Talk:Monty Hall problem/Arguments/Archive 11

The conditional probability proof for Martin's problem does not answer the question asked

Martin laid out a puzzle (see above) and then asked this question: "What is the probability that from that same urn the next door ball you draw will be white?" And in regards to this, another editor posted a formal probability calculation which he says affirms Martin's assertion. But neither of them make any sense for the simple reason that the first pick from the urn does not affect the second pick, other than to finalize the mix of ball colors. To suggest otherwise is akin to the Gambler's fallacy. At any given moment, if there are one black and one white ball in a urn and you randomly pick one of them, the odds of getting either one is 1:2 (or 0, if you are picking from the BB urn). To suggest otherwise is like a person who tries to outguess a flipping coin. The hand reaching into the urn is a distinct random event. Nothing which comes before that pick affects the actual pick itself in any way. The only thing a previous pick does is reduce the total choices down to two. But when it's at two, the answer to the question "What is the probability that from that same urn the next door ball you draw will be white?" is 1:2 (or 0). I think that the reason why Martin's puzzle is false, is that the contents of the two urns form two distinct sample spaces such that the richness of the W population per space can help you calculate which space you are in, but once you've removed that 1st ball, there are only two left per urn, and the positions of the balls inside the urn are fungible - they are in flux. For that reason, there's nothing which carries over any information from the 1st pick. Conditional probability can't be used for this puzzle. After removing a W ball from the urns, you are left with BB and WB - just like I said above. The next pick into either of these sets can not be 2:3 - there are not enough remaining positions in either set to yield that. Once a ball is removed from the urn, it does not effect the odds of the next pick, except, as I said, to set the mix as fixed as either BB or WB. Tweedledee2011 (talk) 05:08, 17 January 2014 (UTC)

Time to put your money where you mouth is

According to you, if we were to play the urn game that I suggested. You would win half the time. I will offer you twice your bet if you win, which should be a good deal for you. If you want to play let me know, otherwise I have nothing more to say.

Several users have tried to explain probability to you but you have resisted all attempts. The fact that no one has agreed with you must surely open in your mind the possibility that it is you who is wrong. The only way you will be convinced is by a demonstration in which you lose money. If you want to do this let me know, otherwise I will not respond. Martin Hogbin (talk) 10:06, 17 January 2014 (UTC)

Martin, your phrasing says "remove a ball". In your problem, is the ball returned to the urn after seeing what color it is? If so, the word "removed" is misleading. Tweedledee2011 (talk) 16:12, 17 January 2014 (UTC)
No, it is not returned. Do you want to play? Martin Hogbin (talk) 17:38, 17 January 2014 (UTC)
True or false - If we flip a perfect coin three times, there are 2x2x2 (8) possible outcomes
True or false - There is only a 1/8 chance that of three flips, three in a row will be heads
True or false - The odds of a 4th flip being heads is 1/2
True or false - The odds of the 3rd flip being heads was 1/2
True or false - The cumulative odds required to get to the 1/8 on the 3rd flip is not the same as the 1/2 chance on each flip
True or false - The 2/3 you are talking about, is akin to the cumulative odds of the 1/8 chance of three heads in a row
If all true, then please explain how following your decision tree and getting 2/3 odds differs from a stranger walking in, picking from your urn and getting 1/2 odds?
I ask in all seriousness because by your example, you could sequentially place all of the available W and B balls in the world into a series of urn #2's (while using up some to populate the urn #1's) and even though the distribution of balls in those #2-modeled urns would be WB (one each), because you set them all in place via the scenario of this problem, your next choice from any of those urn #2's would have a 2/3 chance of being white. But if anyone else in the world were to draw a ball from any of your #2 urns, they would face 1/2 odds of a white. And remember, all the white and black balls in the world are used up by populating urns (in decreasing quantities, by using the removed balls) and your next draw must be from a #2 modeled urn which arrived at a WB configuration because you personally removed a W from it. Please reply to this, before asking me to answer. Tweedledee2011 (talk) 19:49, 17 January 2014 (UTC)
Sorry, I am not interested in trying to explain things to you. I am happy to clarify exactly what the game is so that there are no misunderstandings and we can play the game for real money. Do you want to put your money where your mouth is? Martin Hogbin (talk) 21:02, 17 January 2014 (UTC)
Provided that the computer code we use to test is valid, I am happy to take your money. And, since you are such a good sport giving me odds and such, let's start by you paying off on the first test I ran manually (see below) Tweedledee2011 (talk) 00:24, 18 January 2014 (UTC)
This illustrates sample space size reduction from ball removal

Which urn was it?

Tweedledee, consider that if you knew which urn it was then the answer would be different. For example, if you peek in the urn before picking the second ball, and see that it contains one of each color then, knowing this, you would say the probability of picking a second white ball is 1/2.

Without cheating like this, we don't know which urn it was, not definitely. However, the color of the first ball gives us some evidence about which urn it was. So consider this question about what the evidence tells us:

• Given the evidence of having picked a white ball, what is the probability that this is the urn that started out with more white balls in the first place?
— The answer to this elementary probability question is simple: it is 2/3. If you are not sure how to calculate this then please ask.

The point of calculating this probability is this:  Yes, the mix is either BB or WB but, given the evidence, it is twice as likely to be WB. It is not correct to say "Nothing which comes before that pick affects the actual pick itself in any way": it affects which urn is actually being picked from. This is the whole point of the question asked in Martin's problem. ~ Ningauble (talk) 14:07, 17 January 2014 (UTC)

@Ningauble I do not deny that a pick from a pool of WWB is more apt to yield a W than a pick from a pool of WBB. But you are misquoting me by omitting my proviso. Here is what I am saying: At the point with any urn, where there is one W an on B in it only, the next pick is a discrete random event and as such, is 1:2. There is no difference between two urns identically configured with one W and one B. But for your explanation to be true, if just before the 2nd pick, you took the original urn #2, which now has one W and one B in it and you substituted an urn #3, which has one W and one B in it -even though the urns are identically configured, the results of pick #2 would have different probabilities between those urns. But that's nonsense. Also, the question says that on the fist pick we are to "remove a ball", so you do agree that the first ball is taken out and not returned to the urn, yes?. Also, if Martin is referring to problem, the way he stated it is not the same Tweedledee2011 (talk) 15:32, 17 January 2014 (UTC)
There are several problems with this:
1. I was not saying that a pick from WWB is more apt to yield a W than a pick from WBB, which is true. What I said is that a result of W is more likely to have come from the WWB urn than from the WBB urn. (Please re-read the bulleted question in my post.)
2. My explanation does not imply anything about any third urn. Martin's problem refers to picking a second ball "from that same urn", and that is the urn to which I refer. (Please re-read Martin's problem.)
3. It is an urn with a history, the urn that was first selected by opening a door. The Gambler's fallacy, to which you refer in opening the top level thread, applies to independent events. It is not relevant because selecting balls from the urn depends on which urn was selected. (Please read the "Non-examples of the fallacy" section of the article to which you linked.)
4. Also, I did not misquote you. The post from which I quoted contains no "proviso" indicating your words pertain to a different situation than the one posed in Martin's problem. (Please re-read your post.)
Martin's problem is much simpler than the Monty Hall problem, because it does not involve any "sneaky" selective information. It is a problem of pure random sampling where it is easy to make a statistical inference about the population (urn) from which the sample is drawn. ~ Ningauble (talk) 20:29, 17 January 2014 (UTC)
But the doors are irrelevant. What this problem is saying is that from any pool of 2-1 (2 same, 1 different) of items, the chances of picking two same items in a row are 2/3. That's nonsense because once you remove the first ball, the remaining pool is 1-1. And the next pick is a fully random and distinct pick. Tweedledee2011 (talk) 20:55, 17 January 2014 (UTC)
I already demonstrated in the opening sentence of this sub-thread a full and sufficient reason why it matters which urn is picked: because "if you knew which urn it was then the answer would be different". This goes to the very essence of what it means to say statistical events are dependent. Since you don't seem to understand the concept, let me explain it another way:
1) The door determines which urn is used.
2) The contents of the urns are different: one contains a second white ball and the other does not.
3) Therefore the door is not irrelevant.
Quod erat demonstrandum.
Once you remove the first ball you still don't know which urn was picked, but you can make a statistical inference about the probabilities for it being one urn or the other. ~ Ningauble (talk) 14:15, 18 January 2014 (UTC)
I have no argument with what you are saying. My hang-up was two-fold 1) My sloppy proof of .25 is wrong and b) the original "correct proof" done by Nijdam was even less accurate than mine - he originally claimed it was 2/3. It was that claim of 2/3 which was simply too erroneous for my math-enfeebled mind to accept. And it turns out, my (correct) rejection of 2/3 is what spurred the dialog onward and led to Nijdam finding his error. Even so, I have pulled my math book off the shelf and am re-thinking how best to refrain from trying to intuitively solve these problems. Also, perhaps some of the spatial recognition I used would have helped Nijdam to see his error sooner. Perhaps asking oneself "how would this look" before concluding the math is right, would help. Turns out, I did too much "how would this look" and perhaps Nijdam did not do enough. Tweedledee2011 (talk) 16:19, 18 January 2014 (UTC)

Question for Nijdam

Nijdam - In my manual test, there were 48 picks and of that; 23 were 1st pick W. That's .47 (almost 50%) 1st pick W, which is correct because we start with 3 of 6 balls being W. Next; of those 23, 14 were in urn #2, which is .61, which is also correct because it's supposed to that 2 of 3 come from urn #2. And it's supposed to be that 2 of 3 first pick W's come from the #2 urn because 2 of the 3 W's which exist in the game, are in urn #2. So .61 is close enough, given the number of iterations I ran. Now, as best as I can tell; we can correctly say that if you get a W on a first pick, the odds are 2/3 that you are picking from urn #2. That said, Martin's question starts at this point (that our first pick was a W) and it asks "What is the probability that from that same urn the next ball you draw will be white?". What I am concerned about is this: I think your 1/3 answer (which I agree with) would not be correct if the balls were not removed. I am thinking, (am I wrong?), that if the selected ball is only examined for color (at the first pick) and returned to the urn, that the odds for this are different than with removal. I say this because for the 2nd pick, if it came from urn #1, the probability of a W from a 2nd pick in urn #1 is no longer 0, because the W was not removed. And from the urn #2, the probability of getting a W when the mix is WWB, is greater than when the mix has been reduced to WB. So my question is, would you be so kind as to show the correct calculation for both ways - ball removed and ball returned? Tweedledee2011 (talk) 05:00, 19 January 2014 (UTC)

When drawn balls are replaced the calculation is:
P(W2|W1)=P(W2W1)/P(W1)=[P(W2W1|urn1)P(urn1)+P(W2W1|urn2)P(urn2)]/(1/2) = [2/3x2/3x1/2+1/3x1/3x1/2]/(1/2)=5/9. Nijdam (talk) 11:32, 19 January 2014 (UTC)

In light of the urn problem

I would like to propose that it's indeed beneficial to this talk board to try to talk through the various feeble arguments posted by the weaker math people (such as myself) rather than simply quickly disprove them. And that's what I want to do with MHP. I want help to line-up the correct math such that the gaps between the correct math and the wrongly applied verbal reasoning can be ferreted out. And I think the best place to start is to clarify exactly how big the sample space would be from the perspective of the player, at discrete points in time in the game. The first point in time I want to examine is after the player has made his first choice, but before a door is opened and before he's given a 2nd choice. And FYI: I do agree that switch is better and yields 2/3 wins. But what I am seeking is a better way to explain it to novices. But since my math is weaker than others here, I need collaborative dialog, not rebuttals. Tweedledee2011 (talk) 16:27, 18 January 2014 (UTC)

OK, so let us start with what you would consider the full sample space before the player chooses his first door. Martin Hogbin (talk) 00:55, 19 January 2014 (UTC)
The word 'guess' is a complete irrelevance. Use of that word in describing the problem is misleading and should be avoided. It is not used in any of the well known problem statements.
Whenever you set up a sample space part way through a problem you have to be very careful. For example if you decided to set up a sample space after the host had opened a door you would probably end up answering a different problem (vos Savant's little green man problem). I therefore think that to really answer this problem properly the sample space should be set up at the start, where the player has a choice of three doors.
That being said, with the usual setup and assumptions, there is an obvious symmetry with respect to the player's initial door choice. In other words the player's initial door choice makes no difference to the problem. Traditionally, therefore, the problem is often addressed that way. So, on the basis of the standard setup and assumptions (please say if you are not clear about these), what is your sample space for a player who has just chosen door 1? (I will not be able to respond before Monday evening.) Martin Hogbin (talk) 09:25, 19 January 2014 (UTC)
No, what I am saying is that a player on the game, when he first picks a door, must choose from (1) door from only (1) of (3) possible 3-door configurations - but he doesn't know which configuration he's facing. If the car is at door #1, then by the vernacular of the .png diagrams I posted, the player is facing "layout" #1. And if the car is at #2, then it's layout #2, etc. If there's only three possible places the car can be, then there's only three possible layouts. Quoting the Sample space article: In probability theory, the sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment. Now without getting hung up on the word "outcome", look at the other included meaning: "result" (of an experiment). To a player on the show, until he sees a door open and gets a 2nd choice, he thinks that his "experiment" is to choose a door - and he does so, anticipating it will be opened. However, the result of his experiment is not immediately known because the host throws a curve ball. But until that happens, from the player's perspective, there are only three possible layouts and it's irrelevant to the player which one he's facing. Because no matter what, until a door is opened and a 2nd choice is given, the odds a player faces are static, at 1/3. If a player chooses a door, the player expects that door will be opened, and he hopes that of the three, he's picked the right one. Spread across the three possible layouts there are three cars and six goats. So at that point, there are 9 possibilities. And even though only one layout is actually on the stage, the possibility exists that it could have been any of the three possible layouts. So, it's 3x3, or 9. Tweedledee2011 (talk) 08:11, 20 January 2014 (UTC)
On the basis of the standard setup and assumptions (please say if you are not clear about these), what is your sample space for a player who has just chosen door 1? Or, if you prefer, show your sample space for some other stage in the game. Martin Hogbin (talk) 19:05, 20 January 2014 (UTC)
If the player has chosen a door, and if the host has not yet opened a door or offered a 2nd pick; then it's a 3x3 matrix of possibilities - from the perspective of the player. But not from our vantage point, because unlike the player, we know what happens next. What I want to discuss is the view from the player's perspective. From that vantage point, there are 3x3 possibilities (for the brief moment just after the 1st pick, but just prior to the host opening a door and offering 2nd choice). Tweedledee2011 (talk) 20:17, 20 January 2014 (UTC)

Verbal solution for MHP

Think of the 2-door set of not-chosen-at-first doors as two shoe boxes containing between them 67 cents. The first door you chose has 33 cents (which is 1/3), and the 2-door set has 67 cent between them (which is 2/3). But within the un-chosen 2-box set, the 67 cents is completely contained in one of the two boxes. And when one of the two boxes is opened and proves empty, the 67 cents doesn't partially migrate to the first-choice 33 cent box. Rather, it stays entirely where it was. As a result, it holds 67 cents and is more valuable than the first choice door. So we are certainly better off to choose it. Tweedledee2011 (talk) 22:26, 20 January 2014 (UTC)

MHP solved with verbal reasoning

If you are a player on this game show, your objective will be to win the car. And the sole means by which you can win the car is to choose the correct door. But when the game starts, because you have no information about where the car is, the best you can do is guess. Therefore, with one car and three doors, your best 1st pick will be a blind guess of one in three, or 1/3. However, as soon as you pick a door, the host then opens another door (a losing one) which leaves only two doors. And on top of this, the host then gives you a 2nd choice of “stay or switch”. If you “stay” you will stand pat with your original door selection. If you “switch” you will abandon your original choice, and instead choose the last remaining door. The math of this problem proves why it’s better to switch, but this can also be explained verbally, as follows:

When you first pick a door, regardless of what happens later, your odds for that first pick begin at, and stay at 1/3. That’s because the car does not move and only with only one car in three doors, only 1/3 of your first picks can be a correct guess. However, once the host opens a door, you are no longer merely guessing, because the open door provides you with information. The information provided is that the open door does not have the car. Therefore, here is what you can deduce:
1. Your first pick will be a guess that has a 1/3 certainty of being correct.
2. Between them, the other two doors hold double that amount, or a 2/3 certainty of being correct.
3. When one of the unpicked doors is opened, half of the doubt from the 2-door set of unpicked doors is removed.
4. Of the 2-door set you did not initially choose, the certainty was 2/3 and the doubt was 1/3. But with 50% of that doubt gone, the doubt for the remaining door is only 1/6. And that’s the funny thing about doubt – when it’s exposed, it goes away.
5. When a door was opened, the doubt it held leaves the game. Perhaps on some level it still exists, but because you can’t accidentally choose it, as far as you are concerned, it's gone. So of the original doubt, 2/3 of it stays with your first pick, 1/6 leaves the game and 1/6 is assigned to the remaining door which you did not pick.
6. Because the car does not move, the doubt for your original choice remains at 2/3. That's because even though the total remaining doubt between two doors is 50%; that 50% of doubt is not re-assigned equally to the two remaining doors.
7. The reason for this is simple: The information was developed asymmetrically. No new information was learned about your first pick. But new information was learned about the 2-door set which you did not initially pick from.
8. Instead of thinking about the cars to doors ratio, think about the information sets which develop during the game.
9. Wherever the car is at the start is where it will still be at the end. But by removing doubt, you can get closer to where it actually is.
10. To get away from doubt, it’s better to switch to the door with less of it. And in this situation, the "switch to" door not only has less doubt, but it also has more certainty. That's because even though some doubt left the game, all of the certainty of the 2-door set was was retained. Because we are certain that the car did not leave the game, we can be 2/3 certain that it's behind the "switch to" door. And that's because the state of, and information about, our original door do not change. The new information improves the odds of the "switch to" door.

Tweedledee2011 (talk) 22:28, 20 January 2014 (UTC)

As I said, everybody has their own pet way of seeing the solution. I cannot see this getting into the article though, it is long, complicated, contains non-standard terminology and errors, and it is not supported by any reliable source. Martin Hogbin (talk) 23:00, 20 January 2014 (UTC)
Martin -Everything with you is a naysay. Did I say I wanted this particular text in the article? No. What I said was, this is one way of reasoning it through. I posted another one directly above it. Honestly, everything with you is an argument. Tweedledee2011 (talk) 03:19, 21 January 2014 (UTC)
This line of reasoning is essentially the same as what is presented in the Solutions section of the article under the headings of Carlton and, more fully, under Adams and Devlin. The verbal explanation is accompanied by an illustration for the benefit of readers who are more visually oriented. ~ Ningauble (talk) 15:28, 21 January 2014 (UTC)

Martin's problem is wrong because...

Any initially picked W balls are removed from the sample space after the 1st pick. This means that 1/6 of the time a first pick is made, the entire #1 urn is no longer part of the puzzle as it can't possibly yield 2 whites in a row. And, 1/3 of the time a first pick is made, the possible choices of the 2nd urn are reduced from WWB to WB. This change in the original sample space must be taken into account on the 2nd pick. The proof provided above by Nijdam only works if the W balls picked are looked at to see the color, but not removed from the mix (reinserted into the urn). 50% of the time a 1st pick is made, the test fails because B comes of first and 50% of the time (1/6 + 1/3) the sample space changes (because the mix of available balls is changed by the removed W's) and that must be taken into account Tweedledee2011 (talk) 00:11, 18 January 2014 (UTC)

No, the problem is not wrong, but some if the answers being discussed definitely are. In the problem, as clearly laid out by Martin, you first pick one of the urns. It might be the one with two black balls. You can not un-pick it later. ~ Ningauble (talk) 15:08, 18 January 2014 (UTC)
Since the time I posted this, Nijdam has corrected his original false proof of the answer being 2/3. That was what this argument was all about. I agree it's 1/3 and I agree that my rebuttal was sloppy and in fact wrong. But the others here also need to accept that I was right when I said it's not 2/3 - which is what the issue was. And frankly, my intuitive answer of .25 is indeed a lot closer to the true answer of 1/3, than the original false answer of 2/3 - so the naysayers here should put that in the pipe and smoke it. This entire dialog adds weight to my original point which was: Intuitive thinking isn't as bad as some make it out to be. Rather that being mocked, it should be bolstered with some probability math training. In fact, it was only me here who recognized that the answer simply could not be 2/3 - but it was my weaker math skills which stopped me from proving correctly. Tweedledee2011 (talk) 16:07, 18 January 2014 (UTC)
'...it was only me here who recognized that the answer simply could not be 2/3 '? Come off it. Nijdam quickly corrected his error and I had already given the answer of 1/3. Martin Hogbin (talk) 01:13, 19 January 2014 (UTC)
It took him 2 days, he hid his mistake and he only saw the right solution after the page filled up with arguments. Also, as part of one of my arguments, I was the first one to insert a 1/6 into what I was saying. Why don't you please admit what I am telling you: We could make MHP better if you stopped trying to think that I am trying to defeat the math - I'm not. Rather, I'm trying to find a narrative that helps verbal thinkers see the solutions more easily. And I know what I'm trying to do because I am a verbal thinker and it takes me real effort to break out of that sometimes - especially when things initially look simpler than they are. Tweedledee2011 (talk) 02:19, 19 January 2014 (UTC)
You need to stop fighting with other editors. I understand what you are trying to do and support it but the fact is that you probably will not be able to think of a way to make the correct solution clear to everyone. Countless people have attempted to do this and most regulars here are aware of the wide range of explanations in everything from peer reviewed journals to the popular press and personal web sites. The MHP is one of the world's most unintuitive simple problems and it is absolutely amazing how many people will not accept the correct solution even when presented to them in a variety of different ways.
I do not want to stop you from trying or dishearten you but you might like to look through the long talk page history of this article to see the explanations that have been proposed. Different people seem to see the correct solution differently and no one explanation seems to suit all. Also bear in mind that, for a good quality article, reliable sources should be provided for what you want to say. Good luck! Martin Hogbin (talk) 09:34, 19 January 2014 (UTC)
For the record, it was Tweedledee who posted the first attempt to answer Martin's problem, saying "the probability is .25" before Nijdam's little slip up. In point of fact, as may be seen in this snapshot of the discussion immediately before Nijdam's post, Tweedledee had already posted more than a thousand words about his own answer before anybody mentioned anything about the answer being 2/3. ~ Ningauble (talk) 14:37, 19 January 2014 (UTC)
Yes, and as part of my proof, I said that 50% of all the 2nd picks from the #2 urn would be W - and my test proves that to be true. What's at issue here is that too many editors like you, try to mock others (like me) into silence and push-back develops. And for your information, my answer of 50% for urn #2, is the literally correct answer (2/3rds of the time) for the question: "What is the probability that from that same urn the next door ball you draw will be white?". And the fact that of 14 first-pick W from urn #2, there were 7, 2nd pick W from that same urn (see my manual test) proves this to be true. But at the same time, if Martin had said, "What is the probability that if a first pick is W, the 2nd pick will be W too?", then yes, the Nijdam solution of 1/3 would have been more clearly foreseeable. But for the question which Martin actually asked, as evidenced from my manual test, if the first W pick comes from Urn #2, then the next W pick is indeed 50:50. And the first W found will come from Urn #2 2/3 of the time. So then, what the correct 1/3 represents is the combined probabilities across the whole mix. Where this dialog went south was - and the edits clearly show it - my read from Martin is that he was saying it was 2/3 (which it's not). In any case, please step back from this and see what I am saying: Part of my solution said .50 - and that .50 definitely does show up. But it's the fact that it's not a stand-alone 50% which makes the difference. And that's what I am trying to help explain: The right way to solve these is to take all the facts into account. Tweedledee2011 (talk) 03:54, 20 January 2014 (UTC)
It was certainly not my intention to mock anyone. In the process of mathematical inquiry people do make incorrect deductions, and there are two ways for someone who notices the error to approach it: one can provide an alternative derivation with a different result to rebut the conclusion, or one can attempt to explain why the deduction is wrong. This is not mockery, it is mathematical inquiry.

Providing an alternative derivation leaves people in the position of deciding for themselves which derivation is better, so it does not necessarily settle the question. I chose the second approach when I joined the discussion of this problem by raising the issue of "Which urn was it?". I thought I had identified where your original derivation and subsequent argument went off the rails. In attempting to explain why this matters I was not trying to mock you, I was trying to help you understand the logic.

However, if the error arises from not understanding the meaning of Martin's statement of the problem, or from reading Martin's explanations as saying something other than what he said, then we are not dealing with a problem of mathematics at all. It is at best a problem of semantics.

Yes, by all means do take all the facts into account. Unfortunately, it is not clear that we are talking about the same set of facts because I am at a complete loss to understand how you arrive at the meanings you ascribe to what others have written. I am not saying this to mock you, but to say I don't know how to help.

I am not going to argue about semantics. ~ Ningauble (talk) 16:06, 20 January 2014 (UTC)

@Ningauble - By saying "I am not going to argue about semantics" you have exactly framed this discussion well enough to prove my point: Many math experts think that verbal discussions (semantics) are "arguments". As such, when one tries to verbally reason with a math expert, the typical retort from the expert is a reiteration of the literally correct math and zero discussion about how to adjust one's reasoning to see that applicability of that math. Rather than send me off on a huge rabbit trail by trying to goad me on with a bet, Martin (and you) would do better to say "yes, odds of a pick from each urn matter, but you have to account for the fact that 2/3rd's of 1st pick W's will come from Urn #2. Because of this, the odds of the 2nd pick being W is skewed by the more frequent dips into the #2 urn". And then the math is more obvious. As it was (and seems to be) it's actually semantics which clouds the issue - the semantics of how the question is phrased to begin with. Now that I've been able to re-think what the question is actually asking, here's a verbal way to answer the question: Because 2/3rds of the W balls are in urn #2, when a W is drawn on the first pick, 2/3rd's of the 1st pick W's will come from Urn #2. And, because urn #1 has only one W ball it it, on the 1/3 times that 1st pick W comes from urn #1, a 2nd pick from that urn must be 0 probability for W. That's because 1st W's are removed after picking and taking one out of urn #1 removes the sole W from urn #1. For this same reason (removal of W), after the 1st pick there is now a 50/50 mix of W and B in urn #2. Therefore, here's what has to happen: Of 300 1st pick W's. The yield will be 100 from Urn #1 and 200 Urn #2. There will be ZERO 2nd pick W's from Urn #1 and 100 2nd W's from Urn #2. The totals are 100 2nd pick W's and 300 1st pick W's. 100/300 = 1/3. Suffice it to say, rather than try to mock me into given up on the true parts I did reason out correctly (0% of 2nd picks from urn #1 are W, and 50% of 2nd picks from urn #2 are W), the others on this board should try to be more helpful towards what I am trying to do: I am trying to bridge the gap between verbal reasoning ("semantics"), and the math. Tweedledee2011 (talk) 20:11, 20 January 2014 (UTC)
My point about arguing semantics is this:  You have repeatedly asserted that people said things which they did not say, and that they did not say things which they did say. (In point of fact, Martin's very first response to your solution of this problem made the very point that you now say he should have made, and my very first post about this problem was an amplification of that very same point.) Since it would be foolish for anyone to deny the fact of what was actually written here, I can only assume that you dispute the meaning of what the words say. This is arguing about semantics.

The reason I am not going to argue about semantics is because it leads to an infinite regress of meaninglessness, i.e., there is no point in saying anything about semantic meanings if the semantic meaning of what one says about it is not going to be recognized. ~ Ningauble (talk) 15:26, 21 January 2014 (UTC)

The reason why you think this is mere semantics, is because you and Martin aim your dialog towards shutting the other person up. You don't seek to draw good information from editors like me. Rather, you rebuke them enough that it's hard to keep every dialog straight. The simple fact is that the order of actions was this: Martin posted a question - and asked Nijdam to post the correct math. I posted an incorrect proof, but some of the parts of my proif were correct. Where I went astray was to not factor in the greater frequency of 1st pick W's from urn #2. Then, Nijdam posted the math (which was wrong - it claimed 2/3) And then WITHOUT debunking Nijdam, Martin proceeded to prod me deeper into dialog - all the while, his REQUESTED proof, which I was arguing against, was wrong. Tweedledee2011 (talk) 04:08, 22 January 2014 (UTC)

Results of my manual test of Martin's urn problem

I ran a manual simulation as follows:

1) I made 24, 1 of 3 picks from a BBW urn and 24, 1 of 3 picks from a WWB urn
2) This is accurate because in a big test, you'd randomly select first from either urn an equal number of times
3) Results from the BBW first pick urn are 9W, 15B
4) Each time a W came up first, I removed the W from the BBW mix, leaving only a BB mix
5) All possible 2nd picks from BB are B - no need to test this (but if we did, the number of 2nd pick fails would be 9)
6) I then made 24, 1 of 3 picks from a WWB urn
7) Results from WWB urn first picks are 14W, 10B
8) Each time a W came up first, I removed one W from the WWB mix, leaving only a WB mix
9) I then picked a 2nd pick from this mix of WB
10) I made 2nd picks 14 times
11) Of the 14 2nd picks, exactly 7 were W, which is 50% of the 2nd picks
Martin - do you want to pay by check or paypal?
I had to double-check my counts more than once, but the numbers now posted are accurate. I can scan my notes and post them, if you want
Also, if we had made 2nd picks from the urn #1 (BB at point of 2nd pick), the results would be 9 more failed 2nd picks. The total number of 2nd picks would then be 9+14, which is 23. Of those, 7 were W. This .3043 (7/23) is pretty close to .25 and is close enough for this number of iterations. So then, exactly as I said; the chances of the 2nd pick being W if the first pick is W, is .25

Tweedledee2011 (talk) 00:02, 18 January 2014 (UTC)

Just in short. Check that from the trials 9+14 showed a white ball at the first draw. From these only 7 gave a white ball at the second draw. Conclusion: 7/(9+14)=7/23 is the estimated desired probability (which is 1/3 BTW).Nijdam (talk) 13:53, 18 January 2014 (UTC)
Indeed: Tweedledee has conveniently omitted any samples of W that came from from the BBW urn. ~ Ningauble (talk) 14:31, 18 January 2014 (UTC)
p.s.:  His afterthought including them is quite remarkable in observing that 7/23 is "close" to his prediction of 1/4, when it is actually closer to Nijdam's prediction of 1/3. Of course, on a scale of 0 to 100, both are "close" to 0. ~ Ningauble (talk) 14:52, 18 January 2014 (UTC)
See my comment below - and the quote from Nijdam regarding the correct proof. I was trying to disprove a wrong solution which claimed 2/3 - I do not deny that 1/3 is correct and that my "proof" of .25 is wrong. Tweedledee2011 (talk) 16:10, 18 January 2014 (UTC)
The experiment you must do

You must do the test correctly as described below.

1) You must have two urns. One urn initially contains 9 black balls and one white ball, the other initially contains 1 white balls and 9 black balls.

2) You place a bet that your second ball will be black.

3) Toss a coin to choose an urn and then take a ball from this urn. That ball is not returned until stated below.

4) Note that I said that the first ball that you take proves to be white. If the ball that you have just taken proves to be black we are not interested. The test stops here, that ball is returned to the urn from which it came, your stake is returned, and you go back to step 1. .

5) If the first ball proves to be white you take a second ball from the same urn.

6) If the second ball is black, you win twice your stake. If the second ball is white, I win your stake.

7) Replace all balls in the urns from which they originally came and start again from step 1.

Do you want to play? Do try this at home first. Martin Hogbin (talk) 12:08, 18 January 2014 (UTC)

Martin - please read this quote, which I agree with - found elsewhere on this page. I agreed with Nijdam's correction, but NOT his original assertion that it was 2/3 ("Hi Martin, with shame I have to admit I made a mistake in my analysis. P(W2W1)=1/6 instead of 1/3, as in the complete experiment only half of the times the urn with WWB is chosen. Hence: P(W2|W1)=P(W2W1)/P(W1)= (1/6) / (1/2) = 1/3. Written out in equal likely posibilities: (Urn WWB) WaWb,WbWa, WaB, BWa, WbB, BWb, (urn WBB) WBa,WBb,BaW,BbW,BaBb,BbBa. First time W: WaWb,WbWa, WaB, WbB, WBa,WBb. 2 of these 6 in favour of WW. Nijdam (talk) 13:26, 18 January 2014 (UTC)" Tweedledee2011 (talk) 16:00, 18 January 2014 (UTC)
I cannot see where Nijdam has said that answer to my original question is 2/3. I agree with his answer of 1/3. Your answer was 1/4. Have you changed your mind? What do you say is the probability that you will win the game described above? Martin Hogbin (talk) 19:41, 18 January 2014 (UTC)
Nijdam removed his original post - after he found his error a couple of days later. And BTW, his error of 2/3 is worse than mine of .25, since the real answer is 1/3. It was that 2/3 "proof" of his, which sent me wildly down my mistaken rabbit trail. I was 100% sure the answer was not 2/3 - and in that, I've been proved correct. Tweedledee2011 (talk) 22:52, 18 January 2014 (UTC)
It was a silly mistake which he admitted to. It might have been better to have left it there however embarrassing. As you have changed your mind I guess that you can now see how knowledge gained from subsequent events can allow you to revise an earlier probability. What do you say is the probability that you will win the game described above? Martin Hogbin (talk) 00:39, 19 January 2014 (UTC)
Martin - I have not "changed my mind" because that was never the issue. The bet you offered was if the solution he offered of 2/3 was correct - had I taken that bet, I still would have won. The correct solution is .33 and my answer was .25. Of course, my answer was wrong, but it was less wrong than .667. In any case, you miss the point, which is; I am trying to change the tenor of the board from combat to cooperation. My original .25 solution was offered in cooperation. My subsequent arguments against the 2/3 was to pin you down on a point where you were wrong. You did not notice that the 2/3 solution was wrong and you offered to bet on it before verifying it. That mentality does not lead to cooperative editing. Tweedledee2011 (talk) 02:15, 19 January 2014 (UTC)
No, the bet Martin offered was not whether Nijdam's little slip with 2/3 was correct. It was entirely about whether your solution of 1/4 was correct. For the record, the bet was first offered two hours before Nijdam's post. ~ Ningauble (talk) 14:41, 19 January 2014 (UTC)
Yes Ningauble, that is exactly correct, Tweedledee did originally claim that the answer was 0.25 and, of course that was what I was betting against. Nowhere have I said or implied that the answer was 2/3. Martin Hogbin (talk) 19:13, 20 January 2014 (UTC)
You bet me after the 2/3 solution was posted - and that's the solution I was trying to disprove. Go back and see my posts - I mention the 2/3 many times. You may not have said it, but you were goading me on, trying to make my 0 & .50 (combined for .25) look bad. In fact, you were so intent on making me look dumb, you overlooked the fact that the solution you asked Nijdam to post was itself wrong. And you still haven't admitted that W 2nd picks account for 50% of all #2 urn 2nd picks - which is exactly what my original, "proof" said. That part, at least, was correct. Tweedledee2011 (talk) 19:40, 20 January 2014 (UTC)
This is simply counterfactual. Please examine the time stamps at the links I provided to when the bet was first offered (18:12, 16 January 2014, UTC) and when and when the 2/3 mistake was first posted (20:17, 16 January 2014, UTC). ~ Ningauble (talk) 17:02, 21 January 2014 (UTC)
He kept his bet standing - even after vigorous debate and WHILE the 2/3 error - which Martin failed to notice - was still posted. And the goading from him is a ratification of his bet - while I was arguing against 2/3. Even with the 2 hour overlap, Martin was banking on a false solution. Tweedledee2011 (talk) 05:52, 22 January 2014 (UTC)

Verbal solution for MHP from a simple mind

I always thought this was simple. 3 doors, one hiding a car, two hiding a goat. With any set of people 1/3 have picked right (a car), 2/3 have picked wrong (a goat). Monty takes a goat away (he knows). 2/3 still have a goat. 2/3 should switch. Fountains of Bryn Mawr (talk) 15:52, 30 January 2014 (UTC)

I do not quite follow that. As I said to Tweedldee, everybody seems to have their own way of explaining the correct answer that seems quite obvious to them. There does, however seem to be no explanation that is immediately obvious to everyone. Martin Hogbin (talk) 17:38, 30 January 2014 (UTC)
Solutions and explanations presented in this article are all from reliable sources and home grown ones are generally not accepted. I think there could be a case for a separate article for simple solutions.
The matter is further complicated by an absurd complication to the problem that some editors have insisted must be included for the solution to be correct. Martin Hogbin (talk) 17:38, 30 January 2014 (UTC)
Fountains of Bryn Mawr, you exactly hit the mark, in respecting that in this famous paradox the host's intent is to draw a blank, and to offer a switch to the other non selected door. Correct. Furthermore, as to the intended paradox, the host of this typical "once and never again show” is completely unknown to us. And it is correct that 1/3 of people will arrive in the "lucky guess scenario", having selected the correct door with the prize behind and should stay, whereas 2/3 of people will arrive in the "wrong guess scenario", having selected one of the two goats, so they should switch. Inescapable.
And yes, the actual location of the car is irrevocable, it’s unchangeable. But some mathematicians (Morgan et al.) said something that does not address the intended paradox. Although the chance of having arrived in the lucky guess scenario still remains 1/3, and of having arrived in the wrong guess scenario still remains 2/3, they say that after the host has intentionally shown a goat behind one of the two non selected doors – one could make both still closed doors somewhat transparent, in order to have a peek behind them, similar as to open all doors at once. They said that the host, regarding the actual location of the car, could eventually provide an indication as to the scenario the candidate actually is in. Suppose that the host, in for once exceptionally opening the door that he usually strictly avoids opening, could indicate that actually the prize is behind his other (preferred) door that he now offers to switch on. They say that in this extreme case the host, by opening the door that he usually strictly avoids to open, could indicate that the contestant actually is in the wrong guess scenario indeed with probability 1 (instead of the average 2/3], so that switching actually will win the prize for sure. As said above, this is similar as to open all doors at once.
Or elsewise, in opening his “preferred door”, the host could indicate that the contestant actually is in the lucky guess scenario (in having selected the prize) with probability 1/2 (instead of the average 1/3), so that switching actually will lose/win the prize with probability 1/2.
But others (Ruma Falk) said that such additional hint as to the actual scenario the contestant is in, can only be possible if the host, given that both unselected doors hide goats, is biased somehow indeed to open one special door and not the other one. But this will be without notable effect, unless the host is biased indeed, and we exactly “know” about his given bias, its extent and its direction. Otherwise such "additional indication" cannot and will not occur. For the famous paradox with an unknown host, such airy-fairy alien predictions are meaningless and inoperative. They do not address the famous paradox. Nevertheless, it's the hobby and favourite subject of mathematicians to use such alien and deviationist examples in teaching conditional probability theory. And that's the confusing problem of this article on a famous paradox. Gerhardvalentin (talk) 19:16, 31 January 2014 (UTC)

Question for Nijdam

Good to see you back.

I have a question for you, nothing to do with Tweedledee. I am sure that you will see its connection with the MHP.

There are two doors each with an urn behind it. One urn contains two black balls and one white and the other two white balls and one black. You select a door and from the urn behind it remove a ball (as always with urns, at random) which proves to be white.

What is the probability that from that same urn the next door ball you draw will be white and how do you calculate this? Martin Hogbin (talk) 15:46, 15 January 2014 (UTC)

Martin - I think you meant to say "What is the probability that from that same urn the next ball you draw will be white"Tweedledee2011 (talk) 07:44, 16 January 2014 (UTC)
Yes, thanks, I have corrected it.
And to answer your question, the probability is .25 and here's why:
URN #1 has BBW and if you first choose W from this, the next choice from this urn can only be B, which is 0
URN #2 has WWB and if you first choose W from this, the next choice from this urn is between W or B, which is .50
Since you don't know which urn had what to start, if you repeat this 2 choice series enough times, you will find W on the second choice .25 of the time.
Half the times, you have W at 0 from URN #1 for a possible 2nd choice result and half the times, you have W at .50 from URN #2 for a possible second choice result.
That's why you only get W .25 of the time on your 2nd choice
Tweedledee2011 (talk) 08:21, 16 January 2014 (UTC)
Tweedledee, I did not set this problem as a trap for you but to demonstrate a completely different point to Nijdam. As it happens though you have made a mistake which helps me demonstrate the point that I want to make.
The problem lies with your step, 'Half the times, you have W at 0 from URN #1 ... and half the times, you have W at .50 from URN #2...'.
Let us suppose that you initially toss a coin to decide which door to open. Let us, purely for our own convenience, call the urns A and B. I think we all agree that before you open a door the probability that you have chosen urn A for example is 1/2.
Next you open the door to see the urn. This tells you nothing which might let you revise that probability.
Next you draw a ball from urn A, which proves to be white. This does give you information about which urn you originally chose. One urn contains two white balls but the other contains only one white ball. Given that you have in fact chosen a white ball you must now consider it more likely that you chose from the urn which had two white balls. This is a conditional probability; the probability of an event given that another event has actually occurred. This has been a much discussed subject here in the past.
To get the correct answer we need to start by considering all the events that might have happened.
[Point for Nijdam] To do this, purely for our own convenience, we may choose to label the balls W1 and W2 in one urn and W3 in the other. The other balls may be labelled K1, K2, and K3. Although all the white balls look the same to us and all the black balls look the same and we only actually see two of the balls, we know what balls there are so to get the probability right we should, in principle, label them all.
So our sample space is AW1, AW2, AK1, BW3, BK2, BK3 (if you really want to we could add BW1, BW2, BK1, AW3, AK2, AK3) but I think we can agree that this is not necessary. Each event in this space has a probability of 1/6.
Now we must condition our sample space on the event that we actually drew a white ball. This means considering only the events that conform to our condition, namely AW1, AW2, BW3. As we know for sure that one of these events actually occurred, we must adjust the probabilities so that they total 1. This makes them 1/3 each.
As you can now see, after a white ball has in fact been drawn, it is twice as likely that your original ball came from the WWB urn than from the WBB urn, that is to say twice as likely that you have chosen the urn with two white balls. Martin Hogbin (talk) 09:50, 16 January 2014 (UTC)
Hi Martin, as you see, I'm not gone, but I'm keeping watch. Concerning your experiment, you're right in your analysis. In terms of conditional probabilities: P(W2|W1)=P(W2W1)/P(W1)= (1/6) / (1/2) = 1/3. I do not see the direct connection with the MHP. Nijdam (talk) 20:17, 16 January 2014 (UTC)
Hi Martin, with shame I have to admit I made a mistake in my analysis. P(W2W1)=1/6 instead of 1/3, as in the complete experiment only half of the times the urn with WWB is chosen. Hence:
P(W2|W1)=P(W2W1)/P(W1)= (1/6) / (1/2) = 1/3.
Written out in equal likely posibilities: (Urn WWB) WaWb,WbWa, WaB, BWa, WbB, BWb, (urn WBB) WBa,WBb,BaW,BbW,BaBb,BbBa. First time W: WaWb,WbWa, WaB, WbB, WBa,WBb. 2 of these 6 in favour of WW. Nijdam (talk) 13:26, 18 January 2014 (UTC)
Nijdam, I cannot see where you made a mistake. In your first answer you appear to say the probability of drawing a white ball from the urn that you have just drawn a white ball from is 1/3. In your second answer you seem to say the same thing. Martin Hogbin (talk) 19:17, 18 January 2014 (UTC)
My point is this. We have two black balls in one of the urns. They are, to most intents and purposes, identical. We may never even see one of the black balls if we draw only two balls from that urn. The balls are not numbered, nevertheless it helps us formulate the problem correctly if we do number the two black balls in one urn (and the two white ones in the other).
Now consider the two urns. They may be identical, we do not know because we only see one of them, but again it is useful to label the urns. One is the urn that we chose and the other is the one we did not choose In my example it matters which urn you choose. If each urn had two white and two black balls it would make no difference but it still might be good practice to label them in some way. Do you agree so far? Martin Hogbin (talk) 17:54, 17 January 2014 (UTC)
Yes, okay, but it seems an open door (sic!) to me. Nijdam (talk) 12:28, 18 January 2014 (UTC)
The problem is that when we move on to the MHP you become irrational. The situation with the goats is just the same. There are two different goats and only one of them is revealed. When we set up our sample space for the problem, if we are to do it really properly, we must consider which goat the host reveals. Martin Hogbin (talk) 13:23, 18 January 2014 (UTC)
Martin, please also have a look above, as I made a mistake in my analysis. I don't understand your concern about the identity of the goats. In the problem with the urns, the identity of the balls also do not play a role. For instance you do not mention wich of the white balls you draw. On the other hand, if the host tells the player the revealed goat is called Beeeertha (the other one is Maaaaaary), the result of the analysis is the same; only is each one of the outcomes now split in two. Nijdam (talk) 13:45, 18 January 2014 (UTC)
It is the identity of the urns that is the analogy with the goats in the MHP. There are two urns hidden behind doors. They are not labelled or identified in the problem statement and you only ever get to see one of them. It is possible that it matters which urn is revealed (if they contain a different selection of balls from one another) or possible that it does not matter (if they both contain the same selection of balls).
In the case of the MHP goats, there are two goats hidden behind doors. They are not labelled or identified in the problem statement and you only ever get to see one of them. It is possible that it matters which goat is revealed (if the host has a known goat preference) or possible that it does not (if the host has no goat preference).
For the case where the player has originally selected door 1 which hides the car there are four outcomes. depending on which door the host chooses and which goat is revealed. We have H=2,G=B H=2,G=M, H=3,G=B H=3,G=M. Of course, it is easy to argue that there is a symmetry between the goats (that is to say it makes no difference which goat is revealed) so we can ignore the goats but I would argue that, if we are to have a complete solution to the MHP it must consider which goat has been revealed.
Thus the full solution of the MHP becomes quite complicated. As well as Morgan's door preference parameter we must add a goat preference parameter. How these two parameters are applied is also more complicated because the host's preferred door may not hide his preferred goat.
You have agreed in the past, I think, that a complete solution should also consider the door originally chosen by the player. Martin Hogbin (talk) 14:33, 18 January 2014 (UTC)
Nijdam, any thoughts on this yet? A proper and complete solution to the MHP would be quite interesting to work on. Maybe we could even get another letter in the 'American Statistician'. Martin Hogbin (talk) 19:48, 20 January 2014 (UTC)
Martin, I (and I guess the mathematical world) do not see the relevance of making a distinction between the goats. Mainly because it does not contribute essentially to any aspect of the MHP. I'm also pretty sure any attempt to get such analysis published will fail. If you would like to work on the MHP, we may try to give an overview on the main litterature with a discussion of the errors made in it. Nijdam (talk) 09:48, 22 January 2014 (UTC)
I did fear that the mere mention of goats would cause you to look at the problem in the traditional way rather than as a mathematician. The situation just like the urn problem above. There are two goats, we may never see them both but that, as you seem to have agreed above, is not important. Neither is the fact that the goats have not been given numbers. We know that there are two goats and that one specific goat has been revealed, it is not inconceivable that the goat revealed could give us information as to whether the player originally chose the car. Please explain to me, as a mathematician, how we can ignore this fact. Martin Hogbin (talk) 13:22, 22 January 2014 (UTC)
Martin, you're right, but what's the point. What are you aiming at?Nijdam (talk) 22:39, 22 January 2014 (UTC)
Martin, I see the point you're trying to make which is that "To the player the two urns are indistinguishable but they are in fact different in terms of probability space" or you might use differing terms. The reason this doesn't translate to the Goats, or in fact to the doors is that the Goats and the doors are identical in terms of probability. As someone said above if in your situation both urns had 2white and 2 black balls, you could collaps the problem down to drawing balls from a single urn as the door and urn choice would have no effect on the problem P(W2|W1)=1/3. In your version of the problem the two urns are different. U1{W1, W2, K1} and U2{W3, K2, K3}. However the doors are identical. You don't need to consider D1{U1}&D2{U2} and D1{U2}&D2{U1} because both of those scenarios are identical. just like in the monty hall problem where the actual door number and the individual goat are irrelevant. SPACKlick (talk) 14:03, 28 April 2014 (UTC)
Is it just the goats?

Nijdam, imagine the MHP game is not played with a car and two goats but with three balls. One black, one yellow, and one white. The balls are in small boxes with doors and the game is played to the normal rules. If you end up with the black ball you win a prize, otherwise you get nothing. What would be your sample space now? Martin Hogbin (talk) 14:17, 22 January 2014 (UTC)

Normal rules? What does the opened door show? I suppose it reveals the yellow ball. Sample space: B=door number of black ball, Y=door number of yellow ball, X=chosen door, H=opened door; samples

XBYH
1122
1123
1132
1133*(X=1,Y=H=3,B=1)
1213
1233*(X=1,Y=H=3,B=2)swapping shows the black ball
1312
1322
2123
etc. Nijdam (talk) 22:41, 22 January 2014 (UTC)

I agree. What if the both the non-black balls were white? What if they were slightly different shades of white? Martin Hogbin (talk) 09:25, 23 January 2014 (UTC)
Are you serious? The only point is, do we distinguish between them or not. So???Nijdam (talk) 21:30, 23 January 2014 (UTC)
Yes, of course I am serious. If we wish to be strictly correct we must distinguish between them. All macroscopic objects are in principle distinguishable.
What I cannot understand is why you say that we should distinguish between the two doors that the host might open but not between the two non-winning objects that he might reveal. In the case that the player initially chooses door 1 we know that, in principle, it is possible that the probability that the prize is behind the originally chosen door may depend on whether the host opens door 2 or door 3. It is equally possible, in principle, that the probability that the prize is behind the originally chosen door may depend on which non-prize object the host reveals. You have agreed above that the fact that two objects have similar appearance does not mean that we cannot distinguish between them.
Of, course, as Gerhard points out below, we may note various symmetries in our setup which we may choose to apply to the problem. We can observe that, from the player's state of knowledge, the problem is symmetrical with respect to non-winning object identity and with respect to door number.
Having noted these symmetries we may choose to apply them in two consistent ways. The first way is to use a sample space in which every possible choice mentioned in the problem statement is represented. This includes the player's initial door choice, the host's door choice, and the host's choice of object to reveal. Having set up this sample space we can then use our symmetries to assign probabilities to the various outcomes.
Alternatively, we may apply the known symmetries at the start of the problem and use them to intuitively reduce our sample space to the simplest one which will solve the problem.
Gerhard and I prefer the latter method and you and others prefer the former but, if you do it that way, you must do it properly. Martin Hogbin (talk) 10:25, 24 January 2014 (UTC)
No, the point is that in the problem statement the doors are presented as different, and the goats are not. (As you may well know the standard notion of the problem is equivalent to the problem with the goats removed, leaving one door with a car and the other two empty. I hope you will not start a discussion about whether two empty spaces might be distinguished. There has been people discussing the issue whether an empty set of numbers is the same as an empty set of vectors.)Nijdam (talk) 11:04, 24 January 2014 (UTC)
I do not agree that the standard notion of the problem is equivalent to the problem with the goats removed. We are told quite clearly that there are two goats and that one goat is revealed. To some people, goats may look quite similar but it is quite probable that the host, for example, would have been able to distinguish between them. He might even have had a favourite (in fact it is much more likely that he had a favourite goat than a favourite door).
I might agree that, if this were an exam question, it might well be intended that we should distinguish between doors (because they have been numbered) but not between goats (because they have not been numbered) but this was not a question in a statistics exam it was a puzzle in a popular general interest magazine.
If we are to answer a question of this nature we must do as Seymann suggests and ask ourselves what the intent of the questioner was. As it happens in this case, we do know that Whitaker did not number the doors and that vos Savant added to numbers only to make the problem setup clearer. She has made quite clear that it never was her intention to distinguish between door numbers. Now it might be argued that vos Savant made a bad job describing the problem that she wished solved and that superior mathematicians and statisticians should reformulate her problem properly before attempting to answer it. Fine, if that is what we wish to do let us do it that way; let us cover the most general case that is described by the problem statement and then solve that, taking account of such symmetries that we see fit. Your decision to ignore the goats is your personal choice; there is no mathematical justification for this. You have already agreed that two urns with similar appearance should be treated as different urns.
Historically, the two goat identities have been ignored, although historically speaking many have ignored the door numbers too. If you want to be pedantic you cannot ignore either. Martin Hogbin (talk) 12:41, 24 January 2014 (UTC)
To put my point more simply, are you claiming that 'the host reveals goat A behind door 3' is the same event as 'the host reveals goat B behind door 3'? Martin Hogbin (talk) 10:45, 25 January 2014 (UTC)
To put my answer simply: I do not have a goat A and a goat B. Nijdam (talk) 17:15, 25 January 2014 (UTC)

Martin, are you addressing the famous paradox that MvS attempted to present, with a host of which absolutely nothing is known, except that he intentionally shows a goat, so he impossibly can show the car? Ruma Falk says that the chances of the door first selected by the guest remains 1/3 not only because in any case he can do so, but moreover because any "biased" choice between two goats never may be taken into consideration, unless such one-sidedness host's bias is given indeed, and we absolutely know of such bias, its extent and its direction, otherwise NOT? What idiosyncratic variant are you addressing? Gerhardvalentin (talk) 17:36, 23 January 2014 (UTC)

Gerhard, I hope my response to Nijdam makes it clear what I am doing. I am addressing the MHP as stated by Whitaker/vS with the normal rules. Applying well-founded principles of probability you and I can see that, from the player's state of knowledge, the player's initial door choice, the host's door choice, and the host's choice of object to reveal do not affect the probability that a player who chooses to swap will win the car. We use this fact to reduce the problem to a simple form.
Others prefer to do everything the long way round. That is to list all the possibilities that are implied by the problem statement to create a large sample space and use this to solve the problem, applying the symmetries implied by the player's state of knowledge to assign probabilities to outcomes in this space.
If course we know that these people will come up with exactly the same answer that we got. They might argue that their way is more rigorous, and more general in that it covers variants in which the host may have a preference for one of the goats or doors, for example. We might say that we are not interested in answering variations of the problem so their method is too complicated.
My point to Nijdam is that, if you are going to solve the problem the long way round, you should do it properly and include in your sample space every choice that is implied by the problem statement. Martin Hogbin (talk) 10:25, 24 January 2014 (UTC)
Martin, I see the point you're trying to make which is that "To the player the two urns are indistinguishable but they are in fact different in terms of probability space" or you might use differing terms. The reason this doesn't translate to the Goats, or in fact to the doors is that the Goats and the doors are identical in terms of probability. As someone said above if in your situation both urns had 2white and 2 black balls, you could collaps the problem down to drawing balls from a single urn as the door and urn choice would have no effect on the problem P(W2|W1)=1/3. In your version of the problem the two urns are different. U1{W1, W2, K1} and U2{W3, K2, K3}. However the doors are identical. You don't need to consider D1{U1}&D2{U2} and D1{U2}&D2{U1} because both of those scenarios are identical. just like in the monty hall problem where the actual door number and the individual goat are irrelevant. SPACKlick (talk) 14:03, 28 April 2014 (UTC)

_________________

Martin - see below Tweedledee2011 (talk) 16:42, 16 January 2014 (UTC)

Programming question

I was trying to set up a simulation of the Monty Hall to grab some data for a talk I'm giving and I set it running for 2^20 cycles and it came out with a ratio of close to 3/5 wins rather than 2/3 by switching. Can anyone see where I haven't set the problem up right or is my RNG just too far from random?

```Dim CarDoor as byte
Dim ContestantsDoor as byte
Dim HostsDoor as byte
Dim HostsDoorAllowed as boolean
Dim SwitchDoor as byte
Dim Winner as Boolean
Dim WinCount as Long
Dim TotalCount as long
Dim RunCount as long
Dim Percentage as double

TotalCount = 2^20                              'Total number of runs
For RunCount = 1 to TotalCount                 'For each run

CarDoor = Int(Rnd * 3)+1                       'Randomise the CarDoor between 1 and 3
ContestantsDoor = Int(Rnd*3)+1                 'Randomise the Contestant Pick between 1 and 3

HostsDoorAllowed = false                       'Resest Host Door Test
Do until HostsDoorAllowed                      'Loop until Host picks valid door
HostsDoor = Int(Rnd *3)+1                      'Randomise the host door
HostsDoorAllowed = HostsDoor <> CarDoor _      'Check Hosts door is not car door
and HostsDoor <> ContestantsDoor               'Or contestant door
Loop

SwitchDoor = 6 - (ContestantsDoor + HostsDoor) 'Switch to the other available door
Winner = SwitchDoor = CarDoor                  'You win if you have the car

If Winner then
Wincount = Wincount+1                          'Count the wins
End If

Next Runcount                                  'Start Again

Percentage = WinCount*100/TotalCount           'Check the win percentage
```

Before the comments, I know I could make the selection of the host door more efficient using a library but that didn't feel like it was obviously the problem. I mean if I just wanted a mathematically equivalent program I'd just generate a random number(1-3) and you win if it isn't 3 but that doesn't look like Monty Hall. SPACKlick (talk) 10:39, 1 May 2014 (UTC)

I cannot immediately see a problem but you could try keeping a count of the number of times each door is chosen as car, host and player. Martin Hogbin (talk) 18:06, 1 May 2014 (UTC)
First, thanks for the move, second I've found the problem. My random number generator gives the same number for Cardoor and Contestant door unreasonably often. Usually it's 1 higher than contestant door for first hostdoor. I've made the RNG volatile and it works fine now. — Preceding unsigned comment added by SPACKlick (talkcontribs) 09:07, 2 May 2014 (UTC)

Monty Fail needs more explanation

Monty Fail -- aka

1) The player chooses a door at random 2) The Host randomly chooses a door of the ones remaining. 3) The Host's door is revealed to have a goat behind it. 4) The Player is offered the choice to switch.

The reason that switching when a randomly chosen one has a 50% chance of success, while the original problem where the host chooses one that is a goat is 66% is that there are 2 cases unaccounted for in the Monty Fail scenario. What happens if the Host chooses the car?

If the game is reset, then you are eliminating the 1/3 of the time where the Host chooses the car, and thus you have 1/3 vs 1/3, which is 50%.

If the player would lose, then either the player has a 1/3 chance for the remaining two, or he loses.

If the player would win, then either the player has a 1/3 chance for the remaining two, or he wins.

A variation on the above is that one choice is randomly removed by the host. After the player makes his choice, the removed item can be revealed. This makes it the equivalent to the "Player loses" scenario. And of course if the Host's choice is added to the winnings no matter what, then that gets added to the "Player wins" scenario. And if it is only added back if the player makes the switch, then that is the original scenario.

Please translate this to better language ;) — Preceding unsigned comment added by 64.17.230.98 (talk) 20:07, 22 May 2014 (UTC)

Why do you not have a go yourself? Martin Hogbin (talk) 21:54, 22 May 2014 (UTC)

Month Hall as a limit theorem

I tried to find an equation expressing this problem that avoids simulations. Where the person switches from original choice it is summation from k = 1 to k = infinity and from n = 0 to n = infinity of ((1/n+1)/2)^k. Santacruz1962 (talk) 20:15, 30 May 2014 (UTC)

Simple logical solution to the Monty Hall Problem

The solutions given are too complicated for the average reader. The statements below ombine probability theory with simple logic to reach a solution.

Here are more ways to annalize the Monty Hall problem:

(1) If you can pick only one door your odds of getting the car are 1/3, but if you were allowed to pick two doors it would double your odds to 2/3. You can randomly pick two doors to open if you decide, in advance, to switch. You decide which door will not be opened and choose that door. Then the other two doors are opened, one by the host and the other one by you when you switch. You may say “yes, but the host will always open a door with a goat. That’s OK since if you opened both doors at least one would have a goat. The fact that you randomly got to pick the two doors to open is what sets the odds at 2/3 that you will get the car.

(2) Again, you know that you will switch. Now you choose a door which you hope will be a “goat door”. Your chances of it being a “goat door” are 2/3 because there are two goats, Assuming you are correct, the host is forced by the rules to open the other “goat door” so you switch and open the door with the car. Therefore you had a 2/3 chance of getting the car,

(3) Look at the problem from a different angle. What are your chances of not getting the car if you decide to switch? If you have decided to switch, you will only not get the car if you chose the door with the car behind it since you will switch. But it is only a 1/3 chance that you will choose the door with the car. If your chances of not getting the car are 1/3 then your chances of getting the car are 2/3.

(4) If you have decided not to switch you have only a1/36 chance of getting the car,

Thus, you should always switch!

Sawtooth mountin dude (talk) 22:25, 28 February 2014 (UTC)Sawtooth mountain dude

Here is the problem. Once you can see why the player should switch, it is very hard to see why other people cannot understand this this. Everybody has their own pet solution to the Monty Hall problem. By 'pet solution', I mean solution or explanation that works for them or that made it all clear to them and which they firmly believe should make it instantly clear to anybody else the moment they see it. Unfortunately people who have not yet seen why you should swap are notoriously hard to convince, with any argument, however clear and obvious it may be to the presenter of that argument.
The general consensus here has therefore been to mainly keep away from home grown solutions and add to the article only explanations found in reliable sources. I agree that the article could and should make a greater effort to convince people of the correct solution but the problem is, how to do this without just adding a large number of users' individual solutions like yours. Martin Hogbin (talk) 11:02, 1 March 2014 (UTC)
As many sources mention, another problem is the disconnect between the typical "easy to understand" solutions and the mental image created by the problem statement. These solutions pretty much all contrast the overall results of picking a door (say #1) and staying with it vs. picking a door (say #1) and switching to either of the other two doors. In the former case, you win if the car is behind the one door you pick. In the latter case, you win if the car is behind either of the other two doors. So, obviously, in an overall sense switching must be twice as good as staying. However, in the mental image created by the problem statement you've picked a door (say #1) and are deciding to switch after you've already seen the host open a different door (say #3) - i.e. you win by staying if and only if the car is behind door #1 and by switching if and only if the car is behind #2 (since you know for sure it is not behind #3). With this mental image, the question is not what is the probability the car is behind door #1 (1/3) vs. the combined probabilities of doors #2 and #3 (2/3) - but rather what is the (conditional) probability the car is behind door #1 vs. the (conditional) probability the car is behind door #2 given you've seen the host open door #3. As it turns out, these questions have the same answer (and they must have the same answer if everything is symmetrical between the doors) - but seeing the relationship between these two (different) questions is quite difficult. As one source puts it "the distinction between [these questions] seems to confound many". -- Rick Block (talk) 21:14, 1 March 2014 (UTC)
To further the explanation of the simple solution I propose the following;

(1) It is known in advance that you will ultimately have to choose between the door you choose and the remaining door

(2) It is known in advance than you will ultimately have to choose between the door with the car and the door with the remaining goat

(3) It is axiomatic that the odds of the door you choose being the door with the car are 1/3 whichever door you choose.

(4) From 1-3 The odds of getting the door with the car by switching to the remaining door are 1- the odds in 3 or 2/3.

a lot of people seem to like to overcomplicate the problem, considering sample spaces and probability trees including the choice of door 1-3, the location of car in door 1-3, Whether goat A is in door 1-3, whether goat B is in door 1-3, whether the host opens door 1-3, whether the host reveals goat A or B, whether the player sticks or switches. Yet every time I've seen the problem, the probability tree in my head is just 2 branches
a. Contestant picks the door with the car behind it, Host opens a door revealing a goat, leaving a door with a goat behind it 1/3
or
b. Contestant picks a door with a goat behind it, Host opens the other door with the goat behind it leaving a door with a car behind it 1/3
Every action is determined from the one choice made in the scenario and that choice is a 1/3 choice. — Preceding unsigned comment added by 213.106.233.97 (talkcontribs)
Your numbered steps above include as #3 "It is axiomatic that the odds of the door you choose being the door with the car are 1/3 whichever door you choose.". Yes. But this is axiomatic only at the beginning of the game - specifically before the host opens a door. Looked at in a certain way, the entire question is what effect (if any) the host opening a door has on this probability. It's definitely not axiomatic that this probability remains 1/3 regardless of how the host chooses which door to open. For example, if the host opens a random other door (and happens to reveal a goat), the probability the door you chose now hides the car is 1/2 (not 1/3). Even if you say the host MUST open a door revealing a goat, the probability the originally chosen door hides the car after seeing what door the host opens still might be anything between 1/2 and 1 (depending on how the host decides which door to open if the originally chosen door hides the car, see https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions).

When you say it's axiomatic that the probability of the originally chosen door is 1/3, what you're really saying is if you pick a door and don't switch you have a 1/3 chance of going home with the car - and if you pick a door and switch to whichever door the host doesn't open (meaning there are still two possibilities - i.e. if you pick door 1 and switch you might be switching to door 2 or door 3) you have a 2/3 chance of going home with the car. You've moved the decision point from after the host opens a door to before the host opens a door - so you're not deciding between (say) your original choice of door 1 and door 2 (in the case the host opens door 3), you're deciding between door 1 vs. (door 2 or door 3). With the latter decision, of course you double your chances by deciding to switch. But what about after the host opens a door?

The chances after the host opens a door are a conditional probability, see https://en.wikipedia.org/wiki/Monty_Hall_problem#Conditional_probability_by_direct_calculation for how this is computed. -- Rick Block (talk) 04:06, 30 April 2014 (UTC)

I think you missed my point. In the original problem there are only Two scenarios. "You picked the car first", "You didn't pick the car first". Once that action is taken nothing else effects things. The host cannot pick a door randomly, unless you picked the car, due to Axiom 2.
Considering the set of all the various strategies by which the Host could give information by which goat he picks when he has the choice of 2, such that this choice provides information which changes the probability of "the door you chose" being "the door with the car", that set is necessarily symmetrical (Every strategy can be reversed by simply switching "opens Door 2" with "opens door 3" and vice versa in every case) and each door has even odds of being preferred in the case where the host has a choice as they are indistinguishable except by label. So the host COULD be using a strategy which makes the probability >2/3 but he's equally likely to be using one that makes the probability<2/3. Therefore, given there is no information in the problem about what strategy the host is using, the consideration of all strategies has a net effect of multiplying the probabilities by 1. Leaving you with the odds of finding the car of 2/3 by switching.
If you feel you must consider the case where the host has a specific strategy, about which no information is provided, without considering the possibility of all other strategies, then you are considering a subset of the Monty Hall and may as well consider the problem where the contestant can see through doors (the problem never says he can't). The odds that the contestant can detect the car using various senses are greater than 1 and therefore the odds of getting the car are necessarily greater than 2/3 overall.
Of course, if you limit yourself to simply considering the probability puzzle presented to you by the problem, the answer is There are only 2 scenarios. One is twice as likely as the other. In that scenario you win by switching. Switching is the best tactic at that point in the game.
I'm not denying that the mathematical analysis of the results under various different strategies is interesting, fun, clever or anything else. Simply that it isn't an analysis of the Monty Hall Problem any more. SPACKlick (talk) 16:13, 1 May 2014 (UTC)
Sorry, to slightly clarify the above rambling response by directly quoting from you. "you're not deciding between (say) your original choice of door 1 and door 2 (in the case the host opens door 3), you're deciding between door 1 vs. (door 2 or door 3)." The reason for this is that in the original stated problem there is no information which would allow any distinction between door 2 and door 3 if they were both goats. They are, necessarily indistinguishable.
By analogy to consider the case of a strategy whereby your odds by switching have dropped to 50/50 without expressly limiting yourself to that subset of scenarios, even if just to give the answer "Your odds are between .5 and 1" makes as much sense as giving the answer that "your odds are either 100% or 0" given that the car either is or isn't where you'll end up by switching. Which is always true, no matter the conditions. The interesting answer, the answer almost invariably sought by this sort of question, is in what proportion of scenarios possible under the given information am I 100% and in what proportion of possible scenarios am I 0%. And at the point where the host has opened door 3, the car is in door 2 66.7% of the time. SPACKlick (talk) 16:23, 1 May 2014 (UTC)
In rewriting the article to reduce bloat I've come against the issue that I'm still not certain of the utility of the conditional arguments. They tend to be P(Car behind door x|Contestant chose door y and host revealed a goat behind door z). I have several issues with this.
1) trivial one but still, the host revealed a goat is part of the background set up, it had to happen so that section should be changed to "door z was eliminated"
```P(Car behind x|Picked y and Eliminated z)
```
2) The three doors at the start equiprobably conceal the car and cannot be distinguished in any meaningful sense, so Picked y is an irrelevance because there is no conceivable effect of it.
```P(Car behind x|Eliminated z)
```
3) Some door will always be eliminated. The doors are indistinguishable within the puzzle. The only way "given the host elminated z" can have a probablistic effect on the puzzle is if it conveys some information. But any mechanism for it conveying information would have to be contained in the puzzle. There is none. Conditional probablity is useful in variations where there are competitive strategies by the host but in the vanilla problem I don't see it being any more useful to consider
```P(C2|P1&H3)
```
than to consider
```P(C2|P1&H3&A&B&D&E&F&G)
```
```Where Cn is car behind door n
Pn is Player selects door n
Hn is host reveals a goat behind door n
A is contestant doesn't have x-ray vision
B is host doesn't tell the contestant where the car is
D is none of the doors stick shut
E is there was no earthquake in Taiwan in the 4 hours before the show
F is the contestant is paying attention to the host
G is the game is played in earth gravity.
```
All the givens just seem unnecessary.
Just as it is unspecified in the question that the host chooses at random A-G are unspecified and A, B, D and F could all have significant effects on the probability if their negatives were considered. I just don't get it. SPACKlick (talk) 13:03, 18 June 2014 (UTC)
Spacklick, many people here agree with you that vos Savant's attempt to clarify the original problem by saying 'Say door 1' does not lead to the necessity to take the fact that the player chooses door 1 as a condition of the problem. There was a very long discussion on this subject in which no one from either side changed their mind. In the end there was an Arbcom decision that we should start with the simple solutions, which do not distinguish between door numbers and, later on in the article, we should mention the 'conditional' solutions. That is the principle that we should stick to in the article.
Regarding discussion here, good luck, but everyone seems firmly stuck in one camp or the other and cannot be persuaded to change. Martin Hogbin (talk) 17:38, 18 June 2014 (UTC)
Martin, the Arbcom case did not decide that, they pointedly declined to rule on article content. Rather, it was a separate RfC facilitated by the Mediation Committee that decided to present "simple" solutions first and "conditional" solutions later. ~ Ningauble (talk) 15:22, 19 June 2014 (UTC)
Having spent some more time reading the bayes theorem section I sort of get it. It's proof of the obvious that requires conditional probability, that being that the specific door the host opens conveys no differentiating information. While it seems trivial from reading the problem, proving it adds value. SPACKlick (talk) 12:47, 19 June 2014 (UTC)
SPACKlick, it is not the only way to prove the point, but we need to include it. While I agree with your observation that an analysis of conditional probabilities is not strictly necessary to solve or understand the problem, it is definitely necessary to include it in the article because that is how the overwhelming majority of academic sources derive the result.

My own sense of the situation is that academicians are using the Monty Hall Problem to explain conditional probability, which has some utility as a simple "degenerate case"; but the converse, using conditional probability to explain the Monty Hall Problem, is less useful for precisely the same reason – using an elephant gun to kill a mosquito certainly proves the point, but it does not shed much light on the nature of mosquitoes. Still, we are stuck with it because so many academic sharpshooters apparently like to use a mosquito to demonstrate the power of their big guns. ~ Ningauble (talk) 15:30, 19 June 2014 (UTC)

I agree, 'it is not the only way to prove the point, but we need to include it' is about right, as is, 'so many academic sharpshooters apparently like to use a mosquito to demonstrate the power of their big guns'. We have reached a decision that does not require this argument to be settled so let us stick to that. Martin Hogbin (talk) 17:15, 19 June 2014 (UTC)
The point is not that editors here disagree, but that numerous extremely reliable sources point out the difference between the "simple" and "conditional" solutions, and such sources keep getting published. See https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions. Regardless of how many editors here think it's not useful to consider the impact of the specific door the host opens, since there are a significant number of sources that do so the article must present this approach.

And, for discussion on this page (not related to the article, but rather to understanding of the problem), the claim that the doors are indistinguishable is patently false. They're clearly understood to be physical doors (persistently) numbered 1, 2 and 3 - i.e. they are not inherently indistinguishable as marbles of the same color are in an urn problem. The information that is provided in the problem statement allowing you to distinguish between them is the fact that they are numbered. Given this, then the host may or may not pick randomly when the player's initial choice is the door hiding the car (if they were truly indistinguishable they would be indistinguishable to the host as well, in which case the host cannot have a preference). Since you have no information about this choice, in a Bayesian sense you can choose to consider the doors as indistinguishable - but by doing this you're effectively moving the decision point from after the host opens a door to before. Consider how you must model a simulation, and the difference between tallying every repetition (distinguishing only "switch" and "not switch") vs tallying only those cases where the player picks door 1 and the host opens door 3. In the former case, the ratio of winning by not switching vs switching will approach 1/3:2/3 regardless of how the host decides which door to open if he has a choice. In the latter case, the ratio will approach 1/3:2/3 if and only if the host chooses randomly when the car is behind door 1. -- Rick Block (talk) 15:52, 19 June 2014 (UTC)

Rick, I was trying not to restart the argument again. As I said above, not one person changed their mind in the long argument over this subject. You are welcome to try to persuade me on my talk page if you wish but I doubt that you will get anywhere.
Regarding the article, there was a decision reached and it did not claim that any one side was correct, only that we should deal with the 'simple' solutions (without disclaimers) first, then the 'conditional ones. Not sticking to that agreement would start a major and pointless war. Martin Hogbin (talk) 16:52, 19 June 2014 (UTC)

For the record: the so called simple solution is firmly defended by people who do not fully understand the problem, or in due course did understand it, but then anxiously are trying to find arguments to justify the simple explanation they once have proposed. They follow the line set by vos Savant. She tried to defend the simple explanation by stating that she did not explicitly meant to mention the door numbers 1 and 3. This however is of no help, as the problem demands the naming of the door chosen at first and the door opend by the host, be it doors 1 and 3 or any other combination. Nijdam (talk) 18:49, 19 June 2014 (UTC)

Do the Door numbers matter?

I have been mulling this one over for a little while, and read the sources and can't come to an answer. We make the doors identifiable because in the formation they're referred to as "say 1" and "say 3".

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

If we consider the doors indistinguishable from the outside (with some methodology for them to be distinguished by the shows organisers in order to achieve the algorithm of the game) does that make a gnats fart of difference to the maths? I can't see why it would, except that it makes the conditional probability more obviously a pneumatic drill to crack a nut. SPACKlick (talk) 11:59, 26 June 2014 (UTC)

In my opinion your last comment is sums the situation up perfectly.
There has been endless discussion about this and similar topics in the past. Firstly, it is not clear from the above wording whether the questioner intended to number the doors. In fact, we know that vos Savant added the numbers to the original question from Whitaker, purely to make the explanation of the game easier; something she later said that she regretted doing.
Does it actually matter if the doors are numbered? If you believe that using conditional probability is necessary to solve the problem then it is perfectly possible to justify this even if the doors are not numbered, however it is interesting to note that nearly all the sources that promote the use of conditional probability (starting with Morgan et al, who misquote the question) ignore the ambiguity in the question and take the doors to be definitely numbered. Martin Hogbin (talk) 15:10, 26 June 2014 (UTC)

Tense

There has been some considerable discussion as to whether the actions of the host in opening a door to reveal a goat represent what he did on one occasion or what he always does. I do not think anyone has considered the way tenses are used in English The well known (W/vS) problem statement is:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The problem is written in the [simple present] (present simple) tense. This tense is generally used by English speakers 'to express the idea that an action is repeated or usual. The action can be a habit, a hobby, a daily event, a scheduled event or something that often happens'. For example 'He smokes', He drinks, 'He opens the door to the shop at 8:20'. If the action is not always repeated this is usually indicated, with the word 'usually' for example.

One-off events would be generally indicated by the [simple past] tense, as in 'The host opened a door' or, if currently taking place, by the [present continuous tense], as in 'The host is opening a door'.

I am sure that there must be exceptions to this general rule but at the moment I find it hard to think of one.

In the light of this standard English grammar and usage it is actually quite clear that the actions described are to be repeated, with the word 'say' indicating an element of choice at that point. Many languages do not have a present continuous tense and therefore can use the simple present tense to indicate one-off actions and this may be the cause of some confusion amongst non-native English speakers but to a native English speaker the meaning of the question is quite clear. This is what always happens. Martin Hogbin (talk) 08:25, 2 July 2014 (UTC)

Games using non-standard rules

I use the term 'standard rules' to mean that the host always opens an unchosen door to reveal a goat and always offers the swap. In response to suggestion that other rules might apply and that these might account for the common wrong answer of 1/2, I make the following observations.

The host does not have to open a door at all

The problem, as stated, cannot really apply to the case where he does not open a door for what happens next? Does the host offer the choice of either unchosen door? Does he not offer a swap at all? Even if we consider this a realistic option there is no uncontrived way to get the answer 1/2

The host can open the door chosen by the player

So then what? He asks the player whether they want to swap their car for a goat! Or swap their goat for a chance of getting the car! Not exactly rocket science, or an interesting game show.

The host does not always offer the swap

This allows the host to completely control the game. He has the power to make the player always win or always lose by swapping. That would not be permitted by any regulatory authority. In any case it does not easily lead to an answer of 1/2, unless contrive a specific host policy.

In the actual TV show we know that the host never offered the swap so this is irrelevant to the question.

The host can reveal the car

But there is then no problem to solve. We all know where the car is.

A challenge

I challenge anyone to describe a realistic scenario other than the standard rules and, in particular, one which leads to the answer 1/2. Martin Hogbin (talk) 08:51, 2 July 2014 (UTC)

Simple

1 The host always opens one of the two doors you didn't open, completely at random, if he picks the car, you've lost. In the test case he didn't pick the car. Odds by switching 50/50. The game element is that you have a 1/3 chance to stop him picking the car in the first place.

This is a different scenario still and one well covered in the article and the literature. It is essentially the same thing as vos Savant's little green woman. She and others say things along the lines of, If the host is clueless (as to where the car is) there is no advantage in switching.

2 The host doesn't always open a door and offer the switch. Sometimes you just get your 1/3 shot however, 50% of the times you pick the car and 50% of the times you pick goat 2 (or 25% of the times you pick any goat) the host reveals goat 3 and offers the switch. The test case is one of such times and it's 50/50 whether you have the car or a goat. This is reasonably realistic in that it allows the host to extend tension and runtime in shows where the recording runs short.

I do not agree that this is a realistic scenario. Firstly it is somewhat contrived, you have to assume a specific way in which the host acts and that certainly involves a lot more speculation than assuming the standard rules. Secondly, allowing the host to vary the chances of winning at will would not be permitted for any game show in the US or Europe with a car as the prize. Martin Hogbin (talk) 08:29, 3 July 2014 (UTC)
All this being said, i don't think it's useful, because the real issue at hand is how was the original phrasing to be interpreted and how was the original phrasing interpreted by those readers who reached the 1/2 and 1/3 answers. SPACKlick (talk) 11:02, 2 July 2014 (UTC)
Agreed, but with a little common sense there is really only one possible way to interpret the rules. Combine that with the meaning of plain English, as described above and the standard rules become just the obvious rules. You do not really need to interpret the question at all, just understand English. The simple present tense is used to describe habitual actions or things that are generally true, just like it is in this sentence. Martin Hogbin (talk) 15:48, 2 July 2014 (UTC)

You say “I found hundreds of examples of people disagreeing with the 2/3 answer. Many of these people state, in their reasoning, that Monty Hall will always open a losing door...” How many? And why did they state this? And was their reasoning entirely consistent with this premise? (Obviously not, or they would have gotten the 2/3 answer, right?)

Over 100 probably less than 250, I didn't keep the firm count after the first pass.SPACKlick (talk) 02:52, 3 July 2014 (UTC)

You say “The literal interpretation is naive, in that it doesn't take into account the context being a maths puzzle.” Who says it is a math puzzle? (Vos Savant deals with all kinds of questions, not just math puzzles.) And even if it is assumed to be a math puzzle, why does this imply that the host has no freedom of choice? The mathematics of the problem when the host has freedom of choice is far more interesting and challenging than the trivial little probability problem when the host has no choice.

The problem is presented as a maths puzzle, in Vos savant, in all the papers following, Selvin's original letter refers to it as "A problem in probability". It's a maths puzzle, not a question on real life tactics. The context of it being a maths problem is clear. SPACKlick (talk) 02:52, 3 July 2014 (UTC)

You say “It doesn't interpret the additional information given.” What additional information?

You say: “It simply translates the words, computationally, to predicates. It's not incorrect, but it's wrong, it’s not what’s being asked.” But it IS what’s being asked. I don’t know what you mean when you say “It’s not incorrect, but it’s wrong”. The point is, when solving puzzles whose answers depend on the precise wording (like the Boy or Girl Problem) it is absolutely essential to apply the stated conditions, no more and no less. Otherwise you’re answering a different question. I’ve asked you to tell me how the host restriction is implied by the wording or the context or whatever, but so far your replies haven't helped me to see how it is implied (and I remind you that many sources admit it is NOT clearly implied by the Parade problem statement).

Ok to repeat myself I'll quote myself.
• 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. All explixit so far. Three doors, one door has car, two have goats, you get to pick one.
• You pick a door, say No. 1 You have a free choice of doors
• and the host, who knows what's behind the doors, opens another door So the host, who uses knowledge of what's behind the doors opens the door. This door (another door) cannot be your door by implication (in what sense have you picked a door if the host is unconstrained by that choice and why would you say another door if the host was not constrained?). Here we have the first implication that the host's behaviour is constrained otherwise there would be no mention of his knowledge.
• Say No. 3, which has a goat. Here the implication is that whichever door you picked (Say 1) and he picked (Say 3) the door opened has a goat behind it. And now the restriction has been implied. Not stated but implied.
• He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? And so we have the question, "what tactic wins the car most often?". I still say the question clearly includes the restrictions it simply doesn't explicitly state them.
The verbs are in the present tense for the host, not the past tense implying continue or repeated actions rather than "the host opened a door which had a goat. This all implies the standard reading. The sources all say it's not explicit enough for, say, an exam question but that doesn't deny that the standard reading is intended, implied and inferred from this phrasing. SPACKlick (talk) 02:52, 3 July 2014 (UTC)

You say: “Even ANYA appears to understand the intended set up, because her "host motivation" caveat is itself caveated with " If Monty Hall, the game show host of Let's Make a Deal [was the host]" and "in the actual game.". You must be talking about a different Anya. The Anya I’m referring to didn’t say any of those things. All she said was “You never know Monty’s motivation for deciding to reveal a door, so I would definitely stick with my first choice”. There were no caveats, let alone caveats to the caveats.

To quote your source directly " you never know what motivates Monty to reveal a door in the actual game. I would definitely stick with my first choice." Emphasis mine. And also, for context from the source "Well, let's think. If Monty Hall, the game show host of Let's Make a Deal (remember this game?), opened a door and asked you if you wanted to switch" The statement is caveated.
Also, Am I right in assuming English is not your first language, otherwise I don't know how to explain to you the meaning of something being not incorrect but wrong. SPACKlick (talk) 02:52, 3 July 2014 (UTC)

You say: “The only times you see comments about a ... strategic host who [exercises choice] are when the problem is discussed by people who ALREADY understand the answer.” Anya commented on a strategic host. Are you saying she already understood the answer? And which answer did she understand? She said “I would definitely stick with my first choice”.

Again she said in the actual game, let's make a deal, with Monty Hall she'd definitely stick. That's not the same thing at all.

You say: “You also still seem to be suggesting that a real world game show host would NEVER work the way the MHP's host does, to which I say, tosh. The opening of a losing door builds suspense, extends the show...” Here you’re missing the point entirely. You’re making the case for why the host in a real game show (and remember, Craig’s questions said we are on a game show) would exercise choice in deciding when and whether to show a goat and offer a switch. He wouldn’t do it all the time – that would be ridiculous and boring – but he would do it some of the time, and he would decide when to do it based on his motivations, which presumably are to make the show as entertaining as possible, and not to give people obvious choices. So now you have all the information you need to figure out the answer to Craig’s question. Can you do it?IVLeeg (talk) 00:13, 3 July 2014 (UTC)

I'm not getting further drawn into this irrelevance as it's not craig's or selvin's question. SPACKlick (talk) 02:52, 3 July 2014 (UTC)
I agree. Nearly all reliable sources say that the question in Parade was understood to have the meaning intended by its author my most people who answered the question. Alternative interpretations, and the way that they change the solution are covered in the article already.
The problem is loosely based on Let's make a Deal; we know that on the real show there were never any goats and Monty never offered a door swap. We can also say that, if the MHP were a TV game show, the regulators would not allow the host any freedom to influence the game at will. Good examples of this are Deal or No Deal or Who Wants to Be a Millionaire?, the host can joke around but can never influence the player's chance of winning. This would rule out most alternative scenarios for the MHP. So whatever way you look at it, the standard rules are the ones to use.
The MHP was intended to be, and is, a simple probability problem that most people understand perfectly but still get wrong. That is what makes it remarkable. It is a wonderful example of how human intuition can get things completely wrong; lets not spoil it. Martin Hogbin (talk) 08:46, 3 July 2014 (UTC)
The thing that fascinates me about people, like IVLeeg, who try and defend the naive interpretation of the question, is that they ignore the fact that even when the assumptions are explicit, as in the version below, the success rate is similar or in many cases statistically indistinguishable. The "fascinating" idea within the "veridical paradox" is that our intuitive "folk probability" is not built for this sort of puzzle, we don't instinctively know how to assess the available information and yet the problem seems simple enough that our "folk probability" is confident in the answer of 50:50.
• You are presented with a game. Before you are three [door]s, [behind] one of which is a [car] and [behind] the other two [goat]s. The Gamesmaster instructs you to pick a [door], after which he will reveal a [goat] from [behind] one of the other two [door]s. At this point you will be offered the choice of which remaining [door] you wish to open for your prize. Which [door] should you pick to maximise your odds of winning the [car]? The first [door] you chose or the other unopened [door]?
I imagine many people would offer a different answer to a question specified below which makes the opposite of all assumptions.
• I'm asking for advice because you're my phone a friend on a game show totnight. The final round (I got there) works a lot like the old Monty Hall Problem. There are 3 Containers ((1)A red box, (2)a blue bucket, (3)A yellow satchel), one with a prize (I'm hoping for the keys to the porsche, which I also hope comes with a porsche) two with zoinks. I was given the opportunity to pick a container (I chose the red box) and then sometimes the host reveals a door which sometimes is a zoink and if it is then sometimes I get the chance to switch. Well he showed me a Zoink in the Yellow satchel and offered me the switch. What can I do to maximise my odds of getting the prize, take the red box or the blue bucket?
In this version, similar to what IVLeeg seems to be claiming the original MHP is, it's clear this is 1) a real game show where 2) The host acts with human commercial motivations and not as an agent of a puzzle of chance and 3) He has some choice in whether to open any door and 4) He has some choice in whether to reveal a car and 5) he has some choice in whether to offer the switch and 6) the contestant has chosen (1) and the host has opened (3) to reveal a goat. This problem is the one I contend that you have insufficient information but with a uniform distribution switching wins C.56% of the time and for which I'm drawing a probability tree. That still means switch is more likely to win. [My back of the envelope calculations suggest that for all host preferences of the three objects a uniform distribution cancels those out and that symmetry makes the specific container numbers/designators irrelevant]. SPACKlick (talk) 13:32, 4 July 2014 (UTC)
Right, just to finish off calculating from the probability tree, I've made the assumption that if Monty reveals a door it is eliminated and therefore if he reveals the car you lose. The contestant wins, overall (By strategy always switch if offered), 53/144 . The game appears as per the problem (pick door host opens another door to reveal a goat) 11/36 of the time and of that subset contestant wins by switch 6/11 (54.[54]%) of the time. Pretty close to my 56%. Anyone know a good image hosting site so I can post the tree? SPACKlick (talk) 14:18, 4 July 2014 (UTC)
Yes that does seem to be what is being suggested by some, but even in that scenario the standard MHP assumptions are quite realistic and, where they are not, it is because the stated problem is unrealistic.
Game shows in most countries are quite tightly regulated and the rules do not generally allow the host or producers to vary the players odds of winning at will. 'Sometimes the host reveals a door' would be unlikely to be allowed unless this was the result of some specified process not under the control of the host, like 'if I can throw a ball into the bucket the host will open a door', or 'the computer will now randomly decide if the host will open a door'. In a real game show it would need to be specified at the start whether the host can reveal the prize or not and what then happens if he does do that. (What would happen if the host revealed the car is conspicuously missing from the Parade question, suggesting that it cannot happen). So even in the above scenario my advice would be to switch if you can. Martin Hogbin (talk) 16:34, 4 July 2014 (UTC)
One other thing. The assumption that, even if it were allowed by the regulators, the host would try to stop you winning the car is not necessarily correct. The cost of prizes is comparatively unimportant compared with the benefit getting a large audience and audiences generally like to see people winning. Having a host who conspired to prevent the player from winning would make the show a real turn-off. Martin Hogbin (talk) 16:56, 4 July 2014 (UTC)
A real game with non-standard rules

Just for fun, here is what I think is need to ensure non-standard assumptions about the rules:

I crash landed on the planet Zarg. I was unhurt and the locals seem friendly although the seem different from us in many ways. Bizarrely, they invited me to play on a quiz show. I'm asking for advice because you're my phone a friend that show totnight. The final round (I got there) works a lot like the old Monty Hall Problem. There are 3 Containers ((1)A red box, (2)a blue bucket, (3)A yellow satchel), one with a prize (I'm hoping for the keys to the Porsche, which I also hope comes with a Porsche (and that they will transport it back to earth for me) and two with zoinks. I was given the opportunity to pick a container (I chose the red box) and then sometimes (I have no idea when or why) the host reveals a door which sometimes is a zoink and sometimes not, with these people you can never tell, and if it is then sometimes (who knows when) I get the chance to switch. Well he showed me a Zoink in the Yellow satchel and offered me the switch. What can I do to maximise my odds of getting the prize, take the red box or the blue bucket?

Now you can only apply Bayesian inference to all the choices for a set of unspecified rules and calculate a probability of winning by switching on that basis. Martin Hogbin (talk) 16:51, 4 July 2014 (UTC)

My gut instinct on that problem is that all aspects of the behavior are capable of symmetry as any protocol the host may be using can be adjusted with the container to the right of, etc. Therefore the results are almost certainly 50/50, but I would switch because I pretty much always switch when offered, even in "pick a number between one and ten. Got one? would you like to change your mind?" I always change. SPACKlick (talk) 10:10, 7 July 2014 (UTC)
You may be a natural switcher although most people seem to be natural stickers. I guess the psychology is that if you switch and lose it is your fault but if you stick and lose it is just bad luck.
The point I was really making was that you actually have to try quite hard to get a scenario like the MHP in which there is no advantage in switching. Apart from the planet Zarg the only other place it might be better to stick would be on a street-corner game, where the host will definitely be trying to mislead you, but there the best advice is not to play at all. Martin Hogbin (talk) 11:53, 7 July 2014 (UTC)

A test worth seeing

It would be nice to see the results of a study without door ownership. Similar to Krauss and Wangs Full Intuitive Version, It would put the person being questioned in the role of the host, only specify the first choice of the contestant, ask in terms of frequency, except rather than the contestant "choosing a door, say number 1" the contestant would "Divide the doors into two groups (of 1 and 2 doors), say "Door 1 and Doors 2 & 3". In this way none of the doors becomes the contestants. My Sheet for the test group would read.

There is a game show where a contestant is shown three closed doors. Behind one door is the first prize, a new car; behind each of the other doors is a goat. The contestant must divide these doors into two groups (a group of 1 and a group of 2). According to the rules of the game, the game show host, You, who knows what is behind the doors, now has to open one of the doors in the group of 2 to reveal a goat. After You show a goat to the contestant, You must ask the contestant to decide whether s/he wants to open the door in the group of 1 or the remaining door in the group of 2.

The contestant divides the doors, say into [Door 1] and [Doors 2&3]

Image of the doors with boxes round and uneven spacing, also showing contestant and host

You must then, according to the rules, open one of the doors in group 2 to reveal a goat. Now you ask her whether s/he wants to open to door in group 1 [Door 1] or the remaining door in group 2.

There are three doors behind which the car can be.

In how many of these possible arrangements would the contestant with the car after you reveal a goat?,

-if s/he opens the door in group 1? in____out of 3 cases

-if s/he opens the door in group 2? in____out of 3 cases

What should the contestant therefore do...etc etc etc

I think it'd get some differing results. SPACKlick (talk) 10:34, 7 July 2014 (UTC)

I think it would be a great idea to do some online experiments to continue investigation, along the lines of K&W, into how people understand and try to solve the MHP. I did start something once but did not continue, mainly because it is not proper use of WP's space. We could try an experiment and see if anyone objects. I do have access to a private wiki on which we could set up some tests but we would need a means of driving subjects to it. Maybe an external link from the WP would be acceptable. Martin Hogbin (talk) 10:48, 7 July 2014 (UTC)
I can get my hands on test subjects reasonably easily (a couple of hundred minimum, however there is a selection bias, they're all students/graduates so might be more likely to be familiar with the MHP. SPACKlick (talk) 12:24, 7 July 2014 (UTC)

The full conditional solution

1) The hosts preference may not be uniform across doors or goats

2) If the hosts preference varies, we cannot reduce by symmetry.

We consider all doors and goats distinguishable and labelled prior to the scenario Doors {1,2,3} Prizes {Car, Goat A, Goat B}. We assume the prizes are equiprobably randomly distributed behind the doors.

```C door behind which the car is hidden {1,2,3)
X door chosen by contestant {1,2,3}
H door opened by the host {1,2,3}
G Which goat is revealed {A,B}
```

We analyse the preferences of the Host; There are 6 independent possible arrangements

```CAB, CBA, ACB, BCA, ABC, BAC
```

And the host can have differing preferences in all 6

```t = Preference for Goat A when it is behind door 2 and The car is behind door 1
u = Preference for Goat A when it is behind door 3 and The car is behind door 1
v = Preference for Goat A when it is behind door 1 and The car is behind door 2
x = Preference for Goat A when it is behind door 3 and The car is behind door 2
y = Preference for Goat A when it is behind door 1 and The car is behind door 3
z = Preference for Goat A when it is behind door 2 and The car is behind door 3
```

So, in a fully conditional solution we're looking for

P(C=2|X=1, H=3, G=A) = P(C=2,X=1,H=3,G=A)/P(X=1,H=3,G=A)

X=1 is not a probabilistic event, but a decision so it simply defines the universe

```Doors XCHG, XCHG
CAB   112A, 113B
CBA   112B, 113A
ACB   123B
BCA   123A
ABC   132B
BAC   132A
```

So our possibilities for the 8 outcomes are

```112A t/6
113B (1-t)/6
112B (1-u)/6
113A u/6
123B 1/6
123A 1/6
132B 1/6
132A 1/6
```

For our conditional P( A | B) bold and italics above to show which lines are relevant

```P(X=1, H=3, G=A) = u/6+1/6+=(1+u)/6
P(C=2, X=1, H=3, G=A) = 1/6
P(C=2 | X=1, H=3, G=A) = (1/6)/((1+u)/6) = 1/(1+u)
```

This is essentially the morgan solution with a more specific characteristic for the host.

You can reduce the preference variables from 6 to 4 by using

```m = preference for goat A
r = Preference for Door 1 over Door 2
q = Preference for Door 1 over Door 3
p = Preference for Door 2 over Door 3
```

with a conversion of

```u = m(1-p) / (m(1-p)+(1-m)p) = m(1-p)/(m-mp+p-mp) = m(1-p)/(m+p-mp)
```

Which makes our conditional solution

```P(C=2 | X=1, H=3, G=A) = (m+p-2mp)/(2m+p-3mp)
```

SPACKlick (talk) 12:28, 24 July 2014 (UTC)

Worth noting with the final steps, if m=1 and p=1 or m=0 and p=0 the problem is undefined, this is because this represents a 100% preference for Goat A and Door 2 or 100% preference for Goat B and Door 3 and we're looking at when Goat A is revealed behind Door 3 so for our situation to be possible 0<=p<1, 0<m<=1. Whereas u can occupy the full range 0<=u<=1 SPACKlick (talk) 13:41, 24 July 2014 (UTC)
Worth noting, the double integral from 0 to 1 of the final equation is 70.43%
The integral with m=0.5 or r = 0.5 is 69.31% which is the same as the integral of (1/1+u)
Both of these values are significantly higher than 2/3. So if the host chooses preferences at random rather than choosing a door at random (and the player has that information) he makes the odds even better for the player. Although the increased pop from the u solution to the mpqr solution makes me think the mpqr solution is probably not accurate. It relates t and u (and all the others) to eachother and they are in fact unrelated. SPACKlick (talk) 14:14, 24 July 2014 (UTC)
Firstly, do we agree that the Morgan solution is not complete in that it does not allow for a possible goat preference by the host? Martin Hogbin (talk) 15:46, 24 July 2014 (UTC)
Yes, I fully agree that Morgan adds not all of the conditions he could add making his solution a weird halfway house between "Solving the intended problem" and "complete"SPACKlick (talk) 16:15, 24 July 2014 (UTC)
That was exactly my point.
Regarding your solution above, I think you may have gone wrong at some stage. I will add in comments above but keep them brief to avoid losing the thread. We can discuss here. Martin Hogbin (talk) 08:59, 25 July 2014 (UTC)
Very possible, I'm a philosopher not a mathematician, it's been a while since I've had to do so much maths. SPACKlick (talk) 11:25, 25 July 2014 (UTC)
I am a physicist, so not an expert either. I do not quite follow your notation. Nijdam is an expert on probability and should be able to help if we can persuade him that there is something in the goat choice.
I will have another go at a solution. Martin Hogbin (talk) 14:15, 25 July 2014 (UTC)

The Initial

Martin can you go through this initial information and tell me if there's any you think the contestant doesn't know initially or anything you think the contestant knows that is missing. If we can agree on this it will make further discussion much easier

1. There are three doors, identifiable by their position I arbitrarily label them 1, 2 and 3 Door1(Left, Closed, Unchosen, infront of something}, Door2(Middle, Closed, Unchosen, infront of something}, Door1(Right, Closed, Unchosen, infront of something}
2. There is a car Car{Unchosen, Hidden, behind Doora}
3. There are two distinct goats which may be identifiable by their traits I arbitrarily label them 1 and 2 Goat1{Unchosen, Hidden, behind Doorb, has traits G1} and Goat2{Unchosen, Hidden, behind Doorc, has traits G2}
4. Doors a,b&c are Doors 1,2&3 in some order
5. The contestant chooses a door and changes its property to chosen Doorx{position, Closed, Chosen, infront of something}. The contestant will know whether x=1,2 or 3. and therefore the position of x
6. The above step changes the property to chosen for either the Car, Goat1 or Goat2 but we don't know which
7. The host then opens a door, Doory{position, Open, Unchosen, infront of Goatn} The contestant will know whether y = 1,2 or 3 and therefore the position of y.
8. The revealed goat changes the property to revealed Goatn{Unchosen, revealed, Behind Doory}.
9. We cannot tell if n=1 or n=2.
10. We cannot tell if y=b or y=c but y=/=a
11. This will leave Doorz{position, Closed, Unchosen, infront of something} The contestant will know whether z = 1,2 or 3 and therefore the position of z
12. We can now map Doors 1,2,3 on to Doors x,y,z but neither on to Doors a,b,c

Do you agree with all 12 of these initial statements? Are there any of them you would like to re-word, add to, take away from? Once we've agreed that, we'll move on to whether there are any statements missing. SPACKlick (talk) 12:05, 2 August 2014 (UTC)

1

I agree that there are three doors and that they are (like all doors) distinguishable in principle.

It is not clear what you mean by 'identifiable'. The problem statement tells us nothing at all about the doors themselves; the doors can only be identified by their eventual roles in the problem. We have the door that is originally chosen, the door that the host opens, and the remaining door. Why not label them at the start X, H, and R (or by tradition 1, 3, and 2 respectively)? Any additional or alternative labelling is not only superfluous but distracting.

Although I would point out that it is an assumption I am happy to assume, if you like, that we can see all the doors at the start.

Right, we need to stop you where you say "Why not label them at the start X, H, and R (or by tradition 1, 3, and 2 respectively)? Any additional or alternative labelling is not only superfluous but distracting.". The doors have identities separate from their functions in the game.
The situation where X=1, R=2, H=3 is different from X=1, H=2, R=3 but the doors could be the same. When we first see the doors, before any of the game has been played We can tell which door is which but we don't know which we'll choose, which will be opened or which hides what. So we could label them Doorleft, Doormiddle, Doorright or Door1, Door2, Door3 or similar. We cannot label them by their function yet.
Also, that the doors are identifiable (which would be trivial if you could see them) is not an assumption, the contestant is asked to choose one of the three, an action only possible if they can identify all three doors.

By Identifiable I mean that the contestant can distinguish them. If the three doors differed only by their contents, as in the scenario where the contestant cannot see any of the doors to start with, while they are distinct they are not identifiable or distinguishable. SPACKlick (talk) 14:43, 2 August 2014 (UTC)

2

There is a car behind one of the doors. Initially we assume each door is equally likey to hide the car.

STOP, Stop adding to what we're doing, we're simply trying to list what the contestant knows about each object in the universe and what he knows he will or won't know. So He knows there is a car, which is currently behind a closed unchosen door. Agreed? SPACKlick (talk) 14:44, 2 August 2014 (UTC)
3

There are two distinguishable (like all goats) goats but we know nothing else about them. The goats are not behind the door that hides the car. One goat is eventually revealed, one is not. So, like the doors, we can lavel them R and H (or 1 and 2 respectively if you prefer) Any additional or alternative labelling is not only superfluous but distracting.

We cannot see any of the goats at the start of the game

I think we already disagree enough for me to stop here.

Where we differ is that you want to label the doors with some arbitrary labels that have no connection with their eventual role in the problem. Martin Hogbin (talk) 14:26, 2 August 2014 (UTC)

Where you say " There are two distinguishable (like all goats) goats but we know nothing else about them."" You seem to be using Distinguishable to mean Distinct. They may not be distinguishable, they may look identical and be identical twins.
"So, like the doors, we can label them R and H" No this isn't true, just as with the doors, at this stage we don't know which will be the hosts goat or which will be the remaining goat. And since either goat could be behind any door, labelling the same hides information.
By labelling the objects by their role in the game BEFORE the game you are already missing the fact that we don't know at the start of the game which door the host will reveal but we do know which door is which and that ANY of the three doors could be the one the host reveals.
If, as you are saying, the addition of labels to the doors before they are identified by function is superfluous, it should yield the same results as doing it your way. Since it doesn't would you not agree that these labels are providing some function (even if you later dispute the accuracy of that function)?SPACKlick (talk) 14:49, 2 August 2014 (UTC)
Please consider that up to point 4 is the information the contestant has about the set up without any of the action of the game taking place. That is just the "There are 3 doors" "behind the three doors is a car and 2 goats". Whatever happened next in the problem that information would be the same. From 5- 12 is what he thinks he will know after that point but still being considered before any part of the game has taken place. I haven't moved on to additional information on which we can condition, I'm just trying to get you to understand what information is available in the initial against which we can condition. SPACKlick (talk) 15:00, 2 August 2014 (UTC)

Analysis

You both do rather well, not being experts. I'll try to give you the necessary tool. Most of them you allready have provided. So we have X, C, H and G. G being the door number of goat A. For the sake of simplicity I consider conditional probabilities given X=1. I summarised on my talk page:

XCGH
1122 - 1/6 p
1123 - 1/6 (1-p)
1132 - 1/6 (1-q)
1133 - 1/6 q
1213 - 1/6
1233 - 1/6
1312 - 1/6
1322 - 1/6

but now introduced not only p = P(H=2|C=1,G=2), but also q= P(H=3|C=1,G=3), being the conditional probabilities for the host' s strategy to open the door with goat A in the cases where A is behind 2 or behind 3. Of course car and goats are placed at random, and X is independent from this placement. We now may calculate anything we want to know. For instance:

P(C=2|H=3)=(1/3)/(1/2+1/6(q-p))

This may well differ from 2/3. Martin, is this you're aiming at? Nijdam (talk) 22:15, 25 July 2014 (UTC)

Yes pretty well. It is, I think even simpler.
We three have solutions: the simple solution, Morgans solution, and the one that I want, that takes account of goats.
The simple solution calculates P(C=2|X=1), Morgan calculate P(C=2|X=1.H=3), I want to calculate P(C=2|X=1.H=3,G=3)
In all three cases the probability is given by 1/(1+q).
For the simple solution q = 1/2
For Morgan's solution q = 1 - P(H=3|X=1.C=1) (They do not actially mention X=1)
For my solution q = 1 - P(H=3|X=1.C=1.G=3)
That is it! That probability, which we have both mentioned before, lets us calculate what we want.
The above probabilty can be anything from 0 to 1.If the host cares nothing for goats but does care about doors we have Morgan's solution again. If the host cares about goats but not about doors then P(H=3|X=1.C=1.G=3)=1 only if the preferred goat is behind door 3. If the host has some intermediate preference then we have to calculate, on the basis of his relative preferences for goats and doors, what the above probability is.
So, as SPACKlick has agreed, the Morgan solution, as stated by them, is a strange halway house between the simple solution and the full condional solution. Martin Hogbin (talk) 21:03, 26 July 2014 (UTC)

The simple "solution" in its simpliest form just calculates P(C=X) =1/3 and reasons that getting the car means switching, and "hence" the probabilty must be the complement. The form you mentioned P(C=2|X=1) is not adequate, as P(C=2|X=1) = 1/3. What you call the Morgan solution is indeed P(C=2|X=1, H=3), eventually in its general form, with host preference for a door. Your P(C=2|X=1, H=3, G=3) is something different, as you add the knowledge of goat A being revealed, what we do not know. I calculate, just like Morgan, P(C=2|X=1, H=3, G=3)), but taking into account preferences for goat and door of the host. Nijdam (talk) 08:08, 27 July 2014 (UTC)

OK, I agree that the simple solutions actually just calculate P(C=X) and its complement. The Morgan solution claims to calculate P(C=2|X=1, H=3) but does not actually quite do this as it does not condition on X=1. If course, it is easy to see that if the car and goats are placed uniformly at random this does not matter but, if we are being fussy, we should do it explicitly.
My main point is that we should realy calculate P(C=2|X=1, H=3, G=3) explicitly. We have no more knowledge about the door opened than we do of the goat revealed. In both cases the object is the one chosen (either directly or by proxy) by the host in the instance of the game that we have observed. The Morgan solution is incomplete.
It may appear that doors are essential to state the problem but I do not think this is necessarily so. I will present an alternative solution belew that shows that the role of the doors is no more important than that if the goats. Martin Hogbin (talk) 09:37, 27 July 2014 (UTC)
No, Martin, Morgan correctly calculates P(C=2|X=1, H=3) under the assumption of no preference of the host for a door to open. I had the idea that your argument against Morgan would be the diffrent outcome in case of the divers host preferences. Your point, as you stipulate, as if we are to calculate P(C=2|X=1, H=3, G=3) is not correct; we do not know which goat is revealed. Of course we may change the problem formulation, or we have to find some reasoning that seeing a goat would indicate knowing which goat it is. Nijdam (talk) 07:26, 28 July 2014 (UTC)
We do know which goat was revealed, we see it. We do not need to see the other one for us to know if the host reveals a different goat in a different instance of the game. Martin Hogbin (talk) 19:07, 28 July 2014 (UTC)
Instead of your p I introduced t-z for combined goat/door preferences (later m for goat preference and p,q&r for door preferences) Only t,u,m&p are relevant when x=1 so they are the only ones in the maths, it ends up being nearly identical. Below shows our notations adjacent eachother. To make sure I'm understanding
In mine it's XCHG (where G is which goat was revealed)
in yours it's XCGH (Where G is which door hid a specified goat).
my u is preference for door 3 with goat A behind it when the car is behind door 1
my t is preference for door 2 with goat A behind it when the car is behind door 1
Your p is preference for door 2 with goat A behind it
Your q is preference for door 3 with goat A behind it
Morgan's p is preference for door 3 when the car is behind 1
```112A, 1122, t/6,     1/6 p,   1/6 1-p
113B, 1123, (1-t)/6, 1/6 1-p, 1/6 p
112B, 1132, (1-u)/6, 1/6 1-q, 1/6 1-p
113A, 1133, u/6,     1/6 q,   1/6 p
123B, 1232, 1/6,     1/6,     1/6
123A, 1231, 1/6,     1/6,     1/6
132B, 1323, 1/6,     1/6,     1/6
132A, 1322, 1/6,     1/6,     1/6
```
Firstly, your and my notation are equivalent with G notating the same facts differently. and t=p, u=q. I'll use your notation from now on. Morgan's p is not the same as your p so I'll use Mp for Morgan's p. In the Morgan calculation, because the preference is independent of goat location so P(C=2|X=1, H=3, G=3) = P(C=2|X=1, H=3, G=2) = 1/(1+Mp). Whereas accounting for the effect goat location can have, P(C=2|X=1, H=3, G=3) = 1/(1+q) and P(C=2|X=1, H=3, G=2) = 1/(1+p) with p and q being independent.
This addition of a distinctive goat's location on host preference is more complete because it allows for the fact that 1/(1+p) + k = 1/(1+q), -0.5<=k<=0.5, 0<=p<=1, 0<=q<=1. SPACKlick (talk) 00:35, 26 July 2014 (UTC)
I think thta we have all converged on the same answer. See my response to Nijdam above. Martin Hogbin (talk) 21:04, 26 July 2014 (UTC)

What is the question?

All the above being said, I still do not agree that this approach answers the correct question. Given that in what we are asked we are put in the perspective of the contestant making a decision. And given that we are told in the continuous/simple present tense. The correct formulation of the question is Find which is greater (A) P(C=x|X=x H=h G=g) or (B) P(C=/=x|X=x H=h G=g) across all C,X,H,G. If the answer to that is (A) when H=xmod3 + 1 else (B) that would be acceptable, as it turns out in all cases (B)>=(A). Trivially in the above it can be shown that seeing one goat doesn't provide information to the contestant about which goat as he's only seen 1, so G can be removed. Similarly it is trivial to show that all the doors are equivalent through symmetry so you can show H is irrelevant and that X is irrelevant. It simplifies to which is greater (A) P(C=x) or (B) P(C=/=x). It's further reasonably simple to show that the host revealing a goat from behind a door doesn't change either of those probabilities and therefore on average P(C=x)=1/3 with the information available to the contestant. It's an interesting discussion to show that the host can never make the probability greater than 1/2 but that's an extension to the question. SPACKlick (talk) 12:28, 28 July 2014 (UTC)

Presumably you mean to say: which one is the greater: P(C=x|X=x H=h G=g) or P(C=/=x|X=x H=h G=g) across all c,x,h,g (or across all values of C,X,H,G). But no, the value x of X and h of H are known to the contestant, hence form a condition to the problem. To reformulate: given x and h, which one is the greater: P(C=x|X=x H=h G=g) or P(C=/=x|X=x H=h G=g) = P(C=r|X=x H=h G=g) (with {x,h,r} = {1,2,3}) across all possible g (?) But then, what do you mean by "across all g"?? Nijdam (talk) 12:46, 28 July 2014 (UTC)
Sorry, you're right, I should have said across all values of C,X,H,G. And yes in the individual case the value of x and h are known, but they are not specified. The question as asked demands consideration of the tactic in (X=1,H=2), (X=1,H=3), (X=2,H=1), (X=2,H=3), (X=3,H=1), (X=3,H=2). And in each of those cases you must consider C equal to X or (to use your new notation) R. By "across all g" I mean with the two goats in either order. As I showed beneath my statement g can be trivially removed from the problem as the contestant only becomes aware of 1 goat and so cannot distinguish it from the unknown goat in any meaningful sense. SPACKlick (talk) 13:10, 28 July 2014 (UTC)
Okay, understood. Nijdam (talk) 19:59, 2 August 2014 (UTC)
We are told, are told that there are two goats. One is revealed to us, the other is not. They are therefore distinguished as much as the doors are. Martin Hogbin (talk) 19:04, 28 July 2014 (UTC)
In fact they are not. We (in perspective of the contestant) have information to distinguish the doors (all three doors are available to us). If I told you the problem contained three doors and showed you only one door, you couldn't condition on the identity of that door, only its function as the door revealed. The only information you have about the revealed goat is that it is revealed. You already had the information that there were 2 goats and you already had the information that one would be revealed. The extra information is that "this goat" was revealed but "this goat" is indistinguishable from "a goat" without some information to distinguish the 2. SPACKlick (talk) 09:15, 29 July 2014 (UTC)
Really? – Distinguishable or indistinguishable? See the literature. We do not need numbers or colours in order to distinguish. It's enough to distinguish their specific roles: door initially selected by the contestant (i) and the two other doors; the one opened by the host (o) and the second one that remained unopened (u). As well as the two goats: the one that is shown (s) and the other one that is not shown (n). So of course we are able to condition on i, o, u, s and n, likewise. Gerhardvalentin (talk) 11:54, 29 July 2014 (UTC)
Yes really. You can label the revealed goat as the revealed goat. But it presents no information about goat revealing, see the new section belowSPACKlick (talk) 11:59, 29 July 2014 (UTC)
With regard to what the 'real' question, this has been discussed at some length before without much agreement. It is true though that the correct answer does really depend on exactly what the question is taken to be. Richard Gill wrote a paper entitled something like. ' The Monty hall problem is not a probability problem but a question of mathematical modelling'. My answer what the 'real' question is is that given by Seymann in his comment at the end of the Morgan paper. 'Without an understanding of the precise intent of the questionner, there can be no single correct solution to any problem'.
The point I was making is not that that we must care about the goats, only that, if we chose to care about doors, then we should also care about goats.
I have also made the point that it is hard to see what question Morgan's solution answers. Is it only the case where the player initially chooses door 1 and the host opens door 3? No one seems to think so (I think). The problem is, as you say above, that if we go beyond that and take the described games as an example only then any host preference will cancel out and therefore be irrelevant. I showed this and Richard Gill stated this from a different perspective, that of game theory. If we assume that the player wants to win the prize and the host wants to stop him winning all the player needs to do is pick a door at random and always swap to win the car with 2/3 probability, no matter how the car and goats are placed and what the host's door policy is. Martin Hogbin (talk) 19:04, 28 July 2014 (UTC)
Concerning the two goats, it is not evident they may be distinguished. At least nothing in the problem formulation tells us how we should tell goat 1 from goat 2. In the next instance of the game, we may well notice whether it is again door 3 that's opened or not, but we are in no positoin to tell if it is the same goat being revealed. Nijdam (talk) 19:59, 2 August 2014 (UTC)

An alternative approach

Just to make the point clearer, I have changed the question a bit to:

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, a black one and a white one. You pick a door and the host, who knows what's behind the doors, opens another door which has a goat, say the black one. He then says to you, "Do you want to the change to the other unopened door?'. Is it to your advantage to switch your choice?

We might reasonably answer the problem this way:

We have the variables X, the player's initial choice, which could have any of the values c, w, or b; and H, the host's choice of what to reveal, which could have the values w, and b.

The sample space consists of the following events:

(X=c) (H=b)
(X=c) (H=w)
(X=b) (H=w)
(X=w) (H=b)

with probabilities (usual assumptions):

1/3 P(H=b|X=c)
1/3 P(H=w|X=c)
1/3
1/3

respectively.

Conditioning on the black goat being revealed and setting p=P(H=b|X=c)leaves:

(X=c).(H=b) Probability p/(1+p) loses by switching
(X=w).(H=b) Probability 1/(1+p) wins by switching.

Of couse, we do not allow simple solutions that do not distingush between the goats and condition on the goat revealed. Martin Hogbin (talk) 16:13, 27 July 2014 (UTC)

This is not sufficiently set up, I'll come to it later. Nijdam (talk) 07:29, 28 July 2014 (UTC)
I'm not sure what you want to achieve. Anyway: the variables C,X,H,G remain as I defined them (G being the door of the black goat) with values 1, 2 and 3. When you refer to the event (X=c, H=b), I guess you mean "the car is behind the chosen door, and the host reveals the black goat", this event is {X=C=1 and H=G=2 or X=C=1 and H=G=3 or X=C=2 and H=G=1 or X=C=2 and H=G=3 or X=C=3 and H=G=1 or X=C=3 and H=G=2}. and in the same way for the other events. Is this how you want to describe the prolbem? Nijdam (talk) 12:54, 28 July 2014 (UTC)
No, the doors are ignored, just as the goats are ignored in Morgan's solution. The values b, w, and c refer only to the objects being offered as prizes. Apart for doors themselves, every aspect of the game is contained in the above formulation: the player's origianl choice of object, the host's choice of object to reveal, and the offer of a swap to the remaining object. I would bet that many people would accept that solution a perfectly correct, after being gently guided to treat the goat choice as significant but the door choices as not.
If course, someone on Wikipedia could come along and say, 'What about the doors?'. Clearly there are three doors and we can distinguish between them by ther roles in the described example of the game. You have set p to be P(H=b|X=c), that it is to say the probability that the host will reveal the black goat given that the player has originally chosen the (hidden) car. Maybe the host has a bizarre door preference doors. Let us number the doors ourselves, the door hiding the object that the player originally chooses we will call 1, the door that the host opens to reveal the object of his choice we will call 3, and the door hiding the remaining object we will number 2. Maybe his choice (when he has one) of which goat to reveal depends on which door hides the car and which door hides the black goat. In the example, we know that door 1 hides the car and door 3 hides the black goat, so we should include that information in our conditioning. The value to use for p is therefore P(H=b|X=c|3=b|1=c). That still boils down to how the host chooses which goat to reveal though. Martin Hogbin (talk) 17:52, 28 July 2014 (UTC)
???This leads nowhere as far as the MHP concerns. In the MHP the doors are distinguishable; there is no doubt about this fact.Nijdam (talk) 20:56, 28 July 2014 (UTC)
All doors are distinguishable. Do you not agree though that the above solution is valid for the question stated above?
Simply, no! The player knows the door she picked, the door the host opened, and the color of the goat.Nijdam (talk) 18:41, 29 July 2014 (UTC)
The absurdity of misinterpretation

Excellent Martin, you hit the mark. Anyone knows that the intended paradox is a pure abstraction, no ifs, ands or buts.
All we know about the host is confined to his extremely biased role only, and never anyone will know better, because this so called "show" never occurred nor will occur in real life.

Your example unmasks and exposes the absurdity of Morgan's misinterpretation of the intended abstraction, by your saying:

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, a black one and a white one. You pick a door and the host, who knows what's behind the doors, opens another door which has a goat, say the black one. He then says to you, "Do you want to the change to the other unopened door?'. Is it to your advantage to switch your choice? – Of course, we do not allow simple solutions that do not distinguish between the goats and condition on the goat revealed.

Excellent. You helped to unmask absurd misinterpretations. Morgan et al. have failed in two respects: First, they forgot about the goats (the one we see and the other one we didn't see), and secondly, they failed to emphasize that they never did address the intended paradox "per se" (being a pure abstraction), but just were pondering about educational material for training purpose in probability calculus. Your contribution will help the article to get clearer and more precise. Gerhardvalentin (talk) 20:41, 28 July 2014 (UTC)

I have yet to persuade Nijdam. Martin Hogbin (talk) 22:33, 28 July 2014 (UTC)