Talk:Monty Hall problem/Archive 32

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Is there a consensus to add the new proposal by Coffee2theorems?

Should we add [1] for new text for the "Solution" section? — Preceding unsigned comment added by Martin Hogbin (talk • contribs) 16:35, 5 October 2012

Yes

I'm fine with it (count me as a yes). -- Rick Block (talk) 05:34, 6 October 2012 (UTC)
Yes its nice and simply written.--Salix (talk): 13:37, 6 October 2012 (UTC)
Yes. I think Ningauble's suggesting of also giving prominence to vos Savant's table, and a little more on the difference in outcome being because monty is picking goats is also good, and not mutually exclusive with this. --Elen of the Roads (talk) 22:28, 6 October 2012 (UTC)
That was Gerhardvalentin's post, not mine. I was thinking along similar lines because it seems hard to omit the most widely cited solution (by those who love it and those who hate it); but I decided against because the table is equivalent to and redundant with the diagram with three cases (which is based on it) and I think the latter presentation is easier to apprehend. ~ Ningauble (talk) 23:38, 6 October 2012 (UTC)
Yes. There are still suggestions of improvements to the proposal, and there's the question of how to implement putting it into the article (where should the rest of the material in the "Solutions" section go?). However, if we can agree that the basic approach is good and that it's better than what is there now, then I think it's best to use the normal editing process for improvements. Also, how to put it into the article is not difficult, if we do not seek perfection: one could simply replace the material in the "Solutions" section up to but not including the "Formal solution" subsection with the proposal (plugging in a few citations).

I would however suggest further moving the "Formal solution" and the stuff after into a "Mathematical details" section coming later in the article, and moving the simulations into a "Simulations" section immediately after the "Solution" section. There's more to the MHP than the solution: The general confusion (cognitive psychology, how even smart and knowledgeable people are fooled, ...), history (origins, connections with other problems, the wide publicity vos Savant gave to it, ...), etc. These could come after the simulations. Many might even want to skip the explanations altogether, looking at the answer only and then being more interested in the social aspects. -- Coffee2theorems (talk) 07:49, 7 October 2012 (UTC)

Yes. The aim and the goal of our efforts should tend to draw up a clear structure of that mingle-mangle article. To first of all present the clear paradox, above all helping the reader to decode the tricky story that intends to present this famous veridical clear paradox (full 1:2 if the host intentionally shows a goat, but just only 1:1 if he inevitably should be showing the car in 1 out of 3). This should be shown early. Without vexatious opaque wish-wash that you could know a little bit better if you just "knew" better, hiding that you never will know better, pretending that phantasmal assumptions that forever will be made up out of thin air, tied to door numbers, are the best knowledge that one must base on. And that conditioning on phantasmal assumptions are the only correct solution. Pretending that such suppositious "knowledge" could be of "great advantage" in making the correct decision. And that ignoring this heterodoxy means "not addressing the correct question". In disagreement to the fact that any staying will inescapably reduce the probability to win the prize. IMO all of that should be shown in much later sections on the weird history. No more misusing the article as a lesson in conditional probability theory.
And: To help the reader to decode the paradox, imo redundancy is most welcome. Redundancy is necessary there, see administrator Avanu's comment above of 9 September 14:15, and comments of a lot of other users.
But, as the tricky story is about a one-time problem (resulting in the host's "secrecy") we should try to avoid the impression of "series" of shows. And, just at the beginning, after the table of MvS, should show Bayes' formula in the "posterior odds = prior odds times likelihood" version. Gerhardvalentin (talk) 11:23, 7 October 2012 (UTC)
The table to which you refer is not in the draft being voted on here. Note also that the draft refers to a "series" of shows in several places, explicitly and implicitly, with references to the number of times things happen and the number of players to whom they happen. You don't really have to explain why you support the draft, but your comments seem more like reasons to oppose it.~ Ningauble (talk) 16:23, 7 October 2012 (UTC)

I referred to [take 2]. The text of all sections should avoid the impression that the famous story was on reiterated "game sows", but should try to pay respect to the one-time show the question was and still is about, this fact should explicitly be underlined. Gerhardvalentin (talk) 17:09, 7 October 2012 (UTC)

I don't think there is any impression that the explanation is inapplicable to a one-time show. True, it uses words like "always stick", "happens 1/3 of the time", "On average, in 999,999 out of 1,000,000 times", etc. (not "series of shows", though!) If you object to that, then we can't cover any simulations either, because in the simulations the game always occurs many times and you count what happens on average and the player always switches. A simulation of just one game does not give any useful results.

We could always explain a simulation first and then analyze what happens in that simulation, if you like that better. We could even explain that in general such simulations are still considered applicable to one-time situations by all statisticians and many philosophers even if the game really occurs only once. -- Coffee2theorems (talk) 18:15, 7 October 2012 (UTC)

Exactly what I meant. Just underline that a "frequentists' view" still does not address reiterated game shows, but still addresses the question of one special game show, of "a game show" the question was / is about, but is just applying a frequentist's view. And I would not only say "does not apply" or "this didn't occur", as Nijdam once suggested 16:51, 7 June 2009, but I would prefer to say

"Three in six players see door 3 opened. Two of them win by switching and one wins by staying.
But what happens to them is irrelevant for those who see door 2 opened."
and
"Three in six players see door 2 opened. Two of them win by switching and one wins by staying.
But what happens to them is irrelevant to those who see door 3 opened."
or sth alike. Gerhardvalentin (talk) 22:28, 7 October 2012 (UTC)

Yes. Richard Gill (talk) 07:18, 8 October 2012 (UTC)

No

Martin Hogbin (talk) 16:35, 5 October 2012 (UTC)
No. I could support the proposition that this is a promising approach, but the question in this referendum is whether to insert the draft text, as it stands, into the article. I do not believe it is ready. Even the drafter, Coffee2theorems, acknowledges here that there are still issues to be resolved. I cannot support inserting this text while there are still significant issues of structure and neutrality to be worked out.
I find it a little bizarre that the person who called the question, Martin Hogbin, is opposed to it, and put the question just one day after discussion at Comments on the proposed text had been joined with only a single brief comment. I cannot escape the feeling that my efforts to contribute to improving the article are being preempted by this misfired parliamentary maneuver. I oppose the question itself. ~ Ningauble (talk) 15:37, 7 October 2012 (UTC)

Discussion

I have moved these assessments of other editors' opinions by Rick to here. To establish a consensus editors must reply in person. Martin Hogbin (talk) 08:42, 6 October 2012

It seems to me

Coffee2theorems is obviously a yes.

Sounds to me like Gerhard is a yes [2].

Elen is a yes [3].

Ninguable is a (qualified) yes [4].

Preceding comments added by Rick Block (talk) 05:34, 6 October 2012 (UTC) in the "Yes" section

That is not what I said. Everyone is entitled to disregard what I actually did say, but kindly refrain from ascribing to me things which I did not say. ~ Ningauble (talk) 15:31, 6 October 2012 (UTC)

Sorry. I interpreted "All that said, this draft is a pretty good framework to build upon, and the prose is reasonably clear." as a qualified yes. I linked to your actual comments. -- Rick Block (talk) 20:08, 6 October 2012 (UTC)

Combining purses

Two men together have 20 pounds in their purses. At the cinema one of them opens his purse and sees it's empty. How much is in the other man's purse? Piece of cake. But now the real story. Two men, heading for the cinema, leave home with each one of them 10 pounds in his purse. Together they have 20 pounds with them. At the cinema one of them opens his purse and sees it's empty. Would you believe the other man saying: no problem, together we had 20 pounds, so your 10 pounds must have jumped into my purse, opens his purse and indeed finds 20 pounds????? Nijdam (talk) 09:46, 18 September 2012 (UTC)

To be sure it's understood, notice the similarity with the combining doors solution. Nijdam (talk) 15:54, 18 September 2012 (UTC)

See False analogy. hydnjo (talk) 13:02, 18 September 2012 (UTC)

The "answer" to the problem (should you switch) can be demonstrated easily. And the answer is "even though it doesn't appear to make sense: Yes". Anything else is argument over -why- it makes sense to switch or if there are extra conditions (which door was opened? Is the host lying? Was something not stated in the original problem?) that can be debated. 83.70.170.48 (talk) 14:45, 18 September 2012 (UTC)

Nijdam, since you seem completely obsessed with this, can you expound on which of the published sources actually gives the answer you want. And if the answer is 'none', have you considered getting published yourself, as you so clearly do have what you believe to be the answer, and it could then be the latest addition to the references. --Elen of the Roads (talk) 22:00, 18 September 2012 (UTC)

I do not like your qualifications. You'd better responded substantively. I'm not at home at the moment, but I'll show you a list of reliable sources. Nijdam (talk) 09:30, 19 September 2012 (UTC)

I retracted some reliable sources from the reference list:

Gillman, Leonard (1992). "The Car and the Goats," American Mathematical Monthly 99: 3–7.
Grinstead, Charles M. and Snell, J. Laurie (2006-07-04). Grinstead and Snell’s Introduction to Probability (PDF). Retrieved 2008-04-02.{{cite book}}: CS1 maint: multiple names: authors list (link) Online version of Introduction to Probability, 2nd edition, published by the American Mathematical Society, Copyright (C) 2003 Charles M. Grinstead and J. Laurie Snell.
Morgan, J. P., Chaganty, N. R., Dahiya, R. C., & Doviak, M. J. (1991). "Let's make a deal: The player's dilemma," American Statistician 45: 284-287.
Mueser, Peter R. and Granberg, Donald (May 1999). "The Monty Hall Dilemma Revisited: Understanding the Interaction of Problem Definition and Decision Making", University of Missouri Working Paper 99-06. Retrieved June 10, 2010.
Rosenhouse, Jason: The Monty Hall Problem. Oxford University Press 2009, ISBN 978-0-19-536789-8 (online copy of the first chapter, preprint).

Nijdam (talk) 10:32, 20 September 2012 (UTC)

Not much comment here. Afraid one has to admit the misleading way of arguing in the so called 'combining doors solution'? Nijdam (talk) 10:04, 25 September 2012 (UTC)

We are still awaiting the source (please indicate page) for the lost 10 pounds, saying that any group of two doors having better chance than one single door cannot help to decode the obvious paradox (2/3 and not 1/2). Gerhardvalentin (talk) 23:30, 27 September 2012 (UTC)

Still no sources that confirm your opinion that any "group of two host's doors" never can have better chance than one single door selected? As to the received impression "two still closed doors, and one of them hides the car", one editor said regarding those three doors:
Partitioning the doors between "your door" and "Monty's doors" is the most persuasive explanation, and just suggesting the partition is often enough of a hint.
It is on "decoding the paradox". You call that a false analogy and present your "combining purses". Please cite your source. Gerhardvalentin (talk) 18:37, 11 October 2012 (UTC)

What happened to the original RfC?

We had a RfC in which we had input from a much broader section of the community that the regulars here. I would be the first to agree that there was no clear consensus but I think there was a slight preference for my proposal, perhaps Guy could do a final count and let us know.

We have now gone back to a 'last man standing' method of dispute resolution.

I therefore think that one method of resolving this dispute would be for the article to be structured as per the proposal in the original RfC that got the most votes but only as a starting point and without and assumption of a consensus for that version. From there we might try to proceed with a renewed attempt a cooperative editing. It seems crazy to me to ask the opinions of a large number of editors and the ignore them completely. Martin Hogbin (talk) 17:04, 6 October 2012 (UTC)

Today is 30 days since the RfC was posted. If we agree there is no consensus why would we proceed with either proposal? Are you thinking the text Coffee2theorems has suggested reflects Proposal 2? If so, why would Gerhard and Salix, who both say they support Proposal 1 [5] [6], apparently both also say they're OK with this text [7] [8]? To some extent, I think this reflects the fact that not everyone here is understanding the specific intent of the two proposals the same way. Rather than focusing on abstract descriptions meant to control the article's content (that different people apparently interpret differently), by proceeding with actual proposed text it looks to me like there's a fairly decent chance of making progress. -- Rick Block (talk) 19:57, 6 October 2012 (UTC)

We had a substantial response to the RfC with many outside editors giving their opinions. That is the point of an RfC. Now we are back to the usual suspects, who have been unable to agree for years. Martin Hogbin (talk) 21:51, 6 October 2012 (UTC)

Martin, I think we're getting to the point where everybody is able to discuss actual changes to the text. This has to be an improvement on these endless theoretical arguments. Elen of the Roads (talk) 22:26, 6 October 2012 (UTC)

I completely agree. I feel even the RfC was flawed in being too theoretical. The distinction between the two proposals wasn't made clear enough, and misunderstandings abounded. It would have been better to have two concrete proposals in the RfC. The only problem with doing that is that then it wouldn't be a vote for which faction is right and which wrong, and by extension which faction ought to own this article. There were two concrete proposals afterwards from the two factions; I thought they could both be improved and merged ideas from both, hoping to find something satisfactory to both factions. (I remember you doing the same, so I guess we agree that it's a good thing to try.) -- Coffee2theorems (talk) 09:27, 7 October 2012 (UTC)

There is a specific proposal for text in the Solution section that (so far) no one but Martin has said they object to using as a basis for further editing, and that several people !voting each way in the RfC seem to support (in addition to those supporting Proposal 1 mentioned above, it is supported by Elen [9] who !voted neither [10] and presumably by Coffee2theorems [who is suggesting it] and definitely by myself, who both !voted for Proposal 2 [11] [12]). This seems to indicate support from a broad selection of folks who participated in the RfC, including at least 2 (Elen and Salix) who are definitely not "the usual suspects". This sounds like progress to me. -- Rick Block (talk) 03:24, 7 October 2012 (UTC)

So what is the point of having an RfC is which many editors give their opinion? Are you now saying the we should now decide on the basis of just a handful of users? Martin Hogbin (talk) 08:51, 7 October 2012 (UTC)

Elen, I think we will find that we are no nearer at all to being able to discuss the text because the same theoretical disagreements remain unresolved. Martin Hogbin (talk) 08:55, 7 October 2012 (UTC)

"Theoretical disagreements" sounds as if there were disagreements about the mathematics of the MHP. AFAIK there aren't any. There are disagreements about how to interpret/model the MHP, though. I made an incomplete list of various ways of doing that here. If you expect everyone to someday agree about what is The One True Way, I'm afraid you will be sorely disappointed.

You believe in a One True Way, and so does Nijdam, but the ways are different! Neither of you is willing to budge, and I'm afraid that will forever remain so. Your Platonic ideals of "the MHP", i.e. your ideas of what the Essence(TM) of "the MHP" is, are incompatible. You probably call Nijdam's Platonic ideal an "academic extension to the problem", Nijdam probably calls yours an "alternative setting" that is not "equivalent to the MHP". You both have to realize there are many possible interpretations, and not everyone who thinks differently from you is thereby wrong! -- Coffee2theorems (talk) 16:02, 7 October 2012 (UTC)

Please do not try to charactersise this dispute as one between myself and Rick or Nijdam. Look at the history of this dispute (either the recent RfC or the longer term history) and you will see many editors who have objected to the unnecessary complication of what was obviously intended to be a simple puzzle.

The problem is that these people are eventually driven away and the final decision is made by the small group who have withstood the endless argument. That small group is not representative of the larger body of users. Martin Hogbin (talk) 23:05, 9 October 2012 (UTC)

My proposal can actually be seen largely to conform to both RfC proposals. That makes it a false dilemma - we can have both! I think that's why some people from both sides have said they like the basic approach. Quoting the proposals and explaining any perceived discrepancies:

"Proposal 1 is for the initial sections including 'Solution' and 'Aids to understanding' to be based exclusively on 'simple' solutions (with no disclaimers that they do not solve the right problem or are incomplete) then to follow that, for those interested, with a section at the same heading level giving a full and scholarly exposition of the 'conditional' solutions."

Explanation: The actual section names are missing, but you could imagine them in place: remove the "Simple explanation" name, add "Aids to understanding" name before the many doors explanation, rename "Explanation with door 3 open" to, say, "Solution with door 3 open", and move all these sections to top level, keeping the structure otherwise intact. I think those section names/levels wouldn't be an improvement, and a rose by any other name.. (also, I'd say that all the solutions in the proposal are in fact simple, even if they are not scare-quote 'simple'; that has to count for something!)

True, it's not a full scholarly exposition, but that should only make the preferrers of 'simple' solutions happier! Full scholarly exposition of the 'conditional' stuff fits well in a much later "Mathematical details" section with full scholarly exposition of the non-'conditional' stuff (e.g. game theory) to keep it company.

"Proposal 2 is for the article to include in the initial 'Solution' section both one or more 'simple' solutions and an approachable 'conditional' solution (showing the conditional probability the car is behind Door 2 given the player picks Door 1 and the host opens Door 3 is 2/3) with neither presented as "more correct" than the other, and to include in some later section of the article a discussion of the criticism of the 'simple' solutions."

Explanation: The actual term "conditional probability" (or "conditional" or "given") is missing and it's not hooked to the general theory / general principles in any other way either, but hopefully that is not too much hardship. Again, a rose by any other name.., and instead of appealing to the correctness of a general theory based on axioms justified in certain ways, it cannot be any more wrong to appeal to the same justifications directly (e.g. "what happens to those who see door 2 open is irrelevant to those who see door 3 open"); it's just more accessible to those unfamiliar with the theory and its justifications (and those who are familiar will see it for what it is – a conditional probability proof expressed in layperson-friendly terms). I'm also not proposing any inclusion of criticism in later sections, but as I'm not proposing its omission either, hopefully I can leave hashing that out to a later day and other people.

If you think that this is an accident, think again :) There's a reason I tried to dig new rationales for including the many-doors explanation (K&W say it's very effective; "explanations", not "solutions") and the 'conditional' explanation (K&W say you need it after imagining door 3 open, because there are now only two cases, and simple case counting can only give a 50:50 result, so you need another way to reason your way out of this bag; showing how that's done explains the flaw in the common intuitive 50:50 "solution", and a part of resolving any paradox is to point out the flaw in original reasoning and fix it if possible). -- Coffee2theorems (talk) 10:23, 7 October 2012 (UTC)

Either of these suits me. The key thing for most in the RfC seems to have been that whatever is presented early on as a solution or explanation is in terms that a lay person can understand. One of the things that the later discussion has established is that these explanations in layman's terms are not wrong. The answer is always that it is better to switch, switching wins 2/3 of the time and it is not possible to get better odds (unless the contestant has X ray vision). Consensus is therefore against Nijdam, who wants a statement that the layman's explanation is wrong. By and large folks do not seem to be against a simple layman's version of the conditional solution - to demonstrate that it doesn't matter which door Monty opens, as long as it has a goat behind it. Consensus is therefore against Martin if this is what he remains opposed to (I'm not entirely sure what Martin is opposed to I have to admit).
I believe we are now at the stage discussing which of the simpler explanations is the most compelling to a lay reader. ~~Ningauble~~ Gerhardvalentin and myself prefer vos Savant's table, contrasted with the outcome if Monty picks a door at random. Others prefer the Economist explanation or something on the 'combining doors' lines. Like Coffee, I find the million goats thought experiment helpful, but it's not a solution that gives a 2/3 answer. I also think there is enormous value in referencing the research showing that most people almost instinctively look at the 2 remaining doors and think it's 50/50, which is why I believe that highlighting the effect of Monty always picking a goat is helpful.
If others could weigh in just in respect of this, it might be helpful to see whether there is a version that everyone can live with. Elen of the Roads (talk) 23:04, 7 October 2012 (UTC)

We should be careful: Not "as long as it has a goat behind it", but only as long Monty intentionally shows a goat behind it. Gerhardvalentin (talk) 06:17, 8 October 2012 (UTC)

Yes, I'm a devil for imprecise language, as you have remarked before. And I think I've got you and Ningauble mixed up again Elen of the Roads (talk) 13:41, 8 October 2012 (UTC)

Maybe I'm missing something, but I think the 2 explanations above are both referring to the same suggested text - explaining how this text meets both Proposal 1 and Proposal 2. There's no "either/or" here - it's both "simple first" (like Proposal 1) and includes an accessible ("simple", even) conditional solution (like Proposal 2). -- Rick Block (talk) 04:14, 8 October 2012 (UTC)

That is the intent. It cannot completely literally meet both (nothing can), but the point is that the differences are superficial. -- Coffee2theorems (talk) 05:34, 8 October 2012 (UTC)

I really am beginning to lose interest in this now. We have had an RfC and lots of new editors have taken the time to respond and express a preference, so we ignore them and propose something completely different. If that is how WP works then it needs fixing.

The problem with the new proposal is that it supports the misconception that the simple solutions do not, in some way, answer the actual question asked. Why do we have a 'Simple explanation' and then an 'Explanation with door 3 open'. What is the reader to assume the simple explanation refers to, some bizarre case in which door 3 is not open?

Where is the explanation and discussion of the other thing that most readers are puzzled by, that it matters that the host knows where the car is? Many people are astounded to find out that if the host pick a goat by chance there is no gain in swapping. Instead we give an explanation of something that nobody except probability students care about, what would happen (obviously nothing different) if the host had opened door 2. The solution addresses the need for some editors to demonstrate their 'deep understanding' of the problem rather than the needs of our readers to understand a simple but incredibly unintuitive puzzle. Martin Hogbin (talk) 23:05, 9 October 2012 (UTC)

Hello. Are you reading the same talkpage? Please refer to my previous post, which starts "I now believe we are at the stage of discussing which of the simple solutions is the most compelling...". Elen of the Roads (talk) 00:04, 10 October 2012 (UTC)

I think we must be reading different pages. Can you show me where you said that please. Martin Hogbin (talk) 16:46, 10 October 2012 (UTC)

[13]. --Elen of the Roads (talk) 17:26, 10 October 2012 (UTC)

Yes, thank you, I did miss that. You actually wrote, 'I believe we are now...' rather than 'I now believe...' so when I did a text search I could not find it.

I agree with that sentiment, of course, but we seem to me discussing a proposal which quickly moves on to a 'conditional' solution. See more below. Martin Hogbin (talk) 22:08, 10 October 2012 (UTC)

Would you like the other names I suggested for the two subsections better? That is,

"Why switching wins" / "Why the 50:50 answer is wrong",

respectively; or,

"Switching is the best strategy" / "The two cases are not equally likely"?

There's a lot of possibilities here. Some more off the top of my head:

"The general solution" / "A further look at the confusing case"
"A different way to think" / "Fixing the initial line of reasoning"
"The correct answer" / "The flaw in common reasoning"
"Strategic thinking" / "Belief update"
"Game theory" / "Bayesian inference"
"How to solve the problem" / "How the confusion arises"
"The correct strategy" / "The correct way to update beliefs"
"Taking a step back: Imagine you are at the beginning of the game" / "The confusing situation: Imagine door 3 was just opened"
"Solution: Always switching is twice better than always staying" / "Resolution: The correct way to update probabilities when things change" (and maybe call the whole section "Resolving the paradox" instead of "Solution")

Using some of the above would probably require changes to the text.

The point about the ignorant host case is a good one. I agree including it somehow would be good; probably in both subsections. Actually, if we include that, then "Resolving the paradox" would be even more apt section name than "Solution", as the ignorant host case is not a part of a solution of the MHP (we can find the correct answer without it), but it is a part of the resolution of it (it's a related thing people are confused about).

I'm puzzled why you think only probability students would be interested in why the two cases are not equally probable, or what went wrong in the original reasoning. That's an essential part of the resolution of any paradox! There are also many publications about the MHP written by people who are not students, and they are not written for students, but the authors' peers. Many of them even aren't probabilists, but psychologists. They never even were probability students!

The essential reason why people get it wrong is that they do belief update incorrectly ("new situation, everything must be equally probable again!"). Looking at the conditional illustration should show you why you fundamentally cannot use that principle twice in related situations and how to do belief update correctly. (Or perhaps it needs more explanation? The point is, it is explainable; the paradox is resolvable!) Using strategic reasoning (game theory) instead of a belief update (Bayesian probability theory) quickly gets the correct result, and there's certainly nothing wrong with doing that. It sidesteps the the reasoning problem entirely by using a different way of thinking, but it neither points out nor fixes the flaw in the original way of thinking. There is no conflict between the ways of thinking, doing both correctly gives the same result, and showing that reconciles everything. -- Coffee2theorems (talk) 04:14, 10 October 2012 (UTC)

"Resolving the paradox" is a brilliant label, yes, and the article in the first line should assist to decode the intended clear and clean paradox that the famous tricky story tried to present: Any group of two doors has better chance than one single door. For years now, the article did lay a smoke-screen on that famous veridical paradox, by presenting multifaceted misinterpretations of the narrated story. Misinterpretations, where the paradox never arises. All of that misinterpretation has to be presented in later sections on the queer vast history. No more nebulizing, but assisting in "decoding the paradox" as descriptive as possible. And once more: redundancy is most welcome for this effort, as admin Avanu showed us above: 14:15, 9 September 2012. And imo it is not bad if editors KNOW what they are talking about, or could present explicit and undisputed sources supporting their narrow beliefs. Gerhardvalentin (talk) 11:30, 10 October 2012 (UTC)

I've just been re-reading Kraus and Wang (psychologists) who have made a thorough analysis of why ordinary people jump to the wrong conclusion (50/50). They also present four quite different strategies (if I remember the number correctly) which can be used by ordinary people, and are used by ordinary people, to gain the insight that the wrong conclusion is wrong. One of their main points is that the formulation of the problem deliberately puts people onto the wrong leg by explicitly naming the door opened by the host. One of their four strategies for gaining insight is "less is more". We are misled by being given more information than we need. We forget about the history of the doors being chosen and opened, and just see in front of us doors 1 and 2 closed, door 3 open revealing a goat. We have to de-visualize. Forget the number of the door which was opened to reveal the goat (everyone will agree that it's irrelevant, anyway, if they are arguing intuitively with subjective probabilities determined objectively by symmetry). The host opens a door revealing a goat. He can do that and he's going to do it anyway. So this does not give you any new information, the chance is still 2/3 that your door doesn't have the prize. So: one of their four strategies is essentially the simple solution or the combining doors solution. One of their four is essentially the Economist solution (the decision theoretic solution, strategic thinking). One of their four is essentially the conditional probability solution, but not using the word or the concept. They emphasize that people get probabilities wrong, but tend to get whole numbers - frequencies - right. It seems to me that the first thing any newcomer to MHP needs to understand is that the intuitive answer "doesn't matter, so I'll stay" is wrong. For this purpose the newcomer doesn't need a complete solution or complete analysis, but a simple analysis looking at the problem from a different point of view which shows them that there is more here that meets the eye, literally. They have to stop and think again. Many readers will be happy with that. Others will be interested in understanding why the original wrong intuitive reasoning was wrong, and they will benefit from a more complete more mathematical analysis. Note that different people think in different ways, different people will appreciate using a different one of Kraus and Wang's four techniques, or a different combination of them. Richard Gill (talk) 16:32, 10 October 2012 (UTC)

Coffee2Theorems, I think 'Resolving the paradox' would be an excellent section to have after the 'Simple explanation' but I cannot see where your last illustration fits in. How does this explain any better than the first two why you have a 2/3 chance of winning if you switch to door 2. You seem to be trying to make a point that, as far as I know, no source makes and few editors here agree with. Are you saying that the 'simple' solutions solve the problem but that somehow the 'conditional' solutions explain why it is that people get the wrong answer, or have I misunderstood you? Martin Hogbin (talk) 16:43, 10 October 2012 (UTC)

Oh, I think several editors agree with him. In fact, I would venture to opine that several editors would be quite happy with a 'looser' simple explanation (ie one that does not appear to be maths based) if C2t's illustration is also included. Some would disagree though. I think it would help if you could provide a better (ie shorter) version of what it is that you want, seeing as we have now disposed to the satisfaction of most, of the 'simple answers are wrong and the article must say so' question. Is it just that you don't find this illustration helpful, or is there some deeper philosophical objection? Elen of the Roads (talk) 17:32, 10 October 2012 (UTC)

My only point is that we should avoid undermining the simple solutions with text that suggests that you cannot understand the problem properly unless you use the 'conditional' solutions.

If you are reading an explanation and desperately trying to convince yourself of its veracity, it is unhelpful to have in sight a more complicated solution with rather cryptic comments, like 'Explanation with door 3 open'. What other kind of explanation would you expect??. Do you see what I mean?

Rick and Coffe2theorems seem to believe that the conditional solutions explain things more clearly and are more convincing, claiming K&W as support but see Richard's comments on that above. If there is a truly convincing argument that the conditional solutions, in pictures, will help some readers to understand the basic paradox where the simple solutions have failed then I might be persuaded to accept them at the start but there is little evidence for this IMO. Martin Hogbin (talk) 22:08, 10 October 2012 (UTC)

I'd put it as "unconditional" vs. "conditional", or "strategic thinking" vs. "belief update". People here seem to consider Devlin's combined doors solution "simple" or non-"conditional", because it's on that side of the political line you've drawn. To me, Devlin's solution is conditional. That's one reason I preferred Adams' combined doors solution in the "simple" solution section – his version is unconditional! That makes a neat split for the sections.

The unconditional/strategic viewpoint takes a step back, closing that pesky door 3, and tries to figure out a winning strategy. You may see door 2 opened, or you may see door 3 opened; you don't know yet! You intuitively ignore all other strategies except "stick no matter what" and "switch no matter what" (the others are pretty bizarre). One strategy gets you 2/3 overall chances, the other only 1/3. Obviously 2/3 is better, case closed!

Unconditional solutions do not, however, give you the probabilities for the two doors in question. They give you overall winning rates of the considered strategies. Consider the ignorant host case. In it, this approach says that staying wins with probability 1/3. Hey! Isn't that the same as in the standard MHP? Why yes, yes it is! And quite right, too. You see, switching also wins with probability 1/3. Equally good strategies, case closed! But this is not a good way to go about finding out the probabilities in the two doors situation; those are conditional probabilities, not overall probabilities. If you try, you might reason "staying wins 1/3 of the time so the car must be behind the remaining door 1-1/3 = 2/3 of the time!" Wrong. Neither door's probability is 1/3, they are both 1/2. Doing arithmetic with the winning rates of various strategies of a game does not generally make much sense. If you want those probabilities, what you want to do instead is a belief update.

People do belief update incorrectly: they go "new situation, must be all equally likely!" To fix that "argument", we need to show them how to do it correctly. Now, unlike Adams (who simply reasons that two is better than one, sensibly enough), Devlin does do a belief update. He talks about changing probabilities. He points out that conditioning on a sure event doesn't change probabilities; that is correct. I think he could've worded it better, and contrasted it with the ignorant host case better, but he's clearly after the conditional probabilities, and trying to explain how probabilities change, and laments that not understanding it trips people. If a good version of this kind of argument can be composed and sourced, I wouldn't even be opposed to including it (in the conditional section), provided the current conditional illustration also remains. The current illustration goes a bit further than Devlin does to demonstrate the general principle, can be used to show (already shows?) the inconsistency he "pass[ed] on pursuing [...] here", and nobody thinks it's incomplete, etc. etc. It's pretty much as close to "case counting" as it gets for a conditional solution, which is nice.

Anyhow, both strategic thinking and belief update are important viewpoints, and that seems to me the natural dividing line for the section, not some political line. -- Coffee2theorems (talk) 22:24, 10 October 2012 (UTC)

RDGill wrote out Devlin's argument corrected/completed in one of his MHP publications. Richard Gill (talk) 05:42, 11 October 2012 (UTC)

Yes, I know of that source. It's a good one as far as it goes. I didn't actually have anything specific in mind when I said "a good version"; I was more hoping that if people want it in the article (is it wanted?), there would be more sources than these two, and we could look for the best ones.

My impression was that it's often put this way: No new information is revealed when the host opens a door to show a goat. It was known he would do that at the outset, it's "no news". Devlin, on the other hand, says that a crucial piece of information is revealed! It's a different notion of "information". There's a lot of potential for confusion here, so we should be careful with the wording. Also, what about the ignorant host case? It's not treated. New information is revealed: It was not known in advance that the host would show a goat. Does anyone explicitly point that out? I'm not saying that we can't necessarly put it more clearly even if no source explicitly does it, just that having sources would, obviously, be better, and they might generally have good ideas about wording things. -- Coffee2theorems (talk) 07:38, 11 October 2012 (UTC)

Let's be careful with the wording, yes.
* Devlin addresses quite another story of the "ignorant host" in letting another visitor decide to chooose one of his two doors. "Crucial" insofar as this is quite "another story", quite contrary to what he said just before regarding the clean paradox:
Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C. (Notice that he can always do this because he knows where the prize is located.).

We have to clearly show that the ignorant host is not "part of MvS' story", but is quite another story and has nothing in common with the famous paradox. We should show this fact to help readers to grasp the difference of those two quite different "stories", to grasp that "1:2" is correct in the scenario where the paradox appears, but "1:1" is correct only outside that scenario, outside the paradox, in quite "another scenario". Not to confuse the reader by mingle-mangling, but in keeping differing scenarios very clearly apart. Gerhardvalentin (talk) 08:55, 11 October 2012 (UTC)

p.s.: It is on the scenario where the paradox arises in a one-time story, with an unknown host who, in opening of one of his two doors, is absolutely incapable to and therefore never can give any additional hint regarding the door that actually hides the car (*secrecy*). The flaw of "conditioning on door numbers though" is to possibly expect another "value of probability" to win by staying resp. to win by switching. In knowing in advance that the result of "conditioning on door numbers", if done correctly, in this scenario can and will inevitably give the average probability only. Knowing that in advance, it is unnecessary, useless and futile to recommend to do it, anyway. See the actual sources. Gerhardvalentin (talk) 09:29, 11 October 2012 (UTC)

pps: Above I read "Unconditional solutions do not, however, give you the probabilities for the two doors in question". But for the given scenario, that the paradox arises in, this is not fully correct. In that scenario the "unconditional solution", as well as the "conditional solution", both similarly "inevitably" must give the probabilities "for the two doors in question ( ! )", and – no surprise at all – inevitably for the strategy to always switch as well, as both these probabilities mandatory are exactly to be the overall winning rates. Inevitably. Unavoidably. No way out. You never will be able to "know better". The flaw of the conditional approach is to (untruly) pretend that you "of course will know better". But you never will. You know that from the outset. Never within the scenario where the paradox arises. Accentuate and highlight this given fact. It is necessary to distinguish "unconditional" and "conditional" in this context, to expose and to unmask. For the famous scenario of MvS, where the paradox arises, "door number opened", (be it #2 or #3), once and forever is and remains completely irrelevant: Both is "one and the same" (symmetry).

And even if you do not address the paradox that MvS left behind, but look at the quite different story told by Morgan et al. with some "well-known host" who doesn't observe secrecy regarding the door that actually is hiding the car, helping you by special hints, even such well-known host forever will be completely unable to signalize that actually "staying" could be the wiser decision. Never. Even such talkative host of quite another scenario, isn't of any relevance for the decision asked for, so "conditioning on door numbers" forever is of no relevance whatsoever for the decision asked for. The article should clearly show that, based on actual sources: You never need to condition on door numbers, even not in the Morgan et al. "variant", but of course you may do it if you like, but it forever will remain without any relevance for the decision asked for. All of that should be shown in sections on the quirky "history in misinterpreting" the famous scenario. Gerhardvalentin (talk) 10:37, 11 October 2012 (UTC)

Proposed text for Solution section, take 1

Solution

The player should switch. Switching wins twice as often as staying.

Simple explanation

Switching loses if and only if the player initially picks the car door, which happens 1/3 of the time, so switching must win 2/3 of the time (Carlton 2005).

Case 1: Car behind door 1	Case 2: Car behind door 2	Case 3: Car behind door 3
Player picks door 1	Player picks door 1	Player picks door 1
No matter what the host does, switching loses	Host must open door 3; switching wins	Host must open door 2; switching wins
One case where switching loses	Two cases where switching wins

The situation can also be analyzed by cases, as illustrated above. Switching loses in only one of the three equally likely cases, and wins in the other two (vos Savant 1990b; Krauss and Wang, 2003).

Many people find the situation intuitively easier to understand by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990; Krauss and Wang, 2003). In this case there are 999,999 doors with goats behind them, and the player's chances of picking the car door are 1 out of 1,000,000. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times (i.e. the times the player initially picked a goat) the other door will contain the prize.

Explanation with door 3 open

Most people take it for granted that the host opens door 3, and base their reasoning on that fact. That assumption is incompatible with the simple explanation illustrated above, because in case 3 the host does not open door 3. Once case 3 is excluded as impossible, simple case counting results in the common incorrect answer that only two cases (case 1 with door 3 open and case 2) are possible and therefore the chances are 50:50 either way (Krauss and Wang, 2003).

It is true that when door 3 is open, there are only two possible cases: either the car is behind door 1 or it is behind door 2. These two cases, however, are not equally likely! This is illustrated in the alternative explanation below, with the two relevant cases on the left side of the thick line.

Car behind door 2	Car behind door 1		Car behind door 3
Player picks door 1 This happens to two in six players	Player picks door 1 This happens to two in six players		Player picks door 1 This happens to two in six players
Host must open door 3 (two in six) Switching wins	Host randomly opens door 3 (one in six) Switching loses	Host randomly opens door 2 (one in six)	Host must open door 2 (two in six)
Three in six players see door 3 opened. Two of them win and one loses by switching.		Three in six players see door 2 opened. What happens to them is irrelevant to those who see door 3 opened.

The above illustration shows everything that can happen to players who pick door 1. On average, three in six end up in the exact situation described in the problem statement (door 3 is opened; left of the thick line). Two of the three win and one loses by switching; therefore the probability that switching wins in that situation is 2/3 (Chun 1991; Grinstead and Snell 2006; Morgan et al. 1991).

Comments on the proposed text, take 1

Since I criticized the other proposals, I felt it is only fair if I put one up for criticism, too. This contains material from both, the article, and some my own. The referencing could probably be improved. How do you like it?

It's intentionally somewhat minimalistic, and adding a "Discussion" subsection or something to it where things like the ignorant host variant could be covered is certainly possible. This is only meant to get the "answer to the standard problem is 2/3; switch!" part done with before the article proceeds on to other stuff. -- Coffee2theorems (talk) 17:07, 26 September 2012 (UTC)

I do not like it, for reasons discussed at length above. Martin Hogbin (talk) 18:39, 26 September 2012 (UTC)

The explanation about the most common mental model is pretty unintelligible - I guess I just don't think that way. Can we have a do over on that. And I don't think the million goats belongs in there. Perhaps the next section should be Still having trouble? which could include the million goats.--Elen of the Roads (talk) 18:45, 26 September 2012 (UTC)

I'm not too attached to the million goats thing, but I put it in there because while it isn't a solution it's certainly an explanation, and it seems to be an effective one. It's often mentioned that it helps people "get" the MHP, and K&W had this to say about it:

"Hell and Heinrichs (2000) obtained 65% switch decisions by investigating a variant of the problem with 30 doors, where 28 doors were opened after the first choice.", which is big compared to "Most studies on the Monty Hall problem only report the percentage of switch decisions, usually around 10-15%." It's actually more effective than their carefully constructed "guided intuition" version with three doors (59%).

I'm loath to include any lengthy "aids to understanding" material before the conditional solution (I'm for proposal 2, after all :), but this one was fairly short and the bang for the buck high, so if there was going to be any then that was the obvious candidate. Do you have any specific reasons why it doesn't belong?

Which part of the mental model explanation is unintelligible? The paragraph before the figure? The figure? The paragraph after it? Is there anything there that is understandable? -- Coffee2theorems (talk) 19:37, 26 September 2012 (UTC)

I think it's the way it's written above. You've run a number of things together and my thought chain goes something like

"After reading the standard version of the problem, most people form a mental model I've never formed a mental model that looks anything like that
The solution illustrated above is incompatible with this mental model Why the hell are you showing it to me then
Instead of switching to a compatible mental model, it is also possible to solve the problem in a way consistent with the common one by considering only cases where door 3 is opened (left of the thick line below) OK, I'm giving up at this point

I think that the idea of mental models may go some way to explain why people get confused, and certainly belongs in the article somewhere. Interestingly, it's not the actual reason everybody I ever show this puzzle gives for thinking it makes no difference to switch. They all mentally look at the two remaining doors (their pick and the one Monty hasn't opened) and say "Well, it's fifty fifty isn't it. The car's either behind my door or that door. What's the point of switching." Every time (admittedly small sample). That's why I put the version where Monty picks at random into my layout above - it can indeed go below vos Savant's solution, but it's that which seems to convince everyone - showing that the outcome is differend because Monty won't show the car. Elen of the Roads (talk) 22:15, 26 September 2012 (UTC)

I rewrote that part now. At the same time I ditched the "mental model" wording; it wasn't even used in the same sense as in the K&W paper. Their meaning is one definite arrangement (e.g. "Car / Goat / Open door" is a mental model in their sense), whereas I used it in the sense of a model that you can mentally simulate. Better? -- Coffee2theorems (talk) 00:10, 27 September 2012 (UTC)

I like it. Except for the first paragraph of the "with door 3 open" bit. It can be deleted. Start straight away with "it is true that with door 3 opened there are two cases".

References: Selvin's (1975b) second letter to the editor has this solution. It's the second oldest solution on record. His first letter to the editor (1975a) has the other solution. Since he invented the problem I think these references are good reasons to put both these solutions right at the start of the article. Both give useful insight, probably both will appeal to many readers. Some readers will have no use for one or the other. So what?

I would like to see the Economist solution here as well. That too appeals to many people. Editors have repeatedly put it forward spontaneously. Richard Gill (talk) 07:14, 27 September 2012 (UTC)

Isn't starting with "It is true that when door 3 is open" a bit sudden? The first paragraph ties that section into the bigger picture. I think that the first sentence is fairly good at least, even if the rest of the paragraph could be improved.

What I'm trying to get across to the reader is essentially these points (without necessarily explicitly spending ink on all of them):

Most people (such as you the reader, in all probability) initially reason based on the assumption that door 3 is open, and that leads to this common fallacious argument and this wrong conclusion (two cases, 50:50).
The common fallacious argument is actually salvageable, and here's how to fix it (two cases yes, 50:50 no).
Now that the argument is fixed, it should be obvious what went wrong. (host is sometimes constrained, sometimes not, and that matters)

I think that showing what went wrong, not just this is how to do it right, is an important part of the solution in a paradoxical problem. Certainly there are other reasons to put the conditional solution in, but this is the one that I think has the most immediate relevance to the reader.

Maybe it would be better if the paragraph didn't refer to the previous solution? Maybe something like the following (replacing both paragraphs)?

Most people take it for granted that the host opens door 3, and base their reasoning on that assumption, imagining door 3 as already open. That leads to considering only two cases as possible - the car is either behind door 1 or door 2 - and incorrectly concluding that the chances are 50:50 either way (Krauss and Wang, 2003). The two cases are indeed the only ones possible when door 3 is open, but they are not equally likely! This is illustrated in the alternative explanation below, with the two relevant cases on the left side of the thick line.

Any better? One might also throw in an extra sentence right after the vos Savant solution, saying that

Therefore, the strategy of picking door 1 and then switching no matter what happens wins 2/3 of the time.

so that if the reader goes back and is confused about what the explanation actually does show, then they are gently nudged in the right direction (there exists an interpretation of that sentence which is true, and it's the most natural interpretation). -- Coffee2theorems (talk) 09:21, 27 September 2012 (UTC)

As to the received impression "Two still closed doors, and one of them hides the car", Ningauble said above:
partitioning the doors between "your door" and "Monty's doors" is the most persuasive explanation, and just suggesting the partition is often enough of a hint.

Being able to see two "groups of doors" (1 door + 2 doors), to distinguish and to help keep them apart: Why not saying just in the beginning:

Three doors, and one of them will hide the car. (Instead of "1/3 2/3" we should write "Player Host") and instead of "1 2 3" we should say (or add) "1/3 1/3 1/3"

Gerhardvalentin (talk) 10:27, 27 September 2012 (UTC)

Three doors, and just only one of them will hide the car. Only one.
Note: After the guest first has selected "his" door, the entity of three doors has irrevocably been divided into two opposing groups.
The reader should be "aware" of this fact, because it is useful to examine those two groups. Each group can hide one car at most.

The guest's group can hide one goat at most, and the host's group can hide two goats at most, but the host's group "must" hide one goat at least. (At this stage, the "biased" host who in the million doors variant leaves only #777'777 of his doors closed, because he NEVER uses to open #777'777 if any possible, is still not "necessary" in this step.)

What the article should do, but never does: Helping to "decode the picture" that the tricky "story" and its course / sequence / devolution tells / leaves behind, resulting in a given obvious paradox.

Help to correctly distinguish, to single out, to decompose the elements and to put them together again in an insightful manner.

Help the reader to understand and to "decode" the paradox, imo he should be able to *vary* his first impression from different perspectives. Why rejecting to offer him this support? The article could be much better than it has been for years now. Gerhardvalentin (talk) 11:31, 29 September 2012 (UTC)

Coffee2theorems, you wrote "Most people take it for granted that the host opens door 3, and base their reasoning on that assumption, imagining door 3 as already open". I find this a very odd statement. We were told that the host opened door 3! (Not opens door 3). Of course one imagines door 3 as being open. Yes, because it was opened. That's why I find this intro superfluous but also confusing since there are maybe some subtleties implied by present versus past tense. But if so, they are too subtle to come across (to me, at least). How about: Because door 3 has been opened to reveal a goat and we see two doors left closed, most people automatically imagine that the two remaining possibilities for the location of the car must be equally likely. Richard Gill (talk) 12:11, 27 September 2012 (UTC)

That seems to be what people do. And if Monty had opened a door at random, they would be right. The trick is to show them that by avoiding showing the car, Monty has made it more likely that the other door has the car behind it. Elen of the Roads (talk) 12:52, 27 September 2012 (UTC)

Exactly! And the reason why it is more likely, twice as likely in fact, is because Monty is twice as likely to open door 3 when he has no choice, as when he does have a choice. Richard Gill (talk) 13:36, 27 September 2012 (UTC)

Incidentally, the ignorant host version is easy enough to explain using the "explanation with door 3 open" figure, too. When the host chooses randomly even when the car is behind door 2, one of the two players who would win in the standard setting ends up being shown the car behind door 2. Only two in six are shown a goat behind door 3, one of who wins and another one who loses. -- Coffee2theorems (talk) 14:31, 27 September 2012 (UTC)

The first sentence is that way because it was based on this one in K&W:

"Although, semantically, door 3 in the standard version is merely named as an example ('Monty Hall opens another door, say number 3"), most participants take the opening of door 3 for granted and base their reasoning on this fact."

I just took that as the basis, nothing deep implied (FWIW, the word is "opens" in the standard problem descriptions, not "opened", and the vos Savant version doesn't use past tense anywhere; you can imagine yourself doing the analysis in front of the closed doors if you want). They also say:

"34 out of the remaining 35 participants (97%) indeed drew an open door 3, and only a single participant (3%) indicated that other constellations also remain possible according to the wording of the standard version."

So there are a few who do see it otherwise, too. It might be fine to say "Because door 3 has been opened" despite that, if we started our article with the conditional solution, but we don't! The simple solutions describe things from another point of view (you analyzing cases in front of closed doors), where it is still possible that door 2 is opened. I think the reader should be told when we are changing the point of view back to door 3 being already open, and why. Trying again:

Most people consider only the situation where door 3 is open, leaving only two possibilities: the car is either behind door 1 or door 2. They then incorrectly conclude that the chances are 50:50 either way (Krauss and Wang, 2003). These are indeed the only possibilities when door 3 is open, but they are not equally likely! This is illustrated in the alternative explanation below, with these two possibilities on the left side of the thick line.

I'm not dead set against stating that door 3 has been opened as a matter of fact instead of saying merely that most people analyze the problem from that POV, but that seems slightly presumptuous as it is not the only way to analyze it. -- Coffee2theorems (talk) 13:59, 27 September 2012 (UTC)

This is all interesting stuff an worthy of a place in the article but first we have to convince our readers of the right answer. It seems we are going round in circles and the 'combing doors' explanation is coming back into fashion. I agree with all those above, that it has great explanatory and convincing power. We need it near the start. Martin Hogbin (talk) 17:59, 27 September 2012 (UTC)

Well, I've almost lost track of where we stand, but the 'combining doors' explanation is a very bad one and as I wrote above: highly misleading. For your interest again: There is nothing to combine. Each of the not chosen doors has 1/3 chance on the car. Combining them means adding 1/3+1/3 =2/3. I have not the slightest clue how this may be helpful in any solution. Why not read what I wrote on combining purses? Easy to understand for anyone! Nijdam (talk) 22:18, 27 September 2012 (UTC)

This is about convincing readers of the right answer. I think many have this incorrect argument ("two cases, 50:50") lodged in their brain and some won't listen to anything that doesn't point out the flaw in their reasoning. The conditional explanation addresses this directly - you have the two cases, and it shows why they aren't 50:50. I hope that what looks like going in circles is actually going in spirals - steadily, if slowly, upward. -- Coffee2theorems (talk) 23:02, 27 September 2012 (UTC)

The combining doors argument is a correct argument for what it delivers, namely that switching gives the car 2/3 of the time. For many newcomers to MHP, it's *the* eye-opener. It's used by many sources, both popular and academic. Some academic mathematical sources even use it as a step towards the kind of solution some editors want to see (namely, a conditional probability statement). Because if we add symmetry to the argument, the door numbers are irrelevant, and 2/3 is a conditional as well as an unconditional probability. Equivalently, door numbers are independent of door roles and we can ignore the numbers; the only thing that is relevant is the roles of the doors. Richard Gill (talk) 07:31, 28 September 2012 (UTC)

Nijdam, we discussed this earlier and, I thought, came to some form of understanding. The crucial, sometimes unstated, fact in the 'combining doors' solution is that the host's actions of revealing a goat do not affect the probability that the car is behind the originally chosen door.

My suggestion, which I thought you had agreed to, was to state, initially without proof or justification, that the probability the car is behind door 1 is not changed by the host showing a goat.

Having done this, we can then compare this with the case where the host reveals a goat by chance. It is easy to explain how that action gives us us information which can be used to revise our probability that the car is behind door 1, resulting in no advantage in switching. We can the proceed to justify our original statement that in the standard case no information is revealed and the probability that the car is behind door 1 is unchanged.

If we make this explanation the last of the simple ones we achieve three worthy objectives: we keep the explanations simple at the start, we explain why it matters that the host knows where the car is, and we show how, in some circumstances, the probability the car is behind the originally chosen door can be changed by the host's actions, thus providing a lead into conditional probability and the Morgan solutions. Martin Hogbin (talk) 14:20, 28 September 2012 (UTC)

It seems to me that the others wanted to use the combined doors solution to illustrate the same thing as the other simple solutions do: that 2/3 of all switchers (who choose door 1) win. That is a very different usage from the one where we claim "no information is revealed". What you are talking about is something like this going into the "Explanation with door 3 open" section:

One intuitive argument groups the doors into the player's doors (door 1) and the host's doors (doors 2 and 3), as illustrated below. Initially, the chances of the car being behind the player's door are 1/3, and the chances of it being behind one of the host's doors is 2/3. This remains true after the host opens door 3, because doing so provides the player no new information about what is behind the player's door — regardless of what is behind the player's door, the host's choice appears completely random to the player.

What the others seem to be talking about is something that would go in the beginning of the "Simple explanation" section. For example, one could replace the Carlton blurb with this:

If the player is determined to always stick to their initial choice, they might as well not be offered the chance to switch at all. Sticking wins exactly when the player initially picks the car, which happens 1/3 of the time. Switching wins the other 2/3 of the time. This may be easier to grasp if the doors are grouped into player's doors (door 1) and host's doors (doors 2 and 3), as below:

The player is effectively given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host opens one of his doors to show a goat, but that doesn't change the fact that switchers always get all cars behind the host's two doors, and stickers get all cars behind their one door.

I'm sure both wordings could be improved. I think the latter approach is somewhat better, as the "information" thinking is notorious for causing various kinds of confusion, and that concept is not necessary for solving the problem. The notion of no new information being revealed about what is behind the player's door is also somewhat complicated compared to the simple case counting used in most approaches. -- Coffee2theorems (talk) 16:05, 28 September 2012 (UTC)

Let me make this crystal clear, especially for the defenders of this explanation, i.e. the combining doors. It goes like this:

the chosen door has 1/3 chance on the car
hence the other doors have together 2/3 chance on the car
one of these other doors is opened by the host showing a goat
this door clearly has 0 chance on the car
as both doors together has 2/3 chance, the remaining door must have 2/3 chance on the car

Many will find it very "logically", and may think they now understand the paradox. This explanation is indeed found in several (reliable??) sources. It is impossible to repair this kind of reasoning, as the basic idea, the combining of two chances of 1/3 into 2/3, gives the wrong suggestion, and does not contribute to any understanding. It especially lacks the explanation why the chance of 1/3 for the remaining door magically changes into 2/3.

It is *easy* to complete/repair the reasoning. And the repair job is in the published literature.

@Richard: There is nothing to complete/repair, as the basic idea is wrong. Of course you may see the equivalence with the simple solution, and complete it, but this does not repair the wrong idea of combining the doors. Combining may seem to bring understanding, but what is understood is wrong. The step from 4. to 5. is not logical, and to prove 5. the combining is superfluous and of no help.Tell me what insight the combining brings you. Does it enlighten you that 1/3+1/3 = 2/3 ?! Nijdam (talk) 22:47, 28 September 2012 (UTC)

The competitor chooses door 1. With probability 2/3 there is a goat behind it. The host opens a different door revealing a goat but does not inform you which door he has opened. Probability 2/3 your initial door 1 hides a goat since you have gained no new information. Finally the host informs you that it was door 3 he opened. Still probability 2/3 that your initial door 1 hides a goat because *which* door he opens is independent of whether or not your door hides a goat, by symmetry.

And indeed what effectively is going on is that the host is offering you the possiblity of a car behind doors 2 and 3 in exchange for the possibility of a car behind door 1. Richard Gill (talk) 17:53, 28 September 2012 (UTC)

While the symmetry arguments are neat and all, I do not think they help the general reader. It's just so much mumbo jumbo they could parrot to a mathematician to get them to shut up, and that's about all the mileage they will ever get out of it. You can fill in the details. I can fill in the details. Nijdam can fill in the details. But the general reader can't!

A simple test to check whether the reader understood the argument: "Prove that 2/3 of switchers who pick door 1 and see door 3 opened to show a goat end up getting the car. Show your work." (and maybe same for the ignorant host case) If you have hand-waved something about symmetry, do you seriously believe the reader can turn that into an actual proof showing true understanding? Even assuming a high-school student fairly good at math?

The second illustration in my proposal should impart that level of understanding, at least to that high school student. Much more difficult problems are usually tackled in high school, and that explanation does not even require any understanding of probability theory (much less something like non-trivial statistical independence or conditional probability; are those even covered in high school?). The more complicated alternative mathematical material can well go in a later section (which could cover all non-K&W cases and "reverse mathematics" of MHP investigating necessity of assumptions etc. — all very interesting — to your satisfaction).

Even if you don't require standards of proof that would satisfy a high school math teacher, the "no information" thing is tricky enough. I did try to argue it in the combined doors example above, but it's probably pretty opaque, even though it is the most intuitive approach I can think of (i.e. some way of expressing the line of thought "case 1: car behind your door, host behavior seems random", "case 2: goat behind your door, host behavior seems random"; host behavior indistinguishable in the two cases, he has "perfect poker face", you are none the wiser about what's behind your door). -- Coffee2theorems (talk) 19:23, 28 September 2012 (UTC)

NB. By using random doors, a kind of combining doors solution is possible, but that is not what generally is understood as the combining doors solution.Nijdam (talk) 16:29, 28 September 2012 (UTC)

I agree that in this type of combined doors solution (the first one of the two I illustrated above), the combining of doors is superfluous. Once you show door 3 open, the argument of "taking everything the other guy has" loses its force, because both of you now equally 'own' one closed door! Then we go down the rabbit hole of arguing why it's not 50:50 even though there are only two places the car could be, and I don't see how combining doors helps there at all. It's a "no information revealed" argument in disguise.

If all doors are shown closed, then the host has two closed doors, and that's why taking what he has looks like such a sweet deal. I think that is where the intuitive force of the second type of this argument comes from. Showing an open door ruins the argument; one must instead show them closed and say that "no matter what the host does, you get all cars behind his two doors" or something like that. -- Coffee2theorems (talk) 17:14, 28 September 2012 (UTC)

I think the reason combining doors is *such* an eye-opener is that nobody is taken in by a cardsharp who, accused of hiding the ace of spades in his sleeves, rolls up one sleeve (say, the left one) and declares "see, nothing up my sleeve." Everybody knows this is not evidence about whether he is hiding the purloined card. None whatsoever.

Some professor (or sophomore) might note that if the accused has a tendency to roll up his right sleeve whenever possible then choosing the left one would be highly suspicious, so therefore an objective probability for sleeve preference must be taken into account. To a subjectivist this putative unknown probability is immaterial, and the sleeve chosen does not contribute any evidence about whether he is hiding the card. None. What. So. Ever. However, as a sop to those who think it makes a difference, one could stipulate in advance that the accused is completely ambidextrous and utterly dyslexic. If the professor now wants to plug a 50:50 distribution into some formula then that is his prerogative; however, if he says that one must do so, and that it is illogical to simply use the stipulation as indicating the choice of sleeve makes no difference, then I think it just illustrates what Turing laureate Richard Hamming said in The Art of Probability for Scientists and Engineers (1991), p. 4:

"Probability is too important to be left to the experts. [...] The experts, by their very expert training and practice, often miss the obvious and distort reality seriously. [...] The desire of the experts to publish and gain credit in the eyes of their peers has distorted the development of probability theory from the needs of the average user. The comparatively late rise of the theory of probability shows how hard it is to grasp, and the many paradoxes show clearly that we, as humans, lack a well grounded intuition in the matter. Neither the intuition of the man in the street, nor the sophisticated results of the experts provides a safe basis for important actions in the world we live in." [emphasis in original]

Some mathematicians might prefer adding the words "by symmetry" to the explanation. I doubt many laypersons would find the addition of a five dollar word particularly illuminating. Personally, I think it is a *virtue* of the combined-doors approach that lumping them together points out the symmetry to laypersons without reference to abstract ideas. It is a real eye-opener, and ought not be omitted from the article. ~ Ningauble (talk) 17:43, 28 September 2012 (UTC)

Laypersons omit the door numbers from their explanations because they instinctively and correctly appreciate symmetry, and because they are intuitively using probability in its classical sense. They should be applauded for their sound intuition. I'm all behind Richard Hamming here.

If the word "symmetry" is superfluous for laypersons, no hard done in using it. At least it will keep the pedants among the professionals quiet, and it will delight the others. Alternatively, how about the word "independent"? Whether the host opens door 2 or door 3 is independent of whether there is a car or a goat behind door 1. That is both intuitive and true (and easily proven, using symmetry, if you feel a proof is needed). And it's in the literature, in the discussion of Morgan et al. Richard Gill (talk) 17:58, 28 September 2012 (UTC)

Ninguable, I am with you on this one, 'The experts, by their very expert training and practice, often miss the obvious and distort reality seriously'. Morgan is a classic example of where experts use what is essentially a conjuring trick to ruin a simple puzzle. Martin Hogbin (talk) 22:13, 28 September 2012 (UTC)

I agree that name-dropping a cryptic word like "symmetry" doesn't help. The conditional solution needs to be explained in a way that is actually understood; that is done in the "Explanation with door 3 open" section of the proposal.

The "nothing up my sleeve" analogy is interesting, I didn't think of it like that. Yet if we show one of the doors open or talk about the probability of one door changing, I think we are going to confuse many people. I've often enough seen people say that probabilities don't change and get all confused about that, and explicitly showing two closed doors is just begging for the "two doors, 50:50!" knee-jerk response. Would showing the three closed doors figure and closely following the Adams explanation like this be suggestive enough?

The player is basically given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host is saying in effect: "You can keep your one door or you can have the other two doors, one of which (a goat door) I'll open for you."

This explicitly mentions opening the goat door, but leaves making the connection to sleeve-rolling (or to whatever analogy they think of) to the reader. -- Coffee2theorems (talk) 23:01, 28 September 2012 (UTC)

Once upon a time I went looking for a FA-class reference for the "combining doors" solution. The best I could come up with was Devlin (who, as has been mentioned repeatedly here, botched it and in a follow-up column resorted to a Bayes' Theorem solution). If anyone can come up with a FA-class reference (no blogs, not some random guy with a web page) presenting this solution in a way that is better, please let us know what it is. Perhaps obvious, but I'm fine with the overall approach Coffee2theorems is suggesting here. It sounds like Elen is OK with it, Richard is OK with it, C2T (obviously) is, I am - are Martin and possibly Gerhard the only naysayers? -- Rick Block (talk) 05:06, 29 September 2012 (UTC)

A certain RD Gill fixed K Devlin's solution in print. He did not use any scary words like symmetry. [14], middle of page 5; this is an article in the online statistics and probability encyclopedia of the big national and international statistics societies, and peer review is part of the procedure of publication. The contestant initially chooses door number 1. Initially, his odds that the car is behind this door are 2 to 1 against: it is two times as likely for him that his choice is wrong as that it is right.The host opens one of the other two doors, revealing a goat. Let’s suppose that for the moment, the contestant doesn’t take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind door 1, the information that an unspecified door is opened revealing a goat cannot change the contestant’s odds that the car is indeed behind door 1; they are still 2 to 1 against. Now here comes the further detail which we will take account of in this solution: the contestant also gets informed which specific door was opened by the host – let’s say it was door 3. Does this piece of information influence his odds that the car is behind door 1? No: from the contestant’s point of view, the chance that the car is behind door 1 obviously can’t depend on whether the host opens door 2 or door 3 – the door numbers are arbitrary, exchangeable. Therefore, also knowing that the host opened specifically door 3 to reveal a goat, the contestant’s odds on the car being behind his initially chosen door 1 still remain 2 to 1 against. He had better switch to door 2. Richard Gill (talk) 09:42, 29 September 2012 (UTC)

I could see a section called "Explanation with a door open" going in there, explaining (without an illustration showing door 3 open) how nothing changes for a contestant who does not notice which door was opened - maybe they are blind and are told that a door was opened but not which - and then saying something like:

Many people find it intuitively clear that further being told the door number reveals no new information about what is behind the player's door. However, this is only true when the host chooses randomly among his allowed choices. In a variant where the host even slightly favors one of the doors when possible, new information is revealed by his choice and the probability does change.

Trying to actually show that no new information is revealed to those whom it is not intuitive would get way too complicated, IMO, so leaving it at "many people find it intuitive" seems best. I also feel that pointing out the dependence on the unbiased host assumption is necessary if we go down this route, because readers otherwise probably won't notice and are thus misled to think that this applies to the biased host variant as well. That is entirely analogous to people not noticing the dependence on host's restricted choice assumption and assuming the simple solutions apply to the ignorant host variant as well.

This would be the only part of the "Solution" section that actually explicitly uses belief update and has "changing probabilities". I'm worried that this is unnecessarily confusing with little gain. Many peple are not familiar with the idea that probabilities can change and some resist it. The probability interpretations also rear their thorny head. That said, it's not an impossible idea. -- Coffee2theorems (talk) 10:02, 29 September 2012 (UTC)

Coffee2theorems, I notice that you are not appreciating Devlin's solution because you apparently have a frequentist mind-set. You seem to think that we can only use symmetry if we know the host is unbiased. No. We weren't told anything at all. We don't know whether the host is biased or not. If he's biased we don't know which direction he's biased in. All in all, for us it is equally likely that he'll open Door 2 or Door 3 when he has that choice because we have been given no information to the contrary. And Vos Savant underlined the fact that we are supposed to take the door numbers as mere labels, of no significance at all, in her wording say Door 1 etc.

I notice that many mathematicians have a frequentist mind-set and I suspect this is because they have been exposed to academic probability which is usually taught from a frequentist perspective. The rest of the world is still just as Bayesian as ever.

Long ago Martin said a very wise thing: no one can think seriously about MHP without pondering on the meaning of probability. Probability interpretations raise their head, thorny or not, the moment you start to solve a popular brain-teaser by using probability notions. You had better be aware that there are different interpretations, all of them legitimate, and that there is a big mismatch between the more common (subconscious) popular interpretations (which are the same as those of the founders of probability: Huygens, Pascal, Laplace, Boole ...) and the more common present day academic interpretations.

Like it or not, probability does not belong exclusively to the experts. We need meta-experts here, not experts. Richard Gill (talk) 13:35, 29 September 2012 (UTC)

I'm actually Bayesian; I'm just trying to avoid unnecessarily imposing that on the reader at the outset. If the reader understood common notation, we could bash P(gobbledy) = gook at them with impunity and be interpretation-agnostic, confusing nobody regardless of their personal interpretation of probability. We can't do that, but we can use counting of "equally possible" (sic!) cases and easily understood frequencies ("2 people out of 3") while still remaining basically interpretation-agnostic (it's not like we are using confidence intervals or p-values here). In fact, with K&W assumptions, (almost) everyone agrees regardless of interpretation! That's as close to confusing nobody as we can reasonably expect to get.

You seem to think that laypersons are Bayesians. That is indeed the Bayesian party line, but to be honest, I don't think laypersons actually have a consistent interpretation of probability that a statistician or a philosopher would recognize. It probably also varies quite a bit from person to person. People have different backgrounds, and the use of Bayesian language confuses some people (even if they perhaps think in a quasi-Bayesian way at some level). What little exposure to probability theory people have is also usually explained in frequentist terms, even if nothing inherently frequentist is actually covered. That's one reason I wrote the "However, ..." part in those terms (another is that the K&W assumptions are in those terms as well); it is easily enough re-expressed in Bayesian terms if need be. Having to make a choice of interpretation this way is IMO a major drawback of this explanation. -- Coffee2theorems (talk) 17:27, 29 September 2012 (UTC)

Hear, hear, a certain Nijdam does not consider this a fix of the combining doors "explanation". It's a solution to the MHP, but it does not fix the fundamentally wrong idea behind the combining doors. There simply is nothing to combine in making the paradox better understandable. Nijdam (talk) 08:08, 29 September 2012 (UTC)

Let me explain to dear Nijdam why this is a fix of the combining doors explanation. It focusses on what is behind Door 1. A car or a goat. That means that it lumps together what is behind Doors 2 and 3. We focus on the question: is the car behind Door 1, or is it behind Doors 2 and 3 combined? Secondly it splits up the information which we get from the host's actions into pieces. What information do we get from the host about the question whether or not the car is behind Door 1? The answer: no information. Devlin was a bit careless about this. For those who might still be worried that he did not tell the hold story, my solution discusses what information, about the question whether or not the car is behind Door 1, is given it by being revealed *which* door, Door 2 or Door 3, is opened by the host. Answer: none, by symmetry. Or if you find that word to scary: none, because we are given to understand that the door numbers are just arbitrary labels. There is nothing special about any of the doors. Richard Gill (talk) 09:42, 29 September 2012 (UTC)

So, maybe some editors don't understand some solutions: that doesn't mean those solutions are wrong. It might mean that some editors' powers of imagination, or understanding of common English language, or knowledge of the sources, are too limited. The first important issue is whether or not those solutions are published in reliable sources. The second important issue is whether there is a consensus among editors that they might be useful to our readers. The third important issue is whether or not there is a concensus among editors that thy are notable enough that we have to inform the readers about them. If an editor disagrees with a solution then he or she should point to a reliable source which supports his point of view. Richard Gill (talk) 09:36, 29 September 2012 (UTC)

Well dearest Richard, do you, with all your knowledge of the English language, understand what the fundamentally wrong idea behind the combining doors argumentation is? BTW: what do you think of the combining purses example? Nijdam (talk) 12:16, 29 September 2012 (UTC)

In this early stage of receiving what (picture) the tricky one-time-story of devolution has left in the end, it is not necessary to imply that the "paradox" that it has left can only be viewed under one single slight aspect, and in this early stage it is never necessary to imply the unimportant illusion that the overall chance to win by switching could be "enhanced by staying" in one special "repetition" that was never implied to ever follow. Never necessary in that stage. Gerhardvalentin (talk) 13:23, 29 September 2012 (UTC)

Nijdam: I can see a lot wrong with your version of the combining doors argument, but I see no point whatsoever in writing out an argument which is obviously wrong. This is just like Morgan et al. actually changing Vos Savant's words in order to be able to claim a point and get a publication in a prestigious journal by saying she was wrong. No: Morgan et al showed their stupidity by not having the imagination to see that she was right, if interpreted in a constructive way. As some of the discussants of their paper pointed out.

You say that your version is the combining doors argument, but we don't have to buy your version of it. Especially not, since it is wrong. It's a straw man, a characature.

On the other hand I see nothing wrong with the combining doors solution in the form in which I presented it to you here, quoting moreover from a reliable source, who, I happen to know, wrote it for the express purpose of showing that Devlin was close to home with his argument - he simply made one step without motivation, and when people pressed him on it, he apparently got confused and decided it was safer to go a long way round instead of finishing his short, elegant, route to the summit. But he's not a probabilist or a statistician so I can imagine he lost his self-confidence at that point.

As to the purses, I don't understand your combining purses remarks at all. If you think it has relevance you had better find a reliable source which explains it, and hopefully explains it better. If such a source doesn't exist you'ld better write one. I suspect that you have a somewhat narrow mathematician's notion of probability which doesn't always match well to what ordinary people mean when they use the language of probability in ordinary conversation and non-technical writing. Mathematicians don't have a monopoly on the use of English. Richard Gill (talk) 13:19, 29 September 2012 (UTC)

Happily you admit there is a lot wrong with "my" version of the combining doors solution. Here is a quote from Devlin's article:

Imagine you are the contestant. Suppose the doors are labeled A, B, and C. Let's assume you (the contestant) initially pick door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C. (Notice that he can always do this because he knows where the prize is located.) You (the contestant) now have two relevant pieces of information:

[1] The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3.

[2] The prize is not behind door C.

Combining these two pieces of information, you conclude that the probability that the prize is behind door B is 2/3.

Apparently you admit there is a lot wrong with Devlin's way of reasoning. Nijdam (talk) 14:08, 29 September 2012 (UTC)

There is not a lot wrong in Devlin's way of reasoning. There is one missing step in Devlin's argument. Devlin admitted that there was a step missing. The missing step is easy to fill in. It has been done by yours truly, and no doubt by many others (it's easy to do). What's the point of talking about a wrong argument which is nearly right? Fix it first and then discuss whether or not it is useful in the popular early parts of the article. Let's talk about the article, not about everyone's personal favourite and personal most hated solutions. Richard Gill (talk) 14:25, 29 September 2012 (UTC)

@Richard, please show me where "my" version of the combining doors diffesr, as you stated, from Devlin's. I would not brag much about fixing Devlin's argumentation. It is in fact no more than the complete solution to the problem, which is well known and of a very elementary level. Calling it fixing is something like showing a front tyre and the rear light of a bike and saying it's a bike, but only needs some fixing. Devlin does no more than stating two more or less given facts, coming down to: every door has 1/3 chance on the car and (surprisingly, but apparently not to Devlin), door 3, the opened one, has probability 0 on the car. To me it's a contradiction, instead of a brilliant idea, and I surely hope you do not use this way of reasoning yourself when explaining the MHP. Nijdam (talk) 15:31, 3 October 2012 (UTC)

Nijdam: If I get you right your position is this: even if the door numbers are elided from the problem and we have a "you choose a door, host opens another door, wanna switch to remaining door?" formulation, one should still condition on doors, because they are distinguishable to the player (e.g. left, middle, right door; or top, middle, bottom door). You do not condition on specific goats (e.g. label them A and B yourself and condition on B, just like you label the doors 1, 2, 3 and condition on 3), because the goats are indistinguishable. I am curious: how do you justify that view?

Surely the goats are visible to the player. The goats have a size (say, p times the average goat). If the host chooses randomly, then no matter what continuous size distribution you use for the goats, the revealed goat's size almost surely won't hit the median of the hidden goat's distribution, so "which goat is bigger?" will almost surely not be 50:50. You got information distinguishing the goats; you have the "probably smaller goat" and the "probably bigger goat". Might not goat size therefore matter in general (if you took Freud too seriously, you might believe the host wants to show the bigger goat..?), and need to be conditioned on? -- Coffee2theorems (talk) 07:40, 4 October 2012 (UTC)

I think Devlin's explanation has great convincing power and should be included early on in the article but I agree that it has some problems. To avoid clogging up this page with more discussion I have started a page on the subject where anyone with an opinion is welcome to comment.Martin Hogbin (talk) 11:50, 14 October 2012 (UTC)

Devlin is not a reliable source, and was plainly mistaken in his so called explanation. It may looks convincing, but in an utterly misleading way. It is especially this kind of nonsense that should be banned out once and for all. Combining doors does not in any way contribute to understanding, other than in a false way.Nijdam (talk) 13:57, 14 October 2012 (UTC)

(editing)

@Coffee2theorems: I had the impression you know all the ins and outs of the MHP, so you will understand and support my objection to Devlin's combining doors, as being highly misleading as an explanation. What your question about my position concerns, I do not see the immediate connection with the above discussion about combining doors. If you mean, the audience, being informed about the rules of the game, and unaware of the door the player will choose, is asked in advance whether the player should switch or not after the host will show them a door with a goat, yes, even then the answer "yes, you should witch" will be motivated by considering all the conditional probabilities,m like: if the player chooses door 1, and the host opens door 3, then .... As for the goats, firstly we may just consider them as undistinguishable, or formulate the problem with two empty doors. I think most people accept the goats as no more than "bad luck". But you will know this, so your question must beef a more sophisticated nature. When the goats are distinguishable, we are stuck with two goats, from which we do not know how they are chosen. Clearly if a player wins a goat, she has to be replaced. However I see no difficulties, as long as the host, when applicable, chooses randomly between the door numbers. Is this what you're aiming at? (connection was broken)Nijdam (talk) 12:34, 4 October 2012 (UTC)

I see Devlin's combined doors explanations as variants of the "condition on the event 'door 2 or door 3 was opened'" approach, and I don't see any actually wrong statements there. The question is whether he has given a convincing argument or not.

That leads to the question of whether it is permissible to condition on the event "a door (= door 2 or 3) was opened, showing a goat (= goat A or B)", or whether you should condition also on one of "door 3", "goat B", or both. Essentially, Martin's Billy/Nanny argument; I just translated it into something observable (e.g. goat size), as otherwise it makes no sense to me.

I agree that the problem may well be expressed without goats (I explained my views on that here (01:58, 2 October comment)). What I'm aiming at is that, while the idea of conditioning on goats is surprising (why should that matter! a goat is a goat!), conditioning on doors is just as surprising to others (why should that matter! a door is a door!). I can't help but notice the similarity! Consider that the person you are arguing with (Gill) is also quite competent in this field; to me that is evidence that perhaps the issue is not so cut-and-dried after all.

The MHP can also be formulated without distinguishable doors. One way is to use a bag with a red marble and two blue marbles in it, instead of e.g. shells and peas like vos Savant's (goatless!) simulation. IIRC some years ago Gill opined that the former is still a simulation of "the MHP". I feel that it is not quite satisfactory; but less strongly than I used to. Is that purely a matter of opinion, or a matter with a clear-cut correct/incorrect answer, or something in between? That, I feel, is the core issue. Once you have a formal model instead of a fuzzy word problem, there's not going to be much disagreement among informed people; the issue is in how to get there.

It's true that the "condition on a door" arguments don't work in the case of hosts who favor e.g. the right-hand door, and that that takes people by surprise (oh, door number can matter!). But the "condition on door 3" version also doesn't work for hosts who favor bigger goats, and I bet that would take people even more by surprise (oh, goat size can matter!). Either way, you've taken a real(?) situation, built a simplistic model, and are surprised when reality breaks out of it. If reality doesn't do that (lucky!), was your reasoning nevertheless wrong, because reality could have done so and you would have been surprised if it did - you just got lucky? Can you see how reasonable people might end up having a disagreement here, depending on the details? You might argue that including a door number is a second-order correction and something fuzzy like goat size is a third-order correction (effect limited by observation uncertainty), but by then you have admitted that both are approximations instead of Absolutely Correct One True Solution(s). The first-order model, with indistinguishable doors, already captures the main part of the phenomenon (2/3 vs. 1/2), so it's not entirely unreasonable to argue that it's a legitimate option to stop there ("good enough for physicist work!"). -- Coffee2theorems (talk) 18:59, 4 October 2012 (UTC)

Well, look closer at Devlins (and others') way of reasoning (see above). Try to understand what he means when he says: combining these two pieces of Information. There is nothing to combine.there! And hence nothing is explained. It really is easy to see. Nijdam (talk) 19:44, 4 October 2012 (UTC)

It's about a brain teaser that has been told in a story about a "one-time problem". In this one-time problem, the host either got the car and only one goat in 2 out of 3, being restricted to show his only goat then. – Or, in this one-time problem, the host got no car but both goats in the rest of 1 out of 3. Then – in this one-time problem – he cannot "show" any (for ever unknown) "preference". It is of no avail and therefore unnecessary to "assume" what you never will know, instead of trusting in that what you already know for sure, just from the outset.

And, just not to fall into the dangerous 50:50 trap, it is helping to "see" that in only 1 out of 3 his two doors hide just goats, but in 2 out of 3 he has got the only prize with one of those two goats. Not to be trapped, it is helping to distinguish the door first selected on the one side, and the group of those two host's doors on the other side. There is nothing wrong. And, although it is a one-time problem ("you are in a" game show, and not "you will be in ten game shows"), in only about 33 millions out of 100 millions staying will give you the prize, but in more than 66 millions out of 100 millions switching will give the prize, because of symmetry: "no matter at all, which one" of his two doors the host did open in that 100 millions. Gerhardvalentin (talk) 21:17, 4 October 2012 (UTC)

To Nijdam: Falk says that conditioning on "which door" has been opened makes only sense "if the host IS biased" and that you KNOW about that bias. And I am adding: You of course CAN utilize a mathematically correct phantasmal cure-all, but then you must be aware of it's inexpedience to give any better advice than to "always switch". Thus: no necessity at all to give the correct answer for the MHP, but very suited and helpful to teach and to train conditional probability theory, see the plenitude of available textbooks on teaching conditional probability theory. Gerhardvalentin (talk) 21:51, 4 October 2012 (UTC)

Imagine you are the contestant. You are given the choice of three miniature doors in a bag (like three marbles in a bag, really). Suppose they are labeled A, B, C; the labels are written in a secret compartment in each miniature. You blindly pick one from the bag (secretly, door A). The probability that the car is in the bag (in both senses!) is 2/3. The host looks in the bag and secretly peeks behind the doors, picking a goat door (secretly, door C), and opens it for you. As Devlin says, "he can always do this", where "this" must mean picking a goat door (he can't very well always pick door C, it might be the car door!). In other words, this is a sure event, and so the car is behind one of the host's doors (secretly, door B or C) still with probability 2/3. The car is not behind the opened door (secretly, door C). Combining these two pieces of information, you conclude that the probability the car is behind the door in the bag (secretly, door B) is 2/3. The car is probably in the bag! Switch!

I think this is basically the same thought process, put slightly more clearly, perhaps with some unnecessary wrinkles smoothed out. ("he can always do this" is a dead giveaway, IMO) You can label the doors and talk about the labels without assuming that the contestant knows them or the distinctions they stand for. You can of course argue about clarity of phrasing things and the sensibility of a model with indistinguishable doors. (another possibility is a blind contestant told appropriate things, and that probably fits the story slightly better, but I like bags of marbles as a model better, being a classic and all; also, the A, B, C labeling here is actually entirely superfluous) -- Coffee2theorems (talk) 23:39, 4 October 2012 (UTC)

Sorry, Coffee2theorems, often people come with alternative descriptions as if they would make the problem more understandable. As if the original setting is not simple enough as it is. Anyway, if you like to use an alternative setting, like yours, with tiny doors, etc, you firstly have to show it's equivalent to the MHP, and guess ... it isn't. That's often the point, someone gives an alternative description, from which they believe it is equivalent to the MHP, and then proves their different way of solving with this alternative description. Why are people that reluctant to admit Devlin's combining is nonsense? Nothing particularly difficult there. I will spell it out again. Devlin reasons: 1) the not chosen doors 2 and 3 have together probability 2/3 on the car (what he omits to mention, or simply doesn't realize, is that each has probability 1/3); 2) the opened door 3 has probability 0 on the car. To me this is a contradiction. And to you? Nijdam (talk) 06:53, 5 October 2012 (UTC)

I'm not sure that Gerhard is against this. The use of the closed doors figure alone (with an explanation, to suggest the partition) at the beginning was actually his idea (10:27, 27 September 2012 comment above), I just copied it because it seemed good to me. It may of course be that I somehow misunderstood it, but I don't see how. -- Coffee2theorems (talk) 08:17, 29 September 2012 (UTC)

Proposed text for Solution section, take 2

Solution

The player should switch. Switching wins twice as often as staying. [many citations]

The answer is easily verified experimentally. It has been verified countless times using computer simulations as well as simulations with human participants. [many citations]

Simple explanation

If the player is resolved to always stick to their initial choice, they might as well not be offered the option to switch at all. Sticking wins exactly when the player initially picks the car door, which happens 1/3 of the time. Switching wins the other 2/3 of the time. This may be easier to grasp if the doors are grouped into player's doors (door 1) and host's doors (doors 2 and 3), as illustrated below.

The player is basically given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host is saying in effect: "You can keep your one door or you can have the other two doors, one of which (a goat door) I'll open for you." No matter what the host does, a player who always switches gets all cars behind the host's two doors. (Adams 1990)

The situation can also be analyzed by cases, as illustrated below. Switching loses in only one of the three equally likely cases (case 1), and wins in the other two. A player who always switches wins 2/3 of the time. (vos Savant 1990b; Krauss and Wang, 2003)

Case 1: Car behind door 1	Case 2: Car behind door 2	Case 3: Car behind door 3
Player picks door 1	Player picks door 1	Player picks door 1
No matter what the host does, switching loses	Host must open door 3; switching wins	Host must open door 2; switching wins
One case where switching loses	Two cases where switching wins

Many people find the situation intuitively easier to understand by considering the same problem with a million doors instead of just three. (vos Savant 1990; Krauss and Wang, 2003) In this case there are 999,999 doors with goats behind them, and the player's chances of picking the car door are one in a million. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times (i.e. the times the player initially picked a goat) the other door will contain the prize.

Explanation with door 3 open

When first encountering the problem, most people consider only the situation where door 3 is open, leaving only two possibilities: the car is either behind door 1 or door 2. They then incorrectly conclude that the chances are 50:50 either way (Krauss and Wang, 2003). These are indeed the only two possibilities when door 3 is open, but they are not equally likely! This is illustrated in the alternative explanation below, with these two possibilities on the left side of the thick line.

Car behind door 2	Car behind door 1		Car behind door 3
Player picks door 1 This happens to two in six players	Player picks door 1 This happens to two in six players		Player picks door 1 This happens to two in six players
Host must open door 3 (two in six) Switching wins	Host randomly opens door 3 (one in six) Switching loses	Host randomly opens door 2 (one in six)	Host must open door 2 (two in six)
Three in six players see door 3 opened. Two of them win and one loses by switching.		Three in six players see door 2 opened. What happens to them is irrelevant to those who see door 3 opened.

The above illustration shows everything that can happen to players who pick door 1. On average, three in six end up in the exact situation described in the problem statement (door 3 is opened; left of the thick line). Two of the three win and one loses by switching; therefore the probability that switching wins in that situation is 2/3 (Chun 1991; Grinstead and Snell 2006; Morgan et al. 1991).

Comments on the proposed text, take 2

OK, I modified the proposal again to take some of the above comments in account, as well as other modifications. What do you think? -- Coffee2theorems (talk) 20:15, 29 September 2012 (UTC)

Very well illustrated and professed, really a good assistance. Gerhardvalentin (talk) 21:33, 4 October 2012 (UTC)

I like this choice of three solutions (plus one analogy) for exemplifying basic approaches to the problem, and they seem pretty clear. But I do have some reservations about a few particulars:

The present Solutions section includes a lot of other material, in multiple subsections. You have not indicated how this would be reorganized (or removed?).
I would prefer not to reference simulations in a section on solutions because, as I have mentioned previously, they demonstrate and persuade but they do not solve or explain. I think the immediately following section should be about simulation. That section should begin by noting that many people find the above explanations unconvincing, and give the anecdote about Paul Erdős not being sure of the result until he ran a simulation. (Moving this from the lede, where it is more detail than needed.)
I still don't like the "with door 3 open" gambit. Characterizing it this way and referring to it as "the situation where door 3 is open" [emphasis added] is susceptible to the polemical interpretation that the previous explanations are inapt for this situation. I would prefer to find a less prejudicial heading and to reword the pertinent clause as "... most people consider only that door 3 is open, leaving two possibilities: ..." However, the same observation about people's common error applies equally well to the preceding two explanations, so the sentence might be moved to the main section introduction (in lieu of remarks about simulations).
In the million doors analogy, language about "on average" and "times the player initially picked" uses an expressly frequentist interpretation, but the analogy is equally applicable to a Bayesian interpretation. The same sort of slant occurs in other paragraphs that refer to, e.g., "1/3 of the time" and "always stick" [emphasis added]. I would prefer language that gets the point across while remaining agnostic about interpretations of probability.
In the "six cases" illustration, it might not be immediately obvious that this refers to 3 possible car locations times 2 doors Monty might open = 6 cases. This could be mentioned in the sentence introducing the illustration. I also have some reservations about expressing this in terms of six players because, as I have mentioned previously, replacing "which door do you want?" with "which player do you want to be?" strikes me as an unnecessary multiplication of entities.

All that said, this draft is a pretty good framework to build upon, and the prose is reasonably clear. The general idea I am using in points 3 & 4, and to a certain extent in point 5, is that if we are going to cut to the chase by giving straightforward, accessible solutions without going into different interpretations of the problem and of the meaning of probability, then we need to be agnostic about the differences we are not explaining (at least until it is taken up in a later section, if at all). My recommendations in points 3 and 5 are also influenced by my opinion that the root cause of the naïve 50:50 error lies in missing the distinguished status of the two unchosen doors. ~ Ningauble (talk) 18:27, 5 October 2012 (UTC)

I have indeed not explained what to do about the rest of the material. That's because this proposal is based on the RfC, where the idea of a compact and simple initial Solution section is present in both proposals. I figured that if that part is resolved, it's a big step forward, and trying to propose two things at the same time multiplies already small probabilities of success.

One idea would be to replace everything in the Solutions section before the "Formal solution" subsection with this, and then figure out what to do with the rest. Most of the rest could, for example, be moved into a later section ("Mathematical details"?), and insert other important material (e.g. "Simulation", "Sources of confusion", "History") in between. A lot of that also requires cleanup (look at the list of trivia in the "History" section, ouch!). I think this part ought to be a separate discussion, instead of a proposal from me.

The naming of the subsections was not so easy. There are strong feelings here about the two POVs, and it felt like picking labels for the "pro-choice" and "pro-life" factions, with worries that these would be taken to imply "anti-choice" and "anti-life" by the other factions. The first subsection tells you about the general winning strategy and the second analyzes the confusing "[selected] [remaining] [open]" situation in particular. Certainly it's not intended to imply that a general strategy would somehow fail in this particular case. I suppose "Why switching wins" and "Why the 50:50 answer is wrong" would be possible names for the sections. Would that be better? That could tie into writing a short outline at the beginning, explaining the confusion and that we'll first explain the correct answer using strategic thinking ("always switch" > "always stick", overall probability is 2/3) before tackling the confusion directly.

I don't think the use of frequentist language needs to imply frequentist interpretation. The frequentists don't own the law of large numbers, any more than Bayesians own the Bayes' theorem. AFAIK frequentist language is still commonly used in elementary probability and statistics courses for high schoolers and undergraduates (even those who will be Bayesians!), as well as in newspaper articles and other popular media, because it's easy to understand. The use of Bayesian language also tends to stir the controversy here. I don't really care if this uses frequentist terminology throughout, Bayesian terminology throughout, or mixes and matches willy-nilly, as long as it's written understandably with the general reader in mind and is not downright incorrect.

The use of six players instead of six cases is because otherwise you'd need to explain why you've counted an identical case twice (but moving people around is easy to explain). The explanation that Monty might open a door he's not allowed to open is a bit confusing, especially as it's an important point of confusion in this problem that he may not reveal the car. I thought of using a coin which Monty always tosses and then just enumerating all cases, but that had sourcing problems. -- Coffee2theorems (talk) 16:09, 6 October 2012 (UTC)

Naming the explanations without expressing a point of view might be accomplished by stepping back from characterizing what they mean/answer/entail/imply (or whether they are simplistic/general/particular/abstruse), and just describe the features of the problem that they examine or focus on. The following three subsection headings come to mind:

Two choices: your door or Monty's?
Three possibilities: where's the car?
Six cases: Monty's choice

This is about as neutral as I can make it while giving each a usefully descriptive handle. ~ Ningauble (talk) 17:21, 6 October 2012 (UTC)

Incidentally, regarding the six cases, I am really not sure which is less confusing to people: equal-weighted cases where some are virtual pairs representing a non-choice, or unequal-weighted cases forking only where there are real choices. I.e.:

A forked case: Monty's choice

Virtual cases and unequal-weighted cases are both potential stumbling blocks that take a little thought or explanation. That Monty's choice only arises if the player initially guessed right certainly seems to be part of what makes the problem so baffling. ~ Ningauble (talk) 19:14, 6 October 2012 (UTC)

Having unequal-weighted cases is one of the important points in the conditional explanation. It shows that, even though you have two cases, it's still not 50:50! (that's the paradox!) Now what is the reason for that? Is counting people confusing? I don't think so: if you understand simulations, you should understand this. It is basically the description of a simulation, counting simulated people. But surely having one explanation for the "two cases but not 50:50" fact does not exclude having another! Maybe we should somehow explicitly point out that the reason "you picked the car" case is less likely is that in that case, Monty's (figurative/virtual?) coin had to land in a particular way for you to see door 3 opened (the split in the illustration), but if you picked a goat Monty didn't toss a coin (no split), so "door 3 open" is evidence against "you picked the car"? What would be clearest?

Purely proof-structurally descriptive labels feel like an overkill for this potential POV problem. I much prefer splitting things by theorems, not proofs; most people are much more interested in theorems. Would subsection names "Switching is the best strategy" and "The two cases are not equally likely" do? Maybe with some material at the beginning explaining the common misconception that the two cases are equally likely, and that we will first explain why switching is best and then why the two cases are not equally likely? I can't see any possible way to construe these as somehow POV. -- Coffee2theorems (talk) 19:25, 7 October 2012 (UTC)

Oh dear

It looks like we are back to square one. Coffee2theorems asserts above that, 'Unconditional solutions do not, however, give you the probabilities for the two doors in question', and seems to think that the simple solutions therefore do not actually the question as asked. This is what the years long dispute has been about. I suggest that if C2t wants to continue with this assertion he does so on the /Arguments page. Many (most) editors here believe that the 'simple' solutions are fine (or at least that any deficiencies in them can be initially glossed over).

This page is about improving the article, not showing our mathematical prowess. C2t insists that we must show our readers, early on in the article that the simple solutions are in some way deficient. I thought, and hoped, that we had gone beyond that point.

If we have two section entitled, or implying:

Idiot's solutions/Real solutions
Solutions that do not answer the question/Solutions that answer the question
Incomplete solutions/Complete solutions

we project a lack of confidence in our initial solutions and readers are going to be less convinced by them.

Note that I am not insisting the reverse, that we say the simple solutions are perfectly complete and correct, just that we do not distract our readers with esoteric mathematical issues before we have then on-side. I have absolutely no objection to discussing these issues later on in the article. Martin Hogbin (talk) 08:39, 11 October 2012 (UTC)

You say that Coffe2theorems says "Unconditional solutions do not, however, give you the probabilities for the two doors in question". He's right. But so what? Some people aren't interested in the probabilities for the two doors in question, separately. Others realize that it is a simple step to get there, if you are so inclined. Why can't we just have an initial section with decent *arguments* why it is wise to switch doors? We need to help the reader see the answer isn't 50-50. I don't understand, Martin, why you take Coffee2theorems' remark as criticism of some solutions. It's just a fact. If you find it interesting, great. If you don't, ignore it.

Secondly, why talk about *solutions*? Let's talk about *arguments*. Arguments for switching. The word *solution* implies that this is a maths problem to which there is right solution and many wrong solutions. We are talking about a popular brain-teaser, we want arguments which help people understand what is going on. Richard Gill (talk) 12:18, 11 October 2012 (UTC)

I have been avoiding much interaction on this talk page, but I want to say that I agree with Martin Hogbin's comment completely, and it expresses what I was thinking when I commented in the RFC. Much of the "conditional" solutions appear to me to represent (interesting) academic extensions of the original problem that should be covered after the original problem is discussed. That's the genera pattern all over Wikipedia: begin with the basic case, and cover academic extensions later on. — Carl (CBM · talk) 12:51, 11 October 2012 (UTC)

I didn't say anything about the unconditional solutions not answering the question. They do answer the question: Switch! They even go the extra mile to show that always switching wins 2/3 of the time, and always staying wins 1/3 of the time, so the first strategy is twice as good as the other. You don't even need the K&W assumptions if you randomize the door numbers; randomization guarantees that the unconditional probabilities are objectively correct without any assumptions. If you really doubt that any of the other more bizarre possible strategies could possibly be better, then Gnedin will reassure you.

The problem statement says nothing about epistemic probabilities of the car location in the one door open situation, and they are fundamentally unnecessary for solving the problem (but necessary for resolving the paradox). Of course the unconditional solutions don't give these probabilities; they simply do not condition on anything.

You claim that I somehow insist that we must tell the reader that "simple" solutions (whatever that means) are deficient. I didn't say that anywhere. Your characterizations of the suggested section names are also not in any way apt. For example, how would "Game theory" translate to "Idiot's solutions"? Or "Strategic thinking", "The correct answer", "The general solution", "The correct strategy", "How to solve the problem", or..?? -- Coffee2theorems (talk) 12:58, 11 October 2012 (UTC)

Just to clear some possible points of confusion: I was thinking/suggesting of organization like this for the initial (re)solution section (choosing the exact names is difficult; these are just possibilities):

Section: "Resolving the paradox"

Subsection: "The optimal strategy"

Show that always switching wins 2/3 of the time and always sticking 1/3 of the time.
The solutions would be the same as in the "simple explanation" subsection of my current proposal, or possibly other unconditional solutions. Not e.g. Devlin.
Hopefully we can insert a claim here that by randomization the 2/3 winning rate can be guaranteed no matter what the producer/host does; that way he can't possibly guess your initial pick (K&W is actually unnecessary for the solutions of this subsection!).

Subsection: "A further look at the confusing case"

If we can agree on a Devlin-type solution going in the article, I suggest it go here.
The current illustration of a conditional solution, or equivalent.

There was no intent that it should be implied in any way that the unconditional solutions don't solve the problem (as Martin fears), or that more complicated material would be placed before the simple solutions (as Carl fears).

I'm not also suggesting that the article say anything like "the paradox cannot be resolved without conditional probability". I'm just saying that the essential reason the MHP is paradoxical is that people apply the principle of indifference twice, that's inherently a subjective/epistemic probability belief update issue, and that it's helpful if the subsections are split along the unconditional/conditional line (or strategic / belief update, same thing) for that reason. Solve the problem first by looking at the overall probability of winning of each strategy, then look at the probability of the car in the two doors situation. -- Coffee2theorems (talk) 14:29, 11 October 2012 (UTC)

Your approach is IMO ridiculously academic for what was intended to be a simple puzzle and it is based on the assumption of a bit of dubious mathematical pedantry. Some people may imagine a specific case that the player has chosen door 1 and the host has opened door 3 then they have to choose, others may imagine a more general situation but, with a reasonable assumptions or a Bayesian perspective it really does not matter which situation they imagine, the answer is the same they will double their chances of winning by switching, whether they play once or many times.

To artificially split the problem right from the start is utterly confusing and unhelpful to the general reader. Martin Hogbin (talk) 15:18, 11 October 2012 (UTC)

We should keep clearly in mind that MvS presented a one-time problem. The contestant was not offered to "play many times". The article should carefully never confound the frequentist's view and "often-times repetitions". Gerhardvalentin (talk) 15:42, 11 October 2012 (UTC)

Coffee2theorems is *not* artificially splitting the problem. There are a number of different ways in which ordinary people can br helped to realize that switching doors is smart. Different ways catch on with different people. All of them are natural, all have advantages and disadvantages. See Kraus and Wang (very helpful overview, written by experts in cognition, not mathematics). To every popular solution there corresponds a formal mathematical version. His approach is not academic. If you know the literature, and want *on this talk page* to discuss how to structure the article, then you should use appropriate language to discuss things. It doesn't mean that the article has to be drenched in the same language.

Vos Savant deliberately names the doors Door 1, Door 2 and so on, so that the newcomer to MHP will adopt the visual picture which blocks their understanding and entices them to the wrong answer (see Kraus and Wang). To realize intuitively that 50-50 must be wrong you first have to discard that visual picture. Think strategically, or group the doors, or forget the door numbers. Then you see that switching is wise. Finally, it is interesting to go back to the original situation, the visual picture, and try to understand what was wrong with it. This is not *academic*. This is achieving full understanding. Maybe some readers are just interested in a slick one-liner why switching doors is rather smart. But some readers will want to understand what was wrong with the intuitive but wrong reasoning: Doors 1 and 2 must have equal probabilities. This is not perverse, not a matter of advanced mathematics. It is not an academic extension! Nothing is extended! It can be explained with simple verbal reasoning and needs only common sense. No need to use expensive words like "conditional" or "symmetry". There are absolutely simple arguments why, when you are in this situation (you chose Door 1 and the host opened Door 3) 1/3 of the time the car will be behind Door 1, 2/3 of the time the car will be behind Door 2: e.g., Devlin's argument (with missing step filled in), or a simple calculation from a table, or the observation that the probabilities can't depend on the specific door numbers named in the case at hand combined with what we know from the one-liners. I think something like this will be enlightening for many readers. If it can be said in a few simple words then it belongs in the non-technical, not specialist part of the article. It reinforces the conclusion of the one-liners, it doesn't weaken them. If a reader like Martin is uninterested or finds it too obvious then he skips it. Richard Gill (talk) 10:49, 12 October 2012 (UTC)

Exactly! Artificially splitting the explanations into "simple" and not "simple" (by presenting only "simple" solutions first) is utterly confusing and unhelpful to the general reader. -- Rick Block (talk) 16:44, 12 October 2012 (UTC)

Oh! I didn't realize anyone would think I'd want to drench the article in statisticese. Nothing of the kind! I'm quite simply trying to make a minor modification to my current proposal to satisfy Martin's concerns about section names. This could be as simple as just changing those section names and nothing else. You can see the text there, it's not in statisticese. If changing just the names isn't enough, then we can change more things incrementally, yes?

The reason I brought up Devlin and objective probabilities is simple: I'm trying to figure out how to make the text more palatable to Martin! AFAICS Martin is afraid that if one of the subsections contains the explanations he likes and the other the explanations he dislikes, then the latter might come to be presented more favorably ("separate but 'equal'", with the associated problems). So I brought up the possibility of including a Devlin-type explanation, which I gather he likes! That should make further discussions more stable as well, as Martin would have favorite explanations on both sides of the natural splitline. Other simple conditional explanations acceptable to Martin would serve the same purpose. Having attractive words like "objectively correct" or "hard guarantee" etc. on the unconditional side of the line might also help; hence the mention of objective probabilities (the conditional subsection is far more naturally interpreted as subjective/Bayesian than objective/frequentist).

Unfortunately, I have no idea what Rick is talking about above. Rick, could you clarify? Are you agreeing or disagreeing with me? With Martin? Both? Neither?? Or Richard? None of the above? -- Coffee2theorems (talk) 19:25, 12 October 2012 (UTC)

I'm agreeing with Richard, supporting your presentation, disagreeing with Martin. -- Rick Block (talk) 05:56, 13 October 2012 (UTC)

Step by step

I would like to give the following suggestions for changes in the article:

First step: In the introduction vos Savant's answer could be cited:"Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance." The remaining part of her explanation could be part of a new section labeled "The unconditional solution", for example, because the reader doesn't know her assumptions in this moment, and therefore cannot follow her reasoning.

Second step: The illustration in the section "Solutions" above could have a better place in the section "The unconditional solution". A reader, who has in mind the sentence:"Imagine that you chose Door 1 and the host opens Door 3, which has a goat."(Krauss and Wang), could be disturbed by row 3 of the table with the car behind door 3.

Third step: Vos Savant's assumptions should be pointed out explicitly (as proposed by Rick Block and Elen of the Roads) in the section "The unconditional solution", and it should be explained, that without her assumptions the odds are the same for the remaining doors.

Fourth step: The Extended_description... could be the beginning of a new section "The conditional solution". The conditions are:

1. The contestant picked door 1

2. The host opened the goat door 3

Good night! --213.102.99.51 (talk) 21:21, 17 October 2012 (UTC)

Not sure if I would call it 'the unconditional solution' - it may not be immediately apparent what is meant by 'unconditional'. --Elen of the Roads (talk) 22:49, 19 October 2012 (UTC)

The "unconditional" solution doesn't refer to the special doors, one picked by the contestant and the other opened by the host. The corresponding solution table, presenting only one of the three possible cases (the contestant initially picked door 1), shows the host opening door 3 (row 2) or opening door 2 (row 3). For the "conditional" solution ("Imagine that you chose Door 1 and the host opens Door 3, which has a goat."), row 3 is impossible, and you need another illustration. --213.102.96.29 (talk) 15:16, 20 October 2012 (UTC)

I know what 'unconditional' means in this case. My concern is that since conditional probability hasn't been introduced at this point, people might assume that the normal meaning of unconditional applies. This would misdirect them, since Monty is operating under two conditions (using the normal meaning of the word) - that he has to open a door and that he can't show the car. Elen of the Roads (talk) 15:26, 20 October 2012 (UTC)

The 'rules' of a game are not the same as the 'conditions' occuring during the progress of the game (using the normal meaning of the words). At the beginning of a game you don't know anything but its rules. So, the rules are the game itself and they determine the possible moves of the players. The conditions are the special situations which are created playing the game according to the rules. For example:

Rule: The contestant picks a door.

Move: The contestant picks door No.1.

Condition: The host must not open door No.1

Rule: The host opens another door revealing a goat.

Move: The host opens goat door No.3.

Condition: The car is not behind door No.3.

Rule: The contestant may switch or not.

Move: ?

But if you don't like the term "unconditional" in this context, please remember that I had written above:"...a new section labeled "The unconditional solution", for example..." Feel free to propose another label for this section! --213.102.98.144 (talk) 08:46, 21 October 2012 (UTC)

I am with Elen about not calling such solutions "unconditional", but for a different reason. The word does not mean solutions that do not use or are independent of the methodology of conditional probability, it means something very specific within the methodology of conditional probability. Some writers have characterized vos Savant's solution as unconditional for polemical purposes in the context of addressing the methodology of conditional probability. Whether or not it is appropriate to describe these polemics later in the article, it should not be introduced in the early sections, and Wikipedia should definitely not be taking one side or the other on a matter of dispute.

Regarding step 3: If you are referring to the stipulation that Monty must reveal a goat behind one of the unchosen doors then please note that this is not peculiar to Vos Savant's treatment of the problem, it is common to most treatments. Any sources that do not use this stipulation are really addressing variant problems. Because this issue is such a recurring source of confusion, I am beginning to think it should be pointed out in the lede, immediately after the very first statement of the problem. (Cf. discussion of iterations of the problem definition at Talk:Monty Hall problem/Archive 30#Does this version work.) ~ Ningauble (talk) 20:45, 20 October 2012 (UTC)

I think that it should be distinguished between the answer given by vos Savant, not referring to the specific doors named in the problem text, and the solution with conditional probability given by Krauss and Wang. Vos Svant's answer, her solution table and the simple solutions should be explained in a particular section before presenting the "Extended description...", because her answer doesn't fit to the sentence:"Imagine that you chose Door 1 and the host opens Door 3, which has a goat." --213.102.98.53 (talk) 09:25, 22 October 2012 (UTC)

ASS(Another simple solution)

You have 2/3rds chance of picking a goat. If you pick a goat and switch you win. The reason to change doors is because it effectively gives you another chance to pick a goat at the start(not at the end). If you do not switch then it is 1/3rd chance to win for the obvious reasons. Or alternatively, not switching has 1/3 chance of winning ==> swithcing must have a 2/3rds chance. By casting the problem in terms of switching and not switching instead of door 1, door 2, door 3, makes the problem much easier. Since the permutations of the doors do not change the results it is best not to include them in the analysis. — Preceding unsigned comment added by 65.65.56.243 (talk) 09:31, 19 October 2012 (UTC)

This is no solution at all because you don't take into account the rules of the game for your reasoning. If the host only opens a door if you picked the car, then there is a disadvantage to switch.

The permutations of the doors may change the results, the permutations of the door numbers do not change the results, but we have to know what we are speaking about. The door numbers are a useful enumeration for knowing door 2 being another door than door 3, for example. --213.102.97.190 (talk) 12:41, 19 October 2012 (UTC)

Jerrywickey says:

++ A far simpler and more intuitive explanation ++

for the solution to the Monty Hall Problem might be needed. Readers struggling with understanding probabilities need an explanation that they can "feel" in their gut. The text below is such and if no one objects or if someone encourages me to do so, I will post it to the page. After all helping readers who consult Wikipedia is and should be contributor's goal.

simple clear intuitive explanation

The chance that the prize door will be chosen from three doors at random is 33% (1/3) But more importantly the choice has a 67% (2/3) chance of being the wrong door. Removing one door does not change that 67% chance that the chosen door is wrong. Even if both other doors were not opened but instead removed, there still remains the same 67% chance that opening the chosen door will reveal it to be empty.

When one other door is opened to reveal it to be empty, new information is added to the system that can be exploited to recalculate the odds that the remaining door is hiding the prize. The chosen door still has a 67% chance of being wrong. That can't change. This implies that the remaining door has only a 33% chance of being wrong while it has a 67% chance not 50% of being the prize door.

The intuitive argument against this is that "removal of one door reassigns the odds for both doors to 50% 50%; not just reassigns the chance for the door not chosen without effecting the odds of the chosen door." The error of this false assumption is easily demonstrated if the number of doors is increased.

The chance that the prize door will be chosen from ten doors at random is 10% (1/10) But more importantly the choice has a 90% (9/10) chance of being the wrong door. Removing eight doors does not change the 90% chance that the chosen door is wrong, but since the removal was selective, removing only empty doors, but not removing the chosen door nor the prize door, it becomes easy to see intuitively that which ever door remains after eight empty doors are removed has a far greater than 50% chance of being the prize door. It doesn't make sense that the prize is just as likely to be behind the chosen door as the last remaining door, because the removal was selective. The original choice was not selective. The remover knew which door held the prize, causing the remover to not remove the prize door while the chooser did not have this knowledge and made his or her choice .

This is because when making the original choice with the information available before any doors were removed, the chosen door was very unlikely to be the correct choice, 90% chance of being wrong. That chance isn't changed by the removal of eight other doors. However, if it is known that all eight removed doors were empty, then the chance that the remaining, unchosen door, is the prize door is very high. Much higher than 50%. It has a 90% chance of holding the prize because the chance of the chosen being wrong is 90%.

Jerrywickey (talk) 18:54, 25 May 2012 (UTC)

Jerrywickey, I was a 50% 50% believer, read the whole article and still was. The wording of this comment convinced me. If it has made it to the main page since may its lost in the mix. I would suggest adding it, or raising it to a more prominent location. Zath42 (talk) 14:27, 23 July 2012 (UTC)

Jerrywickey, please start a new sections after earlier discussions. It is a fact that some editors follow Morgan et al. in claiming that the chance of the door first selected by the guest could be changed by the special behavior of the host in opening a losing door. If he should be extremely biased e.g. to open his preferred door if ever possible, then he can do that in 2/3, but if in 1/3 his preferred door hides the car he then would be forced to open his strictly avoided door, showing that the chance by switching to his preferred but still closed door is max. 1 and the chance of the door first selected by the guest could converge to zero. So there is some desire to first of all show by Bayes' formula that "which one" of his two doors the host has actually opened could be of influence on the probability to win by switching. Please read also the archive of this talk page. --Gerhardvalentin (talk) 19:49, 25 May 2012 (UTC)

Jerrywickey, your explanation still does nothing for me. I can't '"feel" it in my gut' at all. You remove the wrong doors... and leave 2, it's therefore down to 50/50. But I'm a linguist, not a mathematician. What do I know? Oh yeah... I'm meant to "feel it in my gut". Malick78 (talk) 22:57, 26 May 2012 (UTC)

Perhaps I can offer some insight into this. For some time I have been incorporating this into some training (on an unrelated topic) that I have been giving to scientists and engineers. Here is what I do:

First I ask anyone who has heard of this problem before to silently watch what I am about to do.

Then I hand out the (fully unambiguous, mathematically explicit version of the standard problem) Krauss and Wang description from this page, in writing, and ask everyone to read it and put their answers on paper (unsigned) and hand it back. I count the answers and write that on a whiteboard. Usually, "no advantage to switching" is way ahead.

Then I open it up for discussion. There is always a spirited debate with much certainty on both sides. I have never seen anyone, ever, change their position based upon hearing arguments from the other side. Ever.

Then I prove who is right using the cups simulation described on the page, moving to ten cups if needed. In my experience, this is the best way to convince engineers. (I use toy cars and toy goats - had to buy ten packages of toy barnyard animals at the 99 cent store to get the goats.) I have also found that playing with me as Monty and someone from the audience as the contestant and having the contestant never switch if he thinks there is no advantage to switching works best. I have never had anyone remain unconvinced that they should switch after choosing cups and keeping score.

My point in the training is that actual data trumps logical argument, no matter how sure you are that you are right. but here is an interesting thing I have observed: a significant number of those who got it wrong and argued vigorously that they were right blame the problem description. and it doesn't matter whether I presented the Vos Savant version or the Krauss and Wang version! Just as something about the human mind makes engineers get the wrong answer and defend it to the end, something about the human mind makes engineers reject the notion that they were wrong and blame the problem description. We tend to "feel" things that are not true. --Guy Macon (talk) 02:48, 27 May 2012 (UTC)

Excellent work Guy, perhaps you should try to get it published somewhere so that we can report it here. It is also a sad fact that not one person has changed their mind on the disputed issues here. Martin Hogbin (talk) 13:24, 27 May 2012 (UTC)

Points well taken.

However, removing only empty doors is an assumption of the riddle. Any examination which explores other possibilities is not an exploration of the Monty Hall Problem. If the prize could be removed, the Mony Hall game makes no sense. It is this selectivity on which the solution must be based. If you didn't "feel" the one before then try this.

When making the original choice one has only a one third chance of choosing the winning door. What is more important to understanding the problem, however, is that also means that the choice has a two thirds chance of being wrong. No later event, removing one door included, changes those odds. After the removal of one door, or any other event aside from exposing the winning door, the chance the chosen door is wrong remains two thirds. Nothing can change that.

With the removal of one door only one other door remains. Since the winning door must be one of the two as an assumption of the riddle, then the chance that the chosen door is wrong is still two thirds, which implies that the chance that the single remaining door could be wrong is only one third. Meaning that the remaining, unchosen door, has a two thirds chance, not fifty fifty, of being the winning door.

Some might argue that "removing one door can not change the odds for one of the remaining doors, but not the other. Just designating one door as chosen doesn't give it preferential treatment." This is an erroneous assumption. The error becomes intuitively obvious if more than three doors are used.

If one were choosing from a hundred doors instead of just three, then the chance that the chosen door is wrong is 99 out of 100. Removing 98 doors does not change that probability. It does not give the chooser any more confidence that his chosen door is correct, but he does intuitively realize that the door he chose still has a ninety nine percent chance of being wrong. He also intuitively realizes that since only wrong doors were removed the one single door that remains has a much greater chance, much greater than 50% 50%, of being the winning door. How much? 99/100 Why? Because the one he chose had and still has a 99/100 chance of being wrong. After all, it was selected from a hundred choices. The selectivity of the removal of 98 wrong doors changes the odds for the one remaining door, but not for the chosen door. — Preceding unsigned comment added by Jerrywickey (talk • contribs) 13:59, 27 May 2012 (UTC)

Curiously, you are both wrong. Jerywickey, you say, 'When making the original choice one has only a one third chance of choosing the winning door. ... No later event, removing one door included, changes those odds'. That is not necessarily true, although under the standard assumptions made about the problem it is true that the odds do not change.

To take a really obvious example first, suppose that Monty tells you that the car is behind door number 2. The odds change then for sure.

Now consider a more interesting and instructive case. Suppose that Monty does not know where the car is and opens one of the two doors that you have not chosen at random and it happens to reveal goat. What are the odds then that the car is behind the door that you originally chose?

The important point to consider is whether any event that occurs after you have chosen your door but before you decide whether to swap or not gives you any information about the whereabouts of the car. Under the standard assumptions you know the Monty will reveal a goat, because he must do under the rules, you also gain no information from his choice of door when you happen to have originally chosen the door hiding the car because the host must choose randomly between the two doors available to him under the standard assumptions in that case. So, the host opening, say door 3 to reveal a goat tells you nothing you do not already know, thus your original odds of having chosen the car cannot change. In the standard version of the problem you have a 2/3 chance of winning if you swap. Martin Hogbin (talk) 16:02, 27 May 2012 (UTC)

Thank you Martin, you are absolutely correct, and you clearly articulate the dilemma of the article: -- Not sure who left this comment

So, what difference does it make if you get a new choice each time a door is opened? Sure, the odds are 1-in-3 when you have three doors. But when you get a new choice from 2 doors, the odds are even. So why do people even talk about the door that isn't part of the problem anymore? I am faced with a new choice that is only about two doors. Why so much ink spilled for a simple choice? -- Avanu (talk) 14:15, 9 September 2012 (UTC)

Well Jerrywickey, I posted a "simpler and more intuitive" solution to this problem which (IMO) was elegant in its simplicity but is now gone. I agree with your basic sentiment that a simpler and more intuitive explanation is indicated, but if contributions stemming from original thinking on ways to explain this problem are deleted outright, then clearly Wikipedia is not the place for them. The fundamental question is whether original solutions or explanations are welcome in this article, or if the article is simply a compilation of solutions proposed by others. I have posted my explanation elsewhere on the web where you're not likely to find it. In the meantime, folks can spin their wheels with overly-complex mathematical explanations which aren't clear to the lay person. Chris319 (talk) 01:53, 16 October 2012 (UTC)

Why "always"?

I don't understand the following phrase: "Vos Savant's response was that the contestant should always switch to the other door." The contestant is a person who plays the game only once. So, she can't switch many times. What is to be said by always? --213.102.99.129 (talk) 17:35, 12 October 2012 (UTC)

It is on a one-time show, yes, but anyone is free to "visualize what happens", though. You also could say "in any case". Gerhardvalentin (talk) 18:12, 12 October 2012 (UTC)

As far as I'm concerned, we're talking about simulations. Any averages or repetitions and so on are repeated trials in simulations. We're just trying to get people to understand probabilities by referring to something concrete, i.e. the average rate of something happening in a simulation of many games (as the simulation length tends to infinity, if you're mathematically inclined). The conditional solutions only apply to a one-time game. If you have multiple games, then it gets really really complicated, and AFAIK we don't have any sources for treating that kind of situation. It's only about a one-time show. Only about imaginary (simulated) repetitions. Would it be clearer if we said that explicitly somehow?

"Always" means a general strategy. It means that no matter what door you pick or what door the host opens, you should always switch, no matter what. Not just in the door 1 chosen, door 3 opened case. And overall, when repeated in simulations, that strategy gets you 2/3 winning rate (given K&W conditions). Further, if you pick your first door randomly, then you get 2/3 guaranteed, no matter what. -- Coffee2theorems (talk) 19:44, 12 October 2012 (UTC)

Asymmetric behaviour of the host in opening one of his two doors (note that in 1/3 he has a "choice"):
The argument of Rick Block so far has always been that "if you follow enough repetitions of this game show you will detect some pattern of the host's behavior...." – tending on some "empirical value" to convince that one "must" base on some "q <> 1/2".
You just asked (regarding simulated repetitions) "Would it be clearer if we said that explicitly somehow?", and my answer therefore is yes, we should.

Falk says that, to applying of any host's bias, the host must be biased and you have TO KNOW about that bias. As we never will have such "necessary evidence", we never "need" to base on unknowns, and therefore all we have is the assumption of symmetry, and nothing else.

But of course we "can" and are free to pay respect to the "best case scenario" and to the "worst case scenario" as well: In 2 out of 3 the odds of the two still closed door (contestant:host) are "at least 1:1" (worst case), and in 1 out of 3 they are "0:1" (best case). We know that for sure, and without applying maths. Let's take the extremely biased host: In 1 out of 3 he got two goats, and he can open his "strictly preferred door", odds being "1:0", and in a second one out of 3 he can also open his "strictly preferred door, having the car and one goat behind that special door, but odds being "0:1". So whenever he should open his strictly preferred door (unknown to us!), then probability to win by switching will be at least 1/2. But if he should open his strictly avoided door (also unknown to us!), because the car being behind his strictly preferred door, then probability to win by switching will be max. "1". So we know that we should switch anyway, and the host is very welcome to be biased as much as he can, giving away additional hints we never can nor ever will be interested in, as confirmed per modern reliable sources. Saying that probability theory never is "the key" to making the correct decision and to giving the correct answer, much less conditional probability theory. Exact "conditional probability to win by switching" once and for all is irrelevant.

Any "actual (!) probability" to win by switching, once and forever is completely irrelevant, as all we can ever "know" is that it will be 2/3. But that is not the point. The point is that we are not interested in "actual" conditional probability, not interested in "probability" at all. As per modern sources, the *point* is never to "know any exact actual (!) probability", but the *point* is that we know for sure, just from the outset, that any "staying" forever will diminish the chance to win, as 2/3 is only guaranteed if any "staying" is definitively excluded, just from the outset, and therefore "staying" never will be any option. Never. Just from the outset! That's what we know for sure, and we do not need to know anything else, as per modern reliable sources. We never need to know "actual probabilities", and we do not need any "actual conditional probability". Asking for "actual probability" is a completely superfluous approach that should be discarded, for making the correct decision and to give the correct answer.

Morgan et al.'s view obviously was in the wrong direction and, originally, they indeed went in that wrong direction, complaining about Savant's wording. – Later on, they had to have been corrected, and they had to officially admit their error, had to admit that they were wrong.
Please underline that the question was not on a series of game shows the contestant is in, but just on "a game show". Gerhardvalentin (talk) 22:17, 12 October 2012 (UTC)

What do we know about "any case"? We only know that the contestant picked door No.1, the host opened door No.3 which has a goat, and offered a switch. We don't know anything about the situation the contestant would have picked door No.2, for example, and about the host's behavior in this case. --213.102.99.209 (talk) 23:10, 12 October 2012 (UTC)

So the article, in its actual form, did not help you to decode the paradox? Please help to make the article clear and intelligible. All we really "know" is that staying forever will diminish the chance to win the prize and therefore "staying" once and forever has to be excluded, just from the outset. You only will have a guaranteed 2/3 chance to get the prize if you switch to the door offered, but never otherwise. Actual "conditional probability" being completely irrelevant. Gerhardvalentin (talk) 00:16, 13 October 2012 (UTC)

Try reading the "Extended description of the standard version" section. Does that settle the issue for you? If the lead is confusing, how would you change it? Would simply removing the word "always" from the lead do?

On a related note, we're currently talking about the possibility of replacing the beginning of the solution section with this new version (visible when you click on "[show]"). That one also has phrasings like "always stick", "happens 1/3 of the time", "on average", etc. I'm wondering whether that is suitably clear, or if we should, at the beginning of the solution section, say something like:

Imagine simulating the game show many times and think about what happens to simulated players on average in the long run. The results of such a simulation can be used as a guide for making the final decision to switch or stay even on a one-time show. Actual simulations that have been done are covered in the simulation section. This section explains what will happen in any long enough simulation, and why.

Do you suppose saying something like this would be helpful to the reader? -- Coffee2theorems (talk) 08:07, 13 October 2012 (UTC)

I agree with 213.102.99.129. 'Always' is unnecessary. In the circumstances described the player should switch, and will double their chances of winning by doing so, just as vos Savant said. 'Always' is a relic of a pointless and pedantic argument by editors here. Martin Hogbin (talk) 08:10, 13 October 2012 (UTC)

I would prefer the proposal of C2t to mention early enough that it's not about an ongoing series of shows, but on a one-time problem that, in exactly this form, never took place in reality, and any necessary simulation addresses the possible outcome of that one-time problem. Just to help other editors here not to continually be hoping for "enough repetitions" in order to detect some host's pattern of behavior.

And I repeat my proposal to help the readers to see that the usual wrong conclusion "1:1" is correct only in quite another scenario, in quite another kind of game (if the host does not "in any case" intentionally show a goat, but just opens one of his two doors, be it the car in 1/3 or be it a goat in "only 2/3" !). To see that 1:1 applies to such quite other scenario like "Monty Fall" or "Ignornt Monty". But doesn't apply to the scenario of MvS and MHP. Gerhardvalentin (talk) 08:44, 13 October 2012 (UTC)

If we say that we are analyzing a simulation as a guide for a possibly one-time-show, I think it's clear enough that nobody is trying to detect any long-run patterns of host behavior. I mean, sure, you could figure out any peculiarities of a simulated host; maybe it's a simulation with real players, with your friend playing the role of the host, and he always tends to pick a particular door when he can. But how would figuring out your friend's quirks help you with the real Monty? :) No, I don't think that would be a common point of confusion among readers, especially as we mention computer simulations as well.

Would the example paragraph I gave above (starting with "Imagine simulating") work for you? I think that things like "never took place in reality" fit better in the extended problem description section than the solution section. Such things are more about clarifying what the problem means (just like the "host is not allowed to open the car door" clarification etc.), than about solutions as such. -- Coffee2theorems (talk) 09:37, 13 October 2012 (UTC)

Yes, whatever you prefer. All I fear is that editors of this article will continue to still insist on being able to detect any given host's bias, in "observing him" on the long run.
Saying that "q" could actually "differ" from its average of "1/2" (symmetry), and consequently any solution "must" only be a cure-all so-called "correct solution" in applying "q" (possible range from 0 to 1, although any exact actual value of "q" forever must remain unknown!) within Bayes' formula.

Mathematically fully correct, yes, a suitable description of the obvious spectrum, yes, but forever completely unnecessary to the contestant, unnecessary in the situation the contestant is in. And as any "actual true q", diverging from 1/2 forever is and forever will be completely unknown, this never can be of any help in the given actual situation the contestant is in. Such cure-all so-called "solution" is not suited and hence unneeded to make the correct decision, for

1.) "q" is and remains totally unknown (but on the other hand no-one "needs" to know any "actual q"), and
2.) as "q" forever must remain within the range of "0 to 1", with symmetry at "1/2", staying forever has definitely to be excluded. Excluded. Period.

Yes, it beautifully describes the full spectrum, yes, interesting for students in maths, yes, but without any relevance for the decision asked for. Completely inappropriate: undue weight. Show all of that unimportant display, unnecessary for the decision asked for, in later sections. Show in the first line why staying never can be an option and therefore is discarded just from the outset, and why switching is the only "correct decision". See all modern sources. You know that the overall probability to win by switching on average is exactly 2/3, from the worst case scenario (at least "1/2" in two out of three) to the best case scenario (full "1" in one out of three). And you know just from the outset that staying never can be any option at all. But they continue to distinguish "before" and "after". Undue weight.

Yes, conditional probability maths is capable to nicely describe the full spectrum, but forever is unnecessary to make the decision asked for. Undue weight. Improper to help to decode the paradox, improper to meet all views from different angles that are indispensable and most necessary to understand and to decode. Nevertheless they say that this narrow approach of just only presenting a cure-all mathematical "description" of the given spectrum, and never more, is "the key" to "making the correct decision" asked for, but it never was and never can be such "key", but just only a description. But they insist on such description to be the only correct approach, and that any other approach is "incorrect", as addressing the wrong question ("before" versus "after"). Not being able to read what the sources really say, and especially what modern reliable sources say. That's the ongoing conflict of the article. Gerhardvalentin (talk) 11:17, 13 October 2012 (UTC)

Has anyone checked what vos Savant actually said? It seems to me that if we're appearing to quote her words, we should check her words. There is far too much "own research" going on here, too much original thinking, If Marilyn said "always", she said always. And hopefully the context would clarify what she meant. Richard Gill (talk) 14:59, 13 October 2012 (UTC)

Vos Savant's response was: "Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?" [15] She didn't use the term "always", and it seems to me her interpretation is that of a one-time-game. So, simulations and repetitions are inadequate to solve the problem.

My conclusion after reading through the article is: Vos Savant's response is the wrong answer to her question but the "Extended description of the standard version" describes the right question to her answer. --213.102.99.248 (talk) 17:03, 13 October 2012 (UTC)

Good, the words "always" shouldn't be there. But this doesn't mean that we can't use simulations to solve the problem, or shouldn't think about repetitions. For many people, "the chance of winning by switching is 2/3" means that if you imagine the situation being repeated many, many times, then in 2/3 of the times, you win by switching. Simulation experiments are an excellent way to verify claims about probabilities. The player plays just once. But the player can figure out what is a smart way to play, by imagining many repetitions.

The player who always switches in every imaginary repetition, wins 2/3 of the times. Moreover, among those occasions on which the player chose door 1 and the host opened door 3, he also wins 2/3 of the times. And the same for any other initial door and/or door opened by the host. As long as the car is equally often hidden behind each door, and the host equally often opens either door when he has a choice. These are imaginary repetitions and in these repetitions, we do things equally often because they are equally likely for us in the single case we are faced with in reality. Richard Gill (talk) 15:01, 14 October 2012 (UTC)

To answer this question: I don't understand the following phrase: "Vos Savant's response was that the contestant should always switch to the other door." The contestant is a person who plays the game only once. So, she can't switch many times. What is to be said by always? Vos Savant said "always" because this game tactic was very frequently used on the show. Actually it was the show in that the quests were always asked if they wanted to switch to a different choice. Gandydancer (talk) 16:07, 14 October 2012 (UTC)

You do realise that the puzzle has nothing to do with the real TV show, don't you. The real Monty Hall may have asked contestants if they wanted to stick with their choice (it's frequent in this kind of game show - Noel Edmonds does it on Deal or No Deal), but this theoretical problem doesn't represent one of the games on the tv show. Elen of the Roads (talk) 19:54, 14 October 2012 (UTC)

Of course I realize that it is a "brain teaser" puzzle. How rude of you to act as though I must be stupid. Doubting that anyone here has an IQ in excess of hers, I assumed that she did not make a mistake...or that she is misquoted in the article - and, as it turns out, she is. Marilyn never said "always". [16]. Gandydancer (talk) 03:16, 15 October 2012 (UTC)

Gandydancer, you might try a little good faith here. If you read through the many reams of talkpages - including this one - you'll find that people do pop up who believe that the puzzle is describing an actual game on Let's Make a Deal (someone above suggests including the results of the actual show in the article). Your post sounded exactly as if you were under the same impression, so I certainly wasn't intending to be rude or suggest you were stupid in having that impression. Elen of the Roads (talk) 10:49, 15 October 2012 (UTC)

I'd suggest you try a little good faith yourself. I'm not going to look for the editor you mention, however I'd bet that s/he has not been here since 2006 and made hundreds of good edits on many difficult articles. Gandydancer (talk) 14:28, 15 October 2012 (UTC)

"Do you know who I am?" is never a particularly good gambit in my experience. Unless you're the Queen. But I've never seen anything that suggested she was interested in maths problems. --Elen of the Roads (talk) 17:45, 15 October 2012 (UTC)

Some people take the time to check those things before they make a rude "you must really be dumb" post - apparently you are not one of them. Gandydancer (talk) 18:49, 15 October 2012 (UTC)

Oh, I'm sorry ma'am, I presumed the Queen of England had flunkeys to edit Wikipedia for her. I have to point out that you are the one saying that making the perfectly reasonable mistake that it represents an actual game on the show is "really dumb." Perhaps you might like to explain to the guy further up the page why you think he's really dumb for making that mistake. Or perhaps you might stop ascribing motives to me that bear no relation to anything I said. Ma'am. Elen of the Roads (talk) 21:20, 15 October 2012 (UTC)

Gandydancer, vos Savant's famous question evidently was about an imaginary one time show, not addressing any of those various TV series with their quite differing very own special rules. Yes, with their quite differing rules. Evidently nothing to do with any one of those various differing special "real TV shows" in reality. The question is on an imaginary "one time problem". I suppose you know that, and I suppose you realize that the puzzle is on a quite imaginary one time problem ("imagine you are in a game show"...). And I am repeating:
So you surely do realize, just like me, that the puzzle has nothing at all to do with any of those real TV shows, don't you. As to me, yes I do. Regards, Gerhardvalentin (talk) 21:24, 15 October 2012 (UTC)

Well Gerhard, your civil reply is much appreciated. As for my edits, until I did the research and found that Marilyn never actually said "always", IMO it actually may have had something to do with the show. I am old enough to remember "Let's Make a Deal" (back then the only show of that sort),Parade Magazine, and Marilyn von Savant...and the actual question and the following Parade articles. Actually, everybody knew about that show and there were a lot of jokes about it...and even a song, "Door Number Three" (I think( by Steve Goodman. Anyway, until I showed that she did not actually say "always", and considering that every word to the puzzle had been questioned, there had to be some answer. Especially so since a quick read of this article did not seem to show that any of the experts were hung up on that word - again, there had to be some reason. So, I thought I'd throw that suggestion out. @ Elen, please pull your claws in as this is getting pretty old. Even if I was a Ma'am, do you really go around calling editors sir and ma'am, or is this just more of your sarcasm? Gandydancer (talk) 22:42, 15 October 2012 (UTC)

Based on the above, I think that your reaction to my original statement may have been based on some 'separated by a common language' problem. Let me assure you again that although I did think that you thought the puzzle was based on the game show, because that's what you appeared to be saying, I do not think that anyone coming fresh to the problem who makes that mistake is 'dumb' and I don't know why you think I am uncivil and Gerhardvalentin (who also thinks you thought the puzzle is based on the tv show) is civil when we have both said exactly the same thing.--Elen of the Roads (talk) 23:46, 15 October 2012 (UTC)

You cannot repeat this game in the same manner without changing the content of the problem. If you think about repetitions you are creating a rule which is not a part of the question. Your (not specified) rule is: the contestants pick a door, the host always opens another door which has a goat, and always offers a switch. But this leads astray, and it generates another problem, not the one-time-game given by vos Savant. --213.102.96.69 (talk) 10:25, 15 October 2012 (UTC)

I believe you are right that as vos Savant's version (which is unfortunately not the only one floating around in the early days) is a one time problem, any explanation should show a one time answer. You can of course calculate odds on a one time event if that is any help. However, the answer to vos Savant's version as she stated it is "it makes no difference", because she only said that Monty opened a door and showed a goat. She never said if he was operating under any particular constraint. If he picks a door at random, the odds of the car being behind either door are 50/50. It is only if there is a rule that he has to open a door and he has to show a goat that the odds change, and it becomes better to switch, because if you chose one of the goats (which the odds are 3/2 (or 2/3-on in bookie-speak) - ie more likely than not that you did), Monty just eliminated the other goat. Elen of the Roads (talk) 11:00, 15 October 2012 (UTC)

Marilyn said "and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat" implying that he *deliberately* opens a goat door, and emphasizing thet you know that he always can. And almost all of the people who wrote to "Parade" understood her question this way (but still disagreed with her answer)d. And as she herself made explict in her next column on the problem. He can do this, and he will do this, whatever door you initially chose. Richard Gill (talk) 12:24, 15 October 2012 (UTC)

I agree that most people probably if they thought about it assumed that this statement was of some significance. But I also think that once you want to explain her answer, you need to highlight the significance in a way that she obviously didn't because if she had it would have tipped people off as to what the answer is (as if, for example, in the "father and son are involved in a car accident" problem, it said "the surgeon, who was a woman." And I think a lot of people who disagreed with her still hadn't realised the significance of Monty knowing what is behind all the doors. Elen of the Roads (talk) 12:42, 15 October 2012 (UTC)

@Elen: Spot-on. Most people evidently do realize that Monty is selectively showing a goat from the two un-chosen doors, but fail to recognize the significance of it. That is what makes it such a wonderful riddle.

Misdirection using selective information is one of the oldest tricks in the book. It is one of the best tricks because people are often fooled even when they know it is selective information. In very simple situations it is obvious (cf. "nothing up my sleeve" at #Proposed text for Solution section above), but people's grasp of selective evidence is so tenuous and weak that even the very slight complication in MHP of a third door, from which a goat is not to be selected, causes whatever insight one might have about selective evidence to evaporate into thin air.

"Nothing has been falsified – except the impression that it gives."
— Darrell Huff, How to Lie with Statistics (1954), Ch. 5

I agree that this needs to be highlighted, i.e., pointed out with emphasis. Several editors and sources express the view that the fundamental reason MHP seems paradoxical is that conditional probability is poorly understood. It is; but I believe people's blindness to the significance of selective evidence is the more fundamental stumbling block. This is why selective evidence is used by magicians, con men, and polemicists of every stripe. It is what makes honest, hardworking statisticians unwitting victims of sample bias, and is why hardworking, dishonest statisticians get away with exploiting it.

MHP is an object lesson about selective evidence for everyone, especially for those who can't make heads nor tails of P(A|B) = P(B|A) • P(A) / P(B). ~ Ningauble (talk) 16:59, 15 October 2012 (UTC)

I think the first 'explanation' up has to show the effect that Monty has in deliberately showing a goat. Because that's what makes the difference, and that's what all the 'combining doors' answers fails to explain. If Monty had picked that goat by chance (or even deliberately, but without the constraint that he has to open a door and show a goat) then it isn't any better to switch than to stick. I think the article has to make that clear. Vos Savant made that clear in her second bite at it, where she came up with the million goats explanation. Elen of the Roads (talk) 17:51, 15 October 2012 (UTC)

Am I wrong? The question is about a one time problem: *secrecy*. And the host, in intentionally having to reveal a goat, did reveal a goat. This leads to the famous "paradox". Quite other scenarios not necessarily will generate the "paradox". Gerhardvalentin (talk) 21:42, 15 October 2012 (UTC)

No I don't think you're wrong. I think we are saying the same thing two different ways. It is a one time problem. The significance of 'Monty knows what's behind all the doors' was not brought out in the original iteration, but vos Savant's answer - always switch - is only correct if Monty has to open a door and can't show the car. Elen of the Roads (talk) 21:55, 15 October 2012 (UTC)

Thank you. But: is the WP article really only on a narrow dissection of the history of the first weeks and days of that brilliant famous puzzle that has been presented as a tricky story? Is the article really only on misinterpreting what the story tried to tell? Marilyn vos Savant responded early enough. My RfC-comment above was: It is not so much important what is *the* original question and its "possible" ambiguity, but it is on the certainly *intended* paradox [...] — It is on helping people, *by modern reliable sources*, to decode and understand the puzzle. – Am I really wrong? Gerhardvalentin (talk) 22:12, 15 October 2012 (UTC)

I see what you are saying. But we have to live with the puzzle as it was allocuted, which was as a tricky story. The only significance of it being one-time is as 213.102 points out - we don't know the rules. Which is where we need the other sources who have established what rules are necessary for MvS's answer to be right. Elen of the Roads (talk) 23:31, 15 October 2012 (UTC)

@213.102.96.69 You are right, it's on a "one time problem", and the host always can show a loser "on purpose", as per Marilyn vos Savant, regardless what the player has chosen. Yes, the host shows a loser "on purpose", and not just "by chance". And Marilyn vos Savant added: "by that" we've learned nothing to allow us to revise the odds of the guest's first choice. Gerhardvalentin (talk) 13:03, 15 October 2012 (UTC)

Probably vos Savant had noticed that her first reasoning (a million doors) was wrong. Therefore in her next column she proposed a repetition of the game, but didn't realize that such a repetition, in the same manner, would imply a rule which wasn't given in her original wording. So, her solution doesn't fit to her statement of the problem.

I agree that the host *deliberately* opens a goat door. But you don't know previously that he will do this, whatever door you initially chose. You only know afterwards that he has opened a goat door when you have picked door No.1. You have to pay attention to the chronological (and logical) progress of the game. Hence you must not use knowledge before appearing in the wording.

Nevertheless, if one door is shown to be a loser, that information changes the odds of the remaining doors to be the winner, in general. If we have no further information, a rule for example (but vos Savant's text doesn't give any rule), we have no reason to assume that one door is more likely the winner. So, the odds for both the remaining doors to be the winner are changing to 1/2. --213.102.97.173 (talk) 18:44, 15 October 2012 (UTC)

Yes to your last sentence, but no to the rest (I think - bear with me and see if this makes sense). Vos Savant said that Monty knows what's behind all the doors, but she doesn't say anything about why he picks the door that he does. You pick a door, and the odds are 1 in 3 that you have the car. Monty opens a door at random, it has a goat. That tells you nothing about what's behind your door, or what's behind the third door. As you say, the odds for both the remaining doors to be the winner are 50/50.

Now if by saying that Monty knows what's behind all the doors, vos Savant meant that Monty won't show the car (standard behaviour for a game show host, spoils the game if he gives the result away too early), then the effect is quite different. The odds of you having the car is still 1 in 3, and if you have picked the car, you'll still lose if you switch. But if you picked a goat - for which the odds are 2 in 3, then Monty is left with one goat and the car. He can't show the car so he must show the goat. If you switch, you'll get the car. So you have 1 in 3 odds to get the car by sticking (because you picked it to start with) and 2 in 3 odds to get the car if you switch (because the 2 in 3 odds were that you picked a goat).

So for vos Savant's answer - you should switch - to be correct, she must have always been intending some rule or rules which were behind Monty's actions. In a way, the puzzle has been rewritten to fit the answer by those who like precision. Elen of the Roads (talk) 21:43, 15 October 2012 (UTC)

MvS promptly responded that the host always can show a loser "on purpose", regardless what the player has chosen.
She promptly responded that the host shows a loser "on purpose", and not just "by chance".
And Marilyn vos Savant added: "by that" we've learned nothing to allow us to revise the odds of the guest's first choice.
And I am adding: She presented a one time problem, audience and host unknown to us. All we know about the host is that he showed (shows) a goat on purpose, as per MvS. Gerhardvalentin (talk) 22:00, 15 October 2012 (UTC)

Ps: The core of the "MHP" is the famous *paradox*. So the basics of the MHP is that the host on purpose was to open a door in order to show a goat, where he can do that in any case, regardless what the player has chosen. I repeat that we should show (as an *eye-opener*) that - if quite outside the MHP! - another host should be known to open a door regardless of what is behind it (in 1 out of 3 the car, and in only 2 out of 3 a goat) the *paradox* will never arise. Not in just only a "variant", but quite outside the "MHP". Gerhardvalentin (talk) 12:59, 25 October 2012 (UTC)

MvS clarified in her later columns: 1) the host has to open a door, 2) the host knows where the car is and must show a goat, 3) the host must make the offer to switch. She didn't clarify how the host chooses which door to open in the case the player originally chooses the car, but in the "standard" problem this is a perfectly (uniform) random choice as well. The setting being a game show implies this is a standard sequence of events, likely to be repeated - but with all of these rules the chance of winning the car by staying (regardless of which door you choose and which door the host opens) is 1/3 and 2/3 if you switch. Gerhard's "it's a one-time offer" notion is simply weird (if he can provide a source for this notion that'd be nice). If it's a one-time offer (and you don't know, for example, whether the host always makes the offer to switch) you don't know whether the host is "evil Monty" (who opens a door and gives you the option to switch only if you originally picked the ~~goat~~car - as Monty Hall himself did in response to being interviewed by the NYT following publication of the Parade columns) - in which case if you switch you lose with 100% certainty. The host not only must open a losing door on purpose (this one time), but also MUST NOT have the option to avoid doing this depending on what the player initially picks (i.e. MUST open a door no matter what door the player picks and MUST offer the trade no matter what). In addition, if you want the chances not to depend on which door the host opens, in the case the player happens to initially pick the car the host must choose between the remaining two doors independently of the player's choice (usually expressed as "chooses randomly" in this case). With all of these rules, the chances are the same whether it's a one-time offer or whether you're going to repeat this 100 or 1,000,000 times. The "Variants" section examines what happens if any of these restraints are relaxed. -- Rick Block (talk) 23:03, 15 October 2012 (UTC)

In just only "a game-show" the host is unable to communicate any "bias", and the "standard" duly pays regard to this fact: *secrecy*, *randomly*.
Rick, it's still the same pattern as for years now. You wrote above that Marilyn vos Savant did say "3) the host must make the offer to switch," and immediately thereafter you have added your ceterum censeo concern "what if you don't know, for example, whether the host always makes the offer to switch". Are you aware that you just depicted the state of this mingle-mangle article. Gerhardvalentin (talk) 13:52, 16 October 2012 (UTC)

It doesn't matter whether it's one time or not, that doesn't change the odds, which are one-time and not cumulative anyway. I think GV's point is something to do with the way it was said. I could be wrong. Elen of the Roads (talk) 00:08, 16 October 2012 (UTC)

Right. The point is not whether it's a one-time event, it's what the rules were by which we arrived at the situation that was described. For example, if the host is not compelled to open a door showing a goat and make the offer to switch (from the outset), then the 1/3:2/3 answer does not follow (whether we do it once, or a million times). -- Rick Block (talk) 02:22, 16 October 2012 (UTC)

Rick, it is still the same pattern as for years now. Confirming the "standard", you wrote above that Marilyn vos Savant said "3) the host must make the offer to switch", and now you you repeated again your permanent ceterum censeo concern: "if the host is not compelled to open a door showing a goat and make the offer to switch" Are you aware that you again depicted the state of this mingle-mangle article. Gerhardvalentin (talk) 13:52, 16 October 2012 (UTC)

What is the article about?

So, as per some few editors here, this article is not to show the strange "famous paradox" with its sole only exclusive solution to "switch" to the other door.
Nor to show the scenario(s) where this paradox can and will crystal-clearly arise.
Nor to clearly show "why" the paradox does arise in an appropriate scenario.
Nor to help the reader to decode the paradox, to decode why it just may "look to him as being a paradox", why his uncalled-for leaving out of fundamental items resp. his uncalled-for adding of inappropriate assumptions can blur and can cloud his / our vision.

Will the article continue to show in the first line that there indeed exists a weird litany of "other" scenarios where the paradox never will arise.
Will the article continue to say that it is impossible to "solve" the paradox but by applying just one special narrow method of approach, resulting in some fixed range of conditional probability (at least 1/2 to 1)? You "can" use conditional probability, but the article always said that any other approach does not address the question asked for. Yes, the conflict will continue.

Rick, you said above that Marilyn said
1) the host has to open a door.
2) the host knows where the car is and must show a goat.
3) the host must make the offer to switch.

But don't forget that the story is saying "you are in a game-show". Nothing is said that you are given the opportunity to detect any special host's behaviour by watching two or twenty or more repetitions. No word about offering such opportunity.

So you need not deplore "but we don't know" regarding the host's choice, if he as a choice. Saying "you still can detect" is incorrect, and so – for the guest himself, and for anyone else except God an the host, there just is no "how the host chooses which door to open in the case the player originally chooses the car". (Btw: it is completely irrelevant "how" he does, because the correct solution is "switch", because any staying will diminish your chance to win the prize.)

As the host's method is and forever remains completely unknown, the "standard" plainly and correctly translates this given fact of our immutable total ignorance to "the host strictly observes secrecy regarding the car hiding door", or in other words "he chooses uniformly at random". Have a look to it.
That's the scenario presented concerning just this one ("a") game-show. Period. And you are right to say that in the scenario as per MvS ("a" show) the chance of winning the car by staying (regardless of which door you choose and which door the host opens) is 1/3 and 2/3 if you switch. Please help the article to be clear and intelligible, by a clear structure. Quite diverging scenarios offside the scenario presented should not be mingle-mangled. Gerhardvalentin (talk) 10:38, 16 October 2012 (UTC)

GV - if I tried to explain this in your first language (German....?) you wouldn't understand a word I was saying, so please don't think this is a criticism of your ability to communicate in a second language. BUT I am having difficulty in getting your point. Marilyn vos Savant gave a version of the Monty Hall problem, in which given the information she provided, the correct answer should have been that it makes no difference whether to stick or switch. However, she said that the answer was that you should switch. The only way her answer was correct was if there was a rule governing the action of the game show host which had not been articulated. Her 'million goats' explanation, and her later comments, suggest that the rule she had in mind was that he had to open a door and not show the car (a loser door, as you describe it). This is nothing to do with 'host bias', this is just a rule of the game, and true for one iteration or one hundred.

So I have to ask - do you understand that if the opened door is chosen uniformly at random, the odds are the same whether you stick or switch. If you do, then I think the point you are making must be just about the choice of words used in the solutions. But the only way you can decide what to do is by guessing (if not given the information) what the host is up to. To get to the right answer in the original vos Savant version, you have to guess that because this is a game show and the host knows what is behind all the doors, the host won't show the prize. And you have to guess that the host had to open a door as part of the game regardless of your choice (because if he only opens a door if you picked the car, then there is a disadvantage to switch). Elen of the Roads (talk) 20:48, 16 October 2012 (UTC)

Elen, it is on the discussion here. Just have a look. Millions of words, and binders full of "But what if the host is biased?" – Binders! – Made up especially by three editors, one of them retired years ago, two others still active. Although we know that the paradox is based on a given scenario as per MvS, she constantly is blamed here of not having given the correct answer to her question regarding doors 1, 3 and 2.

As per MvS, the host has to open a door, the host knows where the car is and must show a goat, the host must make the offer to switch, and – as it is just only "a" show – the host is unable to communicate any additional hint on the actual location of the car. This was adopted as the basis of the "standard".

We should present the paradox and help the readers in decoding and resolving it, as per modern sources that say:

"If all doors are equally likely to hide the car and the host is equally likely to open either door when he has a choice, then the conditional probability of winning by switching, given intial choice of player and given door opened by host, is 2/3".

Showing just in the beginning that "posterior odds = prior odds times likelihood", and then in contrast, as an eye-opener, immediately followed by the contrary "1:1 result" caused by the quite inconsistent *strange scenario* of a forgetful host who is showing the car in 1 out of 3 by just randomly opening one of his two doors, deleting the chance to win by switching in that 1 out of 3, while in the remaining 2 out of 3 (only "goat:car" and "car:goat") the odds of both still closed doors are indeed "1:1" then, as per the most intuitive common appraisal.
Any other various deviant variants, where the paradox never arises should be shown in very later sections. Gerhardvalentin (talk) 09:33, 17 October 2012 (UTC)

She gave the correct answer - the answer to the classically articulated MHP is as you state - and it is not necessary to make any assumption as to whether the game is played once or many times because the answer does not change. The result for any given game is independent of all other games.

It is historically true that when she first articulated the question, she did not say the host knows where the car is and must show a goat, the host must make the offer to switch she only said the host knows what is behind each door and the host shows a goat

I agree with you that there should be no talk of 'biased' or 'forgetful' Monty in these first paragraphs. The classic iteration of the problem has only the rules that you state above. However, I believe - as you do, if I've read you right - that to explain the paradox you must show clearly and simply that it is the rules that create the paradox. Elen of the Roads (talk) 15:14, 17 October 2012 (UTC)

Thank you, yes. But as per some editors here the result for any given game is NOT independent of all other games, because they insist hat one "can detect any host's bias" in opening his door. See their full binders over years of discussion. Although it remains "a" game show, and moreover the host can never be supposed to "show" any bias in repetitions. Giving away hints on the location of the car is completely contrary to his focus of interest. It has nothing to do with the actual question "is it to your advantage". See the standard scenario that pays regard to this fact. But they insisted, referring to MCDD. We just should be aware of that fact.

Once more: We should present the paradox, making way and assisting the reader to decode it. Redundancy welcome for this effort. And, to help the reader to grasp why the paradoxical probability to win by switching pws is 2/3 and not 1/2, we should show the difference very early, of intentionally showing "a goat in any case" – and showing a goat "just only by chance, in only 2 out of 3", where pws is 1/2.

First of all the article should be about the arising paradox, especially based on modern sources. Any sardonic historical misinterpretations are not part of the paradox, but quite another issue. Show in very later sections that there are diverging variants where the paradox does not arise. Gerhardvalentin (talk) 16:08, 17 October 2012 (UTC)

@Gerhardvalentin: If you are right and the "standard" problem has been identified with vos Savant's text, there must have gone something wrong in the discussion here. It is clarified since 1991 (John Tierney) that her original statement is ambiguous:

Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.
Ms. vos Savant acknowledged that the ambiguity did exist in her original statement.
He [Monty Hall] picked up a copy of Ms. vos Savant's original column, read it carefully, saw a loophole and then suggested more trials. On the first, the contestant picked Door 1. "That's too bad," Mr. Hall said, opening Door 1. "You've won a goat." "But you didn't open another door yet or give me a chance to switch." "Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again." --213.102.99.58 (talk) 15:42, 17 October 2012 (UTC)

Indeed it is possible to misread Vos Savant's text. The "standard" problem is Vos Savant's text with the clarification added which rules out this possibility of misunderstanding. Vos Savant pointed out that most of her correspondents had understood her intended meanings just fine, but still got the answer wrong! Richard Gill (talk) 10:12, 23 October 2012 (UTC)

May be that most of vos Savant's correspondents had understood her intended meanings, may be not. We don't know, but in the letters, she published, the words referring to those necessary rules are rare. Nobody should be castigated who insists on the fact that her original statement "is not well-formed" (Gardner). All the pupils and students who are able to give reasons for the 50:50 solution should be praised but not blamed for an alleged misreading of vos Savant's text, and on no conditions they should be designated as psychologically persistent persons.

A brainteaser should be clear and complete concerning its suppositions and should not depend on personal interpretation. In this view vos Savant's text is not a brainteaser but a problem which shows that people are able to start a war for nothing, in this case the war against the 50:50 solution.

If you are a teacher or a professor in mathematics and if you like your students then you should give them the "Bertrand Box Paradox" for exercise, but if you don't like your students and if you want to disturb their minds then you should give them vos Savant's original statement and insist on "switching is better in any case, because the probability of winning by switching is 2/3". --213.102.99.21 (talk) 16:27, 23 October 2012 (UTC)