problem (whether it's technically worded this way or not) as asking whether P(A) is the same as P(A|B) and the issue is not an incorrect mathematical representation of the problem statement but an intuitive failure in evaluating the conditional probability. Your response seems to be saying that you think the one and only "correct" view is to "understand" that the problem is actually asking whether P(A) is the same as P(A). I'm saying there's ANOTHER approach, which is to show them how P(A) is indeed the same as P(A|B). This is Morgan's and Nijdam's point when they say the problem is inherently conditional. In their view, the problem asks whether P(A) is the same as P(A|B), NOT whether P(A) is the same as P(A). I think we keep talking past each other because you keep ignoring or not understanding this point. -- Rick Block (talk) 20:56, 18 January 2010 (UTC)

P(A|B) = P(A) because indeed the car behind door 1 and the host opening door 3 are independent. Thanks for answering. However, that's not what was asked in the problem statement. The question was more about the probability of door 2, which is in fact dependent! Now the trick used in the unconditional method is that the chances of door 2 are easily derived from those of door 1, using the combined door theory, which is possible because of symmetry when assuming equal goats. So it's still a conditional problem solved by unconditional method, or whatever you want to call it. Indeed this all has nothing to do with the paradox. And indeed, you are not saying that the popular solution is wrong. Nijdam is however, but that's a formal issue as far as I'm concerned. I'm glad that we finally or hopefully almost agree on the theory behind. Heptalogos (talk) 21:10, 18 January 2010 (UTC)

The title of this section is: An unconditional solution to the symmetrical problem., but the issue here is: A symmetrical solution to the conditional problem.. I'm glad you admit it is a conditional problem, meaning it has to be solved by calculating a conditional probability. In what way this is done is of no interest to me. You come up with the term "unconditional method", a term you invented yourself. You may use any (correct) method you like, unconditional, unconventional, unlogically, whatever. I never objected this. In fact, if you have followed the ongoing discussion, somewhere I showed how to calculate the desired conditional probability, by the use of the symmetry in the problem. But the popular solution is wrong, as is the combined door solution, the many doors solution, most of the simulations, etc. I.e. as a solution to the "conditional problem", the one I (and not just me) consider the MHP. Nijdam (talk) 23:35, 18 January 2010 (UTC)

When you say the popular solution is wrong what do you mean? It gives the right answer. Sure, it would not apply to a more complex version of the problem (the non-symmetrical case) but this is often the case in mathematics. For any solution to any problem in mathematics it is always possible to find a more general case to which the solution is not valid. (I am not really sure that it is always true but I conjecture that it might be so, it is certainly often true). The simple solution does not give the correct answer just by chance, it gives it because of a symmetry that most people recognise. Martin Hogbin (talk) 00:48, 19 January 2010 (UTC)

@Rick The first point that most people do not spot is that P(getting a car on swapping) = 1 - P(having the car), as you know this is always true. The misdirection of opening a door makes this simple fact not obvious. The combining doors solution makes this fact obvious again.

Next, some people think that P(having the car after the host has opened a door) <> P(originally choosing the car|the host has opened door 3) even when the host is specified to choose randomly when the player has originally chosen the car. Many people think that P(having the car after the host has opened a door) = 1/2 because there are two doors and one car. Again, the opening of a door seems to confuse them. Nobody ever thought that the door opened by the host was important.

Morgan's point is whether P(A|B) = P(A|C), where C is the event that the host opens door 2. If P(B)=P(C)=1/2 I would think that it is obvious to most people that P(A|B) = P(A|C), but most people do not even think about this possible complication. Martin Hogbin (talk) 22:05, 18 January 2010 (UTC)

No, you're completely missing Morgan's main point which is that the question is about the difference between P(A) and P(A|B). As it turns out, because vos Savant treated the question unconditionally she completely overlooked specifying that the host must choose randomly between goats. Since she was answering "does P(A) = P(A)" this makes no difference to her answer. It DOES make a difference if you're addressing whether P(A) = P(A|B), and provides a handy way to show that these are indeed different questions. I will agree that most people don't consider the case where P(B) != P(C), but this is not at all the main point. -- Rick Block (talk) 23:00, 18 January 2010 (UTC)

What I cannot work out is whether you intend to distinguish between P(A|B) and P(A|B or C). Would you describe (PA|B or C) as a conditional probability? Martin Hogbin (talk) 23:29, 18 January 2010 (UTC)

IMO, P(A|B or C) defines the same sample space as P(A) since you're given one of events B or C must happen - this means P(A|B or C) is mathematically the same as P(A). Do you distinguish between these two? If they're not the same (mathematically) what is the difference? P(A|B) on the other hand is NOT the same sample space. If you think the MHP is asking about P(A|B or C) you're saying the question is asking whether P(A) = P(A). Of course it does. This is not paradoxical. The paradox is that P(A) = P(A|B). In words, the paradox is that the prior probability before the host opens a door is the same as the posterior probability after the host opens a door. The paradox is the same no matter which door the host opens, but that doesn't mean the posterior probability is P(A|B or C). The posterior probability is P(A|x) where x is one of B or C. This is entirely different than P(A|B or C), because P(A|B or C) is identical to P(A). -- Rick Block (talk) 01:02, 19 January 2010 (UTC)

In essence, the Morgan argument is that the host, Monty Hall, may be an imperfect human being.

Monty might have a bias. Hard to argue with that. Of course, the contestant (that's us), may be imperfect, too. But, there are producers and legal guys and network guys all around to make sure things are on the up-and-up. No matter, as they are all likely to be imperfect human beings, too.

Late edit by Glkanter: Hey, what about the guy who originally places the car behind a door? He might be imperfect, too. Morgan writes a paper about how the host might be an imperfect human being with a bias, and completely ignores the identical likelihood for the guy who originally places the car? How can they do that?

And so might the guy who places the pea under a shell be an imperfect human being.

Or the factory worker who makes the black and white balls for all these urns. Maybe the imperfect human being removing the balls from the urns can somehow feel the difference between the black and white balls made by this imperfect factory worker.

Do I even need to mention the guy who deals the cards from the deck might be an imperfect human being, or have a bias?

Maybe one of you qualified people could publish this idea somewhere. Then every puzzle could be fouled by this ridiculous line of thinking, as has the MHP. Glkanter (talk) 13:26, 16 January 2010 (UTC)

You have a point, saying that assumptions always need to be made. Otherwise we in fact assume crazyness, chaos, which is also an assumption. (Is this a paradox?) There is no elegant solution that needs no assumptions. There may be elegant solutions making only assumptions which are undisputed, and at the same time treat any undefined information in a consequent manner. These can at least exist in theory. Heptalogos (talk) 13:57, 16 January 2010 (UTC)

Comparative Likelihoods Of Bias: The Host v The Car Placer

Anybody want to tell me the difference?

And then defend, to a Wikipedia reader, how Morgan treats them differently?

That ain't the paradox. Glkanter (talk) 14:20, 16 January 2010 (UTC)

Please try to be economical with new sections. This one could be included in the section above. Also you might ask less and state more yourself. Your implicit statement was already that Morgan are not consequent in assumptions they make. This statement is not new; it is actually already presented as an error in the Morgan paper. Heptalogos (talk) 14:27, 16 January 2010 (UTC)

This is not the actual error in the paper that I mentioned. It is a serious and problematic inconsistency. Martin Hogbin (talk) 14:53, 16 January 2010 (UTC)

The errors can be found on this page in section "Issues about the Morgan paper". Heptalogos (talk) 15:04, 16 January 2010 (UTC)

Combined doors

On the german Wikipedia one of the discussiants gave the following funny version.

When the host offers the player to make her choice, she says: "Let's be practical: open doors 2 and 3, and give me what is behind." The host says he can not do this, but the player tells him she will choose door 1, and then he has to open one of the other doors, where after she will switch and open the remaining door, so in the end door 2 and 3 will be open. So why not immediately open door 2 and 3.

Logic?Nijdam (talk) 11:07, 17 January 2010 (UTC)

This was just meant to amuse you. Did it?Nijdam (talk) 17:54, 18 January 2010 (UTC)

But ... do you see the flaw? Nijdam (talk) 12:03, 20 January 2010 (UTC)

Your example itself lacks logic, as with 2 doors revealed, the placement of the car is always known. The contestant would alter her decision based on what is seen. Besides, the 'Combining Doors' solution shows one open door revealing a goat, then the contestant decides.

Here's a better one: You get one play of the game, with the doors opening exactly as per Whitaker. Whoever gets the car, keeps his life. Regardless of when you are asked, are you going to switch? Glkanter (talk) 12:22, 17 January 2010 (UTC)

Here's the best one: as an honoured scientist, you get two plays of the game, which are broadcasted on live television. Firstly, the host opens door 3 and offers you a choice to switch, after which you demonstrate your sophisticated calculation method, resulting in acceptance of the offer. In the second play, the host opens door 2. Do you recalculate or not? Heptalogos (talk) 12:39, 17 January 2010 (UTC)

No, for symmetry reasons (in the case of random hast behavior, of course) I know the (numerical) answer is the same. But both are conditional probabilities. Nijdam (talk) 16:54, 17 January 2010 (UTC)

Now you confirmed that you actually don't use the information in the number of the door shown. Heptalogos (talk) 18:16, 17 January 2010 (UTC)

How do you know symmetry applies in this case? Is a 'host bias' somehow ruled out? Glkanter (talk) 17:39, 17 January 2010 (UTC) (comment moved to improve clarity of dialog between Nijdam and Heptalogos)

Maybe Rick Block would also answer Heptalogos' question? Glkanter (talk) 18:21, 17 January 2010 (UTC)

Now that you've corrected the anecdote, Nijdam, the answer already exists. You've just described the rationale behind another of the simple solutions. The host is effectively offering the better (or equal) of the 2 doors, as he always reveals a goat. All the business about revealing a goat is just carnival-like misdirection, intended to make the reader think its 50/50. This is old news, Nijdam. Glkanter (talk) 12:49, 17 January 2010 (UTC)

Nijdam, what is your answer to Heptalogos' excellent question above? Martin Hogbin (talk) 12:51, 17 January 2010 (UTC)

He answered: "for reasons of effectiveness and efficiency, I do of course not recalculate, but to keep my honour in the world of science, I would have told the viewers that formally I would have recalculated". In other words, there's no use in recalculating, but these are the rules of science at this moment. Which is not true by the way, because there are no rules, but he simply clampses to his bigger brothers' intuition and experience for using the right method. Heptalogos (talk) 14:03, 18 January 2010 (UTC)

You say, 'he answered'. Where was that, and why not here? Martin Hogbin (talk) 16:49, 18 January 2010 (UTC)

I'm sorry for my sarcasm. I was translating the answer above. Please don't worry about not being civil; nothing bad will happen. Heptalogos (talk) 19:08, 18 January 2010 (UTC)

(outindented)New idea! Combine the chosen door and the door to be opened. The probability that the car is behind this pair of doors is 2/3. Now after the door is opened, the probability is 0 it hides the car, hence the chosen door has chance 2/3 to hide the car. So stick to the initial choice!???Nijdam (talk) 13:10, 21 January 2010 (UTC)

If door "A" is chosen and (from the moderators scope: "B=0+C") he opens "B", then winning chance for "C"=2/3. Your suggestion to "combine chance A+B=0" is based on another game, where door "C" must have been chosen and door "B=0" was opened also. But you forgot the moderator's scope: In your game it would not have been "B=0+C" like in the game before, it must have been "A+B=0" in "your" game. And it is true, in that latter game the chance for "A" would be 2/3, indeed. But it was another game, so I rate your new idea a hoax. -- Gerhardvalentin (talk) 06:42, 25 January 2010 (UTC)

The same mistake again. Maybe you're unable to understand. But, once more, in what seems your terminoligy: chosen door A, before opening B: Ab+Bb+Cb=1; Ab=Bb=Cb=1/3 {Ab means A before etc). Now comes the tricky part: after opening B: Aa+Ba+Ca=1; Ba=0(<>Bb), hence Aa+Ca=1 (Aa means A after opening of B, etc.). But we may not just assume Aa=Ab!! Nor may we reason: Bb+Cb=Ba+Ca. So we really have to calculate Aa, which turns out to be 1/3, and hence Ca=1-Aa=1-1/3=2/3. Got it? Nijdam (talk) 02:04, 26 January 2010 (UTC)

Sorry you do not face the facts and the mistake in your plea. Salu -- Gerhardvalentin (talk) 04:04, 26 January 2010 (UTC)

Gloomy view, Nijdam? Wrong combination. Put on your pair of glasses. The door opened never has been opened within a pair of "one door selected" and "one door refused", neither within a group of three doors. Read the rules. Remember: It always has been opened within the pair of those two refused doors. Put on your glasses, bye. --Gerhardvalentin (talk) 14:19, 21 January 2010 (UTC)

Can't follow you. Every door has chance 1/3 on the car, hence every pair a chance 2/3. Do you agree? Nijdam (talk) 14:56, 21 January 2010 (UTC)

Nijdam what? You can't follow? Please don't wimp out. You are strongly and seriously prompted to follow. Please read the rules. Every pair has a chance of (only) 2/3 as an average. And just ONE door of any pair of doors unevitably has to be a goat. You forgot? Just ONE. No matter which one. But never BOTH doors! And moreover I accuse you to neglect the rule concerning the host's target. The host is always given a "certain" pair of doors. He cannot escape: A "certain" pair of doors is given to the host. And in each and every "pair of doors" there necessarily has to be just ONE unavoidable goat. Agree? So your approach must include, and it positively has to include the following constellation:

The host "is given" (by whom?) a certain pair of doors, one of those two doors priorly having been selected already by the player, and that door selected by the player is a goat indeed. Agree? And the second door "given!" to the host might be the CAR. You forgot? ONLY ONE inevitable goat in any pair of doors. So this possibility that the second door is a car has to be considered. Then what? He cannot show that car. So your unneeded proposal to "combine" is a dead-end street, Nijdam. No more blind alleys, please. Regards, -- Gerhardvalentin (talk) 17:19, 21 January 2010 (UTC)

Nijdam is attempting to show the logical fallacy of the usual "combining doors" solution (he is a professor of mathematics at a university). Perhaps another attempt at this.

1. Each door has a 1/3 chance of hiding the car.
2. Therefore, any pair of doors, not just the two doors the host must choose between, has a combined chance of 2/3.
3. Therefore, door 1 and door 3 have a combined chance of 2/3.
4. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3.

Like Nijdam, I'm not saying this is a valid argument, only that it is just as valid as the usual "combining doors" argument. The reason this argument is NOT valid is the same reason the usual combining doors argument is not valid. -- Rick Block (talk) 19:52, 21 January 2010 (UTC)

The combining doors argument works perfectly well in cases where you can show that the host action does not change the probability of the car being behind door 1. As I have explained on several occasions, there are two methods of doing this if the host chooses a legal door randomly. They are symmetry and information theory. Martin Hogbin (talk) 21:50, 21 January 2010 (UTC)

No Martin, what you are referring to is not the "combined doors argument", as this argument doesn't involve the chosen door. Think. Nijdam (talk) 22:44, 21 January 2010 (UTC)

Back to reality: The group "selected door+open door" has a chance of exactly 1/3, whereas the group "refused door+open door" has a chance of exactly 2/3, and the group "selected door+alternatively offered second closed door" has a chance of exactly 3/3 as a fact. You know about this fact and about the causal chain? Seriously: The reason therefor is clear and evident, curious who can explain it − sorry abt objectionable style, Nijdam, just tried to copycat   :- )   Kind regards -- Gerhardvalentin (talk) 23:39, 21 January 2010 (UTC)

Well, why has the group "refused door+open door" a chance of exactly 2/3? Nijdam (talk) 00:17, 22 January 2010 (UTC)

You can tell it in a hundred ways, e.g. because - after the host showed a goat behind the third door - there are only two remaining doors now, the one originally chosen by the player with its unchanged chance of 1/3 and the still closed "priviledged door" in this game.

Here lies your mistake. Two doors are left to hide the car in this new situation. For neither of them it is clear what the new probability is. You say: the one originally chosen by the player with its unchanged chance of 1/3, but that's what we do not know. We have to calculate this new probability. Although it will turn out to have the same value as the old one, it is a different probability. Study hard, and you may understand it. Nijdam (talk) 17:51, 24 January 2010 (UTC)

"Here lies your mistake?" Sorry Nijdam, you aren't addressing the issue, you go past the issue. You need to see and evaluate the situation and draw the appropriate conclusion, and you have to look straight forward, never backwards and never changing your view. Remember: Those three doors have already definitely been devided in two groups, with a chance 1/3 for the door originally selected and with a chance 2/3 for the denied rest of the doors (it was a pair of two doors). Remember? This rest contains one inavitable goat, at least. It has a chance of 2/3, though. No matter if there are two goats, and no matter if one goat is behind the one door or behind the other door or if there are two goats hidden behind both doors. No point of interest. At least one inevitable goat. This pair has a chance of 2/3 though. If you tell no you're wrong. Everyone knows who can count to two: A pair of two doors, only one car in the game.(Have a look). Whether or not one goat behind this remaining pair of doors is still hidden or will be (randomly) shown later, does'n even matter! Opening one door showing one goat without giving any additional information is NO NEWS in this stage. Proven millionfold. Chances remain 1/3 to 2/3. Repeat: Whether one door showing a goat is still closed or already has been opened: No difference regarding the chances 1/3 to 2/3. Because showing one goat there is absolutely "NO NEWS". Of course, you can try the math, it will bring nothing new, it will only confirm the reality. No "different" chances. "Different probability" is bothering you alone, is not bothering the reality. What counts is the correct result, not the arduous detour which you take to reach the same result. Understanding the paradox is what counts. Math and to read it's history in this MHP is important to know, but not necessary to understand the paradox. You assert "mistake!" without a proof. Respectfully, -- Gerhardvalentin (talk) 22:03, 24 January 2010 (UTC)

Both joined together have a chance of hiding the car of 3/3. Of course you can ignore the situation that has developped and say "two doors, a car and a goat", 50:50. Or you can misinterpret the chance of the door originally selected, forgetting about the correct question and apply a false view, saying "any two doors have a joint chance of 2/3, also the door opened (0/3) and the door originally selected (thus 2/3), so the chance of the door offered as an alternative is 1/3. You can use different false viewing directions, a false line of sight and pretend to ignore your deviation that is so easy to reveal. You can misinterpret anything to an alarming extent, but you will be convicted. -- Gerhardvalentin (talk) 23:16, 23 January 2010 (UTC)

Gerhard - What, exactly, about the reasoning I provide above is in error? I assume you're saying the group "refused door+open door" has a chance of 2/3 because (if the player picks door 1) the probability of door 2 and door 3 sums to 2/3. I'm saying any two doors have a combined chance of 2/3, for the same reason. Actually, this is in fact true (i.e. this is NOT the error in the logic above). There is most certainly a problem because this reasoning can be used to show both (in the the case the player picks door 1) that door 1 has a 2/3 probability and door 2 has a 2/3 probability. To show how similar this is, here's a "combined doors" argument using the same logical steps.

1. Each door has a 1/3 chance of hiding the car.
2. Therefore, the two doors the host must choose between have a combined chance of 2/3.
3. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of door 2 must be 2/3.

I'm guessing that you might think the above is valid reasoning. But if it is, my reasoning about door 1 is valid as well. I don't know if you're saying you don't know what the problem is and you would like someone else to explain it. If this is the case, just ask and I'll be happy to explain what I think the problem is. If you think you know, please explain it. -- Rick Block (talk) 00:34, 22 January 2010 (UTC)

Outintended
Sorry, but I would rather like not playing stupid, believe me. Sorry to find obsolete and untruthful arguments, impossible to agree to those offside embellishments. Listen: Each and every "virgin door" in the standard MHP has a chance of 1/3, two doors have a chance of 2/3, guess what a group of 3 doors will be like. So: If you take a "virgin door", e.g. the door originally selected by the player, it has a chance of 1/3 to hide the car. If you doubt I will give you evidence. And I hear uncaringly saying: "Listen, I am going to combine your virgin door with a demonstrable goat, and believe me, I promise your chance will rise to 2/3. Trust me, I double your chance by adding a goat to your virgin door. No error in the logic." Unbelievable.

You can give numbers to the doors (not necessary) and combine them:
Car behind door 1, player choses door 1, switching hurts
Car behind door 1, player choses door 2, switching wins
Car behind door 1, player choses door 3, switching wins
Car behind door 2, player choses door 1, switching wins
Car behind door 2, player choses door 2, switching hurts
Car behind door 2, player choses door 3, switching wins
Car behind door 3, player choses door 1, switching wins
Car behind door 3, player choses door 2, switching wins
Car behind door 3, player choses door 3, switching hurts

and so on. But you can spare this sortilege by saying: There are three doors: door 1, door 2 and toor 3, and they contain:

Car Goat Goat - Player selects door 1, host opens door 2 or door 3, switching hurts, chance=zero. If you stick: chance=1. Even combined chance with open door 2 or open door 3=1
Car Goat Goat - Player selects door 2, host opens door 3, switching wins, chance=1. If you stick: Chance=zero. Even combined chance with open door 3 showing a goat=zero
Car Goat Goat - Player selects door 3, host opens door 2, switching wins, chance=1. If you stick: Chance=zero. Even combined chance with open door 2 showing a goat=zero

Swithing: 2 of 3 wins the car. Sticking: only 1 of 3 wins the car, even "combined with a goat".
Please understand: I never will follow your suggestion to combine with a "demonstrable goat"! Believe me. - No error in logic!
Why do you try to sell demonstrable goats as "virgin doors"? Suppose soon you will say:
"Any two doors have a chance of 2/3. And even if the host shows two goats, then their chance will be 2/3, also. Believe me. No error in logic."
Looking blank. Regards, -- Gerhardvalentin (talk) 02:02, 22 January 2010 (UTC)

No one is asking anyone to play stupid. The point is to identify the precise problem in the "combining doors" solution. You've switched from combining doors to a different (better!) analysis that correctly shows the probability of winning by switching is 2/3 and the probability of winning by sticking is 1/3. Perhaps paradoxically, by itself this does not mean that if you pick door 1 and see the host open door 3, that door 1's probability is now 1/3 and door 2's probability is now 2/3. If we walk through the false solution that shows the door 1 probability is 2/3, we'll see the same issue in the combining doors solution.

1. Each door has a 1/3 chance of hiding the car. Absolutely true, although to be precise we should say each door has a prior probability of 1/3 of hiding the car, where prior probability means the probability before the host has opened a door.

? If the door will be opened according to the rule, without giving any unauthorised addidional information, the chance of the door originally selected will not change. Whether the door is still closed or if it is open later: No difference, bedause showing a goat there where a goat has to be is absolutely no news in this respect.

2. Therefore, any pair of doors, not just the two doors the host must choose between, has a combined chance of 2/3. Also absolutely true, although these are also prior probabilities.

3. Therefore, door 1 and door 3 have a combined chance of 2/3. Also absolutely true, also referring to prior probabilities.

4. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3. This is where the problem occurs. The probabilities we're talking about here are posterior probabilities, i.e. probabilities after the host has opened a door. It is the sum of the prior probabilities that is 2/3 (by step 3), not the sum of the posterior probabilities. We haven't said anything about the posterior probabilities yet. We know the posterior probability of the car being behind door 3 is clearly 0, but to say the sum of posterior probabilities must be 2/3 because the sum of the prior probabilities is 2/3 does not follow - and is indeed false in this case.

Not correct. 1) the guest has devided the 3 doors in two diffeent groups. You have to look straight forward, not backwards. You never can combine door 3 with door 1 anymore after the doors have been devided in two different groups. Of course the car can be behind the door selected, doors 2 and 3 hiding two goats. But this will happen only in 1/3 of all cases.

Similarly, the problem in the combining doors solution is the 3rd step:

3. Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of door 2 must be 2/3.

This statement is also talking about posterior probabilities where the previous ones were talking about prior probabilities. This one is saying the posterior probabilities of the group "refused door+open door" must be 2/3 because the sum of the prior probabilities is 2/3. The "door 1 must be 2/3" false solution shows that this is invalid reasoning. We know the posterior probability of door 3 is 0, but that doesn't say anything about the posterior probabilities of door 1 and door 2. This is the topic of the 2nd question of the FAQ at the top of this page. The bottom line is that unless you assume or know how the host picks between two goats (if given the chance) all you know is that the posterior probability of door 1 is between 0 and 1/2 (with an average of 1/3), and the posterior probability of door 2 is between 1/2 and 1 (with an average of 2/3). To make these definitely 1/3 and definitely 2/3 you have to say or assume the host picks between two goats randomly (with equal probability), and then use this fact in your reasoning. Combining doors doesn't do it. Even enumerating the cases (as you have) doesn't do it! What this shows is the average probability of winning by switching or staying, but not the posterior probabilities of the chosen door and the refused door. -- Rick Block (talk) 03:20, 22 January 2010 (UTC)

But: (Have a look).

Tank you, Rick, for your efforts. Will be back tomorrow. Until then. My talks in german WP with Nijdam, I really like him very much, esp. for his efforts, had exactly this issue. I showed him millions of "tests", all with the same result. The more tests, the more precisely the results. The chance of the door originally selected remains what it originally was before: Exactly 1/3. And the chance when switching (door open or even before, as soon as once she has selected her original door) will always be 2/3. From the time the guest makes her choice and devides those 3 door into two groups. It has been proven millionfold. Will be back tomorrow. Bye, -- Gerhardvalentin (talk) 03:56, 22 January 2010 (UTC) P.S.: Chances always from the guest's view.

I wrote a simulation a while ago that shows results as well - specifically that the overall probability of winning by switching is 2/3 but this can be true at the same time that the probability of the unchosen door and refused door are not 1/3 and 2/3. See /Archive 1#Excel simulation of difference between "random goat" and "leftmost goat" variants. I'd be happy to discuss this simulation with you if you'd like. -- Rick Block (talk) 04:25, 22 January 2010 (UTC)

Thank you, would really like it! Wanted to go to bed (now 5:30 in the morning), but no, couldn't make a break. Your words "Therefore, if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3. This is where the problem occurs." made me amend my "routine" and I tried: If the door originally selected immediately receives any second "virgin door" as a compagnon (from the group of the two doors originally refused), then the chance will rise from 1/3 to 2/3. But after the host has opened one door and has shown a goat within the pair of the two doors not selected, then the chance will stick on 1/3 in case the goat is given as a compagnon, and the chance will rise to 3/3 in case the still closed door is added as a compagnon. So: Any two doors have the double chance of one door, provided they are "virgin doors", i.e. the compagnon is not a demonstrable goat (shown) nor the other closed door offered as an alternative. So we have to destinguish "virgin doors" and "prestressed doors" to add as a compagnon. Any "virgin door" (1/3) added will double the chance from 1/3 to 2/3. Any "prestressed door" will have other effects: Adding zero in case a prestressed demonstrable goat (0/3) is joined, and adding 2/3 (total now 3/3!) if a prestressed "still closed alternative door" is added. So you are right: Adding a second door will double the chance. But it has to be a "virgin" door also, and it may never be "prestressed". Results will be different ones, then, as shown above. My "routine" clearly showed it. Regards, -- Gerhardvalentin (talk) 04:43, 22 January 2010 (UTC) BTW: Did you see my contrib. above in An interesting result "The rule doesn' say to the guest where the car is, but the rule says to the guest where a GOAT is"? Valid! Bye -- Gerhardvalentin (talk) 05:00, 22 January 2010 (UTC) P.S.: Chances always "from the guest's point of view"

Rick, let me tell you about virgins. I read your words written miles above:

"I think we're on the same page except for the bit about not identifying doors. My view is that the problem is either conditional, i.e. applies to a specific player standing in front of two closed doors and a goat (in which case the doors are inherently identified) or unconditional (applies only in some "average" sort of way). You seem to be thinking there's a third option - "conditional" but without specifying which door." To avoid struggle it could be helpful to correctly name the issues, the items, the circumstances of the case we are mediating.

I hear "unconditional" and "conditional". In MHP only the host does know everything, guess he can peek behind closed doors. If you are the host and the guest selects one door she will do that randomly, and two doors remain unselected/refused. The door she selected has a chance of 1/3. Because it had been selected randomly between three dors. Imagine the host offers her the opportunity to add a second door to her first door, and she would pick out a second door. She also would do this "randomly". Only the host knows what's behind the two doors she originally refused: At least one goat behind one of the doors, and behind the second door, the "priviledged door" in 1/3 a second goat in 2/3 the car. Let's suppose for this example, it was the CAR. The guest did NOT know what's behind the two unselected doors and would select RANDOMLY. In 2/3 of all cases she would pick a goat, in 1/3 of all cases she would pick the car. That's what I mean with "randomly". If she choses the CAR it would help, if she chooses the goat: no effect on her chance. Back to the rules: As the total chance of both unselected doors (randomly) always is 2/3 the chance of each unselected door is 1/3. She could double her chance by adding a second door "randomly": in 50% of the cases the left one, (generally doubling her chance), in 50% of all cases the right one (in general doubling her chance). The host knows what is behind the closed doors. If the host opens one of the two unselected doors now, say the LEFT DOOR showing a goat, the chance of this door is zero. But zero only in this one game. Next game: Other distribution. In the long run she doubles her chances by picking a second door. If it is chosen randomly. This is true "in general". May not apply to one specific case. After the host has opened a door showing a goat the chance of this door is 0 for this specific case. Other times other distribution, chance of this door in general and on the long run: 1/3. We have to be a little more precise to express exactly what we mean. And have to avoid statements like "if the player picks door 1 and the host opens door 3, since the chance of door 3 is now clearly 0 the chance of the player's door must be 2/3" and similar statements. Because it has no effect on the probability of the initially chosen door, and so on. (Chances always from the guests view)

I hear "unconditional" and "conditional". Fine. But if you find precise words to express what you mean you never need maths with "conditional probabilities". Am Austrian, my language is german, not English. Necessary to say exactly what we mean. I am sure we will find excellent names for the phenomenons we want to express, even without mathematics.   Regards   -- Gerhardvalentin (talk) 11:56, 22 January 2010 (UTC)

Rick, if a host bias may exist, we indeed do not know the posterior winning chance of door 1. But why should it be between 0 and 1/2? However, when 'everything' happens randomly and therefore symmetrical, we know for sure that the exact chance of door 1 remains 1/3. Doors 2 and 3 together then have a chance of 2/3. One can of course only compare the prior combinations of doors together (like 1+3). Posterior comparisons like 1+3 don't make sense, no matter if Nijdam makes them.

He is not a professor anyway. If he is the same Nijdam as on Wikibooks, he was a maths lecturer (PhD) at the University of Twente until 2004. Heptalogos (talk) 16:43, 22 January 2010 (UTC)

His contribution to the discussion isn't much more than a repeated "it is a conditional problem, meaning it has to be solved by calculating a conditional probability", which is of course true and done by the combined door solution. The key here is the exact definition of the condition: is it another door with a goat or is it number 3? With the proper assumptions there's no problem with most solutions. And no reliable source says so, but Nijdam does. Without arguments. Heptalogos (talk) 17:43, 22 January 2010 (UTC)

Posterior winning chance of the door originally selected by the guest

Intro: The humor of the host does not matter. He unnecessarily never will lift his left hand nor his right hand high in the air. And he never lifts his right leg nor his left leg noticeably and unnecessaryly high. And - having the choice between two goats behind both doors - he never will prefer neither the rigt nor the left door. He is not grinning broadly and he does not look grumpy, alternately. Useless to say: He never may reveal any unauthorized and irregular information. Take that for granted.

My view:
For the guest, the winning chance of his door originally selected is 1/3 (the host knows better anyway, no point of interest here). But for the guest the winning chance is 1/3.
And the chance of at least ONE goat hiding behind the pair of two doors refused is 3/3 "at least  :-)" The host may see two goats there in 1/3 of all cases: No point of interest here. The guest knows about the only car, so the guest knows: At least ONE goat is hidden behind the two doors not selected.

So: Regardless of WHAT door the Host will open showing ONE goat behind this pair of doors not selected, this will never add any relevant new information for the guest regarding the chance of his originally selected door. The host knows more than the guest, no question. Of course he does. But the chances for the door originally selected by the guest has not changed in any way - from the guest's view. Any assumption that - for the guest's view - this chance could have changed by opening one door is ridiculous.

Am I right or am I wrong? Regards, -- Gerhardvalentin (talk) 17:54, 22 January 2010 (UTC)

If the host is given or is assumed to choose between two goats randomly, then (and only then) it turns out you're right. No one has ever said anything attempting to contradict this. The sources that talk about this are talking about a version of the problem where it is NOT given that the host chooses between two goats randomly. It is by thinking about this sort of version that the weaknesses in many of the arguments used to show the result of the standard version can be exposed. -- Rick Block (talk) 04:54, 23 January 2010 (UTC)

You are right, he chooses randomly if he should have two goats. Therefore I said "Take that for granted". -- Gerhardvalentin (talk) 23:25, 23 January 2010 (UTC)

Is the unconditional problem too easy?

Unless I am misunderstanding his reply, Niijdam at least considers the if the problem is clearly described the solution is obvious.

Can I ask if there is anyone who thinks that the solution to this, I hope clearly unconditional, problem is obvious. It is based on the K&W problem statement.

You are going to be on a game show with the following rules. You’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.

What would you expect to be the chances of winning the car for players whose policy is to always switch? Martin Hogbin (talk) 20:48, 17 January 2010 (UTC)

Before I am starting to study this, is it the exact K&W problem statement? If not, we cannot use it as our problem statement anyway. Heptalogos (talk) 22:49, 17 January 2010 (UTC)

I am not proposing to use it. I am just asking whether editors think the answer is obvious when the question is put like this. Martin Hogbin (talk) 23:05, 17 January 2010 (UTC)

Sorry to maybe change the subject, but I understand your strategy in this section is to find out why we keep disagreeing. IMO it might be helpful to arise from the doors and draw the bigger picture. In a section above I asked Rick about the essence of argue. Apart from the article, he is willing to use unconditional method for solving conditional problems, as long as we make the right assumptions. This is IMO also what Morgan implies. The willingness to agree may not be too big because of possible consequenses for the article (my interpretation), but I agree with them that the sources are leading. I think we actually agree on most probability issues now. Heptalogos (talk) 13:53, 18 January 2010 (UTC)

A simple Bayesian analysis

Situation: door 1 is picked

• Winning by switching.

P(A): car behind door 2 = 1/3.
P(B): door 3 is opened = 1/2.
P(A|B) = P(A and B) / P(B) = (1/3*1) / 1/2 = 2/3.

In this line you seem to use two different values for P(B). Martin Hogbin (talk) 23:04, 17 January 2010 (UTC)

That's correct. You can read the reason in the explanation directly below. Heptalogos (talk) 23:07, 17 January 2010 (UTC)

Explanation for P(A and B): given the car behind door 2, door 3 must be opened, so P=1.

But that is P(B|A). P(A and B) = P(A)P(B) they are not independent. Martin Hogbin (talk) 23:10, 17 January 2010 (UTC)

Looks a bit dodgy to me. See what Nijdam thinks. Martin Hogbin (talk) 23:13, 17 January 2010 (UTC)

• Losing by switching.

P(A): goat behind door 2 = 2/3.
P(B): door 3 is opened = 1/2.
P(A|B) = P(A and B) / P(B) = (1/3*1/2) / 1/2 = 1/3.
Explanation for P(A and B): two goats behind doors 2 and 3 (P=1/3) and door 3 opened (P=1/2).

Is this correct? Heptalogos (talk) 22:54, 17 January 2010 (UTC)

Nijdam's corrections, but now underneath:

Situation: door 1 is picked

A: car behind door 2; P(A)= 1/3
C: car behind door 1; P(C)= 1/3
B: door 3 is opened;

P(B|A)= 1. P(B|C)= 1/2.

P(B)=P(B|A)P(A)+P(B|C)P(C)=1*1/3+1/2*1/3=1/2

P(Winning by switching|B)=P(A|B) = P(A and B) / P(B) =

=P(B|A)P(A)/P(B)=(1*1/3) / 1/2 = 2/3.

No need to calculate "Losing by switching", but if you like:

P(Losing by switching|B)=1-P(Winning by switching|B)=1-2/3=1/3

(That's what Nijdam think) Nijdam (talk) 17:37, 18 January 2010 (UTC) Heptalogos (talk) 18:55, 18 January 2010 (UTC)

Nijdam, what's wrong with my solution? Heptalogos (talk) 18:57, 18 January 2010 (UTC)

I love it.

Situation: door 1 is picked

P(A): car behind door 2 = 1/3.

P(B): door 3 is opened = 1/2.

P(A|B) = P(A and B) / P(B) = 1/3 / 1/2 = 2/3.

A and B are dependent. Together, their chance is 1/3. The overall probability of B = 1/2. It's so logically true. Why would anyone want to watch the theatre below? Heptalogos (talk) 21:37, 18 January 2010 (UTC)

Nothing wrong for the good willing reader. It is clumsy written. That's why I helped you with correct terminology. Instead of "together", which could be interpreted as "united", you better use "simultaneous". And I calculated the (overall) probability of B for you. Glad you begin to show some interest in probability theory. Nijdam (talk) 22:45, 18 January 2010 (UTC)

This is the essence of the conditional solution. Nijdam's solution essentially provides more detailed reasoning. Welcome to the dark side (or maybe welcome out of the darkness). Being picky there are two issues. How do you know P(B) is 1/2, and how do you know P(A and B) is 1/3? Hint: look at the tree diagram or the expanded figure in the Probabilistic solution section. Nijdam's solution derives these from more elemental values, specifically from the assumptions P(A)=1/3, P(C)=1/3, the given that the host opens Door 3, and the "host strategy" rules that P(A|B)=1 (the host must open Door 3 if the player picks Door 1 and the car is behind Door 2), and P(B|C)=1/2 (the host picks randomly between Door 2 and Door 3 if the car is behind Door 1). If these are directly given in the problem statement Nijdam's solution is a fully rigorous proof that P(A|B) = 2/3. -- Rick Block (talk) 22:31, 18 January 2010 (UTC)

Just another straight question for you Rick, with no hidden agenda. By 'conditional' are you just referring to the fact that a door is opened by the host, in which case argument over, or do you mean the condition as to which door has been opened by the host is significant. Martin Hogbin (talk) 22:36, 18 January 2010 (UTC)

I mean a door, but it is inherently a specific door (because the host, and the player, and everyone in the audience can see which door it is). This means if it does (or even might) matter which door, we now know which one it is. We can examine the situation for any specific door we want, and if it doesn't matter which door, then the analysis will presumably have a result that is independent of door. If it does matter, then the result will show the dependency on the individual door. -- Rick Block (talk) 03:45, 19 January 2010 (UTC)

It looks like you are using the term conditional to mean that it matters, or at least that it might matter, which door the host opens. This is of course true, in some cases it does matter.

So, this is essentially what we are arguing about. In the symmetrical case it turns out that the door opened by the host does not matter (change the result). Are we obliged to treat this as a special case of the more general case in which the host may open a legal door non-randomly?

I know of no mathematical principle which states that we must do this, that we must treat any specific problem as a special case of a more general one. For example, you might argue that many the proofs of Pythagoras' theorem are invalid because they do not apply to triangles that are not right-angled. The only true proofs are those that generate relationships between the sides of all triangles of which the right-angled triangle is a special case. Proofs which use the specific properties of right-angled triangles are false.

We can solve the symmetrical MHP by using specific and well-known properties of the uniform distribution of legal host door choice by virtue of either symmetry of the fact that random information is no information. Although learned papers and text books can recommend more general methods as being advantageous, they do not have the power to prevent proofs specific to the symmetrical case from being valid. Martin Hogbin (talk) 17:00, 19 January 2010 (UTC)

Is the sample space in which you're evaluating the prior probabilities (that are in effect before the host "opens a door") the same as the sample space in which you're evaluating the posterior probabilities (those in effect after the host opens a door)? I think this is what is meant by saying the player's initial probability "does not change". If this is what you mean, then I don't think there's any argument. The probability of the player's door is 1/3 and the probability of the other two doors is (combined) 2/3. However, if you want to describe the situation after the host has opened a door then I think you have to be talking about a different sample space. It is only in this different sample space that it makes any sense at all to say the probability of the door the host has opened is now 0 and the probability of the "other" door is 2/3. This sample space cannot be the same as the original sample space since the probability of the door the host opens has changed.

A specific example might help. If the original sample space is 3000 samples of the game (where the player has picked door 1), then in this sample space we'd expect about 1000 of the players to have initially selected the car P(A)=1/3. P(car behind door 2) and P(car behind door 3) are also 1/3 (about 1000 each). After the host opens a door if we're still talking about all 3000 samples, none of these probabilities have changed. However, if we want to talk about one of the cases where the host has opened a door, say Door 3, we're not talking about all 3000 cases any more but (in the symmetric case) only half of them. Now we're talking about 1500 samples, not 3000 samples. The probability the car is behind door 1 in this (new) sample space may numerically equal the probability that the car is behind door 1 in the original sample space but these are different probabilities because they are in different sample spaces. The "unconditional solution" only addresses probabilities in the original sample space - that's what it means to be unconditional. What it is saying is of the 3000 initial samples, whatever the host does we'll still have about 1000 players who've picked the car. The probability of the "other door" is always the probability of BOTH other doors (1/3 + 1/3), not the probability of either individual "other door". The probability of, say door 2, is (can be) different only in some reduced sample space - like the one of 1500 samples where the host opened door 3. -- Rick Block (talk) 19:30, 19 January 2010 (UTC)

A bit more on the above. IMO, the reason the MHP is paradoxical is because the sample space changes asymmetrically with respect to the 3 doors. Half the samples where the player's selected door hides the car end up in each sub-sample space (assuming the host picks equally between goats) but all of the samples where, say door 2, hides the car end up in one and all of the samples where the other door hides the car end up in the other. People are simply not used to this. -- Rick Block (talk) 21:01, 19 January 2010 (UTC)

Rick, you do not need to explain the method of solving the problem by setting up a particular sample space, conditioning it according to the door opened by the host and then using the resulting conditioned sample space to calculate the probability of interest; I already fully understand this method, and have done so for some time. This a perfectly good method of solving the problem, perhaps it is even the best method, but it is not the only method.

For any given mathematical problem there are usually many diverse methods of dealing with it. Sometimes, when the attempted solution of a difficult problem grinds to a halt, a way forward is found using a completely different approach. This new approach may involve a different mathematical discipline and sometimes it may depend on spotting an unrecognised symmetry or connection that enables progress to be made. This is true of the MHP, it can be solved by means other than the the method that you (and Morgan and others) prefer.

You seem fixated of a specific method of solution that involves conditional probability. I agree, it is a good method and it is one that is applicable to the more general case where the host has a known legal door opening policy. I agree also that the method is instructive in showing how conditional probability works, but, it is not the only method of solving the MHP. I do not believe that a mathematician would ever claim that there is only one possible method of solving a given problem. Martin Hogbin (talk) 10:51, 20 January 2010 (UTC)

Conditions again

I strongly have the feeling, you do not understand what it means when we say: the MHP is necessarily conditional. This has nothing to do with the method used, but with the question asked. The question asked is after a conditional probability. The method used to calculate this conditional probability is unimportant, as I already told so many times. On the other hand: an unconditional problem, where one is asked to calculate an unconditional probability, may well be solved through conditional probabilities. What important is, is to understand that the simple solution, as well as the combined doors and similar reasoning, are not sufficient. Nijdam (talk) 12:00, 20 January 2010 (UTC)

I accept what you say in the trivial sense that the question is asked after the host has opened a door. If this is all you mean this fine, let us call the problem conditional, clearly one thing happens than another happens then we calculate the probability. If you want to insist that the term 'conditional probability' is used that is fine with me, so long as it does not determine what methods of solution are valid.

I still have not got an answer to what exactly makes the problem one of conditional probability. Is it because:

Something happens after the player chooses a door?

A door is opened after the player chooses a door?

A specified door is opened after the player chooses a door?

Perhaps you could enlighten me. Martin Hogbin (talk) 16:58, 20 January 2010 (UTC)

Happy to be of service. It seems you don't read thoroughly, as I already have indicated, me nor Rich will insist of the use of the word "conditional probability" in the introductory section. It is however a conditional problem and hence we object the simple solution, etc. without mentioning something about the conditional nature in some way. And we also never insisted on a specific method to be used, as long it is valid! We have discussed conditioning also over and over, so your first possibility is way out of line. Both the other two form a condition. Then it depends on the question asked if this condition is to be taken into effect. Nijdam (talk) 12:41, 21 January 2010 (UTC)

Let me tell you my thinking:

I say that a condition is any event mentioned in the problem statement which might affect the probability of interest. In the MHP we have these events:

1) Monty says the word 'door'. It is not inconceivable that Monty might use the word 'door' only when the player has initially chosen the car, perhaps he says 'this one' otherwise. Such things, however, are not the normal assumptions of mathematical puzzles and we might therefore take it that this event does not affect the probability of interest, even though it is mentioned in the problem statement. It is therefore not a condition.

2) Monty opens door 3. Initially it might be suspected that this event would affect the probability of interest, it therefore might me considered wise to take this event as a condition. If, on the other hand, it can be shown that, when the host chooses a legal door randomly, this event cannot possibly affect the probability of interest then this event need no longer be considered a condition. The problem is no longer, therefore, conditional. Martin Hogbin (talk) 14:22, 21 January 2010 (UTC)

Of course it is not up to you to decide what has to be considered a condition. But I try to follow your line of thinking. The opening of door 3 doesn't influence the probability of the car behind door 1, means, the probability after the opening is (in value) the same as before the opening. We also discussed this several times. The probability after is (formally) the conditional probability. It need not be said with so many words, but it is necessary to mention that it needs proof (may be logical proof) that there is no influence for door 1 (on the value!!). For the other doors there is influence! But for all doors there is influence on the nature of the considered probability. Before: unconditional, afterwards: conditional. Nijdam (talk) 14:51, 21 January 2010 (UTC)

It can only be up to the person answering the question to decide whether the problem is conditional or not, unless the problem statement mentions the word 'conditional' (which the MHP does not). You have never given me a sound basis on which this decision is to be made. My suggestion is that the person answering the question has to consider every event mentioned in the problem statement and decide whether it could possibly affect the probability to be calculated. If an event which might affect the probability of interest exists, then the problem is conditional. If no such event exists then the problem is not conditional. Do you agree?

Regarding the proof that opening door 3 (when the host chooses a legal door randomly) does not affect the probability that the car is behind door 1, we note that, if the car is behind door 1, the host must choose randomly (uniformly) between doors 2 and 3. In this case the door actually chosen gives no information about anything (this is a well-know property of a random choice), in particular the location of the car. Thus we can safely consider this event not to be a condition in our problem. Martin Hogbin (talk) 21:43, 21 January 2010 (UTC)

As soon as you consider the opening of door 3 there is a before and an afterwards, with corresponding probabilities. Whether the opening of door 3 gives info or not is irrelevant for this. It may however be relevant for the calculation of the values. What is it that withholds you from understanding? Nijdam (talk) 22:52, 21 January 2010 (UTC)

What 'prevents me from understanding' is that you cannot tell me what in the problem statement tells us that, "the host says the word 'door'" is not a condition of the problem, with a before and afterwards with corresponding probabilities, but "the host opens door 3 is". What exactly in the problem statement tells me this?

Do you believe that the probability of an event can change if no information about it is revealed? Martin Hogbin (talk) 18:24, 22 January 2010 (UTC)

We had this discussion before. Instead of "the host saying 'door'", you then spoke of "Clapping my hands" or something alike. I explained you that a condition means a reduction of the sample space, do you remember? And just as handclapping is no event in the MHP, so is saying 'door' no event. We only consider events that matters to the problem, and they are about placing the car, choosing a door and opening a door by the host. I showed you the actual reduction, and as I'm recollecting right, Rick also did. Nijdam (talk) 22:53, 22 January 2010 (UTC)

Yes, I remember the discussion well. Your argument that a condition is anything that reduces the sample space does, at first sight, seems reasonable. However, there is nothing about a sample space in the problem statement, we have to set one up based on our understanding of the problem, based on events that matter, as you put it. Thus, before we set up our sample space, we have to decide which events matter, in other words which events might possibly affect the probability that we are trying to calculate. For the MHP, I could choose to set up a sample space including both the set of events in which the host says the word 'door' and the set of events in which the host does not say the word 'door'. According to your first definition, the saying of the word 'door' now becomes a condition of my problem, since considering only the case in which the host says 'door' reduces the sample space. In some circumstances this could be a perfectly reasonable thing to do.

Now, I do agree that in the MHP it is wise to set up a sample space to include the host door choice, as it would, on the face of it, appear that this might affect the probability of interest. In the symmetrical case, we can then do our calculation and discover that it, in fact, does not. That is what you suggest, I believe, but this is not the only way of addressing the problem. If we can positively show that the host choice of door cannot affect the probability of interest then we are free to construct a sample space in which this choice is not represented, and thus, by your definition, is not a condition. In view of the extremely counterintuitive nature of the problem this is probably not a good idea, but we could do it.

I would also add that there may be ways of addressing the problem that do not use a sample space at all but use some other mathematical construct. Maybe there is a geometric way? Who knows? I do not accept that any paper or text book can demand that there is only one possible way to address a problem. This is not normal in mathematics, sometimes solutions come from completely unexpected directions. Martin Hogbin (talk) 11:16, 23 January 2010 (UTC)

Of course you may complicate things by introducing events that are meaningless for the problem. Although the host says the word 'door' somewhere, nowhere in the problem is this considered to be important. So let's stay down to earth. You're constant talking about the symmetrical case, as if your hope is, this can do without conditional probability. Well, it cannot. Calculation, based on all the necessary events(!) may show that some unconditional probability and conditional probability have the same numerical value, but that doesn't imply we may do without some events. Then about your remark on other methods. There you're right, and Gill for instance showed a game theoretical approach. But most people, most sources, MvS herself, speak of probability. So we have to present the MHP firstly as a probability puzzle. Agree? Nijdam (talk) 16:09, 23 January 2010 (UTC)

You talk about events that are meaningless for the problem. The problem statement does not tell us which events are important and which are not. We have do decide which are the important events. To be a little perverse consider this scenario:

Monty Hall is replaced by a new host who tries to be a little more helpful to the player, maybe the show is loosing audience share because the prize is not being won often enough. This new host gives the player a bit of a clue what to do by saying "do you want to change to this door?", with this door being said in a warm and inviting tone, when the player will win by switching. When the player has originally chosen the car, the host says uninvitingly, "or would you like to change your choice?". Now the saying of the word 'door' is an essential condition. A contrived scenario, maybe, but not impossible. One still might chastise Morgan for not including this possibility in their analysis. But, to be a bit more serious, now that you are aware of this possible scenario, do you insist that this 'condition' must be included in every solution of the problem? So it is with the host door number. If we can show that it is unimportant then we are free to propose a solution that does not use it.

I agree that probability is the natural branch of mathematics to address this problem, particularly the extended version which asks for the probability of winning by switching, but, we do not have to include a condition which we can show is not important in our understanding of the problem. Martin Hogbin (talk) 19:58, 23 January 2010 (UTC)

Well Martin, the MHP is a probability problem. The story it is packed in, is just to make it attractive and imaginable. I'm not interested in all strange scenario's that has nothing to do with the problem, and that clarify nothing.

The scenario that I suggest above is not in any way ruled out by the Whitaker's MHP statement. Martin Hogbin (talk) 20:40, 23 January 2010 (UTC)

You continuously seem to be confused about the nature of the problem and the nature of the solution. The latter is unimportant. The problem is described in terms of (A) placing the car, (B) choosing a door and (C) opening a door by the host. This determines the necessary events. It's about time you come to understand that. Nijdam (talk) 20:21, 23 January 2010 (UTC)

I am not confused, you are just repeating your assertion, without supporting argument. Do you not understand my point at all? You have to make a decision at the start of a problem as to what might be important. Suppose you were teaching a student how to decide what to consider a condition upon reading a probability problem. What would you say? Martin Hogbin (talk) 20:40, 23 January 2010 (UTC)

To make my point clear let me ask you this question, supposing the host always uses the two phrases I have described above. Suppose he always uses the word 'door' when the host has not originally chosen the car and never uses it when the host has originally chosen the car. What is the probability of winning by switching, given the Whitaker statement (and that the host opens a legal door randomly)? Clearly it is 1. Now what is the probability given that the host chooses randomly which phrase to use? Do you need to calculate this? Martin Hogbin (talk) 09:08, 24 January 2010 (UTC)

No one is interested in describing the MHP in other terms than my A, B and C. They are necessary and sufficient. Where are you heading at? Nijdam (talk) 17:42, 24 January 2010 (UTC)

You know where I am heading, I am attempting to demonstrate to you that what is a condition and what is not a condition in a probability problem must be decided by the person answering the question. There is no other way in which this can be done. Thus you cannot say that, in any given question, a particular event must be a condition. It may turn out to be irrelevant, like the sneeze. You do not need a calculation to do this. Martin Hogbin (talk) 00:47, 25 January 2010 (UTC)

(outidented)Martin, I really don't know. If it makes you happy, you may formulate a kind of MHP, in which sneezing, handclapping, mentioning the word door, walking your dog, eating a hamburger, etc. playes a role. In my MHP only the mentioned aspect A, B and C do. Before I discuss anything further, please tell me how you describe the MHP. Nijdam (talk) 02:47, 26 January 2010 (UTC)

Nijdam, do you know of any other published conditional solutions to the MHP? I think Chun's table is looking wobbly. Glkanter (talk) 23:01, 21 January 2010 (UTC)

Martin - most of the above was explanatory to make sure we're on the same page with regard to the question I asked you that you didn't respond to. Is the sample space in which you're evaluating the prior probabilities (that are in effect before the host "opens a door") the same as the sample space in which you're evaluating the posterior probabilities (those in effect after the host opens a door)? -- Rick Block (talk) 13:21, 20 January 2010 (UTC)

Yes, in the symmetrical case, for the two reasons that I have given previously. My sample space contains the same two events, both with the same probability, before and after a door is opened. You may want to start with a different sample space. Martin Hogbin (talk) 16:58, 20 January 2010 (UTC)

I was a witness to what happened.

vos Savant's column generated letters that she answered for months. In 1990, there was no e-mail. Parade magazine probably had a printing schedule with 4 - 6 week lead times. Unlike our debates, this all took place in slow motion.

The only question/objection ever raised in Parade was 'It's 50/50'. I was there. That's why I am so adamant that nobody cared at that time about the door numbers. All the furor was about the 50/50 vs 1/3 2/3. Nothing else. When I first read the Wikipedia article, I was floored by the revisionist history I found. You're all entitled to your own opinions, but not your own facts. Glkanter (talk) 04:25, 18 January 2010 (UTC)

Please see WP:V, which is right up there with WP:NPOV (i.e. one of the 3 core content policies that Wikipedia is based on). Your own recollection of anything is completely beside the point as far as Wikipedia is concerned. If it's not in a reliable source it might as well not exist. -- Rick Block (talk) 20:25, 18 January 2010 (UTC)

I started a new section about what the MHP is, purely based on an overview of sources. I do have the impression that Parade is the centre of gravity. Do you have an overview of all Parade editions handling the MHP? Heptalogos (talk) 12:42, 18 January 2010 (UTC)

The four columns are reprinted at [13]. The sections of text in red are Marilyn's responses. I'm not sure exactly how much of the intervening text was actually published (I only have a copy of the first column). -- Rick Block (talk) 20:25, 18 January 2010 (UTC)

Thank you, Rick. That column of letters and vos Savant's responses supports my statement of what transpired 100%. It was all 50/50 and 1/3 v 2/3. There was absolutely nothing about which door or a host bias. She probably had 10 million readers, give or take. How many people read Morgan? Glkanter (talk) 22:02, 18 January 2010 (UTC)

There is one other reference that is important, in addition to those three MvS columns. The one where Marilyn vos Savant said that Morgan's interpretation of the problem was incorrect. I don't know how to link to it (I think it requires special access that I have, but won;t translate to others), but it was published in The American Statistician, Vol. 45, No. 4 (Nov., 1991), pp. 347-348. While Morgan, et al, and Rick Block seem to be of the opinion that MvS was talking about interpretation of her answer, that is clearly not the case. Because there was no such misinterpretation of her answer. The only misinterpretation was of the question.

And while I'm here, let me stress one point: There is no published 'scientific analysis' that concludes the problem is conditional, based on door numbers. Three sources state an opinion that it is, but offer no arguments for why that is so (and one of those even states that sematically, it isn't so). Far more articles that cite the Parade article treat it as unconditional, starting with the original author of the problem (and as MvS has stated she edited the question for content, she has to be considered the original author) and continuing to Leonard Mlodinow. While we can't discount those three dissenting opinions, NPOV guildelines say the two differing viewpoints need to be be described clearly and with balance. That means separating the two viewpoints, and not concluding that "the conditional interpretation" is correct. Since it really isn't a "scientific" issue, but one of interpreting the problem (and then applying "science" to the interpretations), there is nothing NPOV about treating the two problems as separate problems, even in the same article; while it violates NPOV to suggest only one is correct.

You wouldn't consider Selvin "the original author of the problem"? Why not?

There's a Formal Mediation request out there. You'll be the final signee. Glkanter (talk) 17:38, 20 January 2010 (UTC)

And one finale comment, more on editorial style. The article stresses words and phrases like 'scientific', 'probabilititic' and 'correct solution' far too much. The approaches wouldn't be included if those words and phrases didn't apply. As is, the only point of including them seems to be POV on the part of whoever put them in, as if they add weight to their favored interpretation. JeffJor (talk) 17:27, 20 January 2010 (UTC)

Here are some interesting quotes from WP:V.

The appropriateness of any source depends on the context. So which is the best source regarding a question to a popular magazine, the regular writer of the column to whom the question was addressed or a bunch of statistics professors writing years later in a second rate statistics journal?

The original writer who say the full question, and has admitted she condensed it down to the concise statement that reached print. JeffJor (talk) 17:27, 20 January 2010 (UTC)

The most reliable sources are usually peer-reviewed journals; books published by university presses; university-level textbooks; magazines, journals, and books published by respected publishing houses; and mainstream newspapers. Nothing here tells us that one has the power to override the other. Martin Hogbin (talk) 22:58, 18 January 2010 (UTC)

The immediate "peer review," Seymann, contradicts Morgan. The fact that the controversy over the problem - which is the only reason it is famous and included in Wikipedia - had absolutely nothing to do with any conditional/unconditional issues shows that that is a minor sidebar to the article. JeffJor (talk) 17:27, 20 January 2010 (UTC)

Does Seymann contradict Morgan? Please show me where. I only read in his comment that without a clear definition of the problem, it cannot be solved properly. Quite an eye-opener to me. The K&W version BTW is clearly defined.Nijdam (talk) 20:09, 23 January 2010 (UTC)

Don't overlook two sentences later: Academic and peer-reviewed publications are highly valued and usually the most reliable sources in areas where they are available, such as history, medicine, and science. -- Rick Block (talk) 03:25, 19 January 2010 (UTC)

When Chun Divides The Contestant's Door Into 2 Equal Pieces For The Conditional Table, Doesn't He Then Already Know That He's Going To Divide By 1/2?

How did he know to break the 2-goat-door 1/3 into 2 equal halves? Using that same knowledge, can't he use the same 1/2 he used to break the 1/3 out to reconstruct the 1/6 into the post-goat-revealed (1/6)/(1/2) = 1/3? This would be instead of dividing by (1/3 + 1/6 ), which also equals 1/2. Glkanter (talk) 20:52, 21 January 2010 (UTC)

The split of 1/3 reflects the host's decision if he is picking between two goats. In the fully explicit problem as posed by Krauss and Wang, this is given to be a random choice and it is this choice that divides the 1/3 exactly in half. This was not Chun's reason (Kraus and Wang postdates Chun by 12 years or so). If the host strategy is not given in the problem statement, one way or the other this boils down to an assumption (Chun simply assumed random choice in this case). A valid reason for this would be that you're explicitly assuming the problem is symmetrical meaning any chance involving door 2 must be the same as the chance involving door 3. Using this reasoning (symmetry), you can say both the overall chance of the host opening door 2 must be the same as the chance of the host opening door 3 and the chance of the host opening door 2 and door 3 in the case the player has picked the car must be the same as well. Since we're given the host is going to open a door (meaning the sum of the probabilities of the host opening door 2 and door 3 must be 1), the overall probabilities p(host opens door 2) + p(host opens door 3) = 1. By symmetry these are the same, so we have x+x=1, so x=1/2. Similarly, the chances the host opens door 2 and door 3 if the player has picked the car sum to 1/3, so we have y+y=1/3 so y=1/6.

I suspect what you're actually trying to get to is an argument more like this. By symmetry, the player's initial chance of selecting the car is split exactly in half because the host must open one of the other two doors in this case, so the player's total probability regardless of which door the host opens is (1/3)*(1/2). Also by symmetry, since the host must open one of the unchosen doors the total probability of the host opening either one of the two doors must be 1/2. The conditional probability that the player's door hides the car is therefore ((1/3)*(1/2))/(1/2), which is (1/3)*((1/2)/(1/2)) which is (1/3)*1 which is 1/3. -- Rick Block (talk) 01:09, 22 January 2010 (UTC)

Lets follow that path. Only 1 line of the 4 lines needs to be filled in, the 1/6 for when the host opens door 3. But why even bother, if you already know you're just going to divide it right back by 1/2? This demonstrates, as I believe Boris has been saying, that the symmetric conditional solution works, but for what benefit?

Wikipedia is not a college text book. Maybe in a text book, a slightly different problem related to a fun puzzle is a good starting point to learn about conditionality. But it has nothing to do with the paradox. Maybe a new section on 'The MHP In Academia' would give these tangential sources a nice home in the article. Glkanter (talk) 01:33, 22 January 2010 (UTC)

So, it seems to me that Nijdam's 'the condition problem statement is necessary' argument comes down to this:

'1/3' = original door pick, '1/2' = likelihood of either door 2 or door 3

1/3 * 1/2 / 1/2 = 1/3

Looks to me like the '1/3' is the same '1/3' both times. As in, 'nothing changes by Monty's revealing a goat'. Glkanter (talk) 16:58, 22 January 2010 (UTC)

Nijdam is making some progress:

"The opening of door 3..<>..it needs proof (may be logical proof) that there is no influence for door 1 (on the value!!). "

He now needs proof that they are independent.

Before he said:

"Independence doesn't play a big role in the MHP, except the independence of the placement of the car and the first choice of the player of course."

Proving independence is actually disproving all possible dependence. So let's do it the more obvious way: can anyone please come up with one possible case of dependency between the two events, assuming random stuff? Heptalogos (talk) 18:29, 22 January 2010 (UTC)

All I know is that the only conditional solution in the article relies on a tree/table that also solves the problem unconditionally. And the conditional formula is 1/3 * 1/2 / 1/2 = 1/3. And it's the same '1/2' both times. Whatever Nijdam is insisting, it makes no difference in actual practice for the MHP. Glkanter (talk) 21:20, 22 January 2010 (UTC)

Actually, the two 1/2's are different. They're definitely related, and if the first (the host preference between door 2 and door 3) is 1/2 then so is the other one (the likelihood of a player seeing the host open door 2 vs. door 3), but if we call the first one X and the second one Y, then Y=1/3+X(1/3). -- Rick Block (talk) 04:08, 23 January 2010 (UTC)

I don't understand your point above. You claim 'the two 1/2's are different', then you write, 'and if the first (the host preference between door 2 and door 3) is 1/2 then so is the other one (the likelihood of a player seeing the host open door 2 vs. door 3).' That means they are the same. Despite what you personally choose to name them. Glkanter (talk) 13:24, 23 January 2010 (UTC)

What I'm saying is that these are different concepts that both have the same value (in the symmetric case). The value being the same doesn't mean the concepts are the same. -- Rick Block (talk) 19:05, 23 January 2010 (UTC)

The moment you multiply the contestant's door's 1/3 by 1/2, 'Because doors 2 and 3 are equally likely', you know without any further work that you will divide the resulting 1/6 by 1/2, the same 1/2, which is why it's pointless to insist what Rick, and Nijdam and Morgan insist. Glkanter (talk) 13:14, 24 January 2010 (UTC)

Prior and posterior

Quote from Rick:

The player's 1/3 chance at the beginning splits into the player's chances for the case where the host opens door 2 and the case where the host opens door 3. The 1/3 at the end (which is a conditional probability) is only a piece of the original 1/3 (half in the case where p=1/2). p (and 1-p) are the fractions of this 1/3 that end up in each case.

                   1/3           +          1/3         +         1/3    = 1
                   /\                       /\                    /\
                  /  \                     /  \                  /  \
                 /    \                   /    \                /    \
                /      \                 /      \              /      \
host opens:  door 2   door 3          door 2   door 3       door 2   door 3
              /          \             /          \          /          \
             /            \           /            \        /            \
          (1/3)p    +  1/3(1-p)  +   0       +     1/3     1/3     +      0 = 1

If p is 1/2 the two terms at the left are each 1/6, and when the host opens door 3 the (unconditional) chances are (1/6,1/3,0). To express these as conditional probabilities we divide by the sum, i.e. divide each by 1/2, which makes these (1/3, 2/3, 0). If the player is looking at two closed doors and one open door, this split has happened (in accordance with whatever p is). The upshot is that the original (unconditional) 1/3 and the resultant (conditional) 1/3 are never the same 1/3 - and the unconditional solution is talking about the 1/3 on the top line, not the conditional probability you can compute from the bottom line by dividing (1/3)p by ( (1/3)p + 1/3 ). Note that this turns out to be 1/3 if p is 1/2, but depending on p it might be anything between 0 and 1/2. -- Rick Block (talk) 14:55, 23 March 2009 (UTC)

Rick, that's wrong. If no equal goat constraint exists (p is not 1/2), posterior chances of doors 2 and 3 are not necessarily 1/3 or 0 as presented. Let's consider the leftmost door (2) host preference: after opening door 3, the chance of a car behind door 2 is 1. Another host strategy: open door 2 when the car is picked. When door 3 is opened, door 1 has a winning chance of 1, which is more than 1/2. Edit: this chance is of course 0. Heptalogos (talk) 22:05, 23 January 2010 (UTC)

??? The numbers on the bottom line are posterior total probabilities, not posterior conditional probabilities, i.e. they are in the same sample space as the numbers on the top line. In the leftmost door preference case, after opening door 3 the total probabilities are (0,1/3,0) which, expressed as conditional probabilities for this case are (0,1,0). Similarly, if door 3 is opened by such a host the total probabilities are (1/3,0,1/3) which are conditionally (1/2,0,1/2). (continued below) -- Rick Block (talk) 18:46, 23 January 2010 (UTC)

Then what is p and why? You say the probabilities at the bottom/end are not conditional, but above the graphic you say they are. Heptalogos (talk) 20:49, 23 January 2010 (UTC)

p is the probability the host opens one of the doors in the case the player's door hides the car (if he opens one with probability p, he opens the other with probability 1-p). The product of p and 1/3 is the joint probability of the car being behind the player's door and the host opening a particular door (in the case the player's door hides the car). If this is a random choice p is 1/2. If this is not specified in the problem statement it presumably could be anything from 0 to 1. If the player doesn't know what it is, the player might analyze the probabilities assuming it's 1/2, or the player might examine the range of possibilities leaving this as a variable. If you don't like the Morgan et al. source, try the Gillman source. Another one not referenced in the article that makes the same points is [14] (and plenty of others).

The conditional 1/3 I'm referring to above the figure is the posterior conditional probability of the player's door being the one hiding the car. It's the "1/3" people refer to when they say the player's chances have not changed after the host has opened a door. This 1/3 is not shown in the figure - all the probabilities in the figure are total (unconditional) probabilities. -- Rick Block (talk) 21:40, 23 January 2010 (UTC)

Thanks for explaining, I read it wrongly, and also made a mistake. It seems to me as a nice figure. Heptalogos (talk) 22:08, 23 January 2010 (UTC)

When we assume random host behavior, the winning chance of door 1 is always 1/6 / 1/2 = 1/3, which is the conditional probability. Vos Savant skipped the (1/6 / 1/2) part, making it seem like an unconditional probability. She understood logically that because of symmetry the condition of the number was not relevant. It's still a conditional probability, using the opening of "door 2 or 3" as a condition, rather than using one of those numbers as a condition. Heptalogos (talk) 11:41, 23 January 2010 (UTC)

I agree, the opening of a legal door is a valid condition, the door number opened is provably irrelevant in the symmetrical case. Martin Hogbin (talk) 11:58, 23 January 2010 (UTC)

When we assume random host behavior, the total posterior probabilities are either (1/6,1/3,0) [in the case the host has opened door 3], or (1/6,0,1/3) [in the case the host has opened door 2]. If the "event" is that the host opens "door 2 or 3" (keeping the sample space the same as the original sample space) the total posterior probabilities (only one event, right?) are (1/3, 1/3, 1/3) (!), i.e. what this event is saying is we don't know which door the host opens and nothing has logically changed from the original situation. We know only "a door" has been opened. This doesn't affect the probability of the player's door, but since we don't know which door has been opened it hasn't affected the probabilities of the other doors either! This is an exceedingly counterintuitive way to model the problem because it mentally conflicts with the image you certainly have in your brain where the player is standing in front of two closed doors and an open door and the total probability of the open door (in this case) is 0. This image inherently reflects a conditional case in a reduced sample space.

If we contrast this with a problem that is unconditional, like an urn problem with one white ball (you win the car) and two black balls (you win a goat) where the player picks a ball but doesn't look at it and the host removes a black ball and then the player is allowed to switch for the remaining ball, because the two black balls are indistinguishable you can (and pretty much have to) model it the way you're trying to model the MHP. Specifically, the host removing a black ball does not and cannot reduce the sample space because the two black balls are indistinguishable. In this problem there are only two senarios:

1. Player picks the white ball, host is forced to remove a black ball and the player can now switch to the remaining black ball (probability 1/3 because the only decision is the player's initial choice of ball)

2. Player picks a black ball, host is forced to remove the other black ball and the player can now switch to the white ball (probability 2/3)

We can also model this problem with three probabilities, the probability of the white ball being in the player's hand, the probability of the white ball being in the urn, and the probability of the white ball being in the host's hand (after removing it from the urn). After the player picks a ball but before the host removes one these are (1/3,2/3,0), respectively. What happens when the host removes a black ball from the urn? Absolutely nothing. The chance was 0 that the host was holding the white ball before he took a ball out of the urn and remains 0 after he takes out a black ball.

Assuming the problem is symmetrical makes it logically equivalent to the urn problem, but I"ve never seen a source that explicitly takes this approach. The urn problem analogy is definitely WP:OR. -- Rick Block (talk) 18:46, 23 January 2010 (UTC)

I agree, but one aspect mentioned is overlooked too easy. It's the fact that 'another door with a goat is opened' is for sure a condition being addressed by the popular solution, which makes it a solution to a conditional problem, although some sources believe that the wrong condition is used. These sources are being too easy to themselves when they state that the solution is a correct one to this or that unconditional problem. Which is true of course, but that doesn't mean it's not a correct solution to any conditional problem. It's only not correct to the problem in which the specific number 3 is a condition, which is mentioned explicitly by Morgan by the way, but wrongly interpreted by Nijdam who claims that the popular solution is no solution to a conditional problem at all. Heptalogos (talk) 22:25, 23 January 2010 (UTC)

My point is that if the "condition" that "another door with a goat is opened" doesn't affect the player's initial chance of having selected the door, doesn't affect the chance of the car being behind either of the unselected doors either! It does not reduce the sample space, so all original unconditional probabilities remain unchanged. If the player picks door 1 and the host must open a door and opens "door 2 or 3", then the probability of the car being behind each door (not just door 1) is STILL 1/3. P(car behind door 2|host opens door 2 or door 3) is exactly the same as P(car behind door 2) because the event 'host opens door 2 or door 3" has not changed the sample space. If you want to say the probability of one of the unselected doors is 2/3 and the probability of the other one is 0, you have to be talking about a conditional case. This is what Nijdam is saying when he says the problem is conditional. Even if you try not to say which door, like P(car behind the unopened, unselected door|host opens the other unselected door) you're saying something different than "the host opens door 2 or door 3". With this phrasing, it's conditional, but this condition affects the player's door as well; the player's door's total probability is split in half (or in accordance with the probability with which the host has opened "the other unselected door", whichever door it is, in the case the player's door hides the car) and it retains the same numeric value only as a conditional probability. This phrasing is ludicrous given that the player and host and everyone in the audience can see which door the host opens - it's much more clear to use one of the doors as an exemplary case and have the condition be something like "the host opens a door, for example door 3". This says the same thing, but in a way that everyone can understand. -- Rick Block (talk) 01:56, 24 January 2010 (UTC)

Your point is taken. It's the same old beta-perspective in this alpha-beta conflict. There's probably an English term for it, meaning something like:

• Alpha - As a whole; value, purpose, intention, strategical level.
• Beta - Exact details; formula, laws, tactical level.

The elementary, basic discussion we keep having is the phrasing of the problem. When (intentional) language is used, not formula, beta (maths) people must translate it, which is not the biggest talent of beta-science, if I may say so. You have translated my "door 2 or 3" into several options, and you will understand which one comes closest to my intention. When we both agree (we did) that it's conditional, and also symmetrical, you mention that the player's door chances are split, then divided by a sum and finally 1/3 again. Those are details which serve the exact same purpose, and there's really nothing realistic that Vos Savant missed in her logic, neither did I. So yes, the symmetry, the randomness, are all reasonable assumptions which make her logic correct, and they can be made explicit on demand. The popular solution is solving a conditional problem, isn't it? If so, I don't see anything we don't agree about anymore, as far as the "theory behind the article" concerned. Heptalogos (talk) 11:51, 24 January 2010 (UTC)

I think the discussion has been about the intent of the problem (not necessarily the phrasing) and then the logic behind the solution. Martin has argued for a long time (and I think is still arguing this) that the intent of the problem is to ask about the average probability of winning by switching vs. the average probability of winning by sticking. Nijdam is arguing that the intent is to ask about the conditional probabilities faced by a player looking at two closed doors and one open door showing a goat. Although we can, with appropriate assumptions, make the numeric answer of these the same they're still different problems. Even if the intent is for the numeric answer to be the same (which I think is the case here), they're still different problems. For example "assuming n is 2, what is n+n" is a different problem than "assuming n is 2, what is n squared" even though they have and are clearly intended to have the same answer. Is adding 2 plus 2 a logically correct answer to the latter question, or is there quite a bit missing from this solution?

This is probably more appropriate for the article talk page (or even our forthcoming, long awaited mediation) than here, but the point Morgan et al. (and the others who say the problem is conditional) are raising is exactly the same as this. In their (expert) POV, strictly unconditional solutions, in particular those that completely ignore how the host picks between two goats, are just as incomplete as adding 2 plus 2 as a solution to "what is 2 squared".

So, I think we agree about the theory behind the problem. What I don't think we agree on (yet) is:

1. Does the article need to represent this "conditionalist" POV at all? My guess is we'd get at least one no, but the consensus would be yes. My personal opinion is that the answer must be yes due to WP:NPOV regardless of any "consensus" here.

2. Must this conditionalist POV be mentioned with or immediately following any initial unconditional solution? My guess is we'd get at least one yes, but not an insignificant number of no's. Again, my personal opinion is that the answer must be yes due to WP:NPOV regardless of any "consensus" here, but I think presenting a conditional solution as an alternative solution and accurately (in an NPOV fashion) describing the difference but deferring the criticism to a later section would be sufficient. -- Rick Block (talk) 18:55, 24 January 2010 (UTC)

Well, the popular solution section is not mentioning the average probability and Marilyn apparently uses such tests to prove chances are not 50/50. (These actually don't seem unconditional because the sample space is reduced: two doors become one.) Btw, using 'F2' as a solution, she would clearly assume that the door number is not a condition. Furthermore, any 'specific' another door with a goat is opened case has the same probability as the average, when all reasonable assumptions are made. We only need to clarify. But yes, we do agree on the theory I think. I also agree to you article suggestions. Heptalogos (talk) 20:30, 24 January 2010 (UTC)

The puzzle is so easy, and the paradox so profound...

Nobody needs to use probability, or logic, or Bayes, or anything but arithmetic to solve this puzzle.

Selvin wrote out all 9 possible results. 3 car locations by 3 contestant choices.

He proved it with a table. Anything beyond that, unless it helps the puzzle solver understand the paradox, adds no value. Glkanter (talk) 13:33, 23 January 2010 (UTC)

So? Heptalogos (talk) 13:40, 23 January 2010 (UTC)

Why all these other complex or clever solutions in the article, but not Selvin's simple table from which all the others are derived? Glkanter (talk) 14:08, 23 January 2010 (UTC)

Because the article is not a source of what you think is reasonable. You need to change your strategy to really change things. Heptalogos (talk) 14:22, 23 January 2010 (UTC)

Selvin's first solution is a simple non-conditional solution published in a peer-reviewed journal and thus deserves a place in this article. I do not see anything wrong with making the presentation better by adding pictures, if this is considered helpful. The most important thing at the start of the article is keep it simple. I believe that that the maxim keep it simple should be carried through to the 'Aids to understanding' section as well. The last thing that helps understanding is the addition of extra complications.

There seem to be some people who think that the simple non-conditional problem is not, in some way, hard enough. We here have all become so familiar with the problem in its various forms and its many solutions that it is very easy to lose sight of the fact that, even when the problem is presented clearly, unambiguously, and non-conditionally, most people get it wrong. Before we attempt to go into a discussion of which door the host has opened we must make sure that are readers understand the basic problem. Martin Hogbin (talk) 14:29, 23 January 2010 (UTC)

Selvin's letters are primary sources. We could perhaps include them in the "History" section, but including them in the Solution sections is not in keeping with NPOV sourcing policy (see WP:NPOV#Primary, secondary and tertiary sources). -- Rick Block (talk) 19:13, 23 January 2010 (UTC)

I guess that is Morgan's paper out then. Martin Hogbin (talk) 19:36, 23 January 2010 (UTC)

Martin - are you being serious here? Morgan's paper is clearly a secondary source. -- Rick Block (talk) 19:43, 23 January 2010 (UTC)

Morgan is not a review or summary of earlier work, it presents new conclusions, which are the points disputed here. It is certainly a primary source for the contentious issues. Martin Hogbin (talk) 20:01, 23 January 2010 (UTC)

So, you are being serious? The paper is mostly an analysis of previously published unconditional solutions, which they call false solutions. This is actually what you object to, isn't it? You could (IMO, facetiously) call the 1/(1+p) result original - if you'd like to take this route we could, but is this really necessary? -- Rick Block (talk) 04:52, 24 January 2010 (UTC)

There is nothing facetious about Morgan's 1/(1+p) result, but Morgan were the first, I believe, to claim that previous solutions are false. This is also their original contribution to the subject, is it not? Martin Hogbin (talk) 09:02, 24 January 2010 (UTC)

Hey Rick, is Selvin's original unconditional solution to the MHP that he wrote, the 9 row table of all possible outcomes (3 car locations by 3 contestant choices), which he clearly states relies on a random 2-goat host, also false, as per Morgan? Glkanter (talk) 10:19, 24 January 2010 (UTC)

On the face of it, the Morgan paper may look like a secondary source as it considers a number of solutions to the problem. It actually considers six solutions, two of which it attributes to vos Savant and one to Mosteller, the others it seem to have made up (no doubt based on their understanding of popular solutions). One of the solutions, F4 is obviously wrong to anyone familiar with the problem in its standard form. What Morgan does next is to call all the solutions false and then go on to give its own solution. This does not make it a secondary source.

According to WP, '...a secondary source is a document or recording that relates or discusses information originally presented elsewhere. A secondary source contrasts with a primary source, which is an original source of the information being discussed. Secondary sources involve generalization, analysis, synthesis, interpretation, or evaluation of the original information'.

If Morgan were a secondary source it would compare only documents from other sources, say vos Savant with Selvin or with a source that claimed that vos Savant was wrong. At the time of the Morgan paper there were no published sources claiming that vos Savant was wrong, certainly none is mentioned in the paper, this view was originated by Morgan. As this is the only point made by their paper, it is clearly to be regarded as a primary source. Martin Hogbin (talk) 14:19, 24 January 2010 (UTC)

No need to judge it like that. Morgan's paper has two sections. The first section is secondary source, discussing primary sources. The second section is primary source of OR. Heptalogos (talk) 14:30, 24 January 2010 (UTC)

The first part of the paper discusses only vos Savant and Mosteller, whose solutions it calls false. It does not rely on any other (primary) source for this judgment, this it is purely the opinion of the authors. No part of the paper could in any way be called a review of primary sources. Martin Hogbin (talk) 14:40, 24 January 2010 (UTC)

Should it rely on other sources? If not, that judgement part is secondary. Heptalogos (talk) 18:56, 24 January 2010 (UTC)

Is It Agreed, Then?

The Morgan paper is the single worst excuse for a piece of critical writing ever displayed in the English language? Made worse by being in an allegedly peer-reviewed (good work, peers!) professional journal. Glkanter (talk) 17:13, 24 January 2010 (UTC)

Can we back up here? The point I was making was about the suggestion to include Sevlin's solution from his first letter. This is a letter that appeared in a peer-reviewed journal, but was not itself peer reviewed. There is plenty of subsequent material published in more reliable sources we can use. What I'm saying is that, in the grand scheme of things, treating Selvin's letters more or less like primary sources seems appropriate.

There is an extreme undercurrent here of contempt and hatred of the Morgan et al. source. "It has errors!" "It says my beloved solution is false!" "It says vos Savant's solution is false!" "It misinterprets the problem statement!" "I HATE it!" The efforts here to downplay, ignore, or exclude what this source has to say from this article are entirely misguided. If you want to formally challenge whether this source should be considered reliable by Wikipedia's standards, the appropriate forum is Wikipedia:Reliable sources/Noticeboard. I would suggest that this would a ridiculously silly thing to do, since the discussion starts and pretty much stops with "it appears in a peer reviewed academic journal and is cited by scores of other papers in academic journals".

If you want to argue that this paper's POV is not mainstream, please compile a list of sources and lets discuss them (rationally). If you want the article to say anything even remotely like "Morgan et al.'s interpretation that the MHP is inherently conditional is wrong" find a reliable source that says this. Note, in particular, that Seymann does not say this, but claims (incorrectly, based on Morgan et al.'s rejoinder) that while vos Savant considers it a 1-player game with the host acting as an agent of chance, Morgan et al. is considering a 2-player game-theoretic approach where both the player and host might have something to gain. BTW - their rejoinder (to Seymann) explicitly addresses many of the topics we have talked about here. -- Rick Block (talk) 17:45, 24 January 2010 (UTC)

Firstly, I believe that letters to the American Statistician are peer-reviewed. Nijdam may be able to confirm this.

I freely admit to not liking the Morgan paper but I have never suggested excluding it from the article. It should be given due weight along with many other reliable sources on the subject. It should not be allowed to overrule other sources or prevent us from explaining the problem and solution effectively.

The point about the Seymann comment is that it exists, it was published in the same peer-reviewed article as Morgan, and it is politely critical of the Morgan paper. It should also be given due weight in the article. Martin Hogbin (talk) 17:59, 24 January 2010 (UTC)

What is the MHP and what is it not

Some people do misinterpret the MHP as a "developping progress" where somtimes new occurences can take place and new configuration/constellation result and new aspects could eventually occur. This is the false approach. What they overlook is:

• The MHP is an integrated holistic "situation", not a proceeding process.

• What is given in the MHP? Three doors with a chance of 1/3 each and a risk of 2/3 each do evidently appear as two irrevocably separated groups:
  One single door and a separated pair of two doors.

• What is known: The chances and the risks of these two separated groups of doors, including certainty that the pair of two doors must contain at least one goat:
  ONE CERTAIN UNAVOIDABLE GOAT. Position of this given goat does not matter at all (some confusion results from the 1/3 possibility that this group can contain even two goats).

So the problem is already solved, chances and risks are well known, regardless whether you can see one goat within this pair of doors or you still cannot. Showing one goat where there is one certain unavoidable goat, within the pair of doors, is not a necessity to solve the MHP. Showing it or not makes no difference. Don't misdirect. Showing a goat there is no incident, is "no event" at all. Regardless whether a goat is shown or not: The probabilities of those two groups are well known and can impossibly change by "showing one goat" where one goat is rigorously given.

As soon as the guest makes her choice everything is clear. Confusion results from absurd questions and absurd assumptions. "Which one of the two door will be / has been opened" (Irrelevant, including even the question whether it will be or is already opened at all, or not.) "There could be even two goats, which one will be shown then?" - If one door is opened according to the rules and no unauthorised information is given, this question is irrelevant also. As irrelevant as whether the door will be opened at all or not.

You can take a needless approach of course and start maths. Not necessary at all. Please check the phenomenon. Maths is not necessary "to solve the MHP".

What really is important to show: The historical dissent in maths as "a strange map of unnecessary speculation and historical misfire", showing that history was yesteryear.

Any reasonable comments to my unfitting view "MHP is a situation, not a process"?   Regards   -- Gerhardvalentin (talk) 11:00, 25 January 2010 (UTC)

This is an accurate and correct model for a problem that is definitely related to the "standard" MHP, but is simplified in one particular way. The difference is essentially whether the player knows which door the host opens. If you're saying it can't possibly make any difference which door the host opens, what you're effectively saying is the player doesn't know which one is opened or can't distinguish between the unchosen doors. There are several equivalent ways to describe the simplified problem:

• The player is allowed to pick a door, and after picking a door is blindfolded. Now the host says he is opening one of the unpicked doors and does so. The audience shouts out "it's a goat!". The player is asked if she wants to switch to the other closed door (still blindfolded). Only after deciding is the blindfold removed.
• The player is allowed to pick a door. The host says (before opening a door) that he will open a door showing a goat and asks the player to decide (now, before a door is opened) if the player would like to keep what's behind the initially chosen door or the other door that will remain closed.
• Rather than doors, the host uses an urn with three marbles, one white (representing the car) and two black (representing goats) as described above (see #What urn problem is the appropriate model).

In each of these cases, the player does not and cannot use any knowledge that may or may not be given by the specific door the host opens. These may still be difficult problems for most people to solve correctly, however none of them are the "full Monty" where the player is asked to decide looking at two closed doors and one open door showing a goat. -- Rick Block (talk) 16:29, 25 January 2010 (UTC)

Rick, thank you so much for this exact description of the "situation". Everything fits, except the more psychological aspect: I guess that seeing one door open showing a goat - in the final analysis - is not more substantially informative for the guest than knowing that "one door" of the unpicked pair of doors will be (or already has been) opened showing a goat. For the guest it is a less "substantial info", in conclusion no constitutive info with respect to the allocation of chances. His full and doubtless certainty about the rule helps him to allocate them. What I tried is to express is that the "standard" MHP should strictly be detached from the "history of dissent" in interpreting the game and its rules and the behaviour of the host and so on, and from the "history of dissent" in appropriate and matching mathematics. As a subject area for one half permille of people interested in the "paradoxon". To show in first line that the pair of doors have twice the chance of the single door selected, and even because this pair of doors are to contain an inevitable goat that will be disclosed, that even because of this granted goat that will be removed the chance of the remaining closed door, offered as an alternative, necessarily and mandatory unifies the full chance, the full original chance of this pair of two doors. May mathematicians plod in building correct mathematical formulas, using a small subset of info, but that's their delight and not to amalgamate with the clearly laid out paradoxon. In fact rather confusing. -- Gerhardvalentin (talk) 18:20, 25 January 2010 (UTC)

The difference is NOT just psychological. If the player doesn't know or can't tell which door the host opens, the sample space before and after the host opens "a door" is the same so none of the initial (prior) probabilities change - in particular the prior probability of the player's initially selected door is exactly the same as the posterior probability and therefore remains 1/3 (so does the probability of door 2 and door 3!). The probability that remains 2/3 (the one you're focused on) is the sum of the prior (and, in this case, posterior) probabilities of door 2 and door 3. The individual probabilities of these doors don't change (prior to posterior), so their sum doesn't change either. This is what you're actually saying, i.e. that the sum of the prior probabilities of the two unchosen doors is the same as the sum of the posterior porbabilities.

However, if the player does know which door the host opens, the sample spaces before and after the host opens a door are different. The posterior probability of the door the host opens is 0, not 1/3. The posterior probabilities of the other two doors also may or may not be the same as their prior (initial) probabilities. If we are given or assume the host chooses between two goats randomly, this is what causes the player's door's posterior probability to "remain" 1/3 - not that the other two doors can be treated as a "pair" and that we know one of them must be a goat. It is absolutely true that we know one of them must be a goat, but this is not the explanation for why the (posterior) probability of the other one is 2/3 after the host opens a door. -- Rick Block (talk) 04:43, 26 January 2010 (UTC)

Rick, please have a look on the small number of only 100.000 events in reality, indipendent whether one goat is shown or not shown:
 
#events:  Guest's choice:          Guest denied following unselected pair of two doors           switching hurts    switching wins

 33.334           car              goat (not) shown 16.667#   goat (not) shown 16.667#           33.334 times            0 times
 33.333           goat             car                        goat (not) shown 33.333#                0 times       33.333 times
 33.333           goat             goat (not) shown 33.333#   car                                     0 times       33.333 times

100.000 events                     goat (not) shown 50.000#   goat (not) shown 50.000#           33.334 times       66.666 times

"This pictures reality". What has become reality, in the real world?   Tell you what: Probability has become reality.
It's up to mathematic to map this fact onto mathematics. Math doesn't teach reality, but reality teaches mathematics. 
Gerhardvalentin  —Preceding unsigned comment added by Gerhardvalentin (talk • contribs) 16:13, 26 January 2010 (UTC) 

The above example is the result of a simple test (100 milllion runs). There were neither priors nor posteriors, but just random events according to the rule. Each event pictures a given, a special situation. All I had to do was having them counted and to devide by 1000. The results show probabilities, do they? If you try to picture it in mathematics, then "priors" and "posteriors" are (if any necessary) only valid in and for mathematics and do not teach reality. Or do they? Don't think so. Please consider once again what "showing the goat" really does effect, besides a merely psychological implication (and mathematical purpose). Once again: Besides "history", is maths really indispensable as "a proof and evidence" to explain the paradoxon? Regards, -- Gerhardvalentin (talk) 18:17, 26 January 2010 (UTC)

Once again to the pair of two doors denied. You just wrote:
"It is absolutely true that we know one of them must be a goat, but this is not the explanation 
for why the (posterior) probability of the other one is 2/3 after the host opens a door."

Trying to show the obvious explanation, please check yourself. Let's have a look on this pair of two doors, 
on their chances and risks before and after, and on any real allocation of their "real content":

                                              door A risk:  chance:          door B risk:   chance:      sum risk:   sum chance:
Probabilities:                                        2/3      1/3                   2/3       1/3           4/3         2/3

Possible content:                             goat (removed in 50 %)         goat (removed in 50 %)
              or:                             car                            goat (always removed)
              or:                             goat (always removed)          car

Risk   of the remaining and still closed door NOW:   exactly 1/3   (no more 2/3 as before)
Chance of the remaining and still closed door NOW:   exactly 2/3   (no more 1/3 as before)

Facit about the denied pair of two doors:

• Sum risk and sum chance is unchanged, of course, as "no event happened" in this respect. Allocation of risk NOW: 3/3 + 1/3. Before it was 3/3+1/3 OR 1/3+3/3 (only one car!)
• Because never a car, but always a goat (one goat known as "unavoidable" there) has been removed (any second goat being there never is "unavoidable"),
  the chance of the remaining door unifies the consolidated chance of this pair of two doors. Allocation of chance NOW: 0/3 + 2/3. Before it was 0/3+2/3 OR 2/3+0/3 (only 1 car!).
  (Attend to the "known chances" before opening one door: "1/3+1/3" have primary definitively to be expected as "0/3+2/3" OR "2/3+0/3", one goat being rigorousely given there.)
• Allocation of "one given goat" is no point of interest, because it is already shown before this question could even be put, a second goat there never being "unavoidably given".

Consolidation: Chances "1/3+1/3" have primary definitively to be expected as "0/3+2/3(consolidated)" OR "2/3(consolidated)+0/3", because of one inevitable goat being rigorousely given there. Only one inevitable goat. Now allocation of one inevitable goat has been shown there. It is evident that this is exactly the realistic explanation for why the chance of the "other door" definitively is 2/3 after the host opened a door, and for its risk being only 1/3 now. You knew this before, you just did not know the allocation of this one unavoidably given goat. Anyone can check that at just a glance. Has this fact ever been appropriately pictured in maths? If not: Historical mathematical depiction using only a small subset of originally given information never can really help to properly show and understand the paradoxon. Kind regards -- Gerhardvalentin (talk) 23:41, 26 January 2010 (UTC)

There are two logical and self-consistent ways to approach the problem. The first is to use is only on information given in the problem statement, the second is, as the question itself suggests, is from the likely state of knowledge of the player.

If we tackle the problem on the first basis then what we imagine that the player might or might not be able to see is not relevant, either the problem statement tells us the door numbers or it does not. In fact, the (Whitaker's) problem statement is rather ambiguous in this respect. If the problem is tackled in this way, we need to make assumptions about the unstated distributions, namely that of the original car placement, the player's initial door choice, and the host choice of legal door. If we do this consistently, for the problem to be soluble we must take them all to be uniform (random).

If, on the other hand, we tackle the problem from the expected state of knowledge of the player then it is important what the player knows and what they do not know. For example, we might well expect that the player would see what door the host opens to reveal a goat. On the other hand the player would almost certainly not be aware of any significance of this number, it might be significant, for example, because the host prefers one door over the others. Thus, from the player's state of knowledge, she can only take this choice to be random, along with the original car placement.

If the host legal door choice is taken as random, as it must be, then the difference pointed out by Rick becomes somewhat academic, to say the least, and the chance of winning by switching is exactly the same in both cases, 2/3. Martin Hogbin (talk) 18:08, 25 January 2010 (UTC)

Thank you, Martin, for your comments. I do hope that the "MHP" will soon distingush between the "paradoxon" itself and it's "solution" and the historical differences in understanding and interpreting/misinterpreting the rule. Would be beneficial for the article. Thank you. -- Gerhardvalentin (talk) 03:40, 26 January 2010 (UTC)ing

Gerardvalentin, I agree with you. Although I often discus the mathematics and logic of the problem with those that are interested may main point has always been that the MHP is a simple probability puzzle that most people get wrong, where the necessary assumptions to keep the solution simple are made. Martin Hogbin (talk) 11:57, 27 January 2010 (UTC)

Yes Martin, that's my focus. It was so easy to understand if the guest could chose "one door", or alternatively "two doors". In effect that's the fundament of the game. Confusion results just from not opening these two doors simultaneously. That's all. Everyone knows that a goat is unavoidably given in each and every pair of two doors, this pair of doors having a chance of 2/3, though. Double chance! But not opening two doors simultaneously and showing there only the one unavoidably given goat, leaving the privileged "partner"-door still closed, gives birth to confusion. This confusion could easily be rectified, if facts were represented clearly. And if math would stop nebulizing :)   Of course you can show mathematics and its peculiar attempts, but just as an unnecessary historical performance :)   -- Gerhardvalentin (talk) 18:50, 27 January 2010 (UTC)

@Gerardvalentin: I discussed this over and over wirh you. The point is that when I show you where you're mistaken, you do not discuss what I show you, but you turn up with a lot of words, dealing with something different. However, one last effort, and again in your terminology. Consider 18000000 times the game is played. Although the choiced of the player eeds not to be random, it does not influence the type of aalysis, so I assume randomness. Hence:

100 000 times chosen door 1 car behind door 1 and door 2 opened

100 000 times chosen door 1 car behind door 1 and door 3 opened

200 000 times chosen door 1 car behind door 2 and door 3 opened

200 000 times chosen door 1 car behind door 3 and door 2 opened

200 000 times chosen door 2 car behind door 1 and door 3 opened

100 000 times chosen door 2 car behind door 2 and door 1 opened

100 000 times chosen door 2 car behind door 2 and door 3 opened

200 000 times chosen door 2 car behind door 3 and door 1 opened

200 000 times chosen door 3 car behind door 1 and door 2 opened

200 000 times chosen door 3 car behind door 2 and door 1 opened

100 000 times chosen door 3 car behind door 3 and door 1 opened

100 000 times chosen door 3 car behind door 3 and door 2 opened

Notice that in 600000 of the 1800000 times the car is behind door 1, but when door 1 has been chosen and door 3 opened it is in 100000 of the 300000 times there. Both come down to 1/3, but there defiition differs! Of course the same applies for other choices of door and other door opened. Perhaps the main problem for you is the way the MHP has to be considerd.Nijdam (talk) 11:04, 27 January 2010 (UTC)

Of course. Consider the problem as Selvin did, when he originally created and solved the problem, 15 years before Whitaker, vos Savant and Morgan. Rather than 12 lines of outcomes, he only had 9. They were all equally likely. He didn't split out the 'opens door 2 or door 3' individually as you have. Glkanter (talk) 18:05, 27 January 2010 (UTC)

Rick and Nijdam, all your high falutin' theory fails in practice. The only conditional solution in the article, Chun's tree/table shows this.

The moment you split the chosen door from 1/3 to 1/6 'because doors 2 and 3 are equally likely', you know door 2, or door 3 will have a value of 2/3, that is, 1/3 divided by 1/2 'because doors 2 and 3 are equally likely'.

Or, like I said previously, but you choose not to agree with, the 1/3 * 1/2 / 1/2 = 1/3 for the door selected is the same 1/3 at both ends, because it's the same 1/2 in both cases in the middle, again, 'because doors 2 and 3 are equally likely'.

Or, you can use Chun's tree/table to get the 1/6 + 1/6 = 1/3 unconditional solution. You can't solve the conditional without also solving the unconditional at the same time.

This is from sources referenced in the article, not from everybody's, or even my own OR. Glkanter (talk) 14:28, 26 January 2010 (UTC)

Rick, of course the sample space is reduced if another door with a goat is opened. I think no source is making that an issue. All they mention is the number of the door, because they interpret this as a (possible) condition, which may even be part of a bias. That's all. If you interpret reasonably otherwise and assume equal goats, it doesn't make any sense to claim that the 1/3 posterior chance of the player's door is in any way different from the prior chance. Heptalogos (talk) 21:09, 26 January 2010 (UTC)

Quite. This point is made in section 1.3.1 on my Morgan criticism page. Martin Hogbin (talk) 12:02, 27 January 2010 (UTC)

Seemingly, Rick should claim that the faith of the prisoner, hearing the name of another prisoner to be executed, is technically another faith than if he did not hear the name. Should he be able to spell his name, Rick, without knowning any other prisoner, or is any 'sound' valid? Or just the expression on the face of the warden when he thinks about the other prisoner? What's in the flipping coin's name that changes his own faith? Heptalogos (talk) 21:44, 26 January 2010 (UTC)

I don't know whether you're not understanding the point I'm making, or refusing to admit it, or something else - but this is about the point at which there becomes nothing else to say other than that there are reliable sources (not just one or two, and not just three as JeffJor seems to be claiming in some other thread) that say there is a difference between solutions that address the unconditional situation and solutions that address the conditional situation and that these situations are meaningfully different, and that in accordance with the fundamental Wikipedia content policies it really doesn't matter if editors agree with what these sources say or not. Editors are certainly welcome to have whatever personal opinion they want, but the article should say what reliable sources say in a neutral fashion (per WP:NPOV). -- Rick Block (talk) 02:04, 28 January 2010 (UTC)

First of all you keep mixing Wiki-rules and article-discussion with the theoretical discussion on this page. If I want to change something to the article, I will discuss it on the other page. On this one, I am trying to exchange perspectives.

Secondly I am claiming that some of your arguments are actually not (for sure) supported by the sources, but are rather your interpretation of them. Morgan calls F5 wrong because it's not using the right condition; not because it cannot solve any conditional problem.

Thirdly you keep sticking to the idea of 'knowledge of a specific door'. Indeed we cannot be sure about that, but it's not at all reasonable to suggest that the unpicked doors are specific at all. That's why I raised the prisoner issue; the warden could speak out any name, "say Rick". The only reasonable condition is that another prisoner, out of two, left the game, which is really a condition, as an event reducing the sample space.

But we agreed already, and I'm not sure why you change your perspectives. Maybe it's just a matter of implicitly changing assumptions. Heptalogos (talk) 21:36, 28 January 2010 (UTC)

@3. When door No. 3 is opened, this hardly can mean something else than a specific door. It is a translation in term of the problem of the situation the player is in. Concerning the prisoner: for himself it turns out to make no difference, but as I write: it turns out, after the right calculation. For both the other prisoners it differs much, and hence also for the prisoners kowledge of the fate of the others. Nijdam (talk) 01:04, 29 January 2010 (UTC)

• Should you switch fate with any unmentioned prisoner? This is unconditional.
• Should you switch fate with the unmentioned prisoner? This is conditional.
• Does your fate change technically by the mentioning of other prisoners? No.
• Does your fate change technically by the mentioning of another prisoner? You say yes.

Because you use two conflicting realities in the same formula: one in which each of two prisoners may be mentioned, and one in which a specific prisoner is mentioned. Although you know that the first scenario is no reality at all. It doesn't make sense to use Bayes' formula when the cause of the effect is already known. P(E|C) = P(E), definitely. Bayes' formula is useful to define possible causes when the effect is given. Everything else is waisted energy; imaginary realities that create various technical probabilities. Heptalogos (talk) 10:16, 29 January 2010 (UTC)

Your fate changes by the mentioning of another specific prisoner because you can now use knowledge of the warden's selection process. If you don't know which prisoner then you can't use this knowledge. This is the point JeffJor and Martin keep trying to make about the MHP host bias (if it's unknown to the player it must be assumed to be random). However, there's a distinction between structurally unknowable (you don't know which prisoner, or you must decide before the host opens a door) and simply unknown. In the former case, the original sample space is not changed - your original probability is exactly the same in all regards. If you're simulating the problem you count all simulations. In the latter case, the original sample space is changed. You can assume something you don't happen to know is random, however your result won't necessarily match a frequency distribution you observe in a simulation (or reality). Gill made this point somewhere. If you don't explore the range of possibilities in the face of unknowns you can run into extremely surprising results.

With regard to whether it changes your fate or not, if the warden picks randomly (assuming a choice is available) then the posterior conditional probability is numerically the same as the prior probability. If this is what you mean by "fate" then fine - your fate doesn't change. But knowing the specific prisoner means there is a posterior conditional probability (which depends on the warden's selection process) and a prior probability (which doesn't). These can have the same numeric value, but because one depends on the warden's selection process and the other doesn't they don't have to. This means to me that they're different. -- Rick Block (talk) 14:46, 29 January 2010 (UTC)

I do not keep trying to make a point about host bias, I make a point, based on standard practice in statistics, which you have been unable to refute. In the MHP there are three unspecified distributions: the original car placement, the original player choice, and the host door choice. If you take them all to be non-uniform and unknown then the solution is indeterminate, if you take them all to me uniform the answer is always exactly 2/3. Can you give me any reason, based on normal statistical practice, for taking the distribution of the initial car placement to be uniform, but the host choice to be non-uniform? Martin Hogbin (talk) 16:56, 29 January 2010 (UTC)

Martin, while you're point is perfectly correct, it's moot. Selvin, the originator of the puzzle, said the host selects randomly when given the 2 loser situation. Glkanter (talk) 17:59, 29 January 2010 (UTC)

You are right the issue was settled by Selvin but I would still like to see Rick's answer to my point, which is based on standard statistical practice. Martin Hogbin (talk) 18:21, 29 January 2010 (UTC)

I predict: Disappointment. Glkanter (talk) 18:25, 29 January 2010 (UTC)

Martin - I don't know why you keep asking this when it has already been answered innumerable times. You say "in the MHP there are three unspecified distributions". Are there? Or are there none? Or is there one? Or are there two? Perhaps you mean the MHP as it was originally phrased in Parade, but then this left not only these distributions unspecified but also whether the host is required to make the offer to switch and whether the host always opens a goat door. The reason to take the initial distribution as random but not the host choice would be if one is given to be random but the other isn't. I think it's clear Morgan et al. interpret the essence of the problem to be about the difference (if any) between the unconditional prior probabilities and the conditional posterior probabilities. I think it's also clear they analyze a version of the problem matching what they infer vos Savant was addressing based on her subsequent clarifications published in Parade before their paper was published (clearly not including her rejoinder or anything anyone else might have said later), in particular where everything important for the 2/3 answer is specified except for the host's preference. Thus in the problem they're analyzing, it's taken to be given that

• the initial distribution is uniform (as assumed by vos Savant)
• the host always shows a goat (as assumed by vos Savant)
• the host always makes the offer to switch (as assumed by vos Savant)
• but not anything about the host's preference (because vos Savant never mentioned anything about this)

The reason to address this problem is because it is the same problem (they think) vos Savant addressed, and (not coincidentally) it shows the difference between an unconditional and conditional solution. If you accept that the essence of the problem is the difference between the unconditional and conditional situations, then the host preference matters. Selvin knew this. vos Savant apparently did not. They could have said that vos Savant's answer assumes no host bias. Instead they chose to analyze the problem with an unspecified host bias, and (as it turns out) you're still no worse off switching - but whether the host has a bias or not is not the point, the point is that the problem is about the conditional probability so approaching the problem unconditionally is sloppy (at best). -- Rick Block (talk) 20:37, 29 January 2010 (UTC)

That's not correct; nothing was taken to be given by Morgan. This is what they said in the rejoinder: "..while it is quite clear what problem vos Savant wishes to solve, it is not clear that her problem and the reader's are the same. It is the reader's question with which we are primarily concerned, not vos Savant's interpretation of that question." Also, you yourself, Rick, in the discussions here -which are theoretical and not directly about the article-, are repeatedly not willing to take the host behavior for random, while you make all the other assumptions easily, implicitly. Heptalogos (talk) 21:25, 29 January 2010 (UTC)

I read "quite clear what problem vos Savant wishes to solve" and "the reader's question" to mean, respectively, the unconditional probability of winning by switching vs. staying and the conditional probability of winning by switching given which door the host opens - not whether the host behavior is taken to be random. These are as different as "what is n+n" and "what is n squared". They're different questions even if we preface both with "assuming n is 2, ..." (or, "assuming the host picks randomly between two goats, ..."). -- Rick Block (talk) 01:56, 30 January 2010 (UTC)

Quote: "These can have the same numeric value, but because one depends on the warden's selection process and the other doesn't they don't have to." They really have to, because the warden "is flipping a coin to decide which of the remaining names to give". Do you agree that they have to be the same, definitely? Heptalogos (talk) 21:33, 29 January 2010 (UTC)

Yup, just like "assuming n is 2, what is n+n" is EXACTLY the same as "assuming n is 2, what is n squared". Same answer, right? So, therefore, these are the same. Definitely. -- Rick Block (talk) 01:56, 30 January 2010 (UTC)

Rick, you seem terribly muddled in your thinking.

Selvin took the initial car distribution and the host legal door choice to be to be uniform. This is perfectly logical.

Vos Savant, in considereing Whitaker's statement, took the initial car distribution and the host legal door choice to be to be uniform (although she initially omitted to state that she had taken the hosts legal door choice to be so). Again a logical and consistent decision.

Morgan quote the Whitaker statement then claim to have 'an elegant solution that assumes no additional information ', clearly referring to the Whitaker statement that they have just quoted. In fact Morgan make this perfectly clear in their response to vos Savant, as Heptalogos points out above, they say, It is the reader's question with which we are primarily concerned, not vos Savants interpretation of that question. They then proceed to take the initial distribution of the car as uniform but the host door choice distribution as non-uniform (plus settle the rules in the same way as everyone else). There is no logical reason for this inconsistent choice based on Whitaker's question.

K&W later on in their unambiguous formulation take the initial car distribution and the host legal door choice to be to be uniform in a consistent manner.

The only party to treat the problem inconsistently is Morgan, in order to conjure up their confusing complication. Martin Hogbin (talk) 23:20, 29 January 2010 (UTC)

Yup, they're the only ones. Well except for Gillman, and Falk, and Grinstead and Snell, and Rosenthal [15], and Eisenhauer [16], and - well, I've asked this before, but how many would you like? [Rick Block]

Rosenthal calls the 'your original 1/3 chance doesn't change' solution 'actually correct', but he calls it the 'Shaky Solution' as it does not work for certain variants. To make his point, only after going through the random host and the host bias variants does he point this out:

"The original Monty Hall problem implicitly makes an additional assumption: if the host has a choice of which door to open (i.e., if your original selection was correct), then he is equally likely to open either non-selected door. This assumption, callously ignored by the Shaky Solution, is in fact crucial to the conclusion (as the Monty Crawl problem illustrates)."

In no way does he support Morgan's claim that other solutions are 'false', or that the problem must be solved conditionally. Glkanter (talk) 12:17, 30 January 2010 (UTC)

That is very interesting. A reliable source confirming that the original 1/3 chance does not change. Martin Hogbin (talk) 17:57, 30 January 2010 (UTC)

And, if you find it confusing to think about the conditional probability as opposed to the unconditional probability no one is saying you have to. -- Rick Block (talk) 01:56, 30 January 2010 (UTC)

Why is the host choice not random

You are rather straying off the point. You said, 'This is the point JeffJor and Martin keep trying to make about the MHP host bias (if it's unknown to the player it must be assumed to be random)'. You have produced no argument against my point. I am not trying to do a 'source count' neither have I mentioned the word 'conditional'.

Both vos Savant and Morgan address the Whitaker statement of the problem. We all agree that this is the case for vos Savant and it is clearly the case for Morgan, because they say so, as quoted above.

So, having established that both Morgan and vos Savant are addressing the Whitaker question, this is the point that I would like you to answer. What is the justification for taking the initial car placement distribution to be uniform but the host legal door choice to be non-uniform, based on Whitaker's problem statement? Martin Hogbin (talk) 10:41, 30 January 2010 (UTC)

I agree with Martin. Rick, you seem to be hiding after Morgan a lot when we ask your own opinion, while you're supporting Morgan with your own opinion on the other hand, which I find inconsequential behavior. You wrote: "I think it's also clear Morgan analyze a version of the problem matching what they infer vos Savant was addressing." The interesting issue here is that this indeed seems to be true! As you say, they accept every explicit assumption of Marilyn, and not the implicit one. When Marilyn confronts them, they falsely argue that they rather ignore her and limit their view to Whitaker. Many sources after Morgan follow their respected leaders, and so do you, which is not a reasonable argument. Heptalogos (talk) 11:02, 30 January 2010 (UTC)

I already responded to this (above), and (as I said) I believe Martin is reading what they said incorrectly. I believe they're saying vos Savant is answering "what is the average probability of winning by switching" where they interpret to the question to be "what is the player's probability of winning by switching in a specific conditional case, for example where the player has picked door 1 and the host has opened door 3". Martin is interpreting what they say to mean they're focusing on the specific wording of the problem, and (re)interpreting this from scratch. They're granting vos Savant all the assumptions she explicitly made about the problem statement (and there are several) - but disagreeing about the fundamental question the problem asks.

It is quite clear what Morgan wished to do, the say so in their response to vos Savant. "It is the reader's question with which we are primarily concerned, not vos Savants interpretation of that question." I am not sure how they could make it any clearer than this.

If the Morgan paper is perversely to be taken as purely a criticism of vos Savant's work then the whole paper could have been replace with: 'Marilyn, you forgot to specify that the host should choose an unchosen door at random when the player has originally chosen the car'.

My opinion is that Martin's question is misguided. IMO, he's asking why Morgan et al. made the same assumptions vos Savant explicitly made. The answer is obvious. Because vos Savant explicitly made them. Her suggested simulation makes it entirely obvious what she was thinking - and that the host must pick randomly in the case the player's initial choice is correct is NOT one of her conditions. It never crossed her mind that this might matter. I mean, really, she explicitly randomizes BOTH the initial car placement and the initial player choice. She then counts success by "not switching" (200 iterations) and success by "switching" (200 iterations). She's clearly not addressing the conditional chance of a player who sees which door (cup) the host opens (lifts up), she's addressing the unconditional probability of winning by switching vs. winning by staying - exactly the same as her case analysis solution. This solution is insensitive to the host strategy (in the case the player's initial choice is correct) and is exactly the solution Grinstead and Snell says "does not quite solve the problem that Craig posed" and Gillman says "does not address the problem posed" and Rosenthal calls "shaky". She's NOT assuming the host picks equally in the case the player's initial choice is correct - in her solution it doesn't matter. The point Morgan et al. (and all the other sources I'm mentioning) are making is that if you're deciding to switch after the host has opened a door then the host's strategy matters. It's abundantly clear vos Savant overlooked this. -- Rick Block (talk) 05:47, 31 January 2010 (UTC)

You are resorting to Machiavellian contortions to excuse the inexcusable. Morgan published what they claim to be a 'elegant solution' which 'assumes no additional information'. The comment from Morgan that I have quoted above makes perfectly clear that they are concerned with 'the reader's [Whitakers] question' and not 'vos Savants interpretation'.

It is not in question that vos Savant (quite rightly) took the distribution of initial car placement and player initial door choice to be uniform. I am also not arguing that vos Savant did not overlook that she should also have considered the distribution of the host's legal door choice, nobody can ever know for sure, although she later said that she took the host to be acting as 'the agent of chance'.

Whatever the case, that Morgan are answering Whitaker's question, or that they were responding a question based on to vos Savant's assumptions, Morgan still proceeded to address a problem in which the initial car placement distribution is uniform and the host legal door choice distribution is non-uniform; you have still not given me a valid reason for this. Martin Hogbin (talk) 11:10, 31 January 2010 (UTC)

(outidented) What reason is there to be given? It's obvious that the only reasonable distribution of the car is uniform. How the player will act, we do not know. And ... it is unimportant, because all analysis comes down to conditioning on her choice. Instead of refusing to accept Rick's explanations, you better do some effort to understand what he is explaining.Nijdam (talk) 14:29, 31 January 2010 (UTC)

Of course a reason is required. There are three distributions not given in the Whitaker problem statement. Please give me a reason why the producer's choice in placing the car is obviously random but the host's choice in opening a legal door is not random. Here are some reasons why they should be treated in the same way:

1. Neither is given in the problem statement.
2. Both are decisions made by humans who are part of the TV production organisation.
3. It would give a player who studied the show an advantage if either distribution was discovered to be non-uniform.
4. Under standard game show regulations neither would be allowed to be known to the player or the audience.

Apart from an unsupported claim that it is obvious, what reason can you give for treating the two distributions differently? Martin Hogbin (talk) 16:07, 31 January 2010 (UTC)

Like I keep saying, the primary point is to show the difference between unconditional and conditional solutions. Keeping the car placement uniform in both (per vos Savant's assumption) allows the unconditional solution to have a definite answer (2/3) matching one particular conditional case. I really don't understand why you object to this so much. -- Rick Block (talk) 18:55, 31 January 2010 (UTC)

[At this point we have changed the question. I want to know why you think that the host should not be considered to act randomly. I have started a section below on that subject] Martin Hogbin (talk) 10:14, 1 February 2010 (UTC)

Now we are getting closer to agreement. Morgan make the fair point that, if the host is known to choose a legal door non-randomly the solution is necessarily conditional, in other words in might matter which door the host opened (although Selvin must have been aware if this fact because he specified that the host must choose a legal door randomly). Morgan make their point in an interesting and informative way, by starting with a host who chooses any unchosen door randomly (may reveal a car) which results in the commonly given answer of 1/2, they then progress through the case where the host must choose only a door hiding a goat non-randomly, and show that in this case the player can never do worse by switching, and finally they consider the symmetrical case with its answer of 2/3. That is all fine, I have no objection to any of that.

The things that I object to are

1. Morgan's arrogant tone and unpleasant treatment of vos Savant and Mosteller.
2. The claim that any unconditional solution (even to the symmetrical case) is false.
3. The claim that they give an 'elegant solution' that 'assumes no additional information'.
4. The suggestion by any editor here that the Morgan paper should, in any way, be able to disqualify any other source or control the structure of the article.
5. The suggestion that there is any logical reason, based on Whitaker's question, to assume that the host opens a legal door non-randomly, but the producer places the car randomly.

This last point is simply a device used by Morgan to get their point across. Nether you nor Nijdam nor Morgan have given any logical reason why this should be so, and I still challenge you to do so. Starting with Whitaker's question, we must either take the producer's action and the host's action to both be non-random or both random if we are to be consistent in our approach. If Morgan had been clear that they were considering a slight variation to make a point then most of the problems would disappear. The paper would be an interesting demonstration of the need to consider the host's door opening policy and of the fact that the problem becomes necessarily conditional if he is known to choose a legal door non-randomly. Both interesting points that should be included in the article. In fact their storyline, totally random, goat door only non-random, goat door random, might be a good one to include in the article, so long as it is put in a proper context. Martin Hogbin (talk) 20:36, 31 January 2010 (UTC)

Yes, I think we're getting closer to agreement but it seems you're still not quite grasping the point Morgan et al. are making. They are not saying that the solution is necessarily conditional only if the host is known to choose a legal door non-randomly, but that the solution for the conditional question is necessarily conditional (period) - i.e. an unconditional solution is addressing something different. Consider a slightly different progression: host must open a door but does so totally randomly (probability is 1/2), host must open a door and must not reveal the car (unconditional solution is 2/3), host must open a door and must not reveal the car and the player knows which door the host opens (unconditional solution is still 2/3 even though in this case the probability always depends on how the host chooses between doors). You're saying an unconditional solution addresses this last case so long as we also assume (or are given) the host chooses randomly between goats. The way Morgan et al. are looking at it is that an unconditional solution is simply not addressing the conditional case, since it always says the answer is 2/3 whether the host chooses randomly or not.

I cannot imagine what it is that you think I do not grasp about the Morgan paper but I can assure you that I understand it fully.

What you seem not to have understood is what I and several other editors have been trying to explain to you. If the host chooses a legal door randomly (which is the only logical assumption, unless you can show otherwise) there is no condition. Nothing happens that can possibly change the probability that the car is behind the door originally chosen by the player. The unconditional problem is the same as the conditional one because knowing the door opened makes no difference to anything. Whether the players chooses before or after the door is opened makes no difference because there is nothing she can learn from the random choice of door. The result on the condition that door 3 has been opened is identically the same as the result on the condition that door 2 has been opened is identically the same as the unconditional result, not by some fluke but by the application of standard principles of mathematics and logic.

I, Glkanter, Heptalogos, at least have been trying to explain this to you for years. Which bit do you not understand? Martin Hogbin (talk) 00:21, 1 February 2010 (UTC)

I, similarly, cannot imagine what it is you think I'm not understanding. -- Rick Block (talk) 03:11, 1 February 2010 (UTC)

[Nijdam's point moved to section below] Martin Hogbin (talk) 10:14, 1 February 2010 (UTC)

There are two separate subjects here and we seem to have switched subject. I will respond to your point in the new section below on why the symmetrical problem is not conditional. Martin Hogbin (talk) 10:14, 1 February 2010 (UTC)

Andrevan has suggested the difference is like the difference between average and instantaneous velocity. This is close (although there are way more instantaneous velocities than conditional cases), but in this analogy the question is like "A person travels between points A and B that are 100km apart in an hour. What is the velocity at the half way point?" The velocity of interest is clearly the instantaneous velocity, not the average velocity, but the problem (as stated) doesn't give enough information to answer the question. You could assume constant velocity and say "the answer is 100km/hr", but you're not really answering the question. A closer analogy might be "A person travels between points A and B that are 100km apart in an hour and passes through their midpoint C along the way. What is the average velocity from point C to point B?". Again, there's not enough information to answer the question and a response that "the average velocity is 100km/hr" (because the average velocity from A to B is 100km/hr) is simply not addressing the question. This "answer" is a true statement, and it even applies to the question that's asked if the average velocity from A to C is the same as from C to B, but it's only true in this case. -- Rick Block (talk) 21:49, 31 January 2010 (UTC)

If you want to pursue that analogy, which is not particularly apt, then if the velocity is specified to be constant the average and instantaneous velocities must be the same. In the MHP, if the host chooses a legal door randomly, the conditional and unconditional results must be the same. There is enough information to answer the question exactly, nothing is missing, we do not need to know which door the host opens because we know it can make no difference. If the host is known to choose non-randomly that is a different question with a different answer. We seem to have got nowhere. Martin Hogbin (talk) 00:21, 1 February 2010 (UTC)

Nothing has changed, indeed. Including the part where, despite countless reliable sources using unconditional sources, including Selvin, somehow, those editors who don't see the need for the conditional solution are disparaged as not understanding mathematics. Glkanter (talk) 00:37, 1 February 2010 (UTC)

Why a formulation based on only Whitaker's question must assume that the host chooses a legal door randomly

[Continued from above, where the question appeared to have got lost]

I am not sure if you have accepted this assertion or not, but here it is again. Based only on the Whitaker problem statement, there is no logical reason to take it that the producer places the car randomly (thus its distribution is uniform) but the host chooses a legal door non-randomly (thus the distribution of door number opened by the host is non-uniform). We must take both the distributions to be uniform. The only other logically consistent choice is to assume that both distributions are non-uniform, in which case the solution is indeterminate. Does everyone accept this? Martin Hogbin (talk) 10:21, 1 February 2010 (UTC)

For the avoidance of doubt I explain what I mean by a uniform distribution of legal door opened by the host. I mean that the host chooses uniformly between all the doors he is permitted to open under the standard rules (any unchosen door to reveal a goat). Sometimes he has only one door to choose from, sometimes two. Martin Hogbin (talk) 13:22, 1 February 2010 (UTC)

Who is "we" in "we must"? If you are writing your own article that you want to publish on the MHP, you are free to make whatever assumptions you'd like. That's not what Wikipedia editors are doing. Wikipedia articles are summaries of what is published in existing reliable sources. Existing reliable sources say whatever they say, regardless of what we think is logically consistent. -- Rick Block (talk) 14:35, 1 February 2010 (UTC)

You are avoiding the question. There are many reliable sources, saying different things on this subject. We have to use a logical basis to decide how to use these sources in writing an article.

My point is that suppose we, as mathematicians or statisticians, are set the task of providing an answer to the probability of winning by switching, given Whitaker's statement. (Which happens to be the task that Morgan wished to do). We can of course make any assumptions that we like, ranging from the obvious to the bizarre. At one extreme we might simply note that there is insufficient information given to answer the problem. On the other hand we could first assume some reasonable rules (about which there is little disagreement) and then decide to apply the principle of indifference to the problem. I would suggest that a good mathematician would apply that principle consistently to all the unknown distributions.

We might, rather perversely, apply the principle of indifference to the host's legal door choice but not to the initial car position. If we did this, would we be justified in calling anyone who took the car to be uniformly distributed to be wrong?

The only logical and consistent thing to do is to take all unspecified distributions to be uniform. Do you disagree with this? Martin Hogbin (talk) 16:42, 1 February 2010 (UTC)

I'm not avoiding the question. I'm disagreeing with your premise. What I'm hearing you say is that we should decide what we think is the "right" POV and then evaluate sources in the context of that POV. Instead, we need to see what POVs are published and, if they differ from each other (not from whatever "we" think), then we need to assess what the relative prevalence of each is, and then write the article fairly representing each. What "we" think simply doesn't enter into it. -- Rick Block (talk) 17:40, 1 February 2010 (UTC)

Now you really are avoiding the question. I am happy to talk about what the sources say and how they should be incorporated into the article in a different thread but I am puzzled by your sudden reticence to give your own opinion. The question which I want an answer to is, 'Starting with Whitaker's statement, what is the justification for taking the producer's actions in placing the car to be random but the host's action in opening a legal door to be non-random?' I would like to hear what you think. If you really do want to claim that your opinion on the subject is unimportant that is fine, provided that you stick to that view. Martin Hogbin (talk) 22:45, 1 February 2010 (UTC)

No, I'm not avoiding the question. I'm being very clear and direct about why your question has nothing to do with the editing process. As long as you understand it wouldn't mean anything as far as the article is concerned even if we were perfectly unanimous about it, fine. As I've said multiple times, Morgan's justification is clearly that vos Savant treated the problem this way in her columns. She clarified the initial placement and player choice were to be taken as random but never said anything about the host's choice (and, since Gillman interpreted what vos Savant said the same way arguing that this is a ridiculous interpretation seems a little silly). A different justification might be that you'd get an interesting puzzle where the player's initial chance is clearly 1/3 but the resultant chance may not be. It's probably more likely to match how a real world game show would be set up, and may well be a better match to people's expectations given the problem statement as well. Randomizing the initial car placement is obvious. Forcing the host to pick randomly is not (I mean, if the world's smartest person missed it probably lots of others would as well). If I were writing a paper given this problem statement, I'd probably start with both uniform, then initial placement uniform but not host choice, and then both not uniform - although as you say, the last one is really not very interesting. -- Rick Block (talk) 02:26, 2 February 2010 (UTC)

Well, I guess you have answered my question in the end. I agree with your last sentence.

You say, 'Morgan's justification is clearly that vos Savant treated the problem this way', but Morgan do not state this in their paper and specifically deny this in their response to vos Savant.

The status of Morgan as a reliable source is an issue that has been discussed here before and that will no doubt be discussed again. I agree that we cannot avoid mentioning the Morgan paper and its conclusions in the article. I do not accept that it should be allowed to control the structure of the article or declare other reliable sources invalid. It should be given appropriate weight bearing in mind all the factors concerning it, including the opinions of editors here. Martin Hogbin (talk) 12:24, 2 February 2010 (UTC)

Why the symmetrical formulation (host chooses a legal door randomly) is not conditional

@Martin: You seem to clamp to the idea that the symmetrical case is unconditional. I explained you before, it is not. We discussed the difference between the type of problem, which may be unconditional or conditional, and the type of solution. As I explained: the conditional problem, which IMO is the only form of the MHP, may be solved in different ways, but always in calculating a conditional probability, also when using the symmetry. Come to understand this. The example with velocity may be of help. Driving a distance of 100 km in an hour, what is your speed halfway? Nijdam (talk) 08:59, 1 February 2010 (UTC)

Yes, I do indeed assert that in the symmetrical case, it is not important which door the host opens and that the door number of the door opened by the host need not be taken as a condition of the problem.

We have discussed it many times before and your argument that the problem is conditional boils down to your assertion that, 'it is conditional'. I have asked many times for you to give me a definitive way to determine what events are conditions of a problem and you have never done so. You have given me several suggestions in the past but none of them stands up to scrutiny.

The answer is actually simple. An event must be considered a condition of a problem, if it is considered that its occurrence might affect the probability of interest. In the case of the number of the door opened by the host we can show that the, if the host chooses a legal door randomly, the chances of winning by switching are independent of the door number opened by the host. Martin Hogbin (talk) 10:14, 1 February 2010 (UTC)

To reply to your analogy, in general, given only that my average speed is 100 kph, I cannot tell you my speed at the halfway point but, if my speed is known to be constant, then I do know that it is exactly 100 kph. In the MHP if know the host may choose a legal door non-randomly I cannot, without further information, calculate the probability of winning by switching given a specific door opened by the host. If I know the host has chosen a legal door randomly then I can calculate the chances of winning by switch exactly, in the case where a specific door is opened, and in the case where an unknown door is opened. Martin Hogbin (talk) 10:32, 1 February 2010 (UTC)

Right, but you seemingly do not see that the answer you give in the case of constant speed is not my average speed is 100 km/h, but: my speed HALFWAY is 100 km/h. Nijdam (talk) 13:52, 1 February 2010 (UTC)

I give the halfway speed which I know is equal to the average speed. What is the connection to the MHP? I do not see how your analogy helps.

>>Well, you give the conditional speed, under the condition of being halfway. I wonder why you do not understand this.Nijdam (talk) 23:36, 1 February 2010 (UTC)

In the MHP we have the probability of winning by switching at the start of the game given that it is the player's policy to switch. This is, we agree, the unconditional probability. Now various events occur:

The player chooses a door.

If we give an answer now, it it conditional or unconditional?

>>In fact conditioned on the choice, but as the choice is considered idependent of the car, we may treat this as unconditional. (That's why Boris called it "semi-conditional").

The host smiles.

If we give an answer now, it it conditional or unconditional?

>>I'll not respond to such questions. Nijdam (talk) 23:36, 1 February 2010 (UTC)

The host opens a door.

If we give an answer now, it it conditional or unconditional?

>>Definitely conditional. Nijdam (talk) 23:36, 1 February 2010 (UTC)

Martin Hogbin (talk) 16:08, 1 February 2010 (UTC)

Well, there is no need for a response to my second event as you answered my question in your response to the first event. It can be shown by several valid mathematical arguments that the probability that the player has originally chosen the car is independent of the door opened by the host, if the host opens a legal door randomly, thus, as you say above, we may treat the problem as unconditional. Martin Hogbin (talk) 14:33, 2 February 2010 (UTC)

I do not follow you here. THE problem cannot be considered unconditional. And I assume here you also mean the MHP, where the player is asked to change her choice after the host has opend a door. Whatever you want to demonstrate, there is a difference in the independency of the door with the car and the player's choice, which is given in the problem, and some independency you may be able to prove. Nijdam (talk) 17:17, 2 February 2010 (UTC)

The Whitaker problem statement does not actually tell us that the position of the car is independent of the players choice. In the full K&W formulation please tell me exactly what tells you that the car position is independent of the players door choice. Martin Hogbin (talk) 17:46, 2 February 2010 (UTC)

You're right, it is not explicitly stated. But everyone will assume the player has no information about the position of the car. In any analysis this is assumed. But you are avoiding the point I made. Nijdam (talk) 23:36, 2 February 2010 (UTC)

If the host chooses a legal door uniformly at random then there is no difference between the fact that the probability of interest (probability of winning by switching) is independent of the players initial door choice, and the fact that the probability of interest (probability of winning by switching) is independent of the hosts legal door choice. Neither fact is explicitly stated in the problem statement; both facts can be deduced by the application of logic to the problem statement. If you disagree, then please explain what the difference is. Martin Hogbin (talk) 23:44, 2 February 2010 (UTC)

Formulate in sound terminology what you mean. I really do not know what you are talking about. What for example do you suppose to be "the probability of interest"? Nijdam (talk) 12:45, 3 February 2010 (UTC)

[Outdent]I am not quite sure what you are wanting me to do. I want to calculate the probability that the player will win the car if they switch their door choice given:

1. The standard game rules (swap always offered, the host always opens an unchosen door to reveal a goat).
2. The host must choose a door to open, according to the standard rules rules, uniformly at random.
3. The player has chosen door 1.
4. The host has opened door 3.

If you want me to start with a sample set of your choice, I have no need to do this. I can start with any sample set that includes all the events on which the probability of interest, as defined above, is dependent. Martin Hogbin (talk) 13:49, 3 February 2010 (UTC)

I'll hand you the tools, which you are already familiar with. C=number of door with the car, X=number of chosen door, H=number of door opened by host. P(C=1)=P(C=2)=P(C=3)=1/3. P(H=x|X=x)=0; P(H=c|C=c)=0; P(H=h|X=C=x)=1/2 for h!=c. What is the probability of interest? And what is the meaning of "the probability of interest (probability of winning by switching) is independent of the hosts legal door choice"? Nijdam (talk) 01:11, 4 February 2010 (UTC)

You are quite right, I am familiar with the tools you hand me and I am quite capable of producing a conditional solution to the problem using those tools, as we have done on the analysis page, however, I am not in any way obliged to use one particular method to solve the problem. My solution, which uses different tools, is as follows:

• P(I=C)=1/3 and P(I=G)=2/3

Where I is the players initial choice. Note that, although I may know the door numbers, I do not use them as this is not necessary. The probabilities given above are based only on the following assumption, which I forgot to state earlier:

• P(I=G) is independent of the player's initial choice of door. This is true if the car is initially placed uniformly at random. Do you challenge this?

If the host chooses according to rule 2 above the we can also say that

• P(I=G) is independent of the host's choice of door. This is true if the host chooses according to rule 2 above. Do you challenge this?

There is little more to do except observe that if I=G the player will, with certainty, win by switching.

Neither the door initially chosen by the player, nor the door opened by the host are conditions of this problem. What is the error in the above proof? Martin Hogbin (talk) 09:25, 4 February 2010 (UTC)

Terribly sorry, Martin, this leads nowhere. (1) Not important, but why use I instead of the already defined X? (2) I did not oblige you to use any specific method! (3) You do not answer my questions, which you yourself brought up. (4) I see no logic in your reasoning. Nijdam (talk) 11:48, 4 February 2010 (UTC)

1) My 'I' is the initial prize choice of the player, not the door choice. P(I=G) is the probability that the player gas chosen a goat, irrespective of door number. You may insist that this refers to door 1, but I do not care.

Where in the description of the MHP does your "I" turn up? If you want to express the event thet the player has chosen a door with the car behind it, it's simply {X=C}.Nijdam (talk) 14:07, 5 February 2010 (UTC)

Please Martin, is I an event, is it a random variable, what is it? Because everything may be expressed in terms of my X, C and H, express I in thes terms. Nijdam (talk) 00:13, 6 February 2010 (UTC)

2) By giving me probabilities that included door numbers, I presumed that you wanted me to use them. If this was not your intention that is fine.

Everything that happens may be expressed in the terms I gave you.Nijdam (talk) 14:07, 5 February 2010 (UTC)

Yes but I may not want to express it that way. X is the number of the door chosen by the player and C is the number of the door with a car behind it. I do not want to use door numbers. Why must I do this? Why can I not have the event that the player has chosen a goat? Martin Hogbin (talk) 23:19, 5 February 2010 (UTC)

What is your point? I showed you that event, it is the complement of {X=C}. Nijdam (talk) 00:13, 6 February 2010 (UTC)

But you insist on using only the terminology which refers to door numbers. That the player has originally chosen a car is an event in its own right that does not need to use door numbers. If I was asking for information about events, as they happened, I could choose to ask, 'What door number (X) has the player initially chosen?', and 'What door number (C) hides the car?' but I could also ask, 'Has the player initially chosen the car?'. The answer to this could be given without reference to door numbers. This is, I think, similar to what Heptalogos is getting at when he refers to the door numbers as not necessarily being static. Martin Hogbin (talk) 10:01, 6 February 2010 (UTC)

I do not. I challenge you to use other terms, but ... your terms must have the possibility to give expressions for my X, C and H. And the meaning of my terms is: The answer to the question: 'What door number has the player initially chosen?' is: X. And the answer to 'Has the player initially chosen the car?'. is: {X=C}. That's how it works. And not the events themselves are important, but their probabilities. Nijdam (talk) 16:58, 6 February 2010 (UTC)

3) The door number opened by the host is, within the rules of the game, random. It is certain that the host can reveal a goat, thus no information which might affect the probability that the player has a goat is revealed by the observation of the door number opened by the host. There is no new information given which allows the probability I=G to be revised or changed. Thus this probability is independent of the legal door number opened by the host.

If you mean: P(H=h)=1/3 for all h, you need to assume X is uniformly distributed. Or do you mean something else? Please formulate your other sentences in apropriate formulas, I do not know what you mean.Nijdam (talk) 14:07, 5 February 2010 (UTC)

You seem to be insisting that I express everything in terms of door numbers. Is this correct? Martin Hogbin (talk) 23:19, 5 February 2010 (UTC)

If you mean in terms of X, C and H, yes. But if you want to give another description, please go ahead. But my X, C and H are to be expressed in your terms, So why do such effort? Nijdam (talk) 00:13, 6 February 2010 (UTC)

What exactly is X, is it the door number initially chosen by the player? Martin Hogbin (talk) 00:55, 6 February 2010 (UTC)

As I said: X is the door (number) initially chosen by the player. Nijdam (talk) 16:50, 6 February 2010 (UTC)

4)Which line do you object to?

As long as I do not see your calculation, every line. Still you didn't answer my questions. Here they come again: What is the probability of interest? And what is the meaning of "the probability of interest (probability of winning by switching) is independent of the hosts legal door choice"? It is about time you become utterly specific. Nijdam (talk) 14:07, 5 February 2010 (UTC)

I am not being awkward, I do not understand what you are asking for. The probability of interest is the probability that the player will win the car if they switch their door choice given:

1. The standard game rules (swap always offered, the host always opens an unchosen door to reveal a goat, car placed randomly, player chooses randomly).
2. The host must choose a door to open, according to the standard rules rules, uniformly at random.
3. The player has chosen door 1.
4. The host has opened door 3.

Is this not a well-defined probability? Martin Hogbin (talk) 23:19, 5 February 2010 (UTC)

In formula please?! Words are confusing. Nijdam (talk) 00:13, 6 February 2010 (UTC)

Please ask for clarification of any wording that is not clear to you. The problem cannot be put in the form of a formula until it is clear exactly what is to be calculated.

That's why I asked you what you mean by "the probability of interest"etc. Nijdam (talk) 16:50, 6 February 2010 (UTC)

I have already given you a formula. I have chosen a particular sample set that you do not like. My sample set consists of just two events, the player initially chooses a car and the player initially chooses a goat. This sample set conforms to the rules of sample sets. It does not use door numbers because I choose not to address the problem in this way. There is not only one way to address any mathematical problem. Martin Hogbin (talk) 00:55, 6 February 2010 (UTC)

Impossible. The set of events canot just consist of those two events! Nijdam (talk) 16:50, 6 February 2010 (UTC)

[Outdent] The probability of interest is, if you prefer the probability that after the host has opened door 3, the prize originally chosen by the player is a goat. There are only two possible events, the prize initially chosen is a car and the prize initially chosen is a goat. There are no other possibilities. No other events, such as, the player chooses a particular door, the host opens a particular door (according to the rules) are significant, as they cannot affect the probability of the event of interest, thus they do not need to be included in our sample set. The numbers of doors chosen and opened are an unnecessary complication. The only events of interest are the player's initial possible choice of prize. Martin Hogbin (talk) 23:39, 6 February 2010 (UTC)

A short diversion

As neither of us seems to be making much headway in convincing the other let me ask you,Nijdam, for your comments on a problem we have discussed before. An urn contain 6 balls numbered 1 to 6. On ball is removed from the urn (at random as always) which proves not to be a six. Now another is removed, which also turns out not to be a six, How do you calculate the probability that the next ball removed will be a six? Here are two ways the problem might be solved:

A long way

This is a conditional probability problem. The conditions being the ball that was first picked and the second ball that was picked. We set up a sample set with 120 events of the form F=f,S=s,T=t showing the balls picked first second and third. This, in abbreviated form, looks like this:

F=1,S=2,T=3
F=1,S=2,T=4
F=1,S=2,T=5
F=1,S=2,T=6
...
F=2,S=1,T=3
F=2,S=1,T=4
...
F=6,S=5,T=4

with each event having probability 1/120

Next we condition the above sample set by removing all events in which F=6. Then we condition the above sample set by removing all events in which S=6.

We now add up the probabilities of all the remaining events in which T=6 to get a probability of 1/4.

A short way

After the first two picks there are four balls left on of which one is the six and three of which are not the six. Our sample set for the final pick is N N N 6 with each event having a probability of 1/4, thus the probability of interest is 1/4.

Questions

What method would you use?
Which of the above methods is better?
Is the short method above incorrect? Martin Hogbin (talk) 17:53, 7 February 2010 (UTC)

Both methods are okay. Both lead to the value of the coditional probability asked for. It doesn't matter which method I use, just like it doesn't matter how we calculate the conditional probabiltity at stake in the MHP. I told you so may times. You continuously confuse WHAT you calculate and HOW you calculate it. Nijdam (talk) 20:12, 7 February 2010 (UTC)

What you say implies to me that you accept the simple solution as valid for the symmetrical case but you just want to state that the problem is one of conditional probability.

Although I do not agree with you, I would be willing to consider having the word 'conditional' somewhere in the simple solutions provided that it did not confuse the non-expert, maybe something along the lines of, 'The MHP is a well know problem of conditional probability...'. What I would not accept is something having the effect of saying, 'This solution is false/incomplete because it does not address the conditional nature of he problem'.

If we agreed to include the word 'conditional' in the simple solution section, would you be willing to have the aids to understanding section immediately follow the simple solution section. Martin Hogbin (talk) 20:42, 7 February 2010 (UTC)

for editing purposes

Let us suppose the player has initially chosen door 1. The probability of interest as you say, is then the probability that after the host has opened door 3, the prize originally chosen by the player is a goat i.e. the conditional proability

P(C=2|X=1,H=3), see!

Concerning the events you're contradicting yourself. Where is your event "host opes door 3"? —Preceding unsigned comment added by Nijdam (talk • contribs) 17:16, 7 February 2010 (UTC)

The event that the host opens door 3 is just like the event that the host says the word 'door'. It occurs but it is not necessary to include it in our calculation because it cannot possibly affect the probability that the player has chosen a car (actually the event that the host says the word 'door' could affect that probability, as I have explained, but we will ignore that possibility). You continue to insist on using door numbers, this is not necessary to solve the problem. Martin Hogbin (talk) 20:47, 7 February 2010 (UTC)

Is The Morgan, et al Paper A Reliable Source for a Wikipedia Article?

As I read this Wikipedia Guideline, it is acceptable to exclude Morgan from the MHP article.

"Peer review is an important feature of reliable sources that discuss scientific, historical or other academic ideas, but it is not the same as acceptance. It is important that original hypotheses that have gone through peer review do not get presented in Wikipedia as representing scientific consensus or fact. Articles about fringe theories sourced solely from a single primary source (even when it is peer reviewed) may be excluded from Wikipedia on notability grounds. Likewise, exceptional claims in Wikipedia require high-quality reliable sources, and, with clear editorial consensus, unreliable sources for exceptional claims may be rejected due to a lack of quality (see WP:REDFLAG)."

They claim to write about a famous puzzle loosely based on Let's Make A Deal, yet make no mention of the originator of the puzzle, entirely overlooking one of his stated premises. Armed with this lack of information, Morgan calls the work of numerous other reliable sources, including, we are to presume, the person who created and solved the puzzle, 'false'.

We've had an editorial consensus to minimize Morgan for months now. Martin has a user page detailing the weaknesses in their paper. I think the paper is so wrought with errors it approaches the level of inaccuracy of the statement 'The Earth is flat'. Glkanter (talk) 11:50, 30 January 2010 (UTC)

This belongs on the discussion page. Morgan don't claim to write about a famous puzzle. They don't call other reliable sources false (except for Parade). There is no originator of the puzzle. The weakness of the Morgan paper is not relevant. Your arguments go beyond the level of accusation. Heptalogos (talk) 12:56, 30 January 2010 (UTC)

Usually I leave your other-worldly responses unanswered. This one is too much. Your entire response is contrary to the facts. Morgan's paper's title is 'Let's Make A Deal: The Player's Dilemma'. They attack the vos Savant/Whitaker version of the problem that Selvin originally posed. They offer up 6 'false' solutions. That's more than just vos SavantThe weaknesses of Morgan's paper call into question the paper's utility as a reliable source. What does 'beyond the level of accusation' mean? I'll leave this section here, on the arguments page, even though I'm not arguing the underlying math. Glkanter (talk) 13:10, 30 January 2010 (UTC)

Morgan is saying only their solution is correct, all others (including Selvin's which they are ignorant of) are false:

"The intricacies of this simple problem make it an excellent teaching tool, as can be seen from the insights offered by the false solutions F1-F6 and the correct resolution." Glkanter (talk) 13:50, 30 January 2010 (UTC)

I usually wonder why Rick usually doesn't leave you unanswered.

• 1a. Fact: Morgan's paper's title is 'Let's Make A Deal: The Player's Dilemma'.
• 1b. Glkanter: "They claim to write about a famous puzzle loosely based on Let's Make A Deal".

This is correct. Although the title of the paper clearly refers to the TV show, the scenario described in Whitaker's question never actually occurred on any show. Martin Hogbin (talk) 17:23, 30 January 2010 (UTC)

• 2a. Facts: Morgan did not call Selvin at all, neither his solution false. Selvin explained that the basis to his solution is Monty's exact strategy. Vos Savant did not.
• 2b. Glkanter: "Morgan calls the work of numerous other reliable sources, including Selvin, 'false'."

• 3a. Facts: there is no "the puzzle". There are many variations, from 1959 and before. Morgan only commented the Parade statement and several solutions to that.
• 3b. Glkanter: "Morgan make no mention of the originator of the puzzle".

There is little doubt that Selvin and Whitaker based their puzzles on the same show. Selvin was the first to publish. Martin Hogbin (talk) 17:23, 30 January 2010 (UTC)

Morgan responded in 1991 to an article from 1990/91. Should they have sought for mathematical similar publications from the decades before? And mention them too? Martin, does it make any sense to expect this from Morgan? Btw, I would have ended up in 1959. Heptalogos (talk) 20:13, 30 January 2010 (UTC)

• 4a. Fact: Morgan don't make exceptional claims and are not presented as scientific consensus or fact. The 'weakness' of a resource doesn't make it less reliable.
• 4b. Glkanter: "As I read this Wikipedia Guideline, it is acceptable to exclude Morgan from the MHP article."

• 5a. Facts: Morgan calls solutions F1-F6 wrong, as solutions to the stated problem in Parade. The only source they connect to some of it is Vos Savant. They don't mention any (other) solution to any other problem statement.

Based on their statement about his solution, Morgan appear to attribute F6, or something similar, to Mosteller, . Martin Hogbin (talk) 17:23, 30 January 2010 (UTC)

OK, missed that one. Heptalogos (talk) 20:13, 30 January 2010 (UTC)

• 5b. Glkanter: "Morgan is saying only their solution is correct, all others (including Selvin's which they are ignorant of) are false."

Although they do not name any others, Morgan clearly suggest that any solution that does not take into account the door number opened by the host is false. Martin Hogbin (talk) 17:23, 30 January 2010 (UTC)

Only solutions to the Whitaker statement. Where certain assumptions may already account the door number by making it explicitly irrelevant. Selvin has nothing to do with this. Heptalogos (talk) 20:13, 30 January 2010 (UTC)

You obviously have not been reading Rick's interpretations of Morgan's paper as I have, for 15 months. This is exactly how the article has been edited, and only recently and reluctantly allowed to be changed by Rick. Just yesterday, Rick said he interpreted Morgan's paper to include Selvin. Great paper. People can't even agree on what their point is. Glkanter (talk) 11:56, 31 January 2010 (UTC)

I think the point being made here is that the business of door numbers is a distraction. Monty opens one of the remaining doors to reveal a goat. Martin Hogbin (talk) 13:26, 31 January 2010 (UTC)

This is worse than a statement about a flat earth. Heptalogos (talk) 14:32, 30 January 2010 (UTC)

Reasons why Morgan is a good reliable source

It is published in a peer reviewed journal.

It has been cited by several other sources.

Reasons why Morgan is not a good reliable source

It, very unusually, has a critical comment by a respected academic published in the same peer reviewed journal.

Some other sources are highly critical of it.

Many editors here consider its conclusions excessive and that it contains misquotations, errors, and inconsistencies.

It fails to acknowledge and incorporate important details clearly stated by the originator of the problem

Status of the Morgan paper

In my opinion it is a valid source but it is clearly a controversial one. It has a place in the article but it should not be allowed to control the structure of the article or invalidate other reliable sources. Martin Hogbin (talk) 23:56, 2 February 2010 (UTC)

Technical answers

Probabilities may be the same in number, but not in used method.

The following arguments are all used in exactly the same "three prisoners problem". I sometimes cut out a part of the sentence to make it shorter.

Rick used this example: "assuming n is 2, what is n+n" is EXACTLY the same as "assuming n is 2, what is n squared".
Regarding the same problem, he stated: These can have the same numeric value, but they don't have to.
And finally: assuming n is 2, these are the same. Definitely.

The issue here is that n is definitely 2; there is no other possibility. And because n=2, we do not have two realities but rather one, which has two perspectives: n+n and n^2. Both have the same meaning: n twice. You see, they not only have the same outcome, but also the same relevance. Statement:

• When within a certain reality different methods structurally and uniformly have the exact same outcomes, they have the same relevance.

They must! Even if you can't find the logic behind the enforced equivalence, or you don't understand it intuitively, they describe the same overall reality. It is possible that such methods use different detail levels, but these differences definitely have no relevance with respect to the outcomes.

A consequence is that any such method is equally correct and effective, but the simplest method (e.g. with the least details) is most efficient. It also has less error chances.

Finding the simplest method is generally a matter of intuitive intelligence, as is finding the best formula to solve a mathematical problem. The same intelligence is a main determination factor for an IQ-test score. It's nice to see that the highest IQ has a title role in the aftermath of this famous paradox, while at the same time she is hardly present, speaking of efficiency. Heptalogos (talk) 12:25, 30 January 2010 (UTC)

The symmetrical problem is a special case of the non-symmetrical problem, which may be regarded as a special case of the many doors problem, which in turn may be regarded as a special case of the a more general problem still. None of this means that a solution to the symmetrical case is wrong if it only applies to the symmetrical case. There is never an obligation in mathematics or logic to answer a more general question than the question actually asked, although, of course, this might be an interesting thing to do. As I have said before, nobody claims that Pythagoras' theorem is wrong just because it only applies to the special case of right-angled triangles.

There are many solutions that apply only to the symmetrical case. Just because these solutions do not apply to more general cases does not make them wrong. Martin Hogbin (talk) 12:47, 30 January 2010 (UTC)

Taking into account several specific doors that may be opened is not only very special; it is unreal. Bayesian analysis is useful to calculate the possibilities of causes when the effect is given. Not impossibilities! Door 2 cannot be opened. Look at the tree graphic in the article. You know why Rick and Nijdam call the 'unconditional' 1/3 player's door's chance technically different? Because it does not take into account the 1/6 chance when door 2 is opened. Why the heck should it?

Now let me be very special by taking into account the very impossibilities that the player picked door 2 or 3. (Selvin did!) Of course the answers are the same, but now they are technically even more different. Does this prove the given tree wrong? No, it proves Selvin being very inefficient. Why imagine causes that don't exist anyway. Simply reduce your sample space and do an ordinary probability calculation. Bayes did not exist to be misused for such. Heptalogos (talk) 20:43, 30 January 2010 (UTC)

Can you explain what you mean here. You seem to be saying that it does not matter what doors the host could have opened and only the door actually opened matters. Martin Hogbin (talk) 10:26, 31 January 2010 (UTC)

Thank you for this question. I realized quite soon that I was not at all clear in my last post and I want to apologize for that. It was too late to delete it. Here's my next try.

Bayes: P(A|B) =
P(A and B) / [ P(A and B) + P(C and B) ]

Bertrand's box

• Causes (3): choosing box1 or box2 or box3.
• Effect (1): choosing a golden coin from box 2 or 3. (1 has no gold)

P(box2).P(gold2) / [P(box2).P(gold2)] + [P(box3).P(gold3)]
1/3.1 / [1/3.1 + 1/3.1/2] = 2/3

P(box3).P(gold3) / [P(box3).P(gold3)] + [P(box2).P(gold2)]
1/3.1/2 / [1/3.1/2 + 1/3.1] = 1/3

Three prisoners

• Causes (3): pardon prisoner1 or prisoner2 or prisoner3.
• Effect (1): choosing an unpardoned prisoner from prisoner 2 or 3. (1 has asked)

P(pris2).P(unp3) / [P(pris2).P(unp3)] + [P(pris1).P(unp3)]
1/3.1 / [1/3.1 + 1/3.1/2] = 2/3

P(pris1).P(unp3) / [P(pris1).P(unp3)] + [P(pris2).P(unp3)]
1/3.1/2 / [1/3.1/2 + 1/3.1] = 1/3

Three doors

• Causes (3): price door1 or door2 or door3
• Effect (1): choosing an unpriced door from door 2 or 3. (1 is picked)

P(price2).P(unpr3) / [P(price2).P(unpr3)] + [P(price1).P(unpr3)]
1/3.1 / [1/3.1 + 1/3.1/2] = 2/3

P(price1).P(unpr3) / [P(price1).P(unpr3)] + [P(price2).P(unpr3)]
1/3.1/2 / [1/3.1/2 + 1/3.1] = 1/3

P(price1).P(unpr3) = 1/3.1/2: it seems like the probability of door1 has been split into 1/6 for door2 opened and 1/6 for door3. But door 2 is not opened and it's not necessary to make it part of the sample space. For door3 opened, the smallest possible but relevant sample space consists of: 1) door2 priced and 2) door2 not priced (and neither door3). When we compare this to the boxes, situation 1 is similar to the gold box and situation 2 is similar to the mixed box.

1) Gold box: chance to pick a gold coin is 1. <-> Door2 is priced: chance to open door 3 is 1.
2) Mixed box: chance to pick a gold coin is 1/2. <-> Door2 is not priced: chance to open door 3 is 1/2.

The probability that the golden coin is picked from the gold box is full box / (full box + half box) = 2/3. The ratio is askedcause / totalpossiblesamplespace. For the boxes as well as for the doors this is 1 / (1 + 1/2). The probability of door1 priced (1/3) is not at all involved. Let's check by formula.

All 3 causes have the same probability x.
The formulas all have the same structure, simplified as follows:

xa / (xa + xb)
xa / x(a + b)
a / (a + b)
1 / (1 + 1/2)

This is the ratio of 3openedwhen2priced : (3openedwhen2priced + 3openedwhen2notpriced). Nothing else is relevant; door1 does not exist in this reality. Neither does the opening of door2. Heptalogos (talk) 21:57, 31 January 2010 (UTC)

This is a conditional solution! You are considering only the case where the host opens door 3 and explicitly using the fact that door 3 is opened only 1/2 of the time when the prize is not behind door 2. -- Rick Block (talk) 22:16, 31 January 2010 (UTC)

I'm sorry, I didn't mean to. :) Well, of course every solution should address the conditional problem, even if the condition is that 'another door with a goat is opened'. But now I unintendedly numbered the doors. Please read doorO (opened) instead of door3 and doorU (unopened) instead of door2. Heptalogos (talk) 22:42, 31 January 2010 (UTC)

The issue relates IMO very much to quantum theory. Or to go beyond, what is not measured doesn't exist. Let's assume the host doesn't even look at door1. Still it's connected to the other doors; if another door with a car is opened, we know that door1 hides a goat. But making the reasonable assumptions, that's not possible in the player's world. I am interested in that world.

Let's assume the doors are statically numbered. Then it's a world in which door3 is opened, revealing a goat, randomly if possible. It's one of many of those worlds, from which 2/3 have a car behind door2. Are we randomly in one of those worlds? Then our chance is definitely 2/3. Somehow people seem to believe that this is not obvious, in other words, that we cannot assume randomness for where we are. Indeed, no single reality can be random. Because for a single situation no chance at all can be given! Heptalogos (talk) 21:58, 1 February 2010 (UTC)

IMO, the point of probability is to predict what may be observable over a large number of experiments. When talking about a single situation, the Bayesian approach allows a probability to be determined based on the conditions under which that situation occurred. If the conditions are repeated a large number of times (as N approaches infinity), the Bayesian probability will be the frequency of occurrence. I think saying the probability of winning by switching is 2/3 means (should mean) that if you repeat the same conditions a large number of times you WILL observe a convergence to this result. Do you agree with this? -- Rick Block (talk) 20:30, 7 February 2010 (UTC)

Correction

Heptalogos wrote: "If he is the same Nijdam as on Wikibooks, he was a maths lecturer (PhD) at the University of Twente until 2004." Heptalogos (talk) 16:43, 22 January 2010 (UTC)

That should be MSc instead of PhD. Heptalogos (talk) 22:15, 5 February 2010 (UTC)

Table showing the host opening an unnamed door

Thread moved from talk:Monty Hall problem -- Rick Block (talk) 17:53, 7 February 2010 (UTC)

The opened door3 is either A or B, not both. The doors are unique, although we don't identify them statically BEFORE. So AFTER we do have 150 cases instead of 300. If you state door3 may be both A or B, it makes no sense to create 2 columns for them with certain, different distributions. Heptalogos (talk) 22:26, 6 February 2010 (UTC)

BEFORE they are identified vertically; AFTER they are also identified horizontally (statically). Vertically means that each door has a certain distribution (GCG, GGC). Horizontally means that any opened door (last column) is one of those vertical columns. Heptalogos (talk) 23:02, 6 February 2010 (UTC)

Based on these comments, I think Heptalogos means the following is the situation.

Situation BEFORE the host opens a door					Situation AFTER the host opens a door
Door picked	Unpicked door A	Unpicked door B	result if switching	total cases	cases if host opens Door B
Car	Goat	Goat	Goat	100	50
Goat	Car	Goat	Car	100	100
Goat	Goat	Car	Car	100	0

The initial column ordering does NOT mean the columns necessarily correspond to Door 1, Door 2, and Door 3 (in that order).

It seems pretty simple. We're talking about 300 samples where the player has picked a door. The host now opens a door. Now we have 150 samples. In 100 of the 300 samples we started with, the door the host has opened was a goat at the beginning. The other 100 samples where that door has a car (and 50 where the player's initially picked door has a car) now no longer apply. That's what this table shows.

@ Heptologos: You didn't say how you'd describe what's going on in this table, or why it's obvious the top number in the "cases if host opens Door B" column is 50. If this is what you're really talking about, please explain the table. In particular, the words "in 2 out of 3 equally likely cases ..." don't seem to have much to do with it. -- Rick Block (talk) 00:56, 7 February 2010 (UTC)

Unfortunately you stopped using the Arguments page for this.

My reaction above was to your last table above (I copied it back again) in which you don't reduce the sample space after a door is opened. To rephrase: the opened door3 is either A or B, not both. So AFTER we do have 150 cases instead of 300. BEFORE, the doors are only 'identified' vertically, which means that one door has distribution GCG and the other GGC. AFTER, we identify the opened door horizontally as one of those vertical columns.

I did not describe your table because it's not mine. I will use mine again to explain it better. These are the two possible situations, which are symmetrical:

BEFORE							AFTER
door picked	cases	door 'unopened'	opened	door 'opened'	opened	switching result	door opened
Car	100	Goat	50	Goat	50	Goat	50
Goat	100	Car	0	Goat	100	Car	100
Goat	100	Goat	100	Car	0	Car	0

BEFORE							AFTER
door picked	cases	door 'opened'	opened	door 'unopened'	opened	switching result	door opened
Car	100	Goat	50	Goat	50	Goat	50
Goat	100	Car	0	Goat	100	Car	0
Goat	100	Goat	100	Car	0	Car	100

The reason that I don't want to number them BEFORE is that the host may not be able to identify these doors through repeated experiment. So he may not be able to execute a bias. There's an interesting consequence of Morgan's implicit assumption about static doors: their reasoning is that because no assumption is made about 'no host bias', there may be a host bias. But through repeated experiment there may be different hosts and/or different stages. The doors are only identified BEFORE when the cars are placed and a distribution comes into existance. But they may lose their identity to the host.

switching result	AFTER	AFTER	TOTAL
Goat	50	50	100
Car	100	0	100
Car	0	100	100

In 2 out of 3 equally likely cases switching will result in the contestant winning the car. Heptalogos (talk) 12:50, 7 February 2010 (UTC)

Please walk me through your symmetrical tables again. I'm honestly trying to understand what you're saying. Here's one of your tables with my questions

BEFORE							AFTER
door picked	cases	door 'unopened' BEFORE we don't know which door is later opened so this label is confusing	opened what does this column mean?	door 'opened'	opened	switching result	door opened
Car	100	Goat	50	Goat	50	Goat	50
Goat	100	Car	0	Goat	100	Car	100
Goat	100	Goat	100	Car	0	Car	0

I can understand labeling the doors AFTER the host opens one, but (I think) this doesn't reduce the sample set. What I can't understand is labeling the doors after the host opens a door AND reducing the sample set. It would help to talk about an experiment we're repeating 300 times that might exhibit the behavior you're suggesting. Might one be, 1) initial random car placement, 2) player picks a door, 3) the door numbers are now scrambled? -- Rick Block (talk) 18:27, 7 February 2010 (UTC)

I drew two tables that are the same in the BEFORE situation, except for the headers. They show three situations. What you call door A is opened in 50 cases in the first situation, 0 in the second and 100 in the third. AFTER, the only way to identify them is to call them opened or unopened, which is then known. But BEFORE, we already know that these are the two possibilities, so we use both headers, one in each table, and we know that only one of those tables will become reality. Each of them reduces the sample space. Heptalogos (talk) 21:48, 7 February 2010 (UTC)