Talk:Monty Hall problem/Arguments

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. [[1]]

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)

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 [2]. 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 [3] (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 [4], and Eisenhauer [5], 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.

May be I have to be more specific, so you better understand. Which question is asked? If you want to kow the conditional probability of drawing ball 6, after two other balls have been drawn before, both methods are okay. But not if it is a game in which I have drawn the first two balls myself, so I know which number they have. Although the answer will have the same numerical value, it is the value of a different probability.Nijdam (talk) 21:13, 8 February 2010 (UTC)

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)

@Martin. My conclusion: you haven't done your homework well. Firstly, both Rick and I have stated we do not insist on the mentioning of the word "conditional". But we object the presentation of the simple explanation as a solution to what we, and I guess you, consider the MHP, being conditional of nature. Even if we consider the (complete) symmetrical version as the MHP, the simple explanation is not sufficient. Accepting the simple explanation as a true solution is misleading, and for quite a long time it plays this misleading role. I wished you came to understand this. Nijdam (talk) 20:56, 8 February 2010 (UTC)

But you accept the simple solution to my problem above. Although you call the probability to be calculated in my urn problem conditional, the second, simple, solution does not mention or in any way consider the various possible conditions (balls that have been previously chosen, yet you called it OK. Martin Hogbin (talk) 21:42, 8 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)

Here you are mistaken. Nowhere in the problem is the mentioning of the word "door" considered to be part of the problem. But the opening of door 3 is. That's why the problem has to be formulated in principle consisting of 27 elementary events (the combinations of the 3 values of each of X, C and H). As I metioned, the probability of interest will be P(C=2|X=1, H=3) (or similar with other door numbers). Notice how 3 of the 27 events play a role here. Nijdam (talk) 21:22, 8 February 2010 (UTC)

Whitaker's statement itself does not tell us what is and what is not part of the problem. It gives a series of events and asks for a probability (or at least which action is better). We, as the answerers of the problem, have to decide which events are significant. The fact that the host says the word 'door' could easily be significant if he does not always use this word. Martin Hogbin (talk) 21:39, 8 February 2010 (UTC)

This is a kind of childish reasoning. I won't react to it. Nijdam (talk) 11:22, 9 February 2010 (UTC)

I cannot understand what you mean by childish. Morgan point out that, if the host is known to choose a door non-randomly, then the probability of winning by switching may depend on the door that he actually opens. It could equally well be pointed out that if the host does not always say the word 'door' but sometimes says something like 'would you prefer this one', the words actually spoken by the host could change the odds of winning by switching. There is nothing childish about this, it demonstrates an important point about statistics, that any kind of information can change a probability. On the other hand, if the host's choice of door is random then number of the the door actually opened by the host reveals no information and thus cannot possible change the probability that the player's initially chosen door hides the car. These are all well-understood and well-accepted principle of statistics. Martin Hogbin (talk) 14:39, 9 February 2010 (UTC)

I called it childish because it seems to me a kind of reasoning in which one, against better judgement, tries to make one's point. I'have no intention to discuss all the theoretical possibilities of formulating versions of the MHP. As it stands it causes enough trouble. Till now I have found no source, reliable or not, accounting for something else than placing the car, picking a door and opening one. Let us concentrate on this. Nijdam (talk) 16:39, 9 February 2010 (UTC)

I raised the point of the host's words not because I want to include that variant in the article but to demonstrate that what we include as a condition in a probability problem (especially one with a rather unclear problem statement) must be determined by the person who is answering the problem. If you want to be fussy, every probability problem is a conditional one. I am am not even saying that treating the MHP as a conditional problem is a bad idea, I am just objecting to the dogmatic view that the problem must be one of conditional probability (any more than every probability problem is). Martin Hogbin (talk) 19:07, 9 February 2010 (UTC)

Even then: one conditional probability is not the other, although they may have the same value. Nijdam (talk) 22:41, 10 February 2010 (UTC)

Problems in which there are only unimportant conditions usually treated as unconditional. You toss a fair coin on a Wednesday what is the probability of getting a head. Is this a conditional problem? No, because the toss of a fair coin is considered independent of the day of the week, just as the probability that the player's original door hides a goat is independent of the door number opened by a host who chooses a legal door randomly. Martin Hogbin (talk) 22:46, 11 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)

Again, I apologize if I'm seeming dense here but I'm not sure I'm understanding. To clarify this for me it would really help to walk through two examples. In the first example say the distribution is GCG, the player picks door 1 (G) and the host opens door 3 (G). This matches the second line in your first symmetric table (right?). In the second example, say the distribution is GGC and the player picks door 1 and the host opens door 2. This matches the third line in your second table (right?). What I mean by reducing the sample space is that if we run an experiment 300 times, there are some samples we throw out because they don't match the conditions we're interested in. Would you throw either of these samples out? -- Rick Block (talk) 22:13, 7 February 2010 (UTC)

What you call door3 is only opened in the first table. Let me explain the tables. We have horizontal distributions CGG, GCG and GGC, by which we can present the entire sample space. Then we freeze the left vertical column, representing the picked door. Now the (other two) doors can also be presented to have a vertical distribution. We can present the entire sample space by using the vertical distributions, which are in the example GCG and GGC. Of course these are snapshots, because the evenly distributed SS would actually be an endless ..CGGCGGCGGC.. in both cases, but in relation to each other, one of them always has C one step above the other. In any experiment, one of the tables becomes true after a door is opened, which divides the sample space into halves. Heptalogos (talk) 12:28, 13 February 2010 (UTC)

So, if you were running an experiment to verify the 2/3 result, is this what you would do? You would not label the doors beforehand. You would randomly place the car behind any of the three doors. You would have the contestant select a door. You would have the host open an unpicked door. And then you would label the doors "selected door", "unopened door", and "opened door". You would count this as one sample. And you would record for this sample whether switching wins or loses. Is this correct, and what you're meaning to show with your tables? -- Rick Block (talk) 15:30, 13 February 2010 (UTC)

That's correct. When I have enough samples I align then in two columns, 'unopened' and 'opened', both presenting the distribution CGGCGGCGG etc. One of the columns has the car always one row below the car-row of the other column. When all checked, I remove all samples below row 3, because the distribution repeats from there. Now we have one of the tables above, in which switching results in winning the car in 2/3 cases. Heptalogos (talk) 20:10, 13 February 2010 (UTC)

I'm confused about where the columns come from. If you're counting each sample and within that sample only identifying the doors that are selected, unopened, and opened, aren't there only two rows? Specifically, for 300 samples wouldn't your results table look like this?

door picked	door 'unopened'	door 'opened'	cases	switching result
Car	Goat	Goat	100	Goat
Goat	Car	Goat	200	Car

The opened door is always a goat. The picked door and unopened door are either car/goat or goat/car. Do the columns represent right, middle, left, or 1,2,3, or some other persistent identifier of the door? I thought your whole point was to avoid this. If you do, then I can't see how there's anything more than the two rows I've mentioned. And, again, I'm not trying to be difficult - just trying to clarify what your meaning is. -- Rick Block (talk) 20:28, 13 February 2010 (UTC)

No problem, it can be confusing indeed. So is the word label, so let me describe again: I repeat the experiment until the contestant chose the same door 300 times. All other samples are removed. I keep an eye on the doors and write down the distribution for every sample, like 'CGG'. The left character -in this case 'C'- is always representing the same physical door, and so do the middle and the right. I also write down which one is opened. After the experiment, I order all samples vertically like this:

door picked	cases	door	opened	door	opened
Car	100	Goat	50	Goat	50
Goat	100	Car	0	Goat	100
Goat	100	Goat	100	Car	0

Because this pattern keeps repeating vertically, I only show the pattern. But I count all opened doors for all three possible situations and write the totals in the columns 'opened'. The point is that I don't know which of the unpicked doors is 'door3'. The contestant knows, but he cannot relate it to my table. There is no relation between both; that's why I don't want to identify them and why you cannot exclude sample GGC as you do. But what we can clearly see is that switching results in winning 2/3 of the time. Heptalogos (talk) 20:23, 14 February 2010 (UTC)

So, "door picked" is always (say) the leftmost door, and although you know (in order to keep track) you're just not saying which door (middle or right) the other two columns refer to? Is this the same as the following?

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 A	cases if host opens Door B
Car	Goat	Goat	Goat	100	50	50
Goat	Car	Goat	Car	100	0	100
Goat	Goat	Car	Car	100	100	0

With this table, a contestant can't tell whether the host opens Door A or Door B but you know, so you're somewhat artificially creating a difference between what the contestant knows and what you (or the audience, who has more like your perspective) knows. The question "what is the contestant's probability of winning by switching after the host has opened a door" refers to one of the rightmost columns, but not both. Based on your (and the audience's) knowledge of whether the host opened what you're calling Door A or Door B, you can pick the relevant column - but the contestant can't.

Getting back to where we started, my claim was that if you refuse to identify the particular door that is opened (which you're doing here - at least from the contestant's perspective) the probability P(car behind unopened door|host opens a door) involves a condition ("host opens a door") which does not change the sample space, so none of the "before" probabilities are changed. This sort of gets back to what we mean by probability. Is the probability of a fair coin toss being heads 100% or 0%, or is it 50% because that is the expected outcome over (say) 300 trials you could observe? In your experiment, even though you can keep track of the results by physical door, you're saying the contestant cannot. Because of this, the contestant has no choice but to count success/failure with a table like my 2-line table above. You know the conditional probability, given which door the host opens, but you've made this inaccessible to the contestant. -- Rick Block (talk) 22:18, 14 February 2010 (UTC)

No, as I told you, I have no knowledge either which door the contestant opened in the single case. The '300-experiment' and the 'suppose you're on a game show' single event are two different worlds. I only know that one of both columns is true, so the sample space is reduced, but I still don't know which one. Therefore car-goat and goat-car are both possible, next to goat-goat, which adds up to three equal likely samples with a 2/3 chance of winning by switching. Heptalogos (talk) 21:24, 15 February 2010 (UTC)

Well, then I'm still confused. What do you mean when you say you "have no knowledge either which door the contestant opened in the single case"? If anyone knows (who could keep track in the way you're describing) in the 300 sample case, then I think the table is as above (someone knows which door is A and which door is B). If no one knows this, then there are only two rows as I've described. If someone does know, then you're creating an artificial "state of knowledge" difference and changing the question from "what is the probability" to "what is the probability from the perspective of someone who doesn't know which door is which". In either case (whether someone else knows or not), from the perspective of someone who doesn't know which door is which, there are only two possibilities. The door they've originally picked hides the car and the door not opened by the host hides a goat, or their door hides a goat and the other door hides the car. These are the same two possibilities that exist in every sample. They have no way to distinguish cases where the host opens A from B, so the sample space is not reduced (from their perspective).

This is like a coin toss. You know it's heads or tails. But your chance is the average outcome. You may think there's a door A and a door B, but if you don't know and can't tell the difference between A and B your chance of winning the car by switching is not either one of these - it's the average of both, which means it's your chance from the original sample space (not a reduced sample space). -- Rick Block (talk) 22:56, 15 February 2010 (UTC)

Behavior of the Host in 300 games:

(please delete it if not useful, it's just my question)

In case he is given two goats, the Host will open the unpicked doors A and B randomly:

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 A	cases if host opens Door B
Car	Goat	Goat	Goat	50	50	0
Car	Goat	Goat	Goat	50	0	50
Goat	Car	Goat	Car	100	0	100
Goat	Goat	Car	Car	100	100	0

Host always opens door B if ever possible, i.e. if he has a choice between two goats (only if the Car was selected and switching means total loss with zero chance
(Will his behaviour be reducing the chance from zero to "below zero"?)

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 A	cases if host opens Door B
Car	Goat	Goat	Goat	50	0	50
Car	Goat	Goat	Goat	50	0	50
Goat	Car	Goat	Car	100	0	100
Goat	Goat	Car	Car	100	100	0

Will the behaviour of the host reduce the chance from zero to below zero in case the Car had been selected? Sorry, did not get it yet.
(please delete if not useful, thank you) -- Gerhardvalentin (talk) 23:53, 7 February 2010 (UTC)

In your first table a player who sees the host open either Door A or Door B has a 100/150 (2/3) chance of winning the car by switching. This is the same as the chance of a player who has picked door 1 and decides to switch before seeing the host open a door. This player has a 200/300 (2/3) chance of winning the car.

In your second table a player who sees the host open Door A has a 100/100 (100%) chance of winning by switching (there are only 100 cases where the host opens Door A, and in all of these cases switching wins the car) while a player who sees the host open Door B has a 100/200 (50%) chance. A player who decides to switch before seeing the host open a door still has a 200/300 (2/3) chance. -- Rick Block (talk) 00:06, 8 February 2010 (UTC)

Note - to simulate this you need to keep track of the results where the host opens Door A and Door B separately (i.e. the doors have to have a persistent numbering and the host preference has to persist). -- Rick Block (talk) 00:12, 8 February 2010 (UTC)

Thank you Rick, so that's the point why Bayes is constantly being abused in MHP, just only to show this even on one single game. Guess he would forever be rotating in his grave if he knew. Could you show this special Baye's example together with such a simple table, just to put that aspect into perspective and to illustrate its relative irrelevance? Best regards -- Gerhardvalentin (talk) 01:53, 8 February 2010 (UTC)

You seem to be missing the point of this example. If we say the player's choice is Door 1 and the host opens Door 3, this example is showing there's a difference between P(car behind Door 2|player picks Door 1) and P(car behind Door 2|player picks Door 1 and host opens Door 3) - these can actually be different numbers! The sources that present this example take the viewpoint that the MHP is asking about P(car behind Door 2|player picks Door 1 and host opens Door 3), and since this can actually be different from P(car behind Door 2|player picks Door 1) presenting a solution that solves for P(car behind Door 2|player picks Door 1) without saying anything about how the host picks between two goats isn't quite addressing the question. If there were a real show run like this (there wasn't) you could look at 900 episodes and keep track of the results. If all you keep track of is overall results, you'll see (about) 2/3 of the 900 win by switching. If all you keep track of is results by door the player initially picks, you'll see 2/3 of each (whether the players initially pick randomly or not) win by switching regardless of which door is picked. What I mean is if all 900 initially pick Door 1, 2/3 win. If half pick Door 1, 2/3 of these will win. 2/3 of however many pick each door win. However if you keep track of all 6 combinations of player pick and door the host opens, you WILL see something else unless the host picks randomly between two goats - even though 2/3 overall, and 2/3 by initial door choice still win! Rather than rolling over in his grave I suspect Bayes would be smiling. -- Rick Block (talk) 02:57, 8 February 2010 (UTC)

Rick, thank you once again. Smiling? Please help me: In discussion "Simple solution (below: The chances of three doors)", on 19:45, 13 February 2010 I said the following:

... In case he (the host) is given not only one goat there, but the second goat also, the host will open the unpicked doors 2 and 3 randomly, not obscenely flashing additional information. Any other obscene behaviour of the host is not provided for in the simple standard MHP, so any "flashing additional information" would be quite another game, let's call it "The flashing MHP".  -   (btw: Nijdam actually isn't to give an inch, could you occasionally have a look there: again ?)

Please help me: In your discussion here and also above with Heptalogos 22:18, 14 February 2010 (UTC) - are you ALWAYS considering "The flashing MHP"? Will it ever be possible to distinguish between the "MHP not providing for obscene behaviour of the host" and the "The flashing MHP"? Is there any possibility to enunciate this difference, and how can I do that? What "term" should I use? Am curious. Regards, -- Gerhardvalentin (talk) 23:32, 14 February 2010 (UTC)

No, I'm not considering the "flashing MHP" - I'm saying there's a difference between prior probabilities and posterior probabilities that seems to be unclear to you. In the tables above I'm attempting to make this clear. The prior probabilities, the ones that are clearly 1/3:1/3:1/3 are in effect ONLY before the host opens a door. If we know what door the host has opened, then after the host has opened a door we have a whole NEW set of probabilities which are the posterior probabilities. We know the posterior probability of the door the host opens is 0. The entire question is what are the other two posterior probabilities? You keep saying that the two unpicked doors together have a "probability" of 2/3 and that this remains true after the host opens a door (as if this is some universal truism). What makes this true in the "symmetric" problem is that the host picks randomly between two goats, or (equivalently) that the posterior probability of door 2 in the case the host opens door 3 is the same as the posterior probability of door 3 in the case the host opens door 2. If you don't say the host picks between two goats randomly, or that you're assuming the problem is symmetric, or something to this effect, then the posterior probabilities of the two unpicked doors don't have to add up to 2/3 (i.e. the problem could be the "flashing MHP").

Note that the host preference could be way more subtle than "choose the leftmost door if possible". It could be "choose the leftmost door 75% of the time and the rightmost door 25% of the time". With this sort of preference it's not "flashing MHP". If the host opens the leftmost door switching wins with probability 1/(1+3/4) = 4/7 and if the host opens the rightmost door switching wins with preference 1/(1+1/4) = 4/5. There are 3 pairs of player pick/host choice combinations - player picks door 1 and host opens door 2, player picks door 1 and host opens door 3 is one of these pairs. If you run an extended experiment keeping track of the frequency of winning by switching for each of these pairs, unless the host chooses randomly you won't see 2/3:2/3 as the results for each pair. You'll see 1/(1+p) and 1/(2-p) for a value of p (that might be different for each pair!). If p=1/2, these are both 2/3, but this means the host IS choosing randomly. -- Rick Block (talk) 01:02, 15 February 2010 (UTC)

Why the simple solution is correct

The simple solution says: Two doors have the double chance to contain the car than one single door. This is correct. The door selected by the guest has a chance of 1/3, so the pair of two non selected doors have double chance of 2/3, total 3/3.
There are "3 doors, two goats and just one single car", so
3 doors have to hide 2 inevitably given goats
2 doors have to hide 1 inevitably given goat
1 car has no inevitably given goat.
The pair of the two not selected doors has to hide exactly ONE inevitably given goat (the second goat never is "inevitably given there"). So showing one goat there does not change the chance of this pair of two not selected doors amounting to 2/3 to hide the car. By switching the guest will double his chance from 1/3 to 2/3. This is true for every single game and also in large scale. That's the game.

Quite another issue is: It's conceivable that, in case the guest should have selected the door hiding the only car, the host has the choice between two doors to open, i.e. to show either the one goat or the other goat. If he is not choosing randomly, equivalently and symmetric, by doing so he could reveal additional info. If he should prefer only one special door to be opened if any possible e.g., he could do that only in 2/3 of all cases. For in 1/3 of all cases this door will hide the car, preventing this preferred door to be opened. By opening the "other" door then, he would be signalling that the car is behind his preferred door and that switching will guarantee the car, changing the guest's chance from 1/3 to 3/3. In contrary, opening his preferred door is signalling that the still closed door has a chance of 1/2, just like the door selected by the guest. Such behaviour is not provided for in the MHP. And even slightly preferring one door to be opened is not provided for in the MHP. Betraying such additional information is not the scope of the MHP and is not topic of the MHP, but is beyond the scope of the MHP and quite another issue. Such behaviour does not change the probability of two doors having double chance of 2/3 and does not change it in large scale, also. What it does is only giving additional info on the actual status of the three doors: Is the goat actually behind the host's preferred door or is it not. This actual accidental constellation, if it remains secret as it should remain, does not change probability. Showing information about the actual accidental constellation is an entirely different issue, far away from the paradox.

But exactly this is the point mathematicians feel to be vocated to present their calculations using Bayes and conditional probability. As said: This is quite another issue, not concerning the MHP, but only for discussions of mathematicians interested in solving such tasks. They do not relate to the fundamental question of the MHP in any way. Controversial views of mathematicians are historically well documented. But this is to problems of probability theory, but not to the fundamental question raised by the MHP as a well-known paradox. For this well-known paradox the simple solution is the one and only correct answer. Regards -- Gerhardvalentin (talk) 22:55, 15 February 2010 (UTC)

You know that the "flashing MHP" is a possibility (in which case the host still always opens a door to show a goat, and one of the two unpicked doors still MUST be a goat), yet you say "So showing one goat there does not change the chance of this pair of two not selected doors amounting to 2/3 to hide the car."

If the problem is symmetrical or (equivalently) if the host chooses between two goats randomly or (equivalently) if the two unpicked doors are completely indistinguishable (which doesn't match any remotely realistic situation), then showing one goat does not change the chance of this pair of two not selected doors amounting to 2/3 to hide the car. But the fact that one of them must be a goat is simply not sufficient. -- Rick Block (talk) 03:32, 16 February 2010 (UTC)

But you have agreed that the host opens a legal door randomly is a natural assumption in the circumstances thus the probabilty that the originally chosen door hides the car remains at 1/3. Martin Hogbin (talk) 15:33, 16 February 2010 (UTC)

The issue is that the statement "The pair of the two not selected doors has to hide exactly ONE inevitably given goat (the second goat never is "inevitably given there"). So showing one goat there does not change the chance of this pair of two not selected doors amounting to 2/3 to hide the car." is faulty reasoning. What makes the probability not change is that the host chooses between two goats randomly (which is the case if the host opens a legal door randomly, or the problem is symmetrical, or the unpicked doors are indistinguishable, or ...), not that we "already know" one of the two doors is a goat. It's like saying for a right triangle A² + B² = C² because A < C and B < C. A is less than C, and B is less than C, but this is not why A² + B² = C². It is true that one of the two unpicked doors is a goat, but this doesn't (by itself) imply the probability doesn't change when the host shows us a goat. -- Rick Block (talk) 16:41, 16 February 2010 (UTC)

You say, 'What makes the probability not change is that the host chooses between two goats randomly', I agree. But most editors here consider that we should take the host's legal door choice to be random. You agreed that this would be the natural choice given only Whitaker's statement. Martin Hogbin (talk) 18:07, 19 February 2010 (UTC)

Rick: You say that probability does not change, provided that the host's legal door choice is random. If he has the choice between two goats he is not allowed to give illegal further information about the actual constellation, so his door choice always is random, and the simple solution is correct for the MHP. I call it a confusing nonsense to use Bayes and conditional probability without any given and quantified proof that the host indeed already did show additional information by opening his door. Without a given and exactly quantified proof. You can briefly tell about a maximum of abusive and illegal disclosure in a separate section, using Bayes' examples there. And even there a clear table would be more helpful than Bayes, but it's okay to mention Bayes just there. Bayes may be of importance only for the trial, if the host is to be condemned. But for the MHP "in itself" the simple solution is correct and never needs additional "conditional probability" for explaining the paradox. And that's what I insist in: the paradox must be explained, that is exactly what this is about, that's the issue. Regards, -- Gerhardvalentin (talk) 23:56, 19 February 2010 (UTC)

And, what I'm insisting, is that you can't explain the paradox without talking about or at least alluding to conditional probability. What MAKES it a paradox is that you can see that Door 3 is open and it's chances must now be 0. This is a conditional probability. Your "simple" explanation (the host opening a door doesn't change your initial 1/3 chance because you know one of the two doors hides a goat) is faulty reasoning, or more bluntly, wrong. A different simple explanation based on examining the success of switching vs. staying considering all possible locations of the car (vos Savant's solution) is at least not wrong, but it only indirectly addresses the case where the player is deciding to switch after seeing the host open a door. It says your overall chances of winning if you always switch must be 2/3. It's fine to extend this with "... and, assuming any particular case has the same probability as any other particular case, then you have a 2/3 chance of winning by switching if you've picked door 1 and the host has opened door 3". But saying "the host opening a door doesn't change your initial probability because you know one of the two unpicked doors is a goat" is entirely different. -- Rick Block (talk) 00:37, 20 February 2010 (UTC)

Why the simple solution fails

As this is my main concern, and the discussion goes everywhere, I started a new page on this subject here. —Preceding unsigned comment added by Nijdam (talk • contribs) 11:26, 9 February 2010 (UTC)

It's Not 'Unconditional', It's 'Omniconditional'

With all the symmetry, the various simple solutions work for all contestant-door/host-reveal pairings. Chun's tree/table in the Probabilistic solution section uses one of these same symmetry assumptions to divide the contestant's door choice by 2 to get from 1/3 to 1/6. Glkanter (talk) 14:26, 13 February 2010 (UTC)

No. Chun's tree uses joint probability. Each fork shows the probability of that fork. The probability at any node is the product of the probabilities of the forks to arrive at that node, i.e. the joint probability of being at that node. For example, to determine the probability of the car being behind door 1 (in the first "column") you start at the left with probability one and follow one 1/3 fork to get to the top node where the car is behind door 1. The 1/2 forks to the right of this are NOT using symmetry, but the constraint that the host opens door 2 or door 3 randomly if the car is behind door 1. The problem is symmetrical because of this constraint. Another way to put this is that if you're assuming the problem is symmetrical, you're assuming this constraint.

Once again, what the simple solutions mean to be saying is that the average probability of winning by switching is 2/3, and assuming any particular case has the same probability as any other particular case (which must be true if the problem is symmetrical) then in all cases (for example the case where the player has picked Door 1 and the host has opened Door 3) the probability must be the same as the average. The problem is symmetrical if and only if the host chooses between two goats randomly. -- Rick Block (talk) 15:53, 13 February 2010 (UTC)

The meaning of Probability

It may seem rather late in the day to start talking about the meaning of the word 'probability' but it seems to me that much of the argument here is about precisely that.

The main WP article on the subject starts by saying, 'Probability is a way of expressing knowledge or belief that an event will occur or has occurred'. It then talks about two interpretations

1. Frequentists talk about probabilities only when dealing with experiments that are random and well-defined.
2. Bayesians, however, assign probabilities to any statement whatsoever, even when no random process is involved. Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, given the evidence.

Frequentism is not a very helpful concept when applied to the MHP The only logical interpretation of probability is that of a state of knowledge or, 'degree of belief in a statement, given the evidence'. Is ther anyone here who does not agree with this interpretation? Martin Hogbin (talk) 13:59, 20 February 2010 (UTC)

I hope this will not end in discussing the meaning of the english word word. Till now, I think, (almost) anyone would consider probability in the MHP as relative frequency in repetitions. Nijdam (talk) 21:46, 22 February 2010 (UTC)

The two approaches are complementary. Starting with a Bayesian analysis a frequentist should be able to design an experiment showing identical results, and starting with a frequentist's experiment a Bayesian should be able to derive the same outcome. This actually may be a source of conflict here. In particular, if the host's choice between two goats is not specified by the problem statement taking this to be a random choice may not match any particular experiment. This is one way to view what the "conditionalists" are talking about - specifically, what we say is "the probability" of winning in the example case of a player who picks Door 1 and the host opens Door 3 should match what would be observed in a real life experiment where we counted the frequency of winning by switching for players who initially picked Door 1 and saw the host open Door 3. I've said this a number of times, but if we say this probability is 2/3 it will match such an experiment only if the host chooses randomly between two goats. -- Rick Block (talk) 19:02, 20 February 2010 (UTC)

I do not see frequentism as a particularly useful concept here, firstly because of what it says above, 'Frequentists talk about probabilities only when dealing with experiments that are random and well-defined ', and secondly because, as a well-known peer-reviewed paper on the subject puts it, 'The modeling of conditional probabilities through repeated experimentation can be a difficult concept for the novice...'. The outcome of a modeling experiments depends on exactly how the experiment is set up. No problem statement tells us exactly how to do this. It is quite easy to set up an experiment to give any desired result.

We are therefore left with the Bayesian interpretation of probability as the only useful concept here, which may be stated as, ' Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, given the evidence'. So the first question becomes, which individual? Martin Hogbin (talk) 00:03, 21 February 2010 (UTC)

It's quite easy to set up an experiment to verify the probability for the K&W version of the MHP - vos Savant's nationwide experiment was actually quite close. All she needed to add was "The host rolls a die out of sight of the contestant and then lifts up the only losing cup if only one is a loser or the leftmost one if the die was 1-3 or the rightmost one if the die was 4-6. Keep track of the success of winning vs. losing for each combination of initial player choice and cup the host lifts up." This would have showed a 2/3 chance (frequency) of winning by switching for all combinations of player choice and host selection. Her experiment showed an overall 2/3 chance of winning by switching vs. staying, corresponding to the "unconditional" Bayesian analysis.

Addressing the question completely theoretically leaves many people unconvinced. -- Rick Block (talk) 00:51, 21 February 2010 (UTC)

I am surprised that you do not heed the advice of Morgan et al on this matter. The results you get from an experiment depend entirely on how the experiment is set up. This just leaves people arguing about exactly what experiment what represents the MHP. Martin Hogbin (talk) 12:50, 21 February 2010 (UTC)

Are you saying you disagree that the experiment vos Savant suggested in her third column (here) augmented as I suggest above represents the K&W version of the problem statement - or are you arguing just for the fun of it? From the Bayesian perspective, the analogous issue to arguing about exactly what experiment represents the MHP is arguing about exactly what sample set and conditions represent the problem - e.g. does the host opening "another door, say No. 3" constitute a condition? Since we're talking about discrete probabilities here, there is a complete isomorphism between the Bayesian and frequentist perspectives - so it's the same argument, whether you're arguing about an experiment or a Bayesian analysis. Talking about the problem in frequentist terms (IMO) actually makes the conversation less abstract and therefore more accessible to more people. -- Rick Block (talk) 18:29, 21 February 2010 (UTC)

I was being perfectly serious in my comments about frequentism not being a particularly helpful concept in the MHP. This is one area where I do agree with Morgan. As the WP article says frequentism is appropriate 'only when dealing with experiments that are random...'. If you want to assume that the host always opens a legal door randomly that is fine with me. The answer is always 2/3 in that case. If you want to extend the problem to the case where it is known that the host may choose non-randomly frequentism is less helpful.

I agree that your proposed experiment represents the K&W formulation of the problem but I am rather puzzled that you say it only gives and answer of 2/3 for the unconditional problem. What answer would you expect to get if you restricted the results to only the case where the player has chosen door/cup 1 and the host has opened door/cup 3 to reveal a goat, in other word, the conditional case?

In fact, my original question was asking, what do you understand by the term 'probability'? In the end all definitions require us to assume a particular state of knowledge. There is no such thing as absolute probability (except perhaps in QM). Martin Hogbin (talk) 20:40, 21 February 2010 (UTC)

What I meant was vos Savant's original experiment only addresses the unconditional probability. Morgan et al. is not saying an experiment is not particularly helpful - but rather if the question pertains to conditional probability it's relatively easy to end with an experiment that doesn't address the question (which is, in their opinion, what vos Savant did).

The difference between the Bayesian and frequentist definition of probability really doesn't affect the MHP. In both interpretations there's a difference between asking about the probability of winning by switching unconditionally (a frequentist might express this as the probability of winning by switching in all cases) vs. the conditional probability in a specific case, such as the case the player picks door 1 and the host opens door 3. If we're talking about the conditional probability, the Bayesian analysis should result in the same answer as a frequentist's experiment.

The point we keep running up against is that this is true only if the host chooses randomly between two goats. You seem to be wanting to say the probability is 2/3, regardless of any host preference, if the player's state of knowledge doesn't include the host's preference. But if the host does have a preference then an experiment that counts frequency of winning given the player picks door 1 and the host opens door 3 will show this preference, in which case the frequentist's answer and your Bayesian answer will disagree. Since these should agree, what it means is that this experiment is NOT an accurate reflection of this Bayesian analysis. If the player's state of knowledge does not include the host preference, then the experiment must preclude any such knowledge - which means the experiment must address the unconditional probability (e.g. count success of winning by switching in all cases where the player picks door 1). -- Rick Block (talk) 22:54, 21 February 2010 (UTC)

I agree with what you say, right up until the end. You seem to accept that under the normal Bayesian meaning of 'probability', if the player is not aware of any host preference then this preference cannot be used in any probability calculation made from the state of knowledge of the player. This is true even in the conditional case that the player has chosen door 1 and the host has opened door 3. From the player's SoK the probability of winning by switching is 2/3, even in the conditional case where the host may have a door preference.

Where we disagree is on what simulation best represents this case. This is not surprising given Morgan's comments on the difficulty of getting simulations right in cases of conditional probability. I guess we will agree that the way to simulate a known host preference is to do as you suggest in your extension of vos Savant's simulation but have the rule "The host rolls a die out of sight of the contestant and then lifts up the only losing cup if only one is a loser or the leftmost one if the die was 1 or the rightmost one if the die was 2-6". Also only consider cases where the player chooses door 1 and the host opens door 3. This would not give an answer of 2/3. Do you agree that this is a good simulation of the conditional case in which the host has a known door preference?

The disagreement comes in the case that the host door preference is not known. Bayesian probability says that even the conditional probability must be 2/3 here. What is the correct simulation. My answer is not that we consider all door possibilities, that is to say the unconditional case, but that we exclude the use of the hosts door preference. We can only take this to be random and thus apply your original rule for the host's door choice. Thus, even in the conditional case, the answer is 2/3. Martin Hogbin (talk) 23:29, 21 February 2010 (UTC)

Yes, I agree with the "known preference" experiment (in this case, the host has a 1/6 vs. 5/6 preference for the leftmost door). And I think you agree the experiment and the Bayesian analysis should match. The question is what does it actually mean to say the contestant does not know the host's preference? Your experiment is saying it means the host, in fact, is choosing randomly between two goats. I'm saying it means the unpicked doors become indistinguishable, i.e. the problem turns into the unconditional problem. It's difficult to tell which of these is "correct" because they both end up with the 2/3 answer. On the other hand, I think you would also say that the probability without knowing the host's preference is 2/3 even if the host actually has a preference (wouldn't you?). To make the experiment turn out right, then what HAS to happen is we have to count all cases where the host opens either door, i.e. we HAVE to turn the problem into the unconditional problem. -- Rick Block (talk) 01:35, 22 February 2010 (UTC)

Well I guess you could simulate the unknown host preference by saying that the doors are indistinguishable, in fact as many editors have suggested, there is a good case for doing this right from the start, as the player has no idea what the door numbers signify. It has been you who has generally pointed out that the doors are numbered and thus we can distinguish them and that we all know that the host has opened door 3.

A far better way to simulate the unknown host preference is to simply randomise the host preference. We keep the question exactly the same, the player chooses door 1 and the host opens door 3, but the host's door choice preference is unknown and therefore taken as uniform at random.Martin Hogbin (talk) 10:00, 22 February 2010 (UTC)

Substantiated state of knowledge. Who owns what knowledge in the MHP? Some do suspect criminal behaviour of the host and say that everyone knew about. Who knows what? The mathematical aspect of culpable offending sellout is linked to the legal aspect of sellout. Both play outside the MHP and belong to the area of justice and evidence. Illegally given info in signalling that the host has two goats available, to draw conclusions from, may be mentioned in a later section "District court and mathematics on illegal behaviour" where you can tell that the host has been brought to trial. Bayes is inextricably linked to that legal aspect that is outside the MHP and belongs in the field of justice. Yes, the exact distinction is extremely important. -- Gerhardvalentin (talk) 11:45, 22 February 2010 (UTC)

Martin - Would you say that the probability without knowing the host's preference is 2/3 even if the host actually has a preference? To clarify, I mean you run the experiment as you suggested above (host rolls a die, and if he has a choice opens the leftmost door if the die is 1 and the rightmost door if the die is 2-6) but you don't tell the contestant what the host is doing. If the player picks door 1 and the host opens door 3, there's now a difference between the probability from the state of knowledge of the contestant and the state of knowledge of someone who knows how the host is deciding what door to open if there are two goats. If you run this experiment and keep track of the success of winning by switching for players who pick door 1 and see the host open door 3 you will see the effect of the host preference. If you say that the player's chances are 2/3 because the player doesn't know what the host's preference is, the experiment is not showing this probability unless you count cases where the host opens both door 3 and door 2. I think this means you're saying the question "what is the probability of winning by switching given the player picks door 1 and the host opens door 3 but the player doesn't know how the host chooses between two goats" is the same as "what is the probability of winning by switching given the player picks door 1".

You say, "what is the probability of winning by switching given the player picks door 1 and the host opens door 3 but the player doesn't know how the host chooses between two goats" is the same as "what is the probability of winning by switching given the player picks door 1". This is indeed true but it is not what I am saying. I am saying that "what is the probability of winning by switching given the player picks door 1 and the host opens door 3 but the player doesn't know how the host chooses between two goats" is the same as "what is the probability of winning by switching given the player picks door 1 and the host opens door 3, having chosen uniformly at random from the legal possibilities available to him"

This is, BTW, exactly what the Morgan et al. paper is all about and Nijdam's oft repeated point that the problem is inherently conditional. Assuming the player knows which door she's picked and which door the host opens, "the probability" (meaning the frequency an experiment will show for this pair of player pick and door the host opens) is a function of the host's preference between two goats whether or not the player knows what this preference is. The Bayesian analysis based on ignorance of the host preference that says the probability is definitely 2/3 is saying the same thing as saying the player doesn't know which door the host opens. To reflect what's actually happening in this case (meaning to reflect what an experiment will show), the Bayesian analysis must include the host preference, i.e. the solution must be conditional.

What is 'actually happening' is that the car is, in fact, behind one of the doors and the probability of winning by switching is 1 or 0 depending on which door the car is behind. But nobody (who might be answering the question) knows which door the car is behind, thus we must assume that its distribution is uniform. Similarly no one (who might be answering the question) knows what the host door preference might be thus we must also take this as uniform at random from the choices legally available.

What you are doing is exactly what I feared, and exactly what Morgan warn against. You are setting up the wrong experiment in order to represent the case that the player does not know the host door preference and then claiming that it proves the result that you want. As explained above, if the player does not know the host door preference then the experiment to show the probability from the players SoK would have to take the host to choose a legal door uniformly at random and then only consider cases where the host opens door 3. This is all that can ever be done when we have no information on something. From the players SoK the hosts legal door choice is random, even though he is seen to have chosen door 3.

Regardless of host door preference and even when the player chooses door 1 and the host is known to open door 3, from the players SoK the probability is always exactly 2/3. The result is exactly the same if the host has opened door 2 and these results do not require the use of conditional probability, they can be obtained either from symmetry or by noting that random information is no information. We have a condition (host opens door 3 or host opens door 2) which we cam prove makes no difference. You can call that conditional if you like.Martin Hogbin (talk) 16:35, 22 February 2010 (UTC)

Gerhard - a more benign way to look at this is to consider a case where the host is simply not told how he should pick between two goats. In this case, the host might exhibit a preference that could be discovered by watching the show and keeping track of the success of winning by switching given each pair of possible initial player pick and door the host opens. The host might be only accidentally revealing information, not doing so deliberately. And, even in this case, the player is never worse off switching. -- Rick Block (talk) 15:54, 22 February 2010 (UTC)

Simulating the 'player does not know the host preference' case

Martin - you're saying not knowing how the host chooses between two goats means the host is choosing randomly. I'm saying not knowing only means you don't know, and that this is independent of whether the host is actually choosing randomly or not (and, yes, the car actually behind one door, but the host actually picked randomly or not as well). To show the difference between the knowledge of how the host chooses and how the host actually chooses, I'm suggesting the experiment above (the host does not choose randomly, but the contestant does not know). You do claim the contestant's chances in this case (not knowing) are 2/3, right? On the other hand, you do agree the frequency of winning by switching counting cases where the player picks door 1 and the host opens door 3 will not be 2/3. So, what you MUST be saying is the contestant can't tell the difference between the host opening door 2 and the host opening door 3.

There is a difference between something actually being random and not knowing the distribution. Take vos Savant's "little green woman" example. The player picks door 1, the host opens door 3 (choosing randomly between two goats), and now a UFO lands and out pops our little green woman. The player has a 2/3 chance of winning by switching. The little green woman has a 50% chance of picking the winning door. These are simultaneously true. It is also true that if this happens repeatedly, the little green woman will win the car 2/3 of the time she picks door 2. She only has a 50% chance of winning because she has no basis on which to pick so must pick randomly, i.e. she'll pick door 1 50% of the time (and win 1/3 of these times) and she'll pick door 2 50% of the time (and win 2/3 of these times). In total, she wins 1/6 + 1/3 = 1/2 of the time. It's perfectly possible to create an experiment that shows this effect without randomizing the car placement between the two doors after the host opens a door.

Similarly, not knowing the host's preference means you're picking between a 1/(1+p) chance and a 1/(2-p) chance "randomly" (not with equal probability, but in accordance with the probability these cases come up) and therefore have chances equal to the average, which is 2/3. What I'm saying is that the experiment analogous to the little green woman one does not require the host to pick randomly between two goats - but that modeling the "lack of information" about the host's choice means not distinguishing between cases where the host opens door 2 vs. door 3. -- Rick Block (talk) 20:40, 22 February 2010 (UTC)

You might argue that not distinguishing between the doors must give the same result as the host's policy being unknown, and I would not dispute this, but you are making things much more complicated than they need to be. Here is the situation that we wish to simulate. The player has chosen door 1 and the host has then revealed a goat behind door 3. The player has no knowledge of how the car was initially place or the host's door opening policy. To simulate this we place the car randomly, because we have no other basis in which to place it, the player then chooses door 1, and the host opens one of the other doors to reveal a goat, when he has a choice we must from our SoK take this to be random also. Now repeat the experiment but only consider cases where the host has opened door 3.

Just as I predicted, simulation does not resolve the problem, it just turns it into an argument on how it should be simulated.

The Bayesian meaning of probability is much easier to understand and argue about. Starting with the same situation, the player has chosen door 1 and the host has then revealed a goat behind door 3, the probability of winning by switching is dependent on the SoK of the person answering the question, just as the 'little green woman' argument shows.

From the SoK of the producer or the host, who both know where the car is placed, the probability is 0 or 1 depending on where it has, in fact, been placed.

From the SoK of a player who does not know either how the car was initially placed or what the host's legal door opening policy is, the probability is exactly 2/3, even when door 3 has been observed to have been opened.

From the SoK of a person who does not know how the car was initially placed but does know what the host's legal door opening policy is the probability is 1/(1+p).

From the SoK of a person who has arrived and been told only that there is on goat behind one of two doors, the probability is 1/2. Martin Hogbin (talk) 14:43, 23 February 2010 (UTC)

As I said above, Bayesian probability and frequentism are complementary. A Bayesian analysis based on a particular state of knowledge is meant to (does, in fact) exactly match what a frequentist would observe under the same conditions (in the limit as n approaches infinity). I think you agree that if the host is told to roll a die and to open the leftmost door (if given a choice) if the die is 1 and to open the rightmost door if the die is 2-6 that the observed frequency of winning by switching for players who pick door 1 and see the host open door 3 will not be 2/3. My question for you is what does it mean to say, with this set up, that "the probability is 2/3" from the SoK of a player who does not know the host's policy? We know that if we repeat this 3000 times with players who pick door 1 and see the host open door 3 (who don't know the host's policy), we won't see roughly 2/3 of these players win by switching. So what's wrong - does Bayesian analysis not apply here for some reason? My answer is that "not knowing the host's policy" doesn't mean what you think it means. It HAS to mean you can't tell the difference between the host opening door 2 and door 3. If it doesn't mean this, then Bayesian analysis fails (which doesn't seem at all likely to me). The converse of this from the Bayesian perspective must be that knowing which door you open and which door the host opens means your SoK includes the host's policy for choosing between two goats. Whether you "know" the host's policy or not (in the everyday meaning of "know"), you're affected by it - so from a Bayesian perspective knowing the door numbers means you DO know it. -- Rick Block (talk) 01:15, 24 February 2010 (UTC)

Your last sentence is completely wrong. There is only one meaning to the word 'know' and it is the same in probability as in everyday life. Either something is known or it is not. In probability you must answer from a clearly defined state of knowledge, that is pretty well the meaning of the term probability, WP says, 'Probability is a way of expressing knowledge or belief that an event will occur or has occurred'. Obviously, what is not know but is true does affect the outcome, when the player has to make her choice, the car has already been placed so, so we agree, the reality is that the probability of winning by switch is 0 or 1. This is exactly the same as the hosts door preference, in reality, the host may have a preference and the probability of interest will depend on this, but we do not know either of these things.

There are two ways to deal with completely unknown distributions. The stricter way is just to say that they are unknown and nothing can be done. The second is to apply the principle of indifference and take the distribution to be uniform at random.

We have this choice at the start of the MHP. A car is placed behind one of three doors and a player chooses one. What is the probability that the player chooses the car? One answer is to say that we are given no information on how the car was placed or how the player chooses (the car might always be placed behind door 2 and the player might always choose door 1 for all we are told). Some might say that this is the strictly correct answer. If we take this view, then the MHP becomes very simple but rather boring. The answer is an indeterminate value from 0 to 1.

The situation is exactly the same regarding host door preference. The host may well have a door preference and this will affect the probability of winning by switching given that he opens door 3, but we do not know this preference. In this case we can either take it as unknown and produce Morgan's calculation or we can take his choice to be uniform at random.

If we are consistent in how we deal with 'things that have actually happened but the player does not know about when she makes her choice' there are only two possible answers to the probability of winning by switching, indeterminate, or 2/3. Martin Hogbin (talk) 09:34, 24 February 2010 (UTC)

Exactly. Again: The question is "Who knows What". In case the host is given TWO goats and he has any preference, does the guest know about such preference?
Probability: Who evaluates existing probabilitiy after the host showed one goat? It may not be to the host to designate probability, then. Is it to the guest to designate probability? Who does have any "records"? Is it the first show ever, or is it the fifth or the hundredth? What does the guest know about the "behavior" of the Host? Can she only assume: Either 2/3 (no preference) or 1/2 or 1/1 (again 2/3). Once more: WHO claims to be in a position to evaluate probability after the host opened one door? -- Gerhardvalentin (talk) 13:26, 24 February 2010 (UTC)

@Martin - So you're saying Bayesian analysis fails for the experiment as suggested (host is not choosing randomly but the contestant is not aware of this)? What I mean by "fail" is that the analysis says 2/3 but this is not the limit of the frequency as n approaches infinity. -- Rick Block (talk) 14:07, 24 February 2010 (UTC)

@Rick. No the Bayesian and frequentist approaches agree provided that you set up the right experiment.

• To set up the fact that the player is not aware of the initial car position you make this random.
• To represent the fact that the player has originally chosen door 1 you either always choose door 1 or better have the player choose randomly and then only count the cases where the player has chosen door 1.
• To represent that you have no information about the host door opening policy you have the host choose a legal door randomly.
• To represent the fact that the host has opened door 3 you only count the cases where the host has chosen door 3.

The results of this correctly set up experiment agree with the Bayesian approach, as expected. Martin Hogbin (talk) 14:57, 24 February 2010 (UTC)

You're avoiding the question. The scenario is the host does NOT choose randomly, but the player does not know this. What is the Bayesian analysis for the conditional probability given the player has initially picked door 1 and the host has opened door 3? Does this analysis match what we would experimentally see? If not, what's wrong? You seem to be saying this is the wrong experiment. Does Bayesian analysis not apply in this case? -- Rick Block (talk) 15:04, 24 February 2010 (UTC)

Both approaches show that the probability of winning by switching, if the host preference is unknown, is exactly 2/3. Martin Hogbin (talk) 15:12, 24 February 2010 (UTC)

You're still avoiding the question. The scenario is the host does NOT choose randomly, but the player does not know this. Experimentally, if we count the frequency of winning by switching for the case where the player picks door 1 and the host opens door 3 the answer will NOT be 2/3. You're saying the Bayesian analysis says if the player doesn't know the host preference the probability is 2/3. My claim is the experimental results are correct. What's wrong with the Bayesian analysis? -- Rick Block (talk) 15:45, 24 February 2010 (UTC)

I have given a full description of how the correct result might be obtained experimentally in the section below. Martin Hogbin (talk) 16:45, 24 February 2010 (UTC)

You're still avoiding the question. I'll take this to mean you can't or don't want to answer it. I've already said my answer is that the Bayesian analysis based on the player not knowing the host's preference has to mean that the player cannot distinguish the host opening door 2 or door 3, and (equivalently) that knowledge of the doors means that the player knows (in a Bayesian sense) the host's preference. Saying this case defies Bayesian analysis, or that the Bayesian analysis results in a different probability than what you'd measure seems like a completely indefensible stance. -- Rick Block (talk) 17:05, 24 February 2010 (UTC)

I thought I had answered you question. Perhaps you could repeat it for me, I am not deliberately avoiding anything. I have given, at various times a frequentist, Bayesian, and modern probability theory explanation of why the answer is 2/3. The Bayesian approach, as I have said before, has nothing to do with the player not being able to distinguish between doors 2 and 3. The Bayesian approach takes the host door opening policy to be random if it is not known, and is usual in such cases. Martin Hogbin (talk) 17:21, 24 February 2010 (UTC)

Back to the meaning of probability

You're a committed Bayesian, and you have the following question to answer:

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. The car and goats are uniformly distributed. You pick door 1, and the host, who knows what's behind the doors (and has been instructed, unbeknownst to you, to use a specific, non-uniform way of choosing which door to open if both unpicked doors hide goats, e.g. host rolls a die and opens leftmost door if the die is 1 or rightmost door if the die is 2-6), opens door 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" What is the probability that you will win if you switch?

If you do this experimentally, you will not see a 2/3 chance of winning by switching. What is the Bayesian analysis that shows this? In particular, how is the apparent conflict between the non-2/3 experimental result and the 2/3 Bayesian analysis (given the player does not know the host's strategy) resolved? -- Rick Block (talk) 17:39, 24 February 2010 (UTC)

You question is ill posed. Either the host is known to choose non-randomly or he is not. There is no 'he chooses randomly but we do not know this' in probability. To some one who does not know the host chooses non-randomly the answer is 2/3, to someone who know his strategy the answer is not 2/3. Martin Hogbin (talk) 22:35, 24 February 2010 (UTC)

Ill posed? It's a real world experiment I can create. All actors have a well defined state of knowledge. Bayesian analysis obviously applies. I think you just don't like the obvious answer - which is that the contestant's probability is 1/(1+p) for some unknown value of p, i.e. that knowing the specific doors you've picked and the host opens means you "know" about the host preference (from a Bayesian perspective). Not knowing the host preference doesn't mean it's random. It means you don't know what it is. As a Bayesian state of knowledge it means the same thing as not being able to distinguish between the doors the host opens - but this is a technical meaning that doesn't exactly match the common usage of "know". -- Rick Block (talk) 01:24, 25 February 2010 (UTC)

Perhaps we should get the view of someone else on this, Nijdam perhaps. I believe that you have a serious misunderstanding of the meaning of 'probability'. The WP article starts, 'Probability is a way of expressing knowledge or belief that an event will occur or has occurred'. There is no 'technical meaning' to 'knowledge' other than 'information available to the person who is assessing the probability'. Martin Hogbin (talk) 10:11, 25 February 2010 (UTC)

I have asked Nijdam to comment. Martin Hogbin (talk) 10:27, 25 February 2010 (UTC)

The MHP is not primarily meant to discuss the different views on probability. The general probabilistic idea behind the MHP is frequentistic (is this not an english word?), or equivalently based on the idea of symmetry. Let's not complicate the discussion.Nijdam (talk) 11:21, 25 February 2010 (UTC)

I think it is very important to agree on what the word 'probability' means before we argue about what the answer to a specific question on probability is. You are, of course, free not to join in this discussion if you wish. Perhaps you might like to comment on my frequentist analysis of the problem below. Martin Hogbin (talk) 13:06, 25 February 2010 (UTC)

Nijdam - would you say any real world situation involving discrete probability can be perfectly represented using Bayesian analysis, and that the Bayesian analysis will be predictive of the observed frequency? The actual issue here is whether "state of knowledge" has a precise technical meaning that doesn't necessarily match the common understanding of "knowing" something - more specifically, whether knowledge of the specific doors you've picked and the host has opened implies your Bayesian "state of knowledge" necessarily includes the host preference between two goats (whether you "know" the precise value or not). For the question above, it seems to me the Bayesian answer cannot be 2/3 since this is not what will be observed experimentally - meaning the player not "knowing" the host preference must be different from the host preference being in the player's "state of knowledge". -- Rick Block (talk) 15:49, 25 February 2010 (UTC)

Simply, no! The SoK of the Bayesian might well be insufficient to cover the real situation. The Bayesian just uses his analysis, if forced to come up with an answer, but tries to learn from the experiment to update his SoK.Nijdam (talk) 10:39, 26 February 2010 (UTC)

However this distracts from the issue at stake. Most "normal" people, confronted with the MHP, will consider it in a frequentistic way. Nijdam (talk) 10:39, 26 February 2010 (UTC)

The Bayesian concept of probability has been well explained by JeffJor when he gave the well-known example, 'Suppose I draw a card at random. I look at it by myself and see that it is the Queen of Hearts. I tell Ann that it is red, Bob that it is a heart, Carl that it a face card (that means TJQKA), and Dee that its value is even. I ask each what the probability is, that it is the Queen of Hearts. Ann says 1/26, Bob says 1/13, Carl says 1/20, and Dee says 1/24. But I know that this "probability" is actually 1/1. Since all the answers are different, who is wrong?'

'Answer: Nobody. The probability is not about the card, it is about the process. And each person sees a different process, one that leads to the specific piece of information I gave to them. Since each piece is different, the process each is evaluating is different'.

I will add Ed whom I tell nothing, his answer is 1/52. How would you do a frequency experiment to show Carl's probability, and Ed's? Martin Hogbin (talk) 13:45, 26 February 2010 (UTC)

The problem is the question asked. Ann will give as an answer: P(HQ|Red), Bob: P(HQ|H), etc., all different conditional probabilities. Nijdam (talk) 16:45, 26 February 2010 (UTC)

That is a perfectly good way of addressing the question, and it is one which makes doing an experiment easier. As I have said before, every problem can be expressed as a conditional probability asking for P(Even in question|the problem description) with the sample set being all possible events. Alternatively, as you suggest, you can take the set of all cards in a pack as the sample set and condition on the information given to each person. The important thing to note is that the condition in each case is based on the information known by each person. For Ann you condition on the card being red, because that is all she knows. Martin Hogbin (talk) 17:35, 26 February 2010 (UTC)

Well, Martin, I think you only have to admit you're in our "camp". What information has the player, knowing the rules, being on stage, pointing to door 1 and seeing door 3 opened showing a goat? Nijdam (talk) 11:28, 27 February 2010 (UTC)

As for the simulation, the complicating factor here, are the different participants. So draw a card at random. For Ann's answer only count the cases in which it is red, for Bob's answer the cases in which it's hearts, etc. Nijdam (talk) 11:36, 27 February 2010 (UTC)

If you do this 5200 times (making sure you've drawn a face card) and tell Ann the color, Bob the suit, Carl that it's a face card, and Dee that it's even and ask them (randomly, based on what they've been told) to guess the specific card, the results will be close to the predicted probabilities. If you do this an infinite amount of times, the results will exactly match the predicted probabilities.

You make the same mistake here that you make with the MHP. Why must you make sure that you draw a face card? You cannot impose arbitrary conditions on top of the conditions of the problem. Whether the card is a face card or not is not relevant to Anne.

To do the experiment to demonstrate the answers you must draw a card at random then for Anne, for example, only count cases where the card drawn is red (whether it is a face card or not) and then note the proportion of times that the card is the queen of hearts. This is along the lines Nijdam has suggested. Using your method you would get the wrong answer for most of the cases.

I repeat, given persistent door numbering and a constant host preference of p, knowing the door numbers means your Bayesian "state of knowledge" inherently includes the host's preference. Your actual probability of winning by switching (meaning your frequency of winning as N approaches infinity) IS 1/(1+p). Claiming the Bayesian analysis says something different is ludicrous. To ensure in your experiment that the Bayesian "state of knowledge" does not include the host's preference, you have to vary the experiment a bit. A state of knowledge of not knowing "x", means (experimentally) "x" must be independent of the experimental results. So, with regard to the MHP, either you have to count cases where the host opens door 2 or door 3, or you must explicitly randomize the host's choice between two goats. Failing to do one of these means your Bayesian analysis based on not knowing the host preference is wrong because your "state of knowledge" does include the host preference (or, if you'd prefer, means your experiment does not reflect your analysis).

The bottom line is that not knowing the value of p means something different than not knowing the host's preference as a Bayesian state of knowledge. -- Rick Block (talk) 15:27, 26 February 2010 (UTC)

I do not know where you get these strange ideas from but they are very confused. From the player's SoK the host's unknown preference is irrelevant. This is standard stuff. Martin Hogbin (talk) 17:35, 26 February 2010 (UTC)

Martin - Strange as it may seem, I think we're actually saying the same thing here. Let me prove it to you. You've said, given knowledge of the specific doors involved, "the player's state of knowledge does not include the host's preference" implies the host picks randomly between two goats (or, equivalently, the probability of winning by switching is 2/3). I assume you agree if A implies B, not B implies not A. In the experiment I've set up (host does not pick randomly between two goats but the player doesn't know this) the host is not picking randomly (the probability is not 2/3). Thus, it must be true that the player's state of knowledge DOES include the host's preference (not B implies not A), even though the player doesn't know the exact value. This is what I mean by not knowing the value is not the same as not being within the player's "state of knowledge".

But you have set up the experiment wrongly (I describe how to do it correctly and why this is correct in the section below). What you have done is set up the experiment from the our SoK, which is someone who knows the host's door preference but not the producer's car placement preference. Why do we know the host's door preference? Because you have told us. Why do we not know the producer's car placement preference? Because you have not told us. It really is that simple.

This is, BTW, the same reason the problem is inherently conditional. If you solve the problem unconditionally you're ignoring the information afforded by the number of the door the host opens. It is exactly this information (the host's preference) that you're ignoring. Knowing the door number means the host's preference is within your state of knowledge. This may be hard to grasp, but it is absolutely true. -- Rick Block (talk) 19:35, 26 February 2010 (UTC)

It is hard to grasp, because it is incorrect. The host's door preference is not within my state of knowledge. Or, to put this and other way, I do not know what it is. Do you? If so, please tell me what it is. Martin Hogbin (talk) 20:13, 26 February 2010 (UTC)

One obvious point about setting up simulations that I forgot to mention is that you can only simulate information that you know. We cannot set up a simulation involving the host's door preference because we are not told what it is. Martin Hogbin (talk) 20:31, 26 February 2010 (UTC)

Martin - you are completely (deliberately?) missing the point. I'm talking about a Bayesian analysis of the experiment I've suggested. The host strategy is given to the host but we don't tell the player what it is, and we ask what is the probability of winning by switching given the player picks door 1 and the host opens door 3 from the perspective of the player. This is an experiment we can set up. Bayesian analysis does apply, and the Bayesian result should (better!) match the experimental outcome. This experiment shows the difference between not knowing what the host's preference is and the host's preference not being in the player's state of knowledge. Because the host DOES have a preference and we know the doors involved, the Bayesian probability from the perspective of a player who does not know the preference is 1/(1+p) - not 2/3. Knowing the door numbers means you're affected by p - which puts it in your Bayesian state of knowledge (at least as a variable). This is not how you'd set up an experiment to measure the probability where the host's preference is not in the player's state of knowledge - but this is the point I'm making, i.e. that you CAN set up an experiment where "state of knowledge" and "knowing the host's preference" mean slightly different things. This is that experiment.

I have discussed the frequentist approach in the section below. If you set up the experiment correctly the you get the correct result, that the probability of winning by switching is 2/3, given that the player has chosen door 1 and the host has opened door 3. I explain exactly why this is the case when the experiment is set up using consistent assumptions. Your comments are welcome in that section.Martin Hogbin (talk) 10:19, 27 February 2010 (UTC)

I know you're not a probability theorist, so I really couldn't care less how wrong you think I am about this. However, realize that what you're saying is that the Bayesian analysis for this experiment says the probability is 2/3 even though an experiment would show a different result (?!).

No, I am not saying this. The 'experimental' result agrees (as expected) with the Bayesian analysis that the answer is 2/3. Martin Hogbin (talk) 10:19, 27 February 2010 (UTC)

I'm saying this cannot possibly be true, and therefore your Bayesian analysis must be wrong - and the only possible thing that can be wrong is equating "not knowing the host's preference" with "the host preference not being in the state of knowledge of the player". Think carefully about the information imparted by the door numbers. -- Rick Block (talk) 00:46, 27 February 2010 (UTC)

Of course the results should agree, and so they do. Martin Hogbin (talk) 10:19, 27 February 2010 (UTC)

You keep telling me how to experimentally model a particular Bayesian analysis, but I want to go in the other direction. I have an experiment and I want a Bayesian analysis. It's the K&R MHP with the addition that the host is told a specific, non-uniform way to choose between two goats but we don't tell the player this. The question is given the player has picked door 1 and the host has opened door 3 what is the player's chance of winning by switching (from the player's perspective)? More specifically, if we do this 9000 times with random initial player picks, we will throw away all samples except those where the player initially picked door 1 and the host opened door 3, and the question is what is the expected percentage of winning by switching for the remaining samples according to a Bayesian analysis of this situation (from the perspective of the player). I believe you're saying the answer is 2/3. I'm saying if we do this, and count the actual frequency, it will (in the limit) approach a different value, specifically 1/(1+p). I think we agree so far. Where we diverge is what conclusion we draw from this. You're saying the experiment is invalid and is not accurately representing the problem statement, so it's no big shock that the Bayesian analysis and the experimental results don't match. I'm saying the experiment is the experiment and I want a corresponding Bayesian analysis that DOES match.

This is exactly what I do in the section below. I am not modeling a Bayesian analysis, I am modeling the actual situation as described in the problem statement. I give the exact problem statement at the start of the section.

Now, if your problem statement says that the car is placed randomly but the host chooses door 3 with probability q when the car is behind door 1 then a properly set up simulation will, of course, show a probability of winning by switching of 1/(1+q). This agrees with the Bayesian calculation for the same case given the same information.

If your problem statement says that the car is placed randomly but the host chooses door 3 with probability 1/2 when the car is behind door 1 then a properly set up simulation will, of course, then show a probability of winning by switching of 2/3. This agrees with the Bayesian calculation for the same case given that same information.

If, on the other hand the problem statement does not tell us how the car is initially placed or how the host chooses a legal door we are stuck. We cannot do a simulation, or do a Bayesian analysis without making some decisions as to how these things were accomplished, so that in our simulation we can repeat them. For example suppose the problem statement tells us that the car is always placed behind door 1, how would we simulate this? By placing the car always behind door 1 in our simulation. Now suppose the problem statement tells us that the car is placed randomly, how would we simulate this? By placing the car randomly in our simulation. Now suppose the problem statement does not tell us how the car is placed, how would we simulate this? Not so easy. We have to make a decision. The choices are explained in the section below. Martin Hogbin (talk) 19:04, 27 February 2010 (UTC)

The player knows the initial distribution is 1/3:1/3:1/3. The player knows the host must open a door showing a goat. We haven't told the player how the host picks between two goats but the player knows she picked door 1 and the host opened door 3. The player knows her chances of winning in this case depend on how the host chooses between two goats, so has two alternatives - assume the host makes a random choice, or figure out a probability leaving the host preference as a variable. Assuming the host choice is random results (in this experiment) in the wrong answer. Leaving the host preference as a variable ends up with the expression 1/(1+p). The door numbers are clearly within the player's "state of knowledge". So, this expression representing the player's probability of winning is as well. The information in the door that is opened is what allows the player to change her view of the chances of winning by switching from 2/3 to this expression. As Bayesians, we MUST make this change as well or else our Bayesian prediction will not match an experimental frequency. The player's state of knowledge is changed by knowing what door she's picked and what door the host opened. She doesn't know the exact value of the host preference, but she knows she's been affected by it.

And, again, I think this is fundamentally the same thing that you're saying. If our Bayesian analysis says the host's preference is NOT in the player's state of knowledge this experiment does not accurately reflect that analysis. But what this means is that knowing the door numbers requires a slightly different analysis. -- Rick Block (talk) 18:22, 27 February 2010 (UTC)

Any probability problem must be addressed from a defined state of knowledge. In the MHP there are two sensible states we might choose, the expected state of knowledge of a player on a game show (which is what the question seems to ask for), or the information given to us in the problem statement, which is the more formal approach that Morgan, for example, take. There is no way round this, no matter what interpretation of probability you use. You cannot do a repeat show without knowing how the show was set up in the first place. Martin Hogbin (talk) 19:04, 27 February 2010 (UTC)

You are continuing to ignore what I'm saying. Anyone else want to take a shot at this? Nijdam? Gill? -- Rick Block (talk) 19:44, 27 February 2010 (UTC)

Is there anyone here who does not think that any probability problem must be addressed from a defined state of knowledge? Martin Hogbin (talk) 21:06, 27 February 2010 (UTC)

The state of knowledge I'm talking about is that you know the car was uniformly placed, you know the host must open a door, you don't know the host's selection criteria (between two goats), but you know the numbers of the doors you've picked and the host has opened. This is a perfectly well defined state of knowledge and the answer should match any experiment you set up. The issue is treating the host's choice as random means your answer is definitely 2/3 probability of winning by switching when, experimentally, it can be anything between 1/2 and 1, i.e. 1/(1+p). I'm saying because you know the numbers of the doors you know that your probability depends on the unknown value of p, and replacing p with 1/2 in your answer means you are ignoring the information afforded to you by number of the door that was opened - i.e. not knowing p is different from removing the host's preference from your state of knowledge. -- Rick Block (talk) 21:29, 27 February 2010 (UTC)

I have never said that you must treat the host's choice as random. Given the information above you have three choices. Parameterise the host's choice, as you have done above; treat the host's legal door choice as random; use real world information. If we are to use only the information given in your statement we have only two choices: parameterise the host's choice, treat the host's legal door choice as random, both are valid choices, depending on whether you wish to apply the principle of indifference. Agreed? Martin Hogbin (talk) 23:56, 27 February 2010 (UTC)

Rick, thank you for your efforts in explaining: The guest's probability of winning by switching could be affected by the host's selection criteria (when he has the choice between two goats). Probability can be "anything between 1/2 and 1", i.e. "1/(1+p)", where "p" - the host's possible preference - can be from 0 to 1. How do you determin/estimate the host's preference? Do you need additional information - besides the host's actual door choice ("door 3" resp. "door 2")? Where "p=0" will result in probability of 1, "p=1/2" (no preference) will result in probability of 2/3 and "p=1" will result in probability of 1/2. Is this correct?
Please can you tell your result for the guest's probability of winning by switching for the following two different situations: a) guest selects door 1, host opens door 3, and b) guest selects door 1, host opens door 2.
Do you - besides the host's actual opening a specific door - need/use further info about the host's preference? Please help. Thank you so much. -- Gerhardvalentin (talk) 22:58, 27 February 2010 (UTC)

The guest's probability of winning by switching is (not "could be") affected by the host's selection criteria. If this is not given to be random (p=1/2) you have no solid basis to say your probability of winning is 2/3 - if you don't know p its value could be anything from 0 to 1. If you don't know it, you have no way to determine or estimate it. You can ignore it and use its average value (1/2) instead, but by doing this you're actually not using the information provided to you by the specific door the host opens - and this is why the value from the analysis (2/3) can end up not matching the experimentally observed frequency for a specific case (such as player picks door 1 and host opens door 3). If you don't know p, knowing the specific door the host opens (somewhat perversely) means your chances have changed from definitely 2/3 to some unknown value between 1/2 and 1 (with an average of 2/3). What this means is 2/3 of the players who pick door 1 will win by switching, but if you see which door the host opens (and don't know p) your specific chances are some unknown value between 1/2 and 1.

Since the p values come in pairs (if you pick door 1 and the host has a preference p for door 3, the host must have a preference q=1-p for door 2) your probability of winning if the host opens door 3 is 1/(1+p) while your probability if the host opens door 2 is 1/(1+1-p) = 1/(2-p) - which is another expression with values from 1/2 to 1. Not knowing p means there really isn't much difference between the host opening door 3 or door 2 - in either case your probability is some unknown value between 1/2 and 1.

Rolling this back up a bit, the reason we're down this rat hole is Martin's apparent claim that the Bayesian analysis of the conditional situation is not necessarily predictive of the observed frequency. I think it's a rather bad Bayesian who would accept such a discrepancy. -- Rick Block (talk) 00:35, 28 February 2010 (UTC)

Tank you again, Rick, for really helping to clarify this aspect of "p". Regards, -- Gerhardvalentin (talk) 08:54, 28 February 2010 (UTC)

@Rick. I do no make the claim you say above. Please read what I say, it is still there. The Bayesian and frequentist approaches always agree, provided that you use the correct set-up to measure frequency. As Morgan el al point out it is not always easy to se what the correct set-up should be for a frequentist measurement. Martin Hogbin (talk) 09:42, 28 February 2010 (UTC)

Well then, what is your analysis of the experiment I've suggested? You said above it is "ill posed". We're interested in the conditional probability from the perspective of the player given the player picks door 1 and the host opens door 3 in a scenario where the car is uniformly placed but the host has a defined non-uniform, but unknown to the player, preference. The frequentist answer will be whatever value 1/(1+p) works out to. You haven't exactly said it (which I why I said "apparent claim" above), but I believe your "Bayesian" answer is 2/3. -- Rick Block (talk) 17:09, 28 February 2010 (UTC)

Sorry - I missed your reply above Gerhard's. So you agree we must either parameterize or assume random (per the principle of indifference). I agree these are the choices, however if you want to avoid surprises (like the experiment I've suggested) you have to parameterize. If you're going to assume indifference you should really say this is what your doing, which would then give you a hint of where to look if the experimental results and the predicted results don't match. In this case, by assuming indifference what you're doing is ignoring the information provided by the door the host opens. Do you agree with this? -- Rick Block (talk) 17:40, 28 February 2010 (UTC)

OK, let is agree to parameterise if a distribution is not stated in the problem statement. Your experiment is thus correctly set up for the question that you posed. But what about the MHP? Martin Hogbin (talk) 23:22, 28 February 2010 (UTC)

I thought we agreed a long time ago that the "standard form" of the MHP is the K&W version (initial car distribution and host choice are both explicitly uniform). The point of most of these threads seems to be whether considering the host's choice protocol in the solution is necessary, or (putting it another way) whether a simple solution that doesn't mention the host's choice and implicitly assumes it is uniform should be considered complete and correct. With regard to this thread, I think we're agreeing that the frequency of outcome (in the limit) and the Bayesian analysis should match. -- Rick Block (talk) 00:34, 1 March 2010 (UTC)

That frequency of outcome (in the limit) and the Bayesian analysis should match has never been in dispute, but only if the frequency of outcome is based on the correct set-up. This is not always so easy to do.

Yes the K&W is a convenient standard form of the problem but, as you say, the initial car distribution and host choice are both explicitly uniform so there is no need for any parameters. Bayesian analysis and frequentist approach (in which the car is replace uniformly at random and the host chooses uniformly at random each time) both give the answer of exactly 2/3.

You say above, 'The guest's probability of winning by switching is (not "could be") affected by the host's selection criteria. If this is not given to be random (p=1/2) you have no solid basis to say your probability of winning is 2/3' but you have quoted a problem statement where the host's selection policy is given to be random. What notable problem statement do you think requires parameterisation of the host's door policy? Martin Hogbin (talk) 10:10, 1 March 2010 (UTC)

The point is not that the host's selection must be parameterized, but that the conditional probability of winning by switching is always dependent on it. There are a variety of ways to show the 2/3 result that rely on the host selection between two goats being random, but if the solution says nothing at all about how the host picks between two goats the solution is not addressing the conditional probability. Most "simple" solutions address only the overall probability of winning, not the conditional probability in the case the player has picked door 1 and the host has opened door 3 - which is the case nearly anyone reading any version of the problem tries to solve.

I forget what you don't like about the instantaneous vs. average velocity analogy, but the situation is very much the same. A train goes from point A to point C which are 100km apart in an hour. What is the train's velocity at the midpoint B? [some versions say the train goes from A to C at a constant velocity, some don't]

Martin: The train goes 100km in an hour so the velocity is 100km/hr.

Rick: ??? The question is clearly asking for the instantaneous velocity, your answer is the average velocity and doesn't even mention point B.

Martin: <argues endlessly, never admitting that the answer must say something about point B and, if the problem statement doesn't say so, must be explicitly based on the assumption of constant velocity> -- Rick Block (talk) 18:51, 1 March 2010 (UTC)

Let us leave trains out of this. You say that that the conditional probability of winning by switching is always dependent the host's selection policy. Yes, but we are not told what this policy is so we must decide how to deal with that lack of information. One way to deal with this problem is to propose a door choice parameter q and then work out the problem using q as an unknown value. This is what Morgan do. If the host is defined to open a legal door uniformly then we do not need to use a parameter q, we can calculate the answer as 2/3, otherwise we need a parameter to represent the unknown policy. I thought we agreed all this.

My question was asking, for what notable problem statement should we use a parameter (such as q) to get a solution? It is not a trick question. Martin Hogbin (talk) 19:08, 1 March 2010 (UTC)

It's the Morgan et al. and Gillman interpretation of vos Savant's clarification (in her columns) of Whitaker's statement, as described in the variant section of the article. Rosenthal and others address this variant as well. -- Rick Block (talk) 19:36, 1 March 2010 (UTC)

Yes that would be it. That well-known, notable problem statement. Looks like we did not need the trains after all.

This is a complete statement of the required question: There is a game show in which there are 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 player chooses a door randomly and, after the player has chosen a door, the door remains closed for the time being. The game show host, Monty Hall, then has to open any one of the two remaining doors that hides a goat and ask the player to decide whether they want to stay with their original choice or to switch to the remaining door. The host has probability q of choosing door 3 when the car is behind door 1. Given only the above information and that the player has chosen door 1 and that the host has opened door 3. What is the probability of the player winning the car if they switch to door 2? Martin Hogbin (talk) 20:43, 1 March 2010 (UTC)

Would you consider it the same problem without the sentence The host has probability q of choosing door 3 when the car is behind door 1? -- Rick Block (talk) 21:53, 1 March 2010 (UTC)

Not necessarily. Without that statement there is the option of applying the principle of indifference to the host's legal door choice. I would say that, to make the only valid answer 1/(1+q), that statement is needed, otherwise 2/3 is a possible answer. Martin Hogbin (talk) 22:33, 1 March 2010 (UTC)

Do we have agreement?

I may be seen as being a little fussy by some but, given the highly contentious nature of the subject, I think it is justified to be a little pedantic. Do we (Nijdam and Rick) agree that, strictly speaking, the question to which 1/(1+q) is the only valid answer is that given in my statement above (q is defined as P(H=3|C=1)). Martin Hogbin (talk) 13:47, 2 March 2010 (UTC)

I agree 1/(1+q) is the only valid answer to your problem statement above, but I think you're also implying this is the only such problem statement with which I wouldn't agree. I'm not sure why you're so interested in pinning this down exactly - it's like saying the K&W problem statement is the only one for which 2/3 is the only valid answer. Most statements of the MHP are at least somewhat ambiguous. I think we've agreed the "normal" interpretation is consistent with the K&W explicit problem statement, and the answer is the probability of winning by switching is 2/3 whether you view the question as asking about the overall probability or the conditional probability for any pair of initial player pick and door the host opens (such as player picks door 1 and host opens door 3). IMO, we need to present a complete solution for the normal interpretation which means addressing BOTH of these questions. -- Rick Block (talk) 15:44, 2 March 2010 (UTC)

I can see no other problem statement for which the only possible correct answer is 1/(1+q) where q is (P(H=3|C=1). Can you suggest one?

I am trying to take things slowly agreeing as we go along. If you think that there is another unambiguous problem statement to which 1/(1+q) is the only possible valid answer please tell me what it is. Once we have agree this we can move on to consider other problem statements. Martin Hogbin (talk) 15:51, 2 March 2010 (UTC)

Per the extended discussion above, replace

The host has probability q of choosing door 3 when the car is behind door 1. Given only the above information and that the player has chosen door 1 and that the host has opened door 3. What is the probability of the player winning the car if they switch to door 2?

with

The host has probability q of choosing door 3 when the car is behind door 1 but this fact is not known to the contestant. Given only the above information and that the player has chosen door 1 and that the host has opened door 3. What is the probability of the player winning the car if they switch to door 2 from the perspective of the player?

You are right I do disagree strongly with this. I challenge you to find one other person (or reliable source for that matter) that finds this question is well-posed and agrees that the only possible answer to it is 1/(1+q). Martin Hogbin (talk) 16:36, 2 March 2010 (UTC)

or

The probability of the host choosing door 3 when the car is behind door 1 is not known. Given only the above information and that the player has chosen door 1 and that the host has opened door 3. What is the probability of the player winning the car if they switch to door 2?

This one is debatable. It certainly might be argued that is reasonable to apply the principle of indifference to this case. We have no reason suspect that the host prefers any particular door when the car is behind door 1. Alternatively we might imagine, as Morgan do, a population of hosts having values of q uniformly distributed from 0 to 1. In this case the answer is 2/3 (not ln(2) as Morgan claim).

and (I assume) now we disagree. So, now what? I would suggest that if the overall point here is to reach an agreement pertaining to the content of the article that we focus on what sources say (in the context of the ongoing mediation) rather than our own opinions about this (which, I'll remind you yet again, do not matter at all as far as editing is concerned). -- Rick Block (talk) 16:12, 2 March 2010 (UTC)

It is necessary for us to fully understand the subject and the various solutions and their applications to the MHP for us to edit this article. It is neither required nor desirable to just cut and past bits of reliable sources into the article. It should be the work of editors here, supported by reliable sources that those editors understand. Martin Hogbin (talk) 16:36, 2 March 2010 (UTC)

My impression is that you have this backwards, i.e. you want to start with an agreed POV and pick and choose reliable sources supporting this POV as opposed to understanding and neutrally phrasing what the sources say. The latter does not require us to agree. The former does. IMO, this has been your problem for over a year. You don't agree with what Morgan et al. say, so you want it marginalized (put in a section titled "academic extensions"). The POV exemplified by Morgan et al. (and shared by numerous others) is that the problem statement asks about the conditional probability and that the "simple" solutions address only the overall probability. This is not merely an academic point, but fundamental to the understanding of the problem (and its solution). -- Rick Block (talk) 18:11, 2 March 2010 (UTC)

For some reason you say that we need not understand the subject but you want to talk about what is fundamental to understanding the problem. What is fundamental to understanding the problem is the discussion above, which you seem to have abandoned. Martin Hogbin (talk) 19:14, 2 March 2010 (UTC)

What I said was we need to understand what the sources say. We don't need to agree with them. What I'm saying is you are focused way too much on WP:The Truth, and not enough on understanding and saying what the sources say. The fact is we have different sources that say different things. It is not our job to determine which one is correct, or even which one is more correct - other than by determining the prevalence of their views. The operative word is "prevalence", not "correctness". -- Rick Block (talk) 20:42, 2 March 2010 (UTC)

(for editing)

Martin: you often say: the answer is 2/3. You really have to free yourself from this. It is not the value 2/3 that counts, but where it is an answer to. Nijdam (talk) 21:00, 1 March 2010 (UTC)

Not to the question above, that is for sure. The answer to the question above is 1/(1+q). We all agree on that, I think. Martin Hogbin (talk) 21:08, 1 March 2010 (UTC)

Still, where is it an answer to?Nijdam (talk) 21:13, 1 March 2010 (UTC)

When the host chooses a legal door randomly. Martin Hogbin (talk) 23:19, 1 March 2010 (UTC)

Is this supposed to be a question?Nijdam (talk) 08:00, 2 March 2010 (UTC)

No, it is an answer. To say it in full: The probability that the player will win by switching is 2/3 if the host chooses a legal door randomly. Martin Hogbin (talk) 09:53, 2 March 2010 (UTC)

In that case the answer 2/3 is an answer to a question that is not the question of interest.Nijdam (talk) 11:40, 2 March 2010 (UTC)

Frequentist approach

Suppose you are a committed frequentist and you have the following question to answer:

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 No. 2?" What is the probability that you will win if you switch? (Sic)

As a frequentist, the question can be answered by repeating the experimental situation many times and counting what proportion of times you win by switching. The real problem is what exactly to repeat.

Let us start with 'Behind one door is a car; behind the others, goats'. When we place the car and goats each time how do we replace them? The question does not tell us, it might, for example, always be placed behind door 1. We can only do one of the following.

1. We could introduce a parameter x to represent the probability with which the car was is be placed behind door 1 and replace the car, non-uniformly, based on some chosen value of this parameter.
2. We could apply the principle of indifference and place the car uniformly at random behind the three doors.
3. If we are allowed to consider the real world, a good modern example to copy from would be 'Deal or No Deal'. In that show the prizes are clearly stated to be randomly put in the boxes. So we might place the car uniformly at random.

Now 'you pick a door'. How is the door picked? The player might always pick door 1. In fact this generally turns out not to be important but, we have the same choices as for the initial car placement. Whatever method we use we must only consider cases where door 1 has actually been picked.

Next, 'the host, who knows what's behind the doors, opens door 3'. We will take it from the wording that the host never reveals the car. How did the host choose door 3. He might always choose door 3 wherever possible or only choose it if he has to. We have the same choices as before.

1. We could introduce a parameter q to represent the probability with the host chooses door 3 when he legally can, and have a door opened according to this parameter.
2. We could apply the principle of indifference and open a door randomly when there is a choice of goat doors.
3. If we are allowed to consider the real world, a good modern example to copy from would be 'Who wants to be a millionaire'. In that show, when a player chooses 50/50, the computer randomly removes two wrong answers. That is what we must do here. Open a goat door randomly when there is a choice.

Whatever method we use we must only consider cases where door 3 has actually been opened.

So now to run our experiment. If we are consistent in our choices we get three possible results:

1. The answer will turn out to have some value from 0 to 1 depending on what values we chose for x and p.
2. The answer will be 2/3
3. The answer will be 2/3

If we for some bizarre reason choose options 2 or 3 when we replace the car, but choose option 1 when we open a door then we get the result 1/(1+q). But why would we do that? As a well-known peer-reviewed paper says, 'The modeling of conditional probabilities through repeated experimentation can be a difficult concept for the novice...'

Can I assume that nobody disputes that the experimental set-ups described above correctly represent the specified interpretations of the problem statement at the top of this section. Martin Hogbin (talk) 10:17, 1 March 2010 (UTC)

The 'c' word

Before I say anything more, what do you mean by "the answer"?Nijdam (talk) 10:42, 26 February 2010 (UTC)

The probability of winning by switching. Martin Hogbin (talk) 12:48, 26 February 2010 (UTC)

You have to be more specific.Nijdam (talk) 16:36, 26 February 2010 (UTC)

The question is given at the top of this section, I want to know, 'What is the probability that you will win if you switch?'. Martin Hogbin (talk) 17:03, 26 February 2010 (UTC)

If it helps you respond, I can ask this question. What is the probability that the car is behind door 1 given that the player has chosen door 1 and the host has opened door 3? Martin Hogbin (talk) 10:19, 27 February 2010 (UTC)

Now we're talking. Doesn't it strikes you as odd that you ask for a conditional probability?Nijdam (talk) 11:23, 27 February 2010 (UTC)

I am asking for a conditional probability because I include the possibility that the car may not be initially placed randomly and the host may not choose a legal door randomly. In this case the problem is conditional as it clearly might matter which door the player initially chooses and which door the host opens. Let us continue to discuss the subject of conditionality on your page.

Now given that we agree on the question do you agree with my answer above? Martin Hogbin (talk) 12:59, 27 February 2010 (UTC)

Nijdam, you seem reluctant to comment. Do you see an error in what I say above? Martin Hogbin (talk) 09:46, 28 February 2010 (UTC)

I'm not reluctant, just not every moment present here. On the other hand I don't see any sense in stating that the numerical values you mention are okay. They are, but what's important is what they represent. And we have already concluded they represent conditional probabilities. That's the only thing I want to hear. Nijdam (talk) 10:52, 28 February 2010 (UTC)

"Conditional" without any sufficient "condition": Door 1 was selected. Now door 3 opened OR door 2 opened (there's no third possibility). Just "assuming" some "might be possible" condition, not knowing anything about such special condition, not telling anything particular, just enlarging the scope. Curious. Quite another issue was reliably knowing the host's preference and accounting for. But that was not the simple standard MHP any more.
The host's selection policy HAS to be random (p=1/2) in the simple standard MHP. Moreover: The idea that the host's selection policy is NOT given to be random must be clearly stated to be a malignant distortion. Then it is precisely not the paradox, but an entirely different problem, misleading from the pure "Monty-Hall-paradox" to the "Monty-Hall-paradox-problem-problem-problem". And this should clearly be named so. Gerhardvalentin (talk) 17:23, 28 February 2010 (UTC) Gerhardvalentin (talk) 13:57, 1 March 2010 (UTC)

Gerhad, I understand your frustration with this subject. It is all due to a neat little conjuring trick by Morgan et al. It took me a while to work out how they did it, but I can show you exactly how the trick is done. Martin Hogbin (talk) 15:26, 1 March 2010 (UTC)

Sorry, Nijdam. I appreciate that nobody can be expected to answer every question immediately.

I am not that fussed about conditional probability. Let me restate my position here.

1. Every probability problem can be expressed as a conditional problem.
2. What I call the 'academic' MHP, in which the car is initially placed uniformly at random but it is known that the host may open a legal door non-randomly is essentially a problem of conditional probability. As far as I know this is the only way to solve this version of the problem.
3. The 'standard' MHP in which the car is initially placed uniformly at random and the host opens a door uniformly at random (selected from the choice available under the rules) is a special case of the academic case and thus may clearly be dealt with using conditional probability. If you still want to call it a conditional problem that is fine with me.
4. In the 'standard' MHP we can apply either the principle of symmetry, or use information theory to show that the number of the door opened by the host (even if this number is stated in the problem statement) does not affect the probability of winning by switching. If you want to call this conditional that is fine with me also. I would propose the non-standard term 'null-condition' to describe this case but I do not insist that anyone else does or that we do so in WP.
5. Simple solutions, published in reliable sources, to the 'standard' MHP that do not consider which door the host might open, should not be described as wrong or even incomplete. Again, I do not care that much whether the word 'conditional' is used so long as there is no implication, in the 'standard' case, that it might make a difference which door the host opens. We agree that it does not, in fact, matter.

I guess we will agree about 1,2 and 3. We can perhaps agree to differ over terminology for 4. What is your view on 5? Martin Hogbin (talk) 12:20, 28 February 2010 (UTC)

Having moved on from the use of the word 'conditional', perhaps you could confirm that the experimental set-ups described above correctly reflect the stated problem conditions. Martin Hogbin (talk) 12:20, 28 February 2010 (UTC)

(1) You're missing permanently what our discussion is about. The issue at stake is the difference between probability before (may be called unconditional), and after (conditional).

(2) Okay.

(3) It is also essentially about conditional probabilities.

(4) As I explained, not the method of calculation is important, but what is to be calculated is. If the problem is considered a probability problem, it is about a conditional probability. And it's not a good idea to introduce a term of your own.

(5) Me nor Rick insist on the use of the word conditional as we said several times. But we definitely insist on making the difference about before and after. Simple solutions to the standard MHP are wrong, it is not in my power to say they are not. I do not have a clear idea about the meaning of sources (or solutions) that do not consider which door the host might open.

About the experimental set-ups: do you mean what you call the academic and standard versions? Nijdam (talk) 17:48, 28 February 2010 (UTC)

BTW, Martin, it's about time you see the light. I think you're almost there. Study what I wrote on the "combined doors solution", and see why it is wrong. It would be much better if you spend your energy on joining, than on fighting us.

(1) No one is suggesting that we consider the probability before the host opens a door.

Oh no? And what about the random placement of the car? Nijdam (talk) 17:11, 1 March 2010 (UTC)

(3) If you like

It's not whay I like, but what it is.Nijdam (talk) 17:11, 1 March 2010 (UTC)

(4) I do not plan to use any non-standard terms in the article.

Better not use them at all.

(5) This is where we have the problem. Simple solutions give the correct answer to the 'standard' MHP using methods which are correct, even though this may not be that obvious at first sight. A conditional approach may be the better one but it is not the only one.

Too bad for you.Nijdam (talk) 17:11, 1 March 2010 (UTC)

Perhaps you could now confirm that the experimental set-ups described above correctly reflect the stated problem conditions. The exact problem is stated at the top of the section. Martin Hogbin (talk) 23:03, 28 February 2010 (UTC)

Re point #5: we need to say what reliable sources say in an NPOV fashion. Plenty of sources say the simple solutions don't quite address the problem, so we need to say this. It is POV to omit this or push it off into an "academic only" section (or, worse, appendix). I believe this is the crux of our disagreement. -- Rick Block (talk) 19:06, 28 February 2010 (UTC)

It is POV to allow some sources to veto others. We should state what the sources that give simple solutions say about their solutions. None of them say that their solutions are false or incomplete. After we have dealt with this matter we can say what other sources, such as Morgan, say about the solution and the solutions of other sources. ~~

Frequency

Guest pics a door randomly, let's call it "door 1". Host preferes to open always the same other door, let's call it "door 3":
 
#events:  Guest's choice:     Guest denied following unselected pair of two doors                switching hurts     switching wins

 33.334   door 1  car         door 2 goat never shown           door 3 goat shown 33.334            33.334 times            0 times
 33.333   door 1  goat        door 2 car  never shown           door 3 goat shown 33.333                 0 times       33.333 times
 33.333   door 1  goat        door 2 goat shown 33.333          door 3 car  never shown                  0 times       33.333 times

 66.667 events door 1 selected, door 3 opened   chances stay : switch  1/2 : 1/2                    33.334 times       33.333 times
 33.333 events door 1 selected, door 2 opened   chances stay : switch  0/3 : 3/3                         0 times       33.333 times
100.000 events door 1 selected  in total        chances stay : switch  1/3 : 2/3                    33.334 times       66.666 times

Host always opens door 3 if ever possible, i.e. if he has a choice between two goats (only if the Car was selected and switching means total loss with zero chance)

Situation BEFORE the host opens a door					Situation AFTER the host opens a door
Door 1 picked	Unpicked door 2	Unpicked door 3	result if switching	total cases	cases if host opens Door 2	cases if host opens Door 3
Car	Goat	Goat	Goat	100	0	100 Goat
Goat	Car	Goat	Car	100	0	100 Car
Goat	Goat	Car	Car	100	100 Car	0

But: Nobody ever did provide any proof of the host's preference, so that's no evidence of reality.
In case the host should roll a die the guest better was to stay  :-))
-- Gerhardvalentin (talk) 17:46, 24 February 2010 (UTC)

The flip side is nobody provided proof of the host's indifference between the two goat doors, either. -- Rick Block (talk) 18:03, 24 February 2010 (UTC)

That rules out any speculation about any indifference of the host and about any preference of the host. Such speculation is proven to be obsolete and never is part of the paradox. Not subject of the "dilemma" caused by the famous paradox. Such speculations are quite another issue, having nothing at all to do with the evident paradox. A hair-raising marginal historical phenomenon only. Regards,  -- Gerhardvalentin (talk) 18:30, 24 February 2010 (UTC)

Selvin said the host picks randomly in his second letter to The American Statistician in 1975. That's the same source Morgan is published in 16 years later. Glkanter (talk) 18:32, 24 February 2010 (UTC)

What we're talking about here is not whether the problem says the host picks non-randomly, but whether it's important to consider how the host picks when determining the probability of winning by switching. The probability is 2/3 only if the problem says (or you assume) the host picks randomly. -- Rick Block (talk) 01:28, 25 February 2010 (UTC)

6 Plays At 1 Time

The host chooses randomly when faced with 2 goats, as per Selvin.

6 contestants randomly pick each of doors 1 - 3. It turns out each door is selected twice. The host opens different doors for each door # picked.

We are now facing all 6 possible outcomes of contestant selection and host revealing a goat, including Whitaker's selecting door 1 and the host revealing door 3.

Is the contestant in any of these 6 'conditions' more likely to know the location of the car than any of the others? No. And that's why the various Omnicondition Simple solutions answer the appropriate question. Glkanter (talk) 17:27, 27 February 2010 (UTC)

Help in understanding

I'm almost fed up with this discussion, but I'll give it another try with this example.

The possessions of a married couple are described by

m for the male

f for the female

The amount of real estate in the possessions are

rm for the male

rf for the female

For the male the real state part in his possessions is: rm/m, for the female: rf/f and in their total possessions: (rm+rf)/(m+f). They concern different quantities, and with prenuptial agreement the values in general differ. Even when coincidently the value of rm/m equals rf/f (rm/m=rf/f), one wouldn't consider the fractions to represent the same thing.

Now what about a couple married in community of property? Of course as a consequence the values of rm/m, rf/f and (rm+rf)/(m+f) are the same. Yet I wouldn't call them the same "thing". I.e. if someone asks for (the value of) rf/f, it is possible to calculate (rm+rf)/2m, but without mentioning that this gives the right value because of the community of property, I would consider this (as a way of reasoning) wrong.Nijdam (talk) 08:27, 2 March 2010 (UTC)

Firstly, thank you for your patience. I am finding it hard to work out exactly where we disagree but I can assure you that I am not being awkward, it is just that I cannot understand your point.

In your example, I understand that are referring to two different things. In the MHP, I fully understand that: the probability that the car is behind door 1 before the host opens a door, the probability that the car is behind door after the host has opened door 3, and the probability that the car is behind door 1 after the host has opened door 2 are all different things. But consider this analogy. Suppose we have a triangle. It has three sides, all of which are different things, we could make them different colours or number them. Now suppose that we are told that all three angles are the same. We can immediately say, by symmetry, and with no knowledge of geometry, that the three sides must be of equal length. Even though we may have numbered the sides, they now become essentially equal even though we may not know the length of any of them.

It is just the same in the MHP. Once we are told that the host chooses a legal door randomly, we know that the probabilities described above must be equal, without calculating any of them. Thus, in order to calculate any one of them, we may calculate any other. This approach is very common in mathematics and has solved many problems that otherwise might be intractable. Martin Hogbin (talk) 10:50, 2 March 2010 (UTC)

Again, it's not about the calculation, but about what has to be calculated. In your example, we know all sides are equal, but if you need to know the size of let's say AB, you have to calculate AB, and it is not sufficient to say BC = 10. You need to add: AB=BC, and because BC=10, also AC=10.Nijdam (talk) 11:26, 2 March 2010 (UTC)

Nijdam, you must have seen this technique used before. We want to know side AB. We first show that AB=BC=CA by a simple symmetry argument, using the fact that were told that the angles are equal. We can do this without knowing the length of any side. Next we calculate BC (maybe in the given question this is easier than calculating AB). Finally we state that length of AB. Martin Hogbin (talk) 11:58, 2 March 2010 (UTC)

Do you say anything different than what I say?Nijdam (talk) 12:22, 2 March 2010 (UTC)

I am not sure, but if you are happy with what I wrote above, why are you not happy with this? We wish to calculate the probability that the car is behind door 1 after the host has opened door 3, that is to say P(C=1|D=3), a conditional probability. (There I have said the word). If the host chooses a legal door randomly we can say by simple symmetry, without any conditional probability calculation, that P(C=1|D=3) = P(C=1|D=2) = P(C=3). We know P(C=1) = 1/3, thus P(C=1|D=3) = 1/3. Yes we have calculated a conditional probability but without worrying about which door the host has opened, because we know it makes no difference. Martin Hogbin (talk) 13:44, 2 March 2010 (UTC)

Shorter proof of conditional result

Looking back at the original article I find it not so bad at the moment, the only thing that annoys me is the huge section on the Bayesian approach. If one wants to find the conditional probability (under the extra assumptions of uniform randomness which pedantic teachers of Probability 101 add to the statement of the problem so that dull students can solve it routinely with the tools given in the course) then it is much more elegant to do it with Bayes theorem in the "posterior odds = prior odds times likelihood" version.

I suppose the player has picked door 1. I suppose that indepedently of the player's choice, the car is uniformly hidden behind the three possible doors. Suppose the quizmaster opens a door uniformly at random when he has a chance. Let C denote the random doornumber hiding the car, let Q denote the doornumber opened by the quizmaster

Prior odds:

Prob(C=1) : Prob(C=2) : Prob(C=3) = 1 : 1 : 1

Likelihood of location of car C=1,2,3, given data Q=3

Prob(Q=3|C=1) : Prob(Q=3|C=2) : Prob(Q=3|C=3) = 1/2 : 1: 0

Posterior odds:

Prob(C=1|Q=3) : Prob(C=2|Q=3) : Prob(C=3|Q=3) = 1/2 : 1 : 0

The posterior probabilities are therefore 1/3, 2/3, 0 since they must add to 1 and be in the just mentioned ratios.

This computation is easy to generatlize to the situation where the quiz-master may use other probabilities. As long as the initial distribution of the location of the car remains uniform, the posterior odds are

Prob(C=1|Q=3) : Prob(C=2|Q=3) : Prob(C=3|Q=3) = Prob(Q=3|C=1) : 1 : 0

Since Prob(Q=3|C=1) can conceivably be anything between 0 and 1, the posterior odds on the location of the car being door 1 to door 2 is anything between 0 : 1 to 1 : 1. All we can say is "it never hurts to switch".

Gill110951 (talk) 10:25, 2 March 2010 (UTC)

I had the same objections. Nijdam (talk) 11:29, 2 March 2010 (UTC)

Gill, is this correct:
In situation where the quiz-master may use other probabilities but the initial distribution of the location of the car remains uniform, the posterior probabilities are
Prob(C=1|Q=3) : Prob(C=2|Q=3) : Prob(C=3|Q=3)   =   [anything from 0 to 1/2],   [anything from 1 to 1/2],   0   (they always add to 1).
Okay? -- Gerhardvalentin (talk) 14:48, 2 March 2010 (UTC)