Talk:Monty Hall problem/Archive 18

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New Round of Reverts

Rick Block, what do you have against an accurate quotation from a source? Why remove a less-ambiguous sourced statement and replace it with your personal interpretation? Glkanter (talk) 18:18, 27 June 2010 (UTC)

It is stylistically better to paraphrase than to quote and you know full well the main intent of your change was to reintroduce the figure discussed above. Rick Block (talk) 20:42, 27 June 2010 (UTC)

No. I added the actual text because it properly states Carlton's solution. Showing that the new tree is a true representation of Carlton's solution.

Otherwise, you're just a goddam phoney-baloney punk. Glkanter

WP:Civility is a policy which like 3rr can result in violators being blocked from editing. I would strongly suggest you refrain from making any further comments like this. Rick Block (talk) 00:35, 28 June 2010 (UTC)

Glkanter reported for edit warring

Another classic example of misinterpreting unambiguous text:

"== User:Glkanter reported by User:Rick Block (Result: No violation) =="

"* No violation I only count three reverts. Please feel free to open up another request if the user resumes reverting. --B (talk) 21:20, 15 June 2010 (UTC)"

http://en.wikipedia.org/w/index.php?title=Wikipedia:Administrators%27_noticeboard/Edit_warring&diff=368251891&oldid=368249651

—Preceding unsigned comment added by Glkanter (talk • contribs) Glkanter (talk) 00:20, 16 June 2010 (UTC)

Yes, I reported you for edit warring. This is really not the appropriate forum for discussing this but you seem to want to insist. Fine. WP:AN3 is for reporting recent violations of the three-revert rule, and active edit warriors. The admin responding to this is saying you didn't violate wp:3rr (which you didn't, and I never claimed you did) - not that you weren't edit warring (which you were). -- Rick Block (talk) 00:15, 16 June 2010 (UTC)

Just to keep the record straight here, Glkanter continued edit warring and was then blocked for a short time for violating wp:3rr. -- Rick Block (talk) 04:52, 17 June 2010 (UTC)

Rick, your long standing obstruction to various editors' good faith edits has to be addressed by Wikipedia at some point. I'll need to be able to present a concise argument with diffs to support my charges. The situation above is one perfect, simple example, as it demonstrates your violations of all 3 of these WP policies:

Competence

Gaming_the_system

NPOV/Bias

Glkanter (talk) 00:20, 16 June 2010 (UTC)

That bogus RfC/U on me that you conjured up with DickLyon is another perfect example. Now, that was Harassment, and Gaming the system, and... Glkanter (talk) 01:03, 16 June 2010 (UTC)

We are still officially in mediation (although it's on hold pending the assignment of a different mediator). If mediation fails, Wikipedia:Arbitration would be the next step. If you want, feel free to pursue Wikipedia:Requests for comment/User conduct. -- Rick Block (talk) 14:34, 16 June 2010 (UTC)

A stronger approach would be a community ban, per Wikipedia:Banning policy, requiring a consensus of users not involved in the dispute. In fact, come to think of it, why don't we have a straw poll. Would anyone watching this page who has not been involved in this dispute (I think this excludes at least Glkanter, me, Martin, Glopk, Nijdam, Kmhkmh), support a community imposed topic ban against either or both of myself or Glkanter which would ban this user from making edits to articles and talk pages relating to the Monty Hall problem? Please just sign in the appropriate list below. This is in no sense binding, the official way to do this is through a discussion at WP:ANI. If there seems to be significant support, the actual discussion would need to occur at WP:ANI - I'm suggesting a straw poll here just to get a sense for whether there might be some point in opening such a discussion. -- Rick Block (talk) 15:03, 16 June 2010 (UTC)

I would be happy to sign for a ban against Glkanter, but am I not allowed?Nijdam (talk) 15:36, 16 June 2010 (UTC)

In a discussion at WP:ANI about this, you would be welcome to comment, but since you're "involved" I think your "vote" would be more or less ignored. -- Rick Block (talk) 15:47, 16 June 2010 (UTC)

Why support votes only? I don't know if I would be counted as being "involved" in this tempest in a teapot, but I'd oppose topic banning Rick. It would be most counterproductive - he's basically doing a good job, and I think he would even manage this article just fine all by his lonesome. On the other hand, Glkanter seems to be more disruptive than helpful (sorry to say). -- Coffee2theorems (talk) 15:44, 17 June 2010 (UTC)

I also cannot vote below but agree with Coffee2theorems that this is a storm in a teacup. We have had a minor editing skirmish and the bandying around of words like 'harassement'. No animals got hurt and there is no need for anyone to be banned. Martin Hogbin (talk) 23:20, 17 June 2010 (UTC)

I'd support a ban against anyone that's over-complicating this. A few moths ago when I last visited here, there were concise, easy to grasp explanations of what is generally perceived as the MHP. Now it's all muddled. Just like a real encyclopedia, or not. This is why Wikipedia has no real credibility anymore.Feddx (talk) 13:33, 26 June 2010 (UTC)

I think @Rick could not manage this article on his lonesome at the moment, though maybe he could learn to do so in future. The fact that some people think he could, means that he and they people are blinded by their POV to the legitimacy of other POV's. Unfortunately @Glkanter also has a strong POV, but I support him against Rick because I support diversity against dogma. Unfortunately, dogmatism breeds dogmatic anti-dogmatism. @Feddx, if there used to be concise easy to grasp explanations of what is generally perceived as the MHP, that was because it was a concise easy to grasp explanation of one particular perception of the MHP. The whole fun of the MHP is that it is ambiguous and hence that there are many legitimate and fruitful ways to perceive it and to explain it. What is then even more fun and incredibly illuminating is to show how these different perceptions and explanations are so intimately connected. It is important since the application of probability in the real world is non-trivial, often controversial, usually goes badly wrong (in law, medical research, ...). Precisely on unwritten questions about are whether we talking about subjectivistic or frequentistic probability, what are we conditioning on if anything, and should we, whether "random" means "uniformly at random" and when it is a legitimate assumption... It would be so good if the article could actually use the Monty Hall Problem to become a wikipedia FA on how to resolve probability controverses by clarificiation and simplification and making hidden assumptions visible.Gill110951 (talk) 07:05, 29 June 2010 (UTC)

The following users would support a topic ban against Glkanter

• <1st sig here>
• <next sig here>

The following users would support a topic ban against Rick Block

• <1st sig here>
• <next sig here>

The first paragraph of the Solution section.

There's no references, etc., except a mention of Selvin. Otherwise is this paragraph from a source(s)? It reads as OR/editorial commentary. Glkanter (talk) 21:08, 23 June 2010 (UTC)

Besides, The Combined Doors solution does show door 3 being opened to reveal a goat. Nijdam's real complaint with that simple solution actually rests on his argument that a prior 1/3 probability is not the same as a posterior 1/3 probability. The paragraph adds no value whatsoever. Glkanter (talk) 21:14, 23 June 2010 (UTC)

That's f-ing great! Despite it being completely being unreferenced and inaccurate, Coffee2theorems decides to revert my deletion of that crappy editorilaization. Here's his horse crap reasoning:

"Don't throw the baby out with the bathwater. Maybe the previous intro was too detailed, but we do need some kind of introduction here."

And earlier reverted Carlton's support tree for this nonsense:

As discussed on the talk page, this diagram is not suitable for inclusion into this article.

Yeah, right. Because it's too simple for him to grasp, and Rick Block, of course, is against it. It's not OR. I've explained that too many times, already.

So, great. Just what Wikipedia and the MHP article need, another NPOV-violating hypocrite editor. Thanks for nothing, Coffee2theorems. Glkanter (talk) 16:03, 28 June 2010 (UTC)

Generally, articles benefit from having short summaries (introductions) at the beginning of large sections consisting of multiple subsections. If there is no useful summary, then the reader will have to wade through it all to see what the section is about. The introduction you deleted was not exactly of stellar quality, so I didn't feel like reverting the whole edit, but only restoring the bare minimum of the previous intro. The bare minimum is to say that there are two common approaches, and what their main difference is. The intro I restored actually says "multiple approaches" instead of "two common approaches", but that's almost the same thing, and I didn't want to change it right now because Gill is advocating the game-theoretic approach, and that might reasonably be covered as a third solution. I don't think that the intro as it currently stands is the best intro ever, but it's better than no intro. -- Coffee2theorems (talk) 16:30, 28 June 2010 (UTC)

Of course. Rick unilaterally combines all the solutions into a single huge section. Nijdam adds his unreferenced, unsupportable Biased/POV BS, and now you support needing an intro for such a 'large section'. Funny how nobody, including Rick, ever mentioned needing an intro. While I repeatedly ask that all 'controversies' be discussed only after the solutions are provided. More BS phony-baloney hipocracy. You guys are good. Not good for the reader, the article, or Wikipedia, but very good. Glkanter (talk) 17:10, 28 June 2010 (UTC)

AND IT'S STILL WRONG!!!!! HELLO!!!!! "Besides, The Combined Doors solution does show door 3 being opened to reveal a goat. Nijdam's real complaint with that simple solution actually rests on his argument that a prior 1/3 probability is not the same as a posterior 1/3 probability. The paragraph adds no value whatsoever. Glkanter (talk) 21:14, 23 June 2010 (UTC)" Glkanter (talk) 21:45, 28 June 2010 (UTC)

There are actually two "combined doors" solutions in the article. One is by Adams, and I'm not sure if it is supposed to be unconditional or not because it is somewhat ambiguous, but currently it is explained in the article as if it were conditional by amalgamating it with Devlin's solution. The other is by Devlin, and is clearly an informal conditional solution. The figure in the simple solution section showing an open door is also an illustration of a conditional solution, and all the probabilities in that figure are conditional probabilities (note that one of the doors in the figure has the probability of zero - that is only possible in a conditional solution). The introduction says nothing about the unconditional and conditional solutions being neatly compartmentalized into two subsections in the article, and currently, they aren't.

The reason why these conditional solutions seem simple is that they leave key steps unjustified (no wonder many people find the "simple" solutions unconvincing). In the article, that is done in the sentences "the 2/3 chance of hiding the car has not been changed by the opening of one of these doors" and "Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that." Alas, the popular informal conditional solutions are such for the MHP. -- Coffee2theorems (talk) 02:18, 29 June 2010 (UTC)

I don't understand why these various solutions must be labeled using exclusively probability terms. There are other sciences that can solve the problem. Why say anything at all? The sources give their solutions. They're reliably published. Have a 'controversy' section after the solutions. Give the differing viewpoints without judging or editorializing. There is no 'proof' or 'professional consensus' that the simple solutions are faulty. Let the reader decide for him/herself. Selvin used a simple solution, so did vos Savant. Glkanter (talk) 04:26, 29 June 2010 (UTC)

The whole paragraph is unsourced. That's BS. It says there are exactly two methods of solving the problem. That's BS. And various simple solutions, like Selvin's (player chooses box B, host reveals box A) and vos Savant's, and Carlton's claim they *are* answering Whitaker's exact question. Who is this unsourced 'person' saying otherwise at the beginning of the solution section? It's POV, very simply. It has no place in the article. We should provide the varying sources in a way that doesn't confuse the reader or discredit any reliable sources. Why is this such a difficult concept in this article? Glkanter (talk) 04:45, 29 June 2010 (UTC)

The Mathematical Contrivance

In between his 2 letters to The American Statistician, it looks like Selvin heard from a few of his peers. That's likely why he clarified things with:

The host always offers the switch

The host will choose randomly when faced with 2 goats

He offered the more formal conditional solution

Well, as we see from Carlton's solution, it's not the random selection premise that's needed at all. Selvin included it, because his buddies convinced him that he had to state that it was 50/50. Because you can't do the customary doors-based conditional solution without it.

Carlton's car/goat based 'conditional' solution (if you consider 100% 'conditional) shows we simply don't care how he chooses between doors. Because, whatever he does, and however he gets to it, the host is revealing a goat.

Because it's the contestant's State of Knowledge about how the host chooses the doors that would matter. And there is NOTHING in the puzzle about that. You would need a new premise. I don't know what it would be, and I don't care. Because that would be a different puzzle.

All those conditional tree solutions with 50/50 when the host has 2 goats? They're BS. The puzzle doesn't require it. But that popular, flawed, overly-complex conditional solution sure does. It doesn't work without it. Everyone that uses it is wrong. Or maybe just lazy. Selvin was wrong, because he believed them (his original simple solution just lumped the 2 goat doors on a single row of his table, with no % split). It could be any split on the face of the Earth, even 0/100, it doesn't matter. Without a contestant's SoK premise to this affect it makes no difference, and Carlton's solution shows this. And as an added bonus, Carlton's solution proves Nijdam's never-ending '1/3 prior probability <> 1/3 posterior probability' filibuster is BS, too. Any number multiplied by 100% equals the original number. duh.

But, they're reliably published (except Nijdam's 'point', whatever it was), so in the article they go. But the heavy handed editorializing that the Combined Doors and Carlton's simple solutions are flawed should stop right now. Glkanter (talk) 22:16, 26 June 2010 (UTC)

Oh, yeah. And that gang of idiots, Morgan, et al, wrote a entire peer-reviewed article in a professional journal about the host having to choose randomly. And beat up that nice lady for no reason. When it's nothing but a lazy mathematical contrivance. Glkanter (talk) 22:26, 26 June 2010 (UTC)

It's long past time to get the disclaimers, editorialization, caveats, and weasel words out of the solution sections. Yes, various sources have taken shots at various solutions. But none of them is deemed 'right', or 'definitive'. Lets add a 'Controversies' section somewhere after all the solutions, and remove the controversies from the solutions entirely. Glkanter (talk) 08:17, 28 June 2010 (UTC)

Conventional Wisdom

I've been re-reading some past postings. According to Rick, this article has been reviewed on 2 occasions as a 'Featured Article', and that much of what I find inessential actually was a (by)-product of those reviews. Rick is proud of 'shepherding' this article through at least one of those reviews.

So, in some ways, I seem to be arguing against Conventional Wisdom. But I don't feel that way. I have a few college courses on this topic, over 30 years ago, and a lifetime of being a data analyst. My viewpoint is, 'There is no possible way I am wrong about this'. To me, this whole discussion has as much to contribute as a discussion of whether the sun will rise in the east tomorrow morning.

How does a single voice effectively confront the Conventional Wisdom? This is a question not just for Wikipedia, but any societal system. In the US, a swindler set up a Ponzi scheme on Wall Street. Individual investors went to the regulatory agency numerous times, but to no avail. The guy didn't actually get caught. He turned himself in! How does a minority, but important, voice get heard?

Yes, I look at this entire article, excepting maybe 5% of it, as an elaborate hoax. I think everyone went along because they did not want to admit to limited knowledge of the subject matter. Everybody drank the kool-aid. And the emperor is wearing no clothes.

2/3 of the time I will select a goat. Therefore I should switch. Glkanter (talk) 15:32, 7 February 2009 (UTC)

source: http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem&diff=prev&oldid=269139028, 7 February 2009

Glkanter (talk) 08:13, 28 June 2010 (UTC)

@Glkanter, I agree with you and I don't agree with you at the same time. I agree with you that your problem formalization and your solution of it is legitimate, it is mathematically completely respectable. I do notice that you are subjectivistic in your understanding of probability. That's fine but it's not obligatory. It makes life easier, in fact it gives you "no host bias" by definition so that the conditional approach also gives 2/3. But one is not obliged to use subjectivistic probability when trying to solve the problem. Any mathematization which can be argued to reflect the intention of the question-poser, or indeed, which can be argued to be a sensible way that anyone hearing it can understand it, is valid. When you use decision trees to support your solution you must remove the door labels. It's: player chooses a door. Host opens another. Should player switch? Answer *yes* because probability other door has car equals 2/3. One can say that you are ignoring the door numbers. You say they give no extra information. The more involved arguments dealing with the door numbers, either doing the calculatins with them, or assuming them away, finally simply show that the door numbers indeed give no information. So your instinct to ignore them was absolutely correct.

I am annoyed by the attitudes of some other editors. Dogmatically refusing to admit validity of anyone else's opinion is not constructive. It's destructive. It reminds me of religious fundamentalism. It probably means that you don't understand very much so you are very scared when the assumptions on which you have built your life appear to be threatened. You seek safety in numbers and form a gang and go on a jihad to destroy the infidel. You don't argue scientifically against the arguments, but you complain about their style, or try to enlist support of biblical authorities, or threaten to bring in the police. Gill110951 (talk) 06:31, 29 June 2010 (UTC)

One reason for much of the argument

Any statistician seriously considering the MHP is likely to be using the modern definition of probability theory in which we have consider a sample set of probability mass functions. This is a formal mathematical system to allow probabilities to be manipulated to solve probability problems. There is, however, one very important point that may be causing some argument here. To quote directly from probability theory article:

The function f(x), mapping a point in the sample space to the "probability" value is called a probability mass function abbreviated as pmf. The modern definition does not try to answer how probability mass functions are obtained; instead it builds a theory that assumes their existence.

In the MHP there are three undefined (in Whitaker's question) distributions, the producer's original placement of the car and the player's original choice of door, and the host's choice of door. Once these distributions have been determined, a formal calculation is carried out which produces an answer that does not depend on any person's state of knowledge or any real world assumptions. This is why some people have been arguing that that whether the host knows the host's door opening policy is not important. In probability theory, once the relevant distributions have been determined, the answer follows automatically from the mathematical formalism.

As is made clear in the relevant article, probability theory does not tell is how to obtain all the relevant probability mass functions. This is where the basis on which we choose to answer the question comes into play. The player's original choice of door is generally not contentious so I will ignore that here. That leaves two undefined distributions to consider, the initial car placement and the host goat door choice. In order decide these distributions we must use some logical basis. It is in the choice of these distributions that the vague question posed by Whitaker must be formulated into a probability problem.

Consider the producer's initial car placement. How should we decide the probability with which the car is placed behind each door? This may or may not have been done using some random process, but regardless of that, the host and the producer know where the car is. If we decide to answer the question from the point of view, or state of knowledge, of the host, even if the car was actually placed uniformly at random by some process (such as throwing a die) it is not randomly placed from the POV of the host. If we wished to set up the problem from the POV of the host we should not use a uniform distribution of initial car placement. If we had to answer the question from the Host's POV and were not told where the car was placed, our only option would be to parameterise the car placement and say the car was placed behind door 1 with probability C1 etc. This would provide a general answer where, when know, the actual values of the parameters used could be used in a probability calculation to give a numerical answer. The important point to note is that, it is in the setting up of the formal probability mass functions that the state of knowledge from which we wish to answer the question is decided. Even if the car was stated to have been initially placed uniformly at random, if we wanted to answer the question from the host's perspective, we would not use a uniform distribution.

The other distribution that we must consider is that of the host's choice of goat door. The way we must deal with this is in exactly way as we deal with the initial car placement. We have to decide from what state of knowledge we wish to address the problem. Suppose the host has a non-uniform policy of choosing which door to open, say he always chooses door 2 when possible. What distribution should we use in or formal calculation? Exactly as before, that depends on the basis on which we wish to answer the question.

One consistent basis or state of knowledge must be used to determine all unknown distributions in the MHP. Possibilities include:

1. From the POV of the player, as strongly suggested by Whitaker's question.
2. Strictly what is stated in the problem statement.
3. From the POV of the audience
4. From the POV of the host.
5. From our general real world knowledge TV quiz shows.

Options 1, 2, and 3 give us no information whatever about the unknown distributions. In the case of 1 above note that it does matter whether or not the player knows the host's goat door policy. This gives us two further options, parameterise the probability mass functions and give an answer in parametric form into which the real values could be substituted when known which gives us a rather boring and unhelpful answer that probability of winning by switching is anything from 0 to 1 depending on where the car was initially placed and the host goat door policy. The only alternative is to apply the principle of indifference to the distributions and take them all as uniform (regardless of the real, but unknown, basis on which the car was initially placed and the host chooses a goat door). This gives the answer of exactly 2/3. This answer is now agreed by Morgan.

Option 4 is unlikely to be of interest to mist people as the host knows where the car is. Using our real word knowledge of TV shows it is most likely that both distributions would be uniform.

I think it would be more productive for editors to discuss these issues further to determine where exactly any disagreement lies, rather than propose editing blocks.

The cases where the standard assumptions (which say i.a. that all the relevant distributions are uniform) do not apply are discussed in the variants section of the article, and do not seem to be the main point of contention at this time. So it would be more productive to stick to the standard assumptions for now and hash out any issues with the variants section once the stuff preceding that in the article has been dealt with (which has the projected timeframe of "sometime after hell freezes over", so let's not be overly ambitious here..). Accordingly, I will make the standard assumption in the sequel (as usual).

It is of course possible to encode state of knowledge (SoK) into the unconditional probability distributions, and to some extent that is necessary in general, but that's generally how you deal with the fixed part of the SoK. The changing part of the SoK is what you condition on, either because that is the clearest way to proceed or because you want to compute the probabilities for different SoKs. In this case, the knowledge "host opens door 3" is not a part of the player's SoK before the host opens the door, but it is a part of her SoK after the fact, so it is natural (and usual!) to represent that by conditioning. It also allows us to formalize the idea that "the probability that the car is behind door 1 does not change by the opening of door 3" (i.o.w. P(car is behind door 1 | host opens door 3) = P(car is behind door 1)), which is something that is (arguably) important for understanding the MHP.

You can also consider the "fixed" part of the SoK in this case to be a description of the actual behavior of the people involved, and maybe that is the clearest way. The uniform distributions would then mean that the people make their choices literally by throwing dice. Then it is clear that any SoK is not going to change that behavior by one iota, and the unconditional probability distributions must then reflect that reality (i.e. they are uniform). It is by conditioning, then, that you get answers in different kinds of situations.

As an example, if you wanted to look at things from the POV of the host on the game show, then you'd condition on the knowledge that the car is behind door x (we don't know x, but the host does), and the resulting probability would, depending on x, be either 1 or 0 (the host knows everything worth knowing, so there is no uncertainty whatsoever). The resulting solution would be: "If the car is behind door 2, then switching wins with probability P(C=2 | H=3, S=1, C=2) = 1, and if it's not, then switching wins with probability P(C=2 | H=3, S=1, C=1) = 0. Or in other words, the probability is P(C=2 | H=3, S=1, C=x) = x-1, which the host can compute because he knows x." -- Coffee2theorems (talk) 14:22, 18 June 2010 (UTC)

Per Carlton, the condition is not that 'door 3 is opened'. The condition is 'host reveals a goat'. Whitaker, 'say, door #3' uses this as an example. Opening door #3 is not a premise to the puzzle in any published formulation. How can it be a required condition? Glkanter (talk) 14:41, 18 June 2010 (UTC)

Coffee2theorems, I know how to do treat the problem from the host's SoK but this is not what the problem is all about. The question is about how to formulate the problem from the player's SoK. I think it is generally taken that player does not know the host's policy on goat door opening. So it makes no difference what that policy is. When we formulate the problem from the players perspective we must take the initial car placement to be uniform at random by applying the principle of indifference (or else declare the answer to be indeterminate) and also take host's goat door choice as uniform at random and do the subsequent conditioning on that basis. As Morgan now say, if you consider the conditions implicit in the problem (that is to say we are to answer from perspective of the player, who does not know the host's door policy) '...the answer is 2/3, period'. Martin Hogbin (talk) 17:31, 18 June 2010 (UTC)

Another issue

Having established that, regardless of the actual host policy, on any reasonable basis, we must take his goat door choice distribution to be uniform at random (thus making the problem symmetrical) the question arises of whether a simple solution, in which the door opened by the host is not specified, is correct.

Regarding this point I think it is best for neither side to take a hard line. The simple solution gives the correct answer because of an obvious and intuitive symmetry in the problem. Whether it is formally correct could be argued forever. I suggest we use words like formal and informal to apply to the two types of solution (as suggested by someone earlier) to avoid any further conflict. We already give both types of solution in the article. Martin Hogbin (talk) 11:54, 18 June 2010 (UTC)

Martin - if the article is "written from a neutral point of view, representing fairly, proportionately, and as far as possible without bias, all significant views that have been published by reliable sources" (per WP:NPOV), the questions you're raising here are moot. Any editor's opinion about what conditions "we must take" is completely irrelevant. Your lecture about probability theory at best belongs on the /Arguments page, not here. -- Rick Block (talk) 14:10, 18 June 2010 (UTC)

Do you actually disagree with anything I say or are you just objecting on principle? It is all standard probability theory and is an attempt to explain the current heated argument about host bias. I am more than happy to discuss any of the issues involved with you on the /Arguments page if you wish. I think that would be far more productive than trying to get editors who disagree with you blocked.

If you want to discuss what the sources actually say then I am happy to do that also on the /Sources page. Martin Hogbin (talk) 15:15, 18 June 2010 (UTC)

Martin - we've had this discussion repeatedly. I know what you think. I suspect you know what I think. You say above that you think "it is best for neither side to take a hard line about whether a simple solution is correct". I completely agree with this and have been arguing (for months) that the article reflect this. It seems to me that it is you who are completely unwilling for the article NOT to take a hard line about this. If you really think it is best for neither side to take a hard line about this, then why do you keep arguing that the article must take the stance that the simple solution is absolutely correct and that the unconditional approach must be presented as the preferred approach? -- Rick Block (talk) 20:11, 18 June 2010 (UTC)

I don't. Martin Hogbin (talk) 20:13, 18 June 2010 (UTC)

Really? What do you think it means to insist that the simple solutions are presented first, without criticism, without even mentioning that many sources approach the problem conditionally (let alone that some sources explicitly criticize these solutions as not addressing the question), and then following up such solutions with an "aid to understanding" section that further elaborates the "insight" offered by these solutions - and only then in the weakest possible way saying that "some sources interpret the question differently"? I take this to mean the article is saying the simple solutions are absolutely correct and that they are the preferred approach. -- Rick Block (talk) 20:39, 18 June 2010 (UTC)

I disagree that we *have* to take the host choice as uniform at random, though if you are a subjectivist you will naturally and correctly do so. But anyway, without the uniformity, it's easy to see that the conditional probability of switching giving the car is always at least 1/2 and on average 2/3, so it always is wise to switch. Secondly I disagree with calling the unconditional problem and solution "informal" and the conditional problem and solution(s) "formal". The unconditional can be written in unambiguous rigorous maths formulas just as well as the conditional, it is just hardly worth the bother. Please remember that the question is "should you switch?" not "what is the probability"? Gill110951 (talk) 09:06, 29 June 2010 (UTC)

Game theory shows that Monty would be foolish not to choose uniformly at random, and that a player who doesn't switch is stupid. This means that there cannot be any situation where the conditional probability that switching wins is less than 1/2, whatever the host's strategy. The game theoretic approach is exactly meant for the situation where we don't know anything about the host's behaviour. From the mathematical point of view, you ought to treat vos Savant's question (should you switch?) as a question of game theory, since that comes down to solving the problem under arbitrary and unknown host bias. Of course, conditional probability does it for you, too. It's not a difficult problem so it can be solved in many different waysGill110951 (talk) 09:06, 29 June 2010 (UTC)

You don't really have a 'smoking gun', Rick. You're connecting the dots and making a conclusion. In this case, a flawed one. It's the same technique that Sarah Palin used to connect about 5 different aspects of Health Care Reform and conclude that DEATH PANELS are going to euthanize Grandma. Glkanter (talk) 20:53, 18 June 2010 (UTC)

Rick, I have explained my position several times before. I would like to see this article as useful as possible to as wide a range of readers as possible, from those with no interest in the subject who just want the answer to settle a bet for example, through elementary students of statistics, to experts in the subject, who still might get something useful from it.

I am not trying to hide anything or push any POV in the article but just suggesting that we organise the article to present all the information about the subject in a way that works best for our readers. It is logical to present a simple exposition of the subject first then proceed to a more detailed discussion. That is the order used on most text books and WP articles.

The general reader does not want to be distracted by arguments about conditional probability while they are trying to understand one of the world's hardest brain teasers. On the other hand, I doubt that the expert reader will object to skimming through a simplified solution to reach a proper mathematical discussion of the subject, where we can put all the stuff that you claim I am trying to hide.

This continued argument about conditional/unconditional solutions is something of an artifact of discussion here, started by over reliance on a paper by Morgan et al by some editors and still propagated here by some despite a somewhat softened appraoch by the paper's original authors. Martin Hogbin (talk) 10:43, 19 June 2010 (UTC)

Whatever your motivations, are you disagreeing that the structure you keep insisting on has the effect of making the article say the simple solutions are absolutely correct and that they are the preferred approach? -- Rick Block (talk) 14:21, 19 June 2010 (UTC)

Here you go, Rick. This is what a 'smoking gun' looks like:

"Several authors point to the fact that this and some other simple solutions, are not complete, because they do not distinguish properly between the probabilities before and after the opening of a door by the host. (See the probability section for more details.)" Nijdam (talk | contribs) Revision as of 04:05, 18 June 2010 (edit) (undo)

"Whatever your and Nijdam's motivations, Rick, are you disagreeing that the above unsourced commentary he keeps putting back in the Popular Solution section has the effect of making the article say the simple solutions are absolutely wrong and that they are the monority approach?" adapted from Rick's comments above Glkanter (talk) 08:26, 22 June 2010 (UTC)

@Rick, of course I disagree that the structure I keep insisting on has the effect of making the article say the simple solutions are absolutely correct and that they are the preferred approach - because the article says that this is not the case. Martin Hogbin (talk) 17:37, 22 June 2010 (UTC)

I have exactly the same motivations and aims as Martin. But one main difference of opinion. In my opinion the simple solutions are absolutely correct solutions to absolutely legitimate mathematizations of vos Savant's version of Whitaker's questiong which after an introductory rough sketch of the situation, asked "should you switch?" Asking for conditional probabilities is also completely legitimate and interesting. The controversy about host bias and conditioning is entangled with a hidden controversy about "what does probability mean?". For a subjectivist there clearly cannot be host bias. For a frequentist and an economist (game theorist) there obviously could be. But so what: the posterior odds that door 2 hides the car are 1 : q so you should switch! Whether you admit you don't know q, or whether you say that not knowing it means you take q=1/2.Gill110951 (talk) 09:15, 29 June 2010 (UTC)

A rose by any other name

I think the article is looking a lot better. I object to the titles of the two main sections: "popular solution" and "probabilistic solution". There is nothing non-probabilistic about the popular solution. It is good probabilistic thinking. The so-called probabilistic solution is just "the" solution to a more refined and more complex problem. The popular solution is the proper probabilistic solution to a more simple and easy to solve problem. It is a matter of opinion which of the two problems, if either, is THE Monty Hall problem. There are also other options. Economists see THE Monty Hall problem as a problem of game theory or decision theory, not of probability at all. The problem is not "what is this, or that, probability?" The problem is "what should you decide to do?". Decision theory. Game/decision theory is something more complex than probability theory and requires ability to do probability, but goes beyond. Many people see the Monty Hall problem as a problem of psychology or behavioural science, and that is going even further since game theory and decision theory can be thought of as mathematiziations of parts of behavioural science (cf. economics).

I would prefer to refer to "simple problem, simple solution" and "refined problem, refined solution".

Gill110951 (talk) 05:54, 19 June 2010 (UTC)

Right, having said all that, I went ahead and did it. I cleaned up the mathematical solution to the conditional problem, added the game theoretic solution to the game theoretic version, cleaned the solution of the unconditional problem. I hope in a way which demonstrates a totally neutral POV. Still have to add lots of references... please help, if you don't disagree too strongly.

I am trying hard to keep my own Point of View out of this (ie what I think is the right formulation of the problem). I do believe there are three (or more) legitimate mathematizations out there, they can all be justified by recourse to various Authorities or to History or to Good Taste, or whatever, ... so what... They're out there in the wild and wikipedia should talk about them, neutrally, each on its own merits, in length at reasonable proportion to its importance, and go on to summarize the controversy.

Please lets separate maths from controversy. There is clean rigorous nice maths for every interesting variant of the problem. As a mathematician I am going to fight for clean rigorous nice maths. Gill110951 (talk) 07:07, 19 June 2010 (UTC)

I agree with what you say but suggest that there is one case where a slight relaxation in mathematical rigour is justified in the interests of clarity. That is in the symmetrical conditional case, where the player chooses after a specific door has been opened by the host but the host is known to open a door uniformly at random, selected from all unchosen doors hiding a goat - the unambiguous formulation of K&W.

In this case it might be argued that a conditional solution, in which the possible doors that might have been opened by the host are shown, is required. I suggest that an obvious and intuitive symmetry tells us that it is not necessary to distinguish between doors the host might have opened since the answer must be the same whichever door is actually chosen by the host. I believe use of the simple solution in this case is justified initially, although I have no objection to stating the potential problems with this approach later in the article. As Rosenthal says about the simple (unconditional) solution for this case, 'This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem'. Martin Hogbin (talk) 11:32, 19 June 2010 (UTC)

There's no problem with using the unconditional ("simple") solutions for solving the conditional problem, as long as it is mentioned that a symmetry argument or something similar is required. It is not enough to say on the talk page that "an obvious and intuitive symmetry tells us that it is not necessary to distinguish [...]" - a sentence like that must go into the article. It is not reasonable to assume that the reader will come to the talk page to read that sentence. The article should present complete solutions only; or some incomplete solutions, each with a caveat in the article, right next to that solution that the solution is not complete. This is not a question of rigor, but of honesty. Passing incomplete solutions off as complete solutions is dishonest.

In an encyclopedia, there is no need to present any solutions at all to convince the reader. If we want the reader to know that the answer is 2/3, then it is enough to say so and to cite that claim. The citation is enough. The solutions are not there to convince the reader at all costs. They are there because a discussion of the MHP is not complete without presenting a solution, i.e. the reader should be told not only the "what" but also the "why". If we leave many readers with an impression that they understood the "why" when in fact they did not - as an incomplete solution without a caveat right next to it does - then we are doing them a disservice. The task of an encyclopedia is to inform, not to misinform. -- Coffee2theorems (talk) 15:01, 19 June 2010 (UTC)

I agree completely--Kmhkmh (talk) 17:22, 20 June 2010 (UTC)

I would like to put a sentence along the lines that you suggest into the article but I do not know of a source that makes that specific statement (if you do please let me know). On the other hand there are many sources that just give the simple solution as a complete solution to the MHP. To some degree the true, and certainly most notable, MHP is the problem to which the simple solution is the solution. Martin Hogbin (talk) 00:42, 20 June 2010 (UTC)

If there is no host bias which in particular is the case if our probability statements are subjectivistic, the conditional probability is 2/3 by invariance (it is the same for each specific set of choices but averages out at 2/3). There is no relaxiation of mathematical rigour in writing these words. Any mathematician worth their salt can rewrite the english sentence as formal mathematics of similar length and having exactly the same meaning. When Rosenthal says the simple solution is correct but shaky because it fails for variants, he only means that the number 2/3 is the right number for the fully specified conditional problem including no host bias, but that the right number (for the conditional probability) would be 1(1+q) when there is host bias. He is not saying that the simple solution is an incorrect solution to the simple question. There's a simple question which has a simple solution 2/3 and a more subtle question which has a more subtle solution 1/(1+q). The simple solution to the simple question is also the good answer to the more subtle question when q=0.5 which is not a coincidence since q=0.5 makes the problem symmetric hence all conditional probabilities equal to the unconditional.Gill110951 (talk) 09:25, 29 June 2010 (UTC)

Gill's changes

Gill has made some improvements to the readability and factual accuracy of this article as well as adding a section on a game theoretical solution. I hope editors will discuss whatever they feel might be wrong with these changes before blindly 'correcting' them. Martin Hogbin (talk) 15:04, 24 June 2010 (UTC)

I see that I was too late. Martin Hogbin (talk) 15:09, 24 June 2010 (UTC)

The reason for the "correction" (it was actually revert), was that gill replaced old content without seeking consent, i.e. it was by all means a controversial edit without consent. The issue here in particular was not adding new material but removing the old formal solution, which in different was part of the article since it became featured. And the proper order for such a change is to seek consent first and not to make a controversial edits first and ask later.--Kmhkmh (talk) 15:16, 24 June 2010 (UTC)

How about putting the formal solution back in a separate section after the 'Game theoretical' section. Very few people will be interested in it but, if it agreed as a rigorous formal solution, it should have a place. Martin Hogbin (talk) 15:23, 24 June 2010 (UTC)

I have no objection against adding (rather than replacing) the odds-ratio approach and I told Gill explicitly so, nevertheless he insisted on removing it again.--Kmhkmh (talk) 15:32, 24 June 2010 (UTC)

I think it should come later, following the general rule of simple stuff first. We might get something that everyone can agree on. Martin Hogbin (talk) 15:46, 24 June 2010 (UTC)

How about putting the 'Game theoretical' in a separate article. Very few people will be interested in it but, if it agreed as a rigorous formal solution, it should have a place. glopk (talk) 15:37, 24 June 2010 (UTC)

That is going too far. It is far more interesting that a rigorous formal solution, which is only there for completeness. Martin Hogbin (talk) 15:46, 24 June 2010 (UTC)

Martin, who is "it"? To whom it is "far more interesting that a rigorous formal solution"? Are you representing a consensus larger than the number of neurons inside your skull? To me a rigorous formal solution is very interesting and useful in this article. It allows a mathematically conversant reader to cut through the cruft and get a satisfactory grasp for a correct solution.glopk (talk) 15:53, 24 June 2010 (UTC)

Only a small fraction of our readers would understand a rigorous formal solution. They, like you would, no doubt find it interesting. I suspect that more people would be interested in knowing the result if the player and TV show adopt strategies to optimise their own position. But why not have both? Martin Hogbin (talk) 16:01, 24 June 2010 (UTC)

Do you have hard numbers on how "small" a fraction of readers can read math? On the other hand, do you know how many readers would prefer a well-worked out solution to a soup of barely mentioned approaches? Again, please do not stand up on a box as the "advocate" of the "common reader" - you are just an editor like everyone else, pushing your view (one would hope in good faith). Lacking hard numbers, the best one can hope for here is informed editorial consensus. Lacking that, an informed majority. Clearly neither exist in support of Gill's (and now yours) bold changes. glopk (talk) 16:17, 24 June 2010 (UTC)

No, I do not have figures, it is just my estimate. Let us just have both. Martin Hogbin (talk) 08:16, 25 June 2010 (UTC)

The game theoretic approach belongs in the variant sectioni. It is not a solution to the standard problem but a variant where the host has more flexibility. Rick Block (talk) 00:57, 25 June 2010 (UTC)

It is a good solution for vague problem statements such as Whitaker's. There re still three doors, the standard rules still apply, host always offers the swap, and always opens an unchosen door to reveal a goat. Martin Hogbin (talk) 08:16, 25 June 2010 (UTC)

Nice discussion.

In my opinion the odds form of Bayes rule is the simplest rule and the rule which nowadays you will find many authorities using.

The long formal calculation which comes down to proving Bayes rule by hand is superfluous since it exists on another wikipedia page.

The game theoretic version of the story goes back to Nalebuff hence to pre-vos Savant days and "everyone in economics and game theory" knows it, they just are too busy getting Nobel prizes or fat consulting jobs (using game theory to advice governments and multinationals) to waste their time writing up a five second "in your mind" exercise during Game Theory 101. Sorry, that's how they are.

The proof by symmetry, the proof by odds, the proof of the unconditional version by easy logical deduction are all equally mathematically rigorous. The mathematical correctness of a proof does not depend on the number of lines of formula manipulation which the writer feels necessary to write out at length. Mathematicians prefer proofs by using exciting and important concepts. Gill110951 (talk) 09:11, 25 June 2010 (UTC)

My aim by the way is to restore simplicity and unity and clarity to the Month Hall page, and to restore a neutral point of view. An ambiguous sentence like vos Savants' which made the problem famous and is the reason we have this page at all, and the reason these discussions go on endlessly, is open to multiple scientific and in particular mathematical formulations. Anyone who says that their formulation is the only true formulation is a joker or an ignoramus or a terrorist. I suggest all editors of these pages actually read all the literature and also make some effort to understand different points of view, even if they are different from a point of view which they have held strongly for a long time. A true scientist is prepared to change their mind on seeing new evidence. That is what Bayes theorem tells us. A not true scientist makes up their mind and interprets all new information through their prejudiced point of view.

I have changed my mind on Monty Hall many times! I am a mathematician and all the mathematicians I know, care about the unconditional problem. But I know there are lots of folks out there who love the conditional one. It is beautiful, I think, to bring these approaches closer together and show their relations. Finally the whole is more than the sum of the parts. —Preceding unsigned comment added by Gill110951 (talk • contribs) 09:16, 25 June 2010 (UTC)

By the way, the simple verbal solution to the simple unconditional problem and the many alternative and simple verbal solutions to the more complex conditional probability question can all be converted "word for word" into equally short formal algebraic computations, if that is what some of you mean by a rigorous mathematical proof. Real mathematicians like proofs which they can tell their friends while making a walk in the forest. They like to replace calculations by ideas, so that you know why the answer is correct without any calculation. Algebraic manipulations are boring and better left to computers or accountants. There is no way anyone can say that the present tedious lines of formulas are better mathematics than a couple of sentences containing real logical deductions from hard ideas. Just try to digest the ideas in the sentences. Now try to digest the ideas in the formula manipulation. Which meal ends up making you feel you have learnt something, that you understand the problem better now, at a deeper level?

By the way the suggestion some of you made earlier that I am here to push my Own Research is such an incredible joke. Being a senior professional mathematician my professional standing is harmed, not improved, by spending my time discussing elementary probability on wikipedia and writing up educational/amusement maths articles for obscure journals about what I have learnt from all you great guys here. Seriously, I love a debate, I love a fight. That is how science progresses. I am putting in an acknowledgement to all wikipedia editors for the fantastic ideas and fantastic stimulus which comes from this thing. I love it. Gill110951 (talk) 13:22, 25 June 2010 (UTC)

Well this seems to show that you're misunderstanding nature and goals of WP. Encyclopedic articles (in particular WP) are somewhat boring by nature, as they simply (conventionally) summarize the most important and most common aspects of a particular object.

• They are not about what seems currently hip to a particular author
• They are not scientific polemics or opinion essays,
• They are not a place or starting point for intellectual fights or debates (but they rather "boringly" only report on the state and result of such debates).
• Science does not progress in WP but outside WP and WP just reports on the result.
• WP is usually no place for describing a new or different aspect of a problem, because the other aspects can be in literature already. WP is not supposed to augment and improve published literature/knowledge but to reflect it.

There is a difference between WP articles and articles/presentations for math journals, congresses or colloquia. For the latter the listed subject above (without the negation) can be appropriate or even desired but for the former, i.e. in WP, they are not.--Kmhkmh (talk) 08:29, 28 June 2010 (UTC)

Well I beg to differ @Kmhkmh.

• I am not talking about what seems currently hip to *me*. (The game theory approach to Monty Hall was around before vos Savant and is still mentioned in economics texts).
• I am not into polemics or essays on wikipedia pages, but of course we are doing polemics on these talk pages and you in particular are using various debating tricks to defeat an opponent whom you cannot defeat by substantive arguments.
• Talking about probability: science has progressed and wikipedia had better follow. There is no point in wikipedia reflecting probability ideas from 50 year old textbooks when today's readers especially the young ones are learning probability from todays' textbooks.
• There exists in the real world a polemic about the MHP and the wikipedia page has to neutrally report the polemic.
• The page is going to include probability calculations and probability terms, and wikipedia readers would be more helped if that math is written in a modern conventional style, and is written so as to make things easy and simple and not to make things obscure. Editors should take the opportunity to learn some (for them) new tricks, and if they turn out useful help in popularizing them.

Gill110951 (talk) 07:26, 29 June 2010 (UTC)

Well actually polemics don't have of a place on discussion pages either nor should the discussion page be turned into the MHP debate club. If you are looking for "intellectual fights" or polemics over MHP, please look outside WP or at least keep it to the arguments page. There are plenty of math related web fori, blogs or math newsgroups on the internet were you can debate or polemicize over MHP all day long, but WP is not the place for that.

Just to pick a concrete example of how you should not argue from your posting above:

Talking about probability: science has progressed and wikipedia had better follow. There is no point in wikipedia reflecting probability ideas from 50 year old textbooks when today's readers especially the young ones are learning probability from todays' textbooks.

What's the point of such a comment? Did anybody here ever suggest to use outdated material? What's the relation to the MHP article? All the probability textbooks used there are from the 1990s and 2000s. If you expect other editors to take you seriously, I suggest you refrain from such an arguing style.

--Kmhkmh (talk) 10:44, 29 June 2010 (UTC)

Interesting point of view, @Kmhkmh. I suspect a cultural difference here. I deliberately say things in an arguing style thinking that anybody who doesn't like what I say or do is free to ignore it anyway. And thinking that clever people should be able to distinguish what I believe are facts and what are my emotions about those believed facts. When people say crazy things about mathematicsor logic I say crazy things back. Need to let off steam occasionally. I'm deliberately provocative since a real discussion is needed about some basic issues on this page. In particular, people are going round and round in circles because different editors are making different semi-hidden assumptions and refusing to allow debate about them. The page has been stuck for a long long time, mediation has failed, who knows when mediation will start again?, wouldn't it be useful if all active editors on these pages got to understand the fundamental difference of PoV's and start working a) for a neutral PoV and b) for clean elegant accessible logic and mathematics, as far as mathematics is needed at all? It seems to me that the discussion page on the article is the best place to do this, as well as making experimental changes and additions to the page itself. Nobody owns it. Just because it was a FA several times doesn't mean that it can now become a mess and never become a FA again, but maybe it will be necessary for some editors to broaden their opinions about what the page is about, first. An article won't be a FA if it merely is a boring compendium of all stupid opinions and outdated methodolgy which can be found "out there". Some selection, focussing, ... intelligence .. style ... taste ... is needed too. Gill110951 (talk) 14:35, 29 June 2010 (UTC)

Bayes vs. Gill's argument

IMO, the mathematical formulation should be the classic Bayes expansion (as it was before Gill's changes). This expansion appears in dozens of sources and should be included. The simpler conditional argument (equivalent to Rosenthal's) could be included as well but it is distinctly rare (in sources). -- Rick Block (talk) 15:12, 24 June 2010 (UTC)

I agree I see zero reason to remove it and zero reason to assume any consent in that matter. Hence I reverted it originally, however to avoid triggering an edit I refrained from further reverts after Gill insisted on making the change without consent.--Kmhkmh (talk) 15:19, 24 June 2010 (UTC)

How about a separate section again? We then have several solutions, simplest first moving to the more complex. Martin Hogbin (talk) 15:27, 24 June 2010 (UTC)

Essentially that would be fine with me at least as long as no other editors have additional issues regarding the new material.--Kmhkmh (talk) 15:38, 24 June 2010 (UTC)

I have issues both general and specific. General, in regard to the fact that the page is already very long. Specifically, in regard to Gill's verbose and basically unreadable formal treatment - really, I thought the argument on whether math notation is more informative than verbiage had been resolved two centuries ago.

As for the "Game theoretic" section added by Gill, I object to its being little more than an expanded list of references, short on explanation for why it is a genuinely relevant approach. Anyone worth their mathematical salt can come up with a new reformulation of the MHP and call it "original" and "novel". The job of the editors of an encyclopedic article is to both report AND select sources for relevance. glopk (talk) 15:49, 24 June 2010 (UTC)

Maybe the section could be improved but it does represent an alternative approach based different assumptions. Martin Hogbin (talk) 15:53, 24 June 2010 (UTC)

Regarding the article length, the variants section would be the best thing to move if the article is considered too long. Martin Hogbin (talk) 15:56, 24 June 2010 (UTC)

Gill's argument is Bayes argument in a short verbal form which requires less computation and is moreover highly illuminating for understanding the problem.

Translation into English: "It's my own proof, I like it better, everybody else must like it too!"

It is not my own proof. Please try and understand it and see if you like it.

The long mathematical proof is a sequence of formula manipulation steps, suitable for computers, not for humans. To understand it, requires hard work and expansion.

The "long" mathematical proof consists of 1 definition and 6 equations, all of them commented at lenght. As far as symoblic proofs go it's very very short. IMO, A reader who cannot do the "hard work" of reading is obviously not mathematically conversant enough to read any math, so won't likely be interested in any formal proof, either in symbols or in words.

Come off it. One line is better than six.

Moreover the formal Bayes computation is contained elsewhere on wikipedia so need not be duplicated here. However to give all you sceptics a chance, I have simply added the "odds form" proof as an alternative proof, and added two more short alternative proofs. I am astounded that half a page of formal formula manipulations is thought to be more illuminating and useful for the general reader than a couple of sentences based on "posterior odds equals prior odds times Bayes factor".

You must be new to editing WP articles (or interacting with any online community). There are many many thing you'll find astounding along the way, but worry not, you'll learn to be humble and listen rather than just speak,

No I'm not new and I do listen and learn as well as speak, don't worry, I learn by seeking controversy.

I agree, you have to start thinking about odds instead of probabilities. That is difficult for non Anglo-Saxons but still it is useful to get used to the idea.

For example, one thing you will quickly learn (under penalty of ridicule or ban) is that en.wikipedia.org is read and edited by many many English-fluent people who are not Anglo-Saxons, and that Anglo-Saxon people do not have any special status therein. Statements like yours above are seen as condescending at best, offensive at worst. Being seen as condescending or offensive does not add to your reliability as a wikipedia editor.

I'm afraid that Anglo-Saxons often sound condescending or superior. I used to be ashamed of it but now I just be myself. I do not take myself very seriously.

And now you managed to stereotype Anglo-Saxons too. Nice going, old fellow - did you read negotiation theory and practice at Cambridge too, or was it in Attila the Hun's yurt? —Preceding unsigned comment added by 216.239.45.4 (talk) 16:46, 29 June 2010 (UTC)

Of course my writing may not be optimal so please go ahead and improve, cut verbiage, clean and smooth... but first read and try to understand. Then we can discuss again.

That works well in general. However, when a WP article has been peer reviewed enough and deemed of high enough quality to be "Featured", prudence must be exercised in editing. Exactly because the content of WP is not OR, it is rare that anything new is dicovered, that makes a featured article suddenly obsolete. Therefore radical edits to a featured article tend to be seen with skepticism, and the onus is on the editor (YOU) to show to the community that a radical re-edit is worthwhile.

Yes but the situation is now that a long controversy has been going on turning it into a mess and it's time to take a fresh look.

I repeat that the long section with the formal manipulation is totally out of place on this page and moreover superfluous since you can find it elsewhere on wikipedia.

Believe me, I can read, and understand that this is YOUR opinion. Now, open your eyes and read: there are other editors (your peers) that think you are wrong. This is not black and white, it is not a matter of truth of falsity of a theorem. It's an editorial decision, a matter of opinion that intelligent and informed people can compromise upon. Your diktats don't work here.

"diktat" that sounds like a nazi word. I am not a nazi and I am not dictating anything. I'm making a suggestion: the page will be more balanced by removing that superfluous derivation.

I am stunned that anyone objects to the short proof based on symmetry. It is beautiful and entirely correct.

I am stunned that you cannot engage in a constructive conversation about your edits. You have so far used a steamroller approach. I do believe there is some merit in some of the material you have offered, as far as the MHP article. However, your way of offering them has so far only managed to get you a few reverts. Don't you think there is a more constructive way?

Ha Ha! I have fun on wikipedia. There are different ways. Being constructive is to be open to alternative points of view and to read what people write carefully and try to understand what they are doing. It helps if you are not wedded to a particular point of view which makes you defensive, and makes you suspect ill motives of others, and hence makes you behave offensively to others.

I am stunned that noone appreciates learning that Monty Hall is not the exclusive ownership of statisticians. Read the literature, learn Game Theory 101; it can be as important for your intellectual development as Probability 101. Gill110951 (talk) 05:30, 25 June 2010 (UTC)

You get stunned easily, perhaps WP editing is not for you after all. glopk (talk) 22:08, 25 June 2010 (UTC)

I enjoy WP eidting! Especially this page.

BTW the recent and very nice Rosenthal book and Rosenthal article expound the odds version of Bayes in clarifying Monty Hall conditional versus unconditional. It makes Monty Hall so much better understandable. No formulaic mumbo jumbo. Just a simple intuitively appealing and extremely power correct principle which everyone would benefit from knowing. Gill110951 (talk) 13:38, 25 June 2010 (UTC)

This comes from the Conditional Solution section: Gill's response

There are two levels of *assumptions*.

For the time being I don't discuss what the word "probability" means but simply suppose that the basic rules and definitions of probability theory can be used.

Basic assumption) The car is initially equally likely to be behind any of the three doors.

Supplementary assumption) The probability that the host opens door 3 given the player has chosen door 1 and the car is behind door 1 is exactly 1/2.

Under the basic assumption, the *odds* that the car is behind the "other" door given the player chose door 1 and the host opened a door revealing a goat are 2:1. By Bayes theorem, the *odds* that the car is behind door 2 given the player chose 1 and the host opened 3 are 1:Prob(Host opens 3|Player chose 1 and car is behind 1). That latter conditional probability must lie between 0 and 1. The conditional odds therefore lie between 1:0 and 1:1 with them being 1:1/2 under the supplementary condition. The conditional probability therefore lies between 1 and 1/2 while it is 2/3 under the supplementary condition. Switching is never bad and often good. The player should switch.

Now which assumptions you use depends on how you interpret vos Savant's words, and what you mean by probability.

One common meaning of probability is the *ontological*, it is a property of the things or the situation, and can be determined by imagining many many repetitions of the situation with all essential things kept fixed and what you think of as being random allowed to vary freely.

Another common meaning of probability is the *epistemological*, it is defined relative to a particular person's knowledge/information, and reflects that particular person's "reasonable opinion", and can be determined by finding out what betting odds that person would accept.

Commonly these two extremes are called frequentistic and subjectivistic respectively. There are intermediate positions possible, and many subtle distinctions, and philosophers have been worrying their little heads about this for 200 years and will no doubt continue to do so as long as humans play games of chance.

If you are subjectivist then as GKanter correctly observes there is nothing in the problem which suggests a bias for one door or the other. For instance as I wrote in the proposed additional section "forcing symmetry", the subjectivist might well imagine that the host chooses his door with a probability q and that q lies between 0 and 1. This is an known unknown. Since the subjectivist is indifferent to the door labels he will have equal doubts about q, as about 1-q (one gets from one to the other by relabelling doors). The problem is symmetric, and by symmetry the average probability 1/2 is the one to use giving a *conditional* probability of 2/3.

If you read Craig Whitaker's original question to Marilyn and Marilyn's restatement of it, and her later explanation of what she meant, and her perferred solution, you must conclude that for her the door numbers are arbitrary labels which we put into the question as an extra explanatory embellishment.

"Say Door 1" means "We'll name the door which the player has chosen, Door 1".

"Say Door 3" means, "We'll name the Door which the host left closed Door 2, and the one which he opened Door 3".

Originally there were just three doors - say red green and blue, or left middle and right (as seen from the audience). Nobody painted any numerals on those doors!!!! Where did that idea come from? Or think of three boxes which to the player are indistinguishable but to the host not. Initially he hides a euro in one box and nothing in the other two. He shuffles the three boxes and asks you to pick one. He shows that one of the other two boxes is empty. Do you stick to your original choice or do you switch?

Thus the conditional probability we want is the conditional probability that the "other door" hides the car given we chose a door and the host opened another showing a goat. We are conditioning on an event of probability 1 so the odds don't change. The initial odds of player's door versus other doors are 1:2 and these prior odds aren't changed (see Bayes formula, odds form) by conditioning on an event of probability 1 under either hypothesis.

If you read the problem without making a choice of meaning of probability, but you do want to respect the implication that the names of the doors are supposed to be arbitrary or exchangeable (as far as the player is concerned) I think the neatest solution is to let the player's and host's choices determine the names of the doors.

Note that as soon as we've made the problem symmetric in re-numbering of the doors, the conditional probabilities are all equal and equal to their average, the unconditional probability, 2/3

Basicly it is all *very simple* very beautiful and all interconnected.

Psychologists and behavioural scientists and economists see the problem as a problem of game theory. Monty might be trying to cheat the Guest, right? If he knows that he knows that I think that... This is also the way ordinary people think about the problem when they first hear of it. The mathematics of choice and behaviour is called game theory and economics (freakonomics). A game theoretic model recognises this and sees the host's moves as a strategy, and the player's moves as a strategy. Game theory tells us what both parties would do if they were smart enough to follow Game Theory 101 and if the Host intends to keep the car, the Player wants to get it. It's a finite zero sum two person game and it has a solution. Which you can figure out in five minutes in your head once you have grasped the basic probability (above). The host uses symmetric (uniform) probabilities for hiding and opening. The player uses symmetric for initial choice and later uses the most extreme probability, always switching. This is exactly what you expect for a simple game theoretic problem with all these symmetries.

The game theoretic approach has been around since long before Morgan's paper and is still alive and kicking. The behiouralists, neuroscientists and psychologists are still thinking about it this way.

Gill110951 (talk) 17:34, 27 June 2010 (UTC)

I think the door numbers are a red herring, and that the idea of conditioning on a certain event makes no sense from a subjectivist POV (note: discrete problem, so prob. 1 event = certain event). The question seems quite clear. You are on a game show. You see one door open and there's a goat there. The probability that the car is behind that door is zero at that point. If you were to make a bet with someone about the car being behind the door that you see wide open there in all its goat-ful glory, what would you think are the fair betting odds? Right. But the corresponding unconditional probability - the probability before the door was opened - was not zero, but 1/3 (let's say that you are a subjectivist who believes that every agent involved follows their minimax strategy; that fixes your priors). Again, you can refer to the fair betting odds of the car being behind that door before it was opened. So the conditional and the unconditional probabilities are not equal. That means that you conditioned on an event that is not certain. When you compute the probability that the car is behind the door you initially picked, you will of course condition on the same not-certain event (that event represents your state of knowledge, after all, and you use it for all decision-making in the same situation). Note that this argument makes no use of any door numbers.

Actually, you should look at the solution in the "Mathematical formulation" section of the article. It makes it pretty clear that the door numbering is arbitrary. If you replaced every door number "1" in it with "The leftmost door", every "2" in it with "The middle door" and every "3" in it with "The rightmost door", it would still all work unchanged. When you're actually on the game show, you would then take the final probability equation of that section, and substitute into it the values which you have actually observed on the show (i.e. you condition on everything you know in the decision-making situation, as is standard practice). If you see the host open the leftmost door and you initially picked the middle door, then you compute P(C=rightmost door | H=leftmost door, S=middle door) to see if you should switch to the rightmost door, and so on. You can repeat the same exercise with the door labels "red", "green" and "blue" if you so wish, it's not going to make any difference. -- Coffee2theorems (talk) 20:13, 27 June 2010 (UTC)

Come off it, of course I realize that you could equally well talk about left, middle and right door. That's not the point. I'm trying to make a more subtle point. Suppose we play the game in a pub and the doors are replaced by three boxes. There's a Euro in one and nothing in the other two. You can see which box is which but I can't. You shuffle the three boxes and let me choose one. Then you open one and show it's empty. Should I take the other box? This is actually how it is played in real simulations in pubs!Gill110951 (talk) 08:07, 29 June 2010 (UTC)

@Gill - As to the question of where the idea the doors are persistently numbered came from, see the photograph of the Let's Make a Deal stage [1]. Treating the MHP as an urn problem (two indistinguishable black balls representing goats, and one otherwise indistinguishable white ball representing the car in an urn, the player withdraws one without looking at it, the host withdraws a black one and offers the player the opportunity to switch with the remaining ball in the run) is a simplification consistent with treating the problem unconditionally. However, doors are at least usually distinguishable. As Coffee2theorems says above, even if they aren't labeled 1, 2, 3 they're still distinguishable positionally. Part of the difficulty people have with the MHP is that it puts them squarely in a conditional situation - standing in front of door 1 with door 2 still closed and door 3 open and showing a goat. The obvious probability of door 3 is now 0, but this is only the case if the probability we're talking about is the conditional probability - which only exists if the doors are distinguishable. Is the urn problem as difficult to solve correctly? I don't actually know, but I doubt it since treating the doors as if they are indistinguishable (i.e. avoiding addressing the conditional situation) is the basis of most of the simple solutions. -- Rick Block (talk) 13:06, 28 June 2010 (UTC)

See my aswer above to Coffee2theorems. And read my contributions about forcing symmetry. And remember that most readers of vos Savant didn't have access to a photo of the stage. And read the history and read what Marilyn vos Savant actually meant (*say* Door 1: vos Savant names the door chosen by the player "Door 1", whatever the actual numeral written on it, or other distinguishing mark it initially had) and read what Whitaker actually asked (no door numbers!). Your urn problem is beautiful! Two black and one white ball in an urn. You pull out a ball but don't look. The host looks pulls out a black ball. Do you want to switch your ball with the one in the urn? YES!!!! The probability that the remaining ball in the urn is white is 2/3. What's wrong with that? If we use conditional probability with three distinguishable doors and we are not subjectivistic, then the conditional odds are in favour of the car being behind the closed door, but they are unknown. Conditional probability says "switch". If you are subjectivistic in your use of probability then for you, the host's choice of door to open is uniform at random, so the conditional odds are 2:1, the same as the unconditional odds, which means that the door numbers are indeed totally irrelevant. Why do you think the cars problem is different from the balls in a vase? Does it make a difference that the three balls are in three different boxes rather than one? Above all, just remember: the question is not "what is the probability" but "should you switch"./Gill110951 (talk) 08:07, 29 June 2010 (UTC)

The cars problem is different from the urn problem since the odds in the urn problem really are exactly 2/3. Since the two black balls are defined to be indistinguishable, the host CANNOT have a preference between them. There really are only two possibilities - you pick a white ball (with probability 1/3) and the host withdraws a black ball, or you pick a black ball (with probability 2/3) and the host withdraws a black ball. An equivalent urn problem requires 3 different balls, perhaps white, red, and black where red and black are both losers, and an example case - you withdraw a ball, the host looks at the remaining balls and withdraws the black one, do you want to switch? Now the host CAN have a preference between the red and black balls. There is more information available to you. Ignoring it may be unwise. A true subjectivist not knowing the host's preference would realize her chances are unknown, but must be between 1/2 and 1 with an average value of 2/3. The answer to "switch or not" is still switch, but the reasoning now includes all known information. -- Rick Block (talk) 16:54, 29 June 2010 (UTC)

The cars problem is different? source? Switching wins in 2/3 of all cases. punctum. Sure, the host has all information on hand, but he never can change the actual constellation. All he can (eventually??) do, is giving additional info on the actual constellation: e.g., if opening his "unwanted" door (q=0 or 1) in 1/3 of all cases, showing then that his preferred door DOES hide the car (as if he was opening it, too) and, when opening his preferred door, the chances of both still closed doors are 1/2 each. But this is not part of the dilemma. This is a totally different issue. Yes, knowing the Host's bias would be great, but the guest does NOT know it and she should switch anyway! Only a "known and quantified door preference q (off-size 1/2)" could generate a possible "condition" in opening any door, giving a closer hint on the actual constellation. In the absence of such exact knowledge any semi-hidden assumption is unhelpful, confusing, distracting and real bad obfuscation. So-called "Conditional" in the absence of any evident condition cannot give any closer "conditional hint" on the respective actual and given situation. To the guest, the absence of a known and evident condition makes "conditional" solutions cryptical and worthless. No more semi-hidden and confusing assumptions, please. Claiming "conditional" without any evident condition. Such puzzlements are not part of the famous dilemma. Or name them frankly what they are and put them in a separate section, at least. Such meander do not address the famous dilemma. They are a quite different issue. --Gerhardvalentin (talk) 18:05, 29 June 2010 (UTC)

Good discussion.

The doors are distinguishable to the player, and to the host, but the host knows more about them than the player, since he knows which one hides a car.

Two of my contributions (about symmetry) give good reasons to make the extra assumption of 50:50 for the chance the host opens door 2 given player has chosen 1 and it has the car.

I don't understand why supporters of the conditional solution should object to them. Nor can I understand why supporters of the unconditional problem should object, since the conclusion is that *both solutions are the same and both are right*. Let me repeat the argument.

The solution is especially nice if we take a subjectivistic view of probability. I notice that many of the editors actually are perhaps without realising it, subjectivists. There is nothing wrong with that. Nothing at all. It is legitimate, we just have to be clear that that is the kind of probability we are talking about.

Remember, the question is whether the player should switch, given his knowledge ... and therefore given also his ignorance! We want to compute the probability for the player that switching will give him the car.

If we take a subjectivistic approach then probabilities are meant to be expressions, for the subject - in this case the player - of his knowledge. Odds are betting odds, fair stakes. Odds of 4 to 1 mean that you are prepared to wager 4 dollars against my 1 dollar. It means that you expect to lose with probabilty 1/(4+1), so your expected loss is one fifth of 4 dollars. Your expect to win with probability 4/(4+1) you your expected gain is four fifths of 1 dollar. Expected loss equals expected gain. The odds are fair (that is to say, you find them fair, since you think this is a fair bet).

For the player, doors 1 and 2 are initially equally likely to hide the car. *His* prior odds are 1 to 1. He knows that the host certainly will open door 3 if door 2 hides the car, so for him P(3 opened| car at 2) =1. For him, the host is equally likely to open doors 2 and 3 if the host has that choice. So for him. P(3 opened|car at 1) = 0.5. By Bayes *his* posterior odds are therefore initial odds 1:1 times Bayes factor 1:0.5 equals 1:0.5 equals 2:1 means probabilities 2/3, 1/3. Now note that for him, this means that P(switching is good) = P(switching is good| specific door numbers of initial choice and host-goat). So for him, whether or not he should switch is independent of the door numbers.

The fact that for him, the door numbering is arbitrary means that his probabilities do not depend on the door numbering. He can use Laplace's principle of indifference to assign equal probabilities when he has no idea. This interpretation is also supported by the fact that vos Savant added parenthetically in her explanation "*say* Door 1", "*say* Door 3"; while Whitaker didn't talk about door numbers at all in his letter to her! "You choose a door, the host opens another. Should you switch?". The door numbers are irrelevant. The host is therefore equally likely, *from the point of view of the player*, to open either door when he has a choice. The unconditional probability that switching is good is the same as the conditional probability, because of the invariance. In fact the door numbers are independent of whether "switching is good". Everyone is right and everyone has the same opinion, they just don't realise it!

If we don't want to use subjectivistic probability, but be died in the wool frequentists, we have more difficulty in how to incorporate the "arbitrariness" of the door numbers into a formal probability statement and solution. This is where the game theoretic solution tells us why we should always switch.

We can also just postpone facing up to the difficulty and just use Bayes anyway and see what it gives us. The posterior odds in favour of switching are 1 : P(host opens 3|car is behind 1 and we chose 1). Those odds *always* support switching!!!!.

There might be a situation when the odds are even, but then there is certainly another situation when the odds are overwhelming ... so we the player with this frequentist interpretation (where we imagine the host choosing by using a possibly biased coin whose possible bias we do not know) will say "we don't know the conditional probability but it doesn't matter, we are going to switch, since the conditional probability is at least 50% and might be much more, that the car is behing the other door". Game theory will make us feel really happy since it tells us it really is the most sensible thing we can do. It also tells us, by the way, that the most *malevolent* strategy of the host is to hide the car uniformly at random and to open a door uniformly at random when having a choice. The meanest host does not use a biased coin! The sensible player can behave as if the coin is unbiased!

We the mathematicians can also say: let us make the problem invariant under door re-numberings by letting the player initially number the doors himself, uniformly at random. Now it is objectively true that initially all doors are equally likely to hide the car, and it is objectively true that the host is equally likely to open either door when he has a choice. So this mathematical device makes the extended assumptions valid. Why should anyone object to that? It is moreover a possibly reading of vos Savant's wording. For the player, the door numbering or labeling is irrelevant.

We the nit-picking semantist can also say: suppose the doors initially were coloured red, green and blue or that they have spatial locations left, middle, right. There are no numbers on the doors, in my game show. vos Savant meant by saying "say" that "Door 1" is a name given by the player to the door which the player chooses to open. "Door 2" is a name given by the player to the other door which the host left closed. Now the conditional solution is unconditional since we condition only on "host opens another door" which has probability 1; the odds which we are talking about are not of one specific door to another specific door, but are the odds of the initially chosen door to the other two doors. Initial odds are 1:2 against for the car being behind the door initially chosen by the player. The probability that a door is opened revealing a goat is 1 under both alternatives. The Bayes factor is therefore 1. The posterior odds are therefore the same as the prior odds. 1:2 against the initially chosen door, for the other two doors.

Let me do it without door numbers. Instead of "say, Door 1" let's have "say, door Bert". The player gives the name "Bert" to the door he first chooses (which he did uniformly at random). He gives the name "Ernie" to the door opened by the host. He gives the name "Cooky monster" to the other door. The probability that Cooky Monster has the car is 2/3. (There is nothing to condition on)

Or: he gives the names Bert, Ernie and Cooky Monser to the three doors completely at random, and chooses Bert himself. The host opens Ernie. The conditional probability that Cooky Monster has the car is 2/3. (Now we do have something to condition on, and the host opens a particularly named door uniformly at random when he has a choice since we gave the names uniformly at random)

All the solutions *are* finally all essentially the same. The maths is *all* elementary, since Bayes in the odds form is so simple, direct, illuminating. *All* of the controversy is *only* about the meaning of probability (an issue on which philosophers have differed for hundreds of years), and about how to mathematize the notion that the door numbers are arbitrary, irrelevant... Good mathematics illuminates and proves the essential one-ness of all the different solutions and approaches. The controversy goes on and on because the solutions are all so close.

Gill110951 (talk) 05:54, 29 June 2010 (UTC)

Most of what you say above I never disagreed with (and didn't get the impression that most others do either, but obviously I can't speak for them). The assumptions that the car is placed uniformly and that the host picks uniformly among the choices he is allowed to make are standard and can be justified in various different ways, and that's not really disputed. Your calculation of the posterior odds is just the same argument as in the "mathematical formulation" section of the article, just wearing slightly different clothes, so nothing too controversial there either. I agree that the game theoretic view is also of interest, that the player should switch, that the "died in the wool frequentist" (i.e. 1/(1+q) ≥ 1/2) argument is convincing, that it is useful to look at the MHP from multiple POVs, and that all of these approaches are elementary. That covers most of what's said above.

What I don't follow is your arguments that make no use of conditioning, or condition on a certain event. Assume without loss of generality that you have initially picked the leftmost door and the host will open the rightmost door (this, is what I think the "say, door x" things mean; they could equally well be omitted from the problem statement, and you could still use this assumption wlog in the solution, see e.g. this example, and this for a proper conditional solution using your style of naming). At this point you believe that P(car is behind leftmost door) = P(car is behind middle door) = P(car is behind rightmost door) = 1/3. The host opens the rightmost door. Now, if you use a certain event to represent your state of knowledge, P(car is behind rightmost door | what you know) = 1/3. But you see a goat there, so conditioning on this event produces nonsense.

In the case where the doors are called Bert, Cooky Monster and Ernie, you say that P(Cooky Monster has the car) = 2/3. Clearly, that is a part of a prior distribution because it doesn't condition on anything, but it is not the uniform prior, which most people choose when they directly place a prior on the two unopened doors. How did you choose your prior? The only thing you said about choosing it is that you did not choose it by computing a conditional probability, but that's not saying much! -- Coffee2theorems (talk) 01:35, 30 June 2010 (UTC)

He's assigning names randomly chosen by the guest that are unknown to the host, so any "real world" host preference is obscured. This is not a variant I've seen published anywhere, but it does force the conditional solution to use p=1/2 as the host preference. It's simpler just to make the contestant choose before the host opens a door (or blindfold the contestant after she picks a door and have the host loudly announce "I'm opening a door" followed by a collective shout from the audience "it's a goat!"). The point is to avoid a persistent naming known both to the contestant and the host, which logically makes the host preference accessible to the contestant (and could be demonstrated by a frequentist's experiment). The "unconditionalists" are simply going to great lengths to avoid the obvious fact that doors on a gameshow would be spatially persistent because they would be part of the gameshow's set. They seem to truly believe doors can be indistinguishable. -- Rick Block (talk) 04:16, 30 June 2010 (UTC)

Gill's changes/arguments and focus on what matters

After looking over many of the recently posted comments and explanations, I'm getting the feeling the discussion is getting a bit off track. So let's focus on the actual issues of Gill's contested edit and suggestions and how to resolve then. So let me restate, what the issue are and what they are not.

The following points were problematic to some of the other editors:

• p1) large editorial changes without consent in highly contested article (something that ideally should be avoided
• p2) deletion of formal Bayes solution in a 3 step experiment
• p3) lack of citations and possible WP:OR
• p4) style/language which might not be optimal for an encyclopedic article (imho at times it reads a bit like polemic or essay which is ok for journal articles but usually not appropriate for WP, see the use of adjectives for instance)

The following things however were not really problematic, since so far there wasn't any serious disagreement:

• n1) (mathematical) correctness of the odds approach (nobody suggested it is not correct)
• n2) odds approach in the article (nobody really claims that the odds approach cannot be in the article, the issue was with using it as a replacement for formal bayes solution, but that can be discussed in p2)
• n3) game theoretic aspects in the article (nobody really claims it cannot be a part of the article)

So I suggest to leave n1,n2,n3 alone and focus on p1, p2, p3, p4, which for the most part can easily be fixed.--Kmhkmh (talk) 09:24, 29 June 2010 (UTC)

Splendid! Gill110951 (talk) 09:39, 29 June 2010 (UTC)

Clarification of motives: I am a professional mathematician and I do a lot of work on statistics in law, medicine, and so on; I talk to journalists... I am always trying to learn how to communicate probability to ordinary people. Here I try to communicate to you, wikipedia editors, what I think are valuable probability insights. This does not mean that I consider myself *above* the other wikipedia editors, quite the contrary, I consider myself *below* since you are peers of one another but I am an elephant in a porcelain shop. I try different ways to get some subtle ideas across. Some ways turn out to work, some don't. Some ideas turn out to be duds, some grow through the interaction with they guys here. I learn from my experience. And I get new ideas. When people don't understand my way of thinking that is at first puzzling but later one can understand why. Someone is missing something, usually *both* parties are missing something *different*. In the long run this should result in a synthesis. Science proceeds by conflict. Keep up the fight!Gill110951 (talk) 09:52, 29 June 2010 (UTC)

Yes, Speaking Of Stopped Clocks

The common conditional solution only works with the host choosing randomly when faced with 2 goats.

The simple solutions, as demonstrated by Carlton's car/goat based decision tree work regardless of the host's preference when faced with 2 goats. It's the contestant's SoK that matters.

Therefore, those conditional solutions are, like Rick Block likes to say, no better than a stopped clock which is correct twice a day.

Bad news for you Bayesians... that goes for your solution, too. Glkanter (talk) 13:48, 29 June 2010 (UTC)

@Glkanter, this is nothing to do with being Bayesian or not. The conditional odds that switching will give the car are 1:q where q is the probability that the host opens Door 3 when he can choose between 2 and 3. Whatever q might be these odds favour switching. Especially if you don't know q you'ld better switch. I think if you want to do this with conditional probability you should not assume "no host bias" (q=0.5). Whitaker's question is "should you switch", not "what is the probability switching will give you the car". Anyway, whose probability, when? A subjective probability (of whom?) or an objective probabilty (conditional on what?)?

Since the player knows nothing about host bias, if he solves his problem by using subjective probability, and by computing his subjective conditional probability, he'll take q=0.5 since this probability is not the hosts's actual probability in some physical randomizer, it's the player's state of knowledge about the hosts behaviour, which is symmetric in door numbering. Alternatively, whether he prefers subjective or objective probabilities, since he realises that he cannot do anything with the door numbers anyway, he could use your approach and calculate the probability that the remaining door has the car given he selected one and the host showed a goat behind another. Answer 2/3. If the player is a sophisticated mathematician he will realise that the problem is an easy and standard problem of game theory in which there is a minimax solution which provides the optimal strategy, answer, always switch. For me, the question is "should I switch?". You can use a myriad of different approaches to determine the right answer: "yes". What you musn't do is start adding assumptions which you only need in order to make your preferred method of solution applicable. That is called solution driven science. The problems should drive the solutions, not the other way round. Fortunately in this case it is easy to find the solution, and you can argue the correctness of the solution ("yes, switch") using many different approaches. The only assumptions which are needed for any of the approaches we talk about to find the solution "yes, switch" are: initially all doors are equally likely, the host always opens a door revealing a goat, and always makes the offer to switch doors. No-one owns the Monty Hall Problem. vos Savant made it famous and that's why this page is here. Her question is ambiguous. We do know she meant a couple of assumptions to be made, since she used them herself in her "solution". And you don't need more assumptions to get that solution ("yes, switch"). If only people could try to understand this and try to bring out the unity and interrelations, instead of fighting like children whether we are going to do it one person's way or another. All reasonable formalizations are legitimate. All logically correct solutions to a specified formalization are legitimate. The better solutions are the one's which don't make use of unwarrented extra assumptions. The better solutions give enlightenment, explain links, give insight, help the reader to get new tools by seeing smart tools in action (odds, invariance, symmetry, randomization). The best problems are the ones which have multiple and apparently quite different solutions. Anyone who wants to solve this problem using probability theory must make clear what kind of probability he is using. The calculus is the same, Bayes formula is true for all probabilities. The question is what is it about. Some kind of semi-imaginary long run of repetitions? Or one person's reasonable but personal state of knowledge? Some assumptions make more sense with certain kinds of probability than others.Gill110951 (talk) 14:15, 29 June 2010 (UTC)

Rose?

I really do not like the way Gill is intervening, acting like the elephant in the porcelain shop. I'm also very disappointed by him still defending the simple problem. A door (I do not care which one) has been opened before the player is given the offer to switch! Just the "simple" reader of the article will and should have this picture in mind. There is nothing to compromise. So, from now on, I suggest we start with the correct problem and the correct solution, be it in simple wording. Further on in the article we may mention some authors consider simplified versions of the problem! I'm through. Nijdam (talk) 08:07, 19 June 2010 (UTC)

May be someone can explain to me where from the problem formulation it may be deduced the player is given the offer to switch BEFORE the door with a goat is opened?Nijdam (talk) 08:09, 19 June 2010 (UTC)

Moreover, giving the right numerical answer on basis of wrong reasoning, is as bad as giving the 50-50 answer.Nijdam (talk) 10:55, 19 June 2010 (UTC)

It is not entirely clear where any precise formulation of the MHP comes from. The most well known statement, Whitaker's question, is so vague that no method can solve it without some additional assumptions. The simple solutions fail if the host chooses a goat door non-uniformly, the conditional solution fails if the car is initially placed non-uniformly, and even the game theoretical solution fails if the TV station want to boost their viewing figures by engineering a win for the player.

You seemingly care not to comment on the opened door. How vague the formulation may be, it's definitely not vague about the host opening a door and then offering the player to switch. Agree?

No I do not agree. It is far from clear that Whitaker intended to specify a specific door or even that he intended to specify that the player must choose after the host has opened a door. I have tried to discuss this several times before but nobody seems interested. I am happy to talk about it again in a new section if you wish. Martin Hogbin (talk) 13:36, 19 June 2010 (UTC)

Oh please. From Whitaker's problem statement (emphasis added)

You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

If you do want to discuss the important subject of what exact question Whitaker actually wanted to know the answer to I suggest we do so in a new section. I have tried to raise this issue before, only to be hit with 'what do the sources say?'. They actually say very little on that subject. Martin Hogbin (talk) 18:38, 20 June 2010 (UTC)

Seems abundantly clear that the player can see which door the host opens (the door the host opens is not obscured in any way, for example the player is not blindfolded) and the host is offering the chance to switch after he opens a door. The desired mental image is clearly that the player is standing in front of two specific doors (not in front of an "unknown" pair of doors in a Heisenberg uncertainty principle sense) and can see a specific third open door. If the problem is symmetrical the answer will be the same regardless of which specific pair of doors the player is standing in front of, but arguing that the player can't tell which pair of doors are involved is plainly ridiculous. As far as I know, there isn't any published solution (not even a single one - out of at least 50 that I've personally read) that asserts the player must decide before the host opens a door or asserts that the player can't tell which door the host opened. -- Rick Block (talk) 14:15, 19 June 2010 (UTC)

@Martin: why are you so reluctant to accept that the player is standing in front of two closed and one opened door showing a goat. Why? This is where the "paradox" stems from. Nijdam (talk) 16:37, 19 June 2010 (UTC)

Because there may well be a difference between the question that Whitaker intended to ask and the actual words that he used. Any decent statistician would have tried to do what Seymann suggested (and vos Savant pretty well did) and answer the question that Whitaker probably meant to ask. It is extremely unhelpful for a professional to stick rigidly to the exact words used when an amateur asks a question. Martin Hogbin (talk) 17:44, 22 June 2010 (UTC)

The simple solutions fail in all cases. This has nothing to do with the underlying distributions.Nijdam (talk) 11:28, 19 June 2010 (UTC)

In the symmetrical case, the simple solutions necessarily give the correct answer because of an obvious and intuitive symmetry. Spotting symmetries and using them to simplify mathematical problems is a well used and respected mathematical technique. You have no right to demand that a problem can only be solved by a method of your choosing. Martin Hogbin (talk) 11:38, 19 June 2010 (UTC)

Right. And if an unconditional solution actually says this, then it is no longer an unconditional solution but a conditional solution. The point you seem to keep ignoring is that YOU are adding this argument to the published simple solutions. Since they don't argue this themselves we can't argue this on their behalf - that would be WP:OR, even though is it perfectly true. -- Rick Block (talk) 14:28, 19 June 2010 (UTC)

Once again, Martin, do not repeat any more that the simple solutions produce the correct answer. I gather you mean by this they mention a value 1/3. It is incorrect as long as it is produced by false reasoning. Think logic. Nijdam (talk) 16:37, 19 June 2010 (UTC)

We need to cover all the angles in the article, in a way that us useful and informative to our readers. We should, and do, point out in the article the limitations of the simple solution. This should be the case for all solutions. Martin Hogbin (talk) 11:11, 19 June 2010 (UTC)

Right, and we definitely should pass to any reader the warning of several authors about the flaw in the simple solutions.Nijdam (talk) 11:28, 19 June 2010 (UTC)

Indeed, in the appropriate place. Martin Hogbin (talk) 11:38, 19 June 2010 (UTC)

There is nothing whatsoever wrong with the simple solution when seen as solution to a simple problem. The same argument which you guys are using to disqualify the simple solution to the simple question (because you like Bayes theorem), can be used to disqualify your solution in favour of the game theoretic solution. It is yet more sophisticated. It is even more true to the original question. You have to make an extra and unwarranted assumption so that your solution works. The game theorist explicitly takes account of the fact that we don't know the host's behaviour. Remember, vos Savant asked "should you switch"? Who said "probability"? There is no law saying that you have to solve this problem by computing a conditional probability. There can equally well be law saying that you have to solve this problem by computing the minimax strategies (which is what every economist does instinctively and rapidly and easily). In fact, arguably the game theoretic point of view tackles the real problem; the conditional probabilty question doesn't face up to it, but fortunately does get the right answer. It's a simple solution.Gill110951 (talk) 09:35, 29 June 2010 (UTC)

Also there is nothing wrong whatsoever with the proofs of Pythagoras' theorem as seen as proof of the Pythagoras' theorem! The fundamental question above was: Does the player see a door opened? To me it is no question at all. My question more is: Why do some people not want to accept that a door is opened, showing a goat to the player, who then is asked whether to stick or to switch? The Parade version says so, and what may be the charm of another interpretation?Nijdam (talk) 07:20, 1 July 2010 (UTC)

Bayes rule

Bayes rule is "posterior odds equals prior odds times likelihood ratio". Let me write out the proof for you. I talk about two hypothesis H and H', and about some data D.

The prior odds on the two hypotheses are P(H) : P(H')

The Bayes factor or likelihood ratio is the ratio of the probabilities of the data under the two hypotheses; P(D|H) : P(D|H')

The posterior odds are P(H|D) : P(H'|D)

Bayes theorem says

      P(H|D) : P(H'|D)     =     P(H) : P(H')     *   P(D|H) : P(D|H')

Proof: by definition of conditional probability, twice

     P(H|D) = P(H&D) / P(D)  = P(D|H) P(H) / P(D)

     P(H'|D) = P(H'&D) / P(D)  = P(D|H') P(H') / P(D)

Divide first line by second (and skipping the intermediate equality)

    P(H|D)                                 P(D|H)                        P(H) 
    ______                                ______                         _____
                             =                                   *                       
    P(H'|D)                                P(D|H')                        P(H') 

Gill110951 (talk) 06:45, 29 June 2010 (UTC)

This is the same as Rosenthal's "likelihood ratio", which he proves in an appendix to his paper and uses to solve the conditional probability of various variants (including the standard one) of the MHP. Can anyone suggest a better reference for this version of Bayes rule? Although clearly this approach is correct, from an editing standpoint since the (slightly more laborious) standard Bayes approach is presented much more commonly it is the one that should be more prominent in the article. Mentioning this as an alternative (conditional) approach is fine. Rick Block (talk) 16:13, 29 June 2010 (UTC)

It is also in Rosenhouse's book, but in a somewhat less prominent manner than the "standard approach". I'd rather see it as an addition rather than a replacement of the current Bayes description. There are pro and cons for both approaches.--Kmhkmh (talk) 16:39, 29 June 2010 (UTC)

By a better reference, do you mean a reference to something like Gill's proof above? The one that immediately springs to my mind is this one (it's for densities, but clearly the same proof and terminology as Gill's). I think this kind of stuff would be a better fit to the Bayes's rule article, where currently the only odds form makes the assumption that (in Gill's notation) H' = H^c, instead of allowing two arbitrary events. Likewise, in the odds article, there is no mention of this generalized form of odds, only P(H) : P(H^c) = p/(1-p). If these things were added to the Bayes's rule and the odds articles, do we really need to explain them here? We could just give the argument itself and the reader could refer to those two articles if this is new to them. -- Coffee2theorems (talk) 18:01, 29 June 2010 (UTC)

Agree, please let's stay focused. There's no need to put a proof in this article that the denominator (or partition function) in Bayes' P(H|C,S) P(C|S) / P(H|S) goes away when, instead of computing the probability P(C|H,S), one is interested instead in the ratio of posterior probabilities (a.k.a. odds) P(C=2|H,S) / P(C=1|H,S) . Apart from the fact that the "proof" is trivial (just do the division and observe that P(H|S) does not depend on C), it really would belong to the page on Bayes' theorem, if anywhere. But we could add the result (not the above "proof" of Gill's), as an alternative way to see the result in the "mathematical Formulation" section, referring Rosenthal. If the other editors (as well as the esteemed Anglo-Saxon Professional Mathematician) agree, I could add that later today. However, I don't have the Rosenthal reference with me, so someone else who does it should check it. glopk (talk) 19:31, 29 June 2010 (UTC)

Glopk - the Rosenthal reference is [2] (should actually cite the version published in Math Horizons, September 2008 - but I assume the content is identical). -- Rick Block (talk) 03:47, 30 June 2010 (UTC)

Nobody needs to get any permission from any Anglo-Saxon (and in this case, English) professor, whether esteemed or hated. I tell you guys what I think, or at least I try to, and I try to learn from the discussions here, and inspired by them I convert some of them - with the aid of coffee of course - into theorems which I publish elsewhere, and use them in talks which I give to lawyers and medics about statistics fallacies and to my own students ... I'm here to have fun and to contribute my "experience" and ahum "wisdom". I try to promote a neutral POV and good pedagogics. When we are talking about elementary mathematics we shouldn't need the authority of published papers in statistical or mathematical journals. Anyone should be able to see when an argument is correct and when it is incorrect. Anyone should be able to see what assumptions the argument uses, and what it doesn't use. I have the feeling that some people here are going on about no own research and about only use of reliable sources just in order to push a particular point of view. Or because they find it too difficult to learn some new tricks. BTW I agree that the wikipedia pages on odds, Bayes rule in terms of odds, and Bayes theorem generally, are not very good quality. I will certainly be going back to them later. Gill110951 (talk) 15:52, 30 June 2010 (UTC)

What good is the MHP article...

What good is the article if every reader has to read this accompanying talk page to interpret exactly how Rick has chosen to paraphrase/parse/vette each and every source?

Further, why must any changes to the article be specifically approved by Rick under the guise of 'a consensus'? Glkanter (talk) 07:36, 30 June 2010 (UTC)

Hear hear! I vote that Martin Hogbin is made chief editor. He's a non-pushy guy who looks for concensus and who understands subtle issues of mathematics and logic. Gill110951 (talk) 14:24, 30 June 2010 (UTC)

Why the popular solution is [almost] complete

The question is, should you switch?

Everyone agrees that a "switcher" (a player who always switches) wins the car with probability 2/3, precisely because a "stayer" (a player who never switches) wins with probability 1/3.

The logic fails here in the case a door has been opened! Back to school?Nijdam (talk) 20:37, 30 June 2010 (UTC)

Nobody has ever exhibited a strategy with a better than 2/3 *overall* chance of winning. It could only exist if there is some situation (particular initial door chosen by player, particular door opened by host) for which it would be advantageous to stay. Still, given the initial choice, in 2/3 of all cases (cases = many many imaginary repetitions, in which the relative frequencies are very very close to the theoretical probabilities) the initial choice is incorrect. 1/3 of *all* cases are removed when the host opens a particular door. In general, some of these cases could be cases where the initial choice is right, some could be cases where the initial choice is wrong. Most in favour of the player's staying would be that *all* the removed cases were cases where the initial choice was wrong. That leaves at best, for staying, 1/3 where staying is right, 1/3 where staying is wrong. But in this most extreme situation it makes no difference at all to the success rate whether one stays or switches. Therefore there is no situation whatsoever where it would strictly improve the succesrate by staying instead of switching.

In conclusion: the strategy of always switching achieves a succesrate of 2/3 [we already knew that], and moreover no strategy can give a higher successrate [because the conditional probability that staying is good is never bigger than 1/2].

The *only* probabilistic assumption made was that initially all doors were equally likely to hide the car.

I talk about probabilities in terms of relative frequencies in many repetitions. Even if you are a subjectivist, it is always legitimate to do calculations by pretending many repetitions, since the probability calculus is indifferent to the meaning of probability. And any probability model whether subjective of objective can be simulated, i.e., copied to a situation where the probabilities are objective.

I avoid mathematical formalization and reference to probability calculus in this argument but in fact I am using the law of total probability and doing a posterior odds calculation, where I don't need to compute the posterior odds for staying explicitly, I just want to show they are never better than even. So deep down I am using Bayes theorem. The new information that Door 3 hides a goat might alter the balance of the odds to some extent, but it never sways it so far that it points in the opposite direction.

Gill110951 (talk) 14:24, 30 June 2010 (UTC)

Math formulation in odds form.

I propose rewriting the Mathematical Formulation as below, adding a paragraph the end describing the odds form of the argument as an alternative view. Comments? glopk (talk) 14:58, 30 June 2010 (UTC)

* Mathematical formulation *

The conditional solution to the standard problem can be formally proven using Bayes' theorem, similar to Gill, 2002, Henze, 1997, and many others. Different authors use different formal notations, but the one below may be regarded as typical. Consider the discrete random variables:

$C\in \{1,2,3\}$: the number of the door hiding the Car,

$S\in \{1,2,3\}$: the number of the door Selected by the player, and

$H\in \{1,2,3\}$: the number of the door opened by the Host.

As the host's placement of the car is random, all values of C are equally likely. The initial (unconditional) probability of C is then

$P(C)\,={\tfrac {1}{3}}$, for every value of C.

Further, as the initial choice of the player is independent of the placement of the car, variables C and S are independent. Hence the conditional probability of C given S is

$P(C|S)\,=P(C)={\tfrac {1}{3}}$, for every value of C and S.

The host's behavior is reflected by the values of the conditional probability of H given C and S:

$P(H|C,S)\,\,=\,{\begin{cases}\,\\\,\\\,\\\,\end{cases}}$	$\,0\,$		if H = S, (the host cannot open the door picked by the player)
	$\,0\,$		if H = C, (the host cannot open a door with a car behind it)
	$\,1/2\,$		if S = C, (the two doors with no car are equally likely to be opened)
	$\,1\,$		if H $\neq$C and S $\neq$ C, (there is only one door available to open)

The player can then use Bayes' rule to compute the probability of finding the car behind any door, after the initial selection and the host's opening of one. This is the conditional probability of C given H and S:

$P(C|H,S)\,={\frac {P(H|C,S)P(C|S)}{P(H|S)}}={\tfrac {1}{3}}\times {\frac {P(H|C,S)}{P(H|S)}}$,

where the denominator is computed as the marginal probability

$P(H|S)\,=\sum _{C=1}^{3}P(H,C|S)=\sum _{C=1}^{3}P(H|C,S)P(C|S)={\tfrac {1}{3}}\times \sum _{C=1}^{3}P(H|C,S)$.

Thus, if the player initially selects Door 1, and the host opens Door 3, the probability of winning by switching is

$P(C_{=2}|H_{=3},S_{=1})={\frac {{\frac {1}{3}}\times 1}{{\frac {1}{3}}\times ({\frac {1}{2}}+1+0)}}={\tfrac {2}{3}}.$

An alternative proof, equivalent but perhaps simpler (Rosenthal, 2008), considers the odds of player's winning the car by switching doors versus sticking with the original choice, a posteriori of the host's opening of a door. The posterior odds $\,{\textrm {O}}(C_{=i}:C_{=j}|H,S)$ is, by definition, the ratio of the conditional probabilities of the two events $\,C=i$ and $\,C=j$. For this formulation it is not necessary to compute the marginal probability $\,P(H|S)$, since it divides equally both the numerator and the denominator of the odds ratio, when the conditional probabilities are expressed using the Bayes' rule as above. It is then:

${\textrm {O}}(C_{=2}:C_{=1}|H_{=3},S_{=1})={\frac {P(C_{=2}|H_{=3},S_{=1})}{P(C_{=1}|H_{=3},S_{=1})}}={\frac {P(H_{=3}|C_{=2},S_{=1})}{P(H_{=3}|C_{=1},S_{=1})}}={\frac {1}{\tfrac {1}{2}}}=2$.

As you say perhaps simpler, but actually not. I do not see what makes this simpler for our readers. Nijdam (talk) 20:47, 30 June 2010 (UTC)

No need to compute the marginal. glopk (talk) 21:47, 30 June 2010 (UTC)

The current formulation in the article indeed fails to note that P(C | S) is a constant. But why not cut it at its root by putting the P(C | H, S) equation this way:

$P(C|H,S)={\frac {P(H|C,S)\overbrace {\cancel {P(C|S)}} ^{\text{constant}}}{\displaystyle \sum _{C}P(H|C,S){\cancel {P(C|S)}}}}={\frac {P(H|C,S)}{\displaystyle \sum _{C}P(H|C,S)}}.$

????

The formula should read in full:

$\scriptstyle P(C=c|H=h,S=s)={\frac {P(H=h|C=c,S=s)P(C=c|S=s)}{\sum _{k}P(H=h|C=k,S=s)P(C=k|S=s)}}.$

The given form suggests canceling out P(C|S) is possible, even without being constant.Nijdam (talk) 07:02, 1 July 2010 (UTC)

This is just a difference in notation. I used the one already in the article (and the above proposal) with the minimum of changes. I put the note that P(C | S) is a constant there exactly because if it is not mentioned, then the reader's first response to seeing the cancelling is likely to be "hey! that's not legit!"; this way, there's a prominent note to the reader that this time, it is. Apparently that wasn't suggestive enough.. How about the following, then?

$P(C|H,S)={\frac {P(H|C,S)\overbrace {\cancel {P(C|S)}} ^{\text{1/3}}}{\displaystyle \sum _{C}P(H|C,S)\underbrace {\cancel {P(C|S)}} _{1/3}}}={\frac {P(H|C,S)}{\displaystyle \sum _{C}P(H|C,S)}}.$

Adding the "=..." things everywhere clutters things up unnecessarily, IMO. I can see the point of using "C=c" and "C=k" in there to make it absolutely clear that the sum "steals" the variable C, but is it really that confusing? Do you also object to notation like $\scriptstyle p(\theta |y)=p(y|\theta )p(\theta )\,/\,\int p(y|\theta )p(\theta ){\text{d}}\theta$, or is the issue here just the cancelling, or perhaps some scruple about upper case / lower case distinction, or something else..? -- Coffee2theorems (talk) 13:08, 1 July 2010 (UTC)

We may well start in the situation where the player has chosen door No. 1, and skip conditioning on this event. This is done in many textbooks. Further with a less space consuming notation, we get:

$\!P(C_{2}|H_{3})=$

$={\frac {P(H_{3}|C_{2})P(C_{2})}{P(H_{3}|C_{1})P(C_{1})+P(H_{3}|C_{2})P(C_{2})+P(H_{3}|C_{3})P(C_{3})}}=$

$={\frac {1\times {\frac {1}{3}}}{{\frac {1}{2}}\times {\frac {1}{3}}+1\times {\frac {1}{3}}+0\times {\frac {1}{3}}}}={\tfrac {2}{3}}.$

I do not see any advantage in canceling out the 1/3 on forehand. Nijdam (talk) 14:58, 1 July 2010 (UTC)

The first equality above is a common enough way to state Bayes's rule, so there's no need to derive the expression for the denominator separately anyway. The final calculation then becomes:

$P(C=2|H=3,S=1)={\frac {1}{{\frac {1}{2}}+1+0}}={\tfrac {2}{3}}.$

Also, I do not see why you define the posterior odds function O(...). Even if you use odds terminology, it doesn't really need extra notation here, as it's only used once. FWIW I think that "ratio of probabilities" is better than "odds" for these things, and saying e.g. "it's twice as probable that the car is behind door 2" is clearer than "the odds of the car being behind doors 2 and 1 are two to one" (or whatever the proper oddities-terminology is), but I can see some odds fans may beg to differ. Anyhow, you could put the odds equation this way:

${\frac {P(C=2|H=3,S=1)}{P(C=1|H=3,S=1)}}={\frac {P(H=3|C=2,S=1)}{P(H=3|C=1,S=1)}}={\frac {1}{1/2}}=2,$

and wouldn't have to do that ugly subscript "=x" thing just to get things fit in one display formula. -- Coffee2theorems (talk) 02:15, 1 July 2010 (UTC)

OK, here's a list of rationales for my proposal (sorry, I should have written them alongside the text at the very beginning):

1 - Only incremental changes to the current formulation allowed.

2 - Add some expansions to help the mathematically less conversant (think high-school student). In particular:

(a) show explicitly that P(C|S) = P(C) implies P(C|S) = 1/3

(b) rewrite the expansion of the marginal taking out the constant P(C|S) term, so to make it obvious later that it cancels.

3 - Add a formulation in odds form, since it's sourced (Rosenthal, allegedly, but see below) and one editor (Gill) has made a forceful case for its usefulness.

4 - Add an explanation for why the odds form is useful - no need to compute the marginal.

Given the above, here are my responses to the above (welcome) comments from Nijdam and CoffeeToTheorems:

a - @CoffeeToTheorems. I do prefer to write the first presentation of the Bayes rule in the standard simple form, with the marginal not yet expanded. Again, think mathematically-inclined high-school student who has just started reading prob. theory as your target reader (there's a reason why Jeff Gill's textbook presents the MHP in the exercises to the first chapter).

b - @Nijdam. We have already argued about whether the more appropriate formulation is with three-variables or two. Both are common and sourced (Jeff Gill: 2, Henze, 3). I think we can agree that having both would be silly. Personally, I prefer the 3 variable form since it more closely follows the steps in the problem description: a car is placed somewhere, the player selects a door, the host opens another door. It also makes it obvious that the door numbering is irrelevant (but necessary to write down a particular prediction), thus nipping that stray argument in the bud.

c - Given that the marginal needs expanding, it makes sense to show the elision of the 1/3 term in an intermediate step in the final derivation of P(C=2|H=3, S=1).

d - I am not in love with the subscript "=x", e.g. $\,C_{=1}$. However,

It is used in the literature.

I prefer it to a simple subscript ($\,C_{1}$) because it makes it obvious that it is an assigment, rather than the introduction of an indexed variable.

It takes less space than without the subscript ($\,C=1$) or the alternative "right bar" notation ($\,C|_{=1}$). Less space makes it easier to read (think cellphone or netbook screen).

e - @CoffeeToTheorems: Agree that the symbol for the posterior odds could be removed, and a sentence added at the end to the effect that the "odds in favor of switching are two to one".

f - I am not entirely sold on Gill's (the A.S.P.M., not Jeff Gill the author) argument for inclusion of an odds form. This proposal was made in response to his repeated and forceful argument (three reversions and keep on going is forceful enough for me), and his claim that it is well sourced. I don't have the Rosenthal book, and after reading the online article I agree that the odds form is not presented therein. Therefore, unless someone (Hello Gill, that's YOU) gives a proper source, I'd consider the whole argument for inclusion moot. However, I'd keep the other proposed changes in place, as they are minor and useful.

Again, thanks for the comments. glopk (talk) 16:16, 1 July 2010 (UTC)

Gill's initial comment

Interesting. A proof is a proof. IMHO, formal proofs are not necessarily better than verbal proofs. A good proof can be expressed equally well in words as in formulas. A proof which requires blind formula manipuation is not a very good proof since it does not enlighten you as to why the answer is true. It's a proof which can be verified by a computer. A good proof is one which can be easily generalized. It is not necessary to assume Prob(host opens 3| he may choose between 2 and 3) = 0.5 in order to show that switching is favourable. Why prove a special case for which one has to make a special assumption which no-one here except me seems to try to justify?

This proof is not a proof *by* Bayes theorem, but more or less from first principles. Essentially you are above proving Bayes theorem in a rather special case. I think it would be more enlightening to prove the odds form of Bayes theorem in the general case, as I did somewhere above (not very difficult, not very long!) and then use it to prove that switching is never unfavourable. No need to make the assumption that q, the probability that host opens door 3 when he must choose between 2 and 3, is 1/2. The posterior odds on Door 2 against Door 1 are 1:q which its always better than even.

If you are specially interested in the case q=1/2 the proof by symmetry and using that the unconditional probability that switching gives car is 2/3, is enlightening and lightening fast. It also leads up to a meaningful discussion about when this assumption is reasonable (i.e., on which interpretations of vos Savant's ambiguous formulation, and on which interpretation of probability).

My new proof (small addition to usual popular solution, for those who want mathematical proof that you can't beat 2/3 hence the strategy of always switching cannot be beaten), just posted above, can be converted into a "formal proof" of equal length. It does not need any assumption about host bias. It's very very simple, I think. Just common sense.

The advantage of the alternatives I mention here over a formal derivation from first principles is that they build on the popular solution and that they give the reader insight and they teach useful probability ideas. Remember, there already is a formal derivation from first principles on another page. And the odds form of Bayes is already on another page too.

I guess that formal derivations from first principles make some people feel safe. They make me uneasy. I'd rather see a proof using ideas which I could communicate to a lawyer or to a medical doctor, not a proof with formulas which make most people switch off, if not run away screaming.

Gill110951 (talk) 15:14, 30 June 2010 (UTC)

@glopk: That approach is fine with me. In general I very much favor to keep the formal proof in the article, since it useful complementation of the other somewhat less formal approaches. A complete formalization and using Bayes over a 3 step argument is something people often struggle with, so to spell it out in detail is imho beneficial for readers (not for all but for some).--Kmhkmh (talk) 15:46, 30 June 2010 (UTC)

Rosenthal doesn't actually use odds in his solution. His argument is essentially this:

Given a uniform prior, the posterior is proportional to the likelihood. Probabilities must also sum to one. Because of these facts and the usual MHP assumptions, the following table is correct.

	C=1	C=2	C=3
Likelihood	1/2	1	0
Posterior	1/3	2/3	0

He doesn't divide (unnormalized) posterior probabilities with each other (i.e. no odds) in the MHP solution, only unnormalized posterior probabilities with their sum. -- Coffee2theorems (talk) 23:57, 30 June 2010 (UTC)

Much ado about almost nothing

It seems no-one has digested my verbal solution of vos Savant's question "should you switch?" using *only* that initially all doors are equally likely (except that @Nijdam choked on one of the initial sentences - but he needs to go back to school. You can come by my home or office, Wietze, and we'll discuss this over a glass of beer. I'll treat you).

I may well turn up, hope to teach you something.Nijdam (talk) 18:40, 1 July 2010 (UTC) You did learn, as I discovered below. Nijdam (talk) 18:43, 1 July 2010 (UTC)

I love learning! Looking forward to the beer with you! Gill110951 (talk) 08:26, 2 July 2010 (UTC)

My solution builds on the simple solution to the simple problem. My point is that the simple solution is mathematically completed into a solution of the conditional problem, as soon as you can prove that you can't do better - unconditionally - than 2/3. Because to do better would mean to be able to identify a situation when the *conditional* probability that the initially-chosen door has the car is strictly bigger than 50%.

Everyone agrees that a "switcher" (a player who always switches) wins the car with unconditional probability 2/3, precisely because a "stayer" (a player who never switches) wins with unconditional probability 1/3. (Happy, Wietze? Yes, very!Nijdam (talk) 18:41, 1 July 2010 (UTC)). Now, no one has ever shown me or any of you a strategy with a better than 2/3 *unconditional* chance of winning. Some of you just "know" it can't exist (@Glkanter?). Some need to see a mathematical proof.

Well, I don't consider Carlton's solution 'just knowing'. Thanks for the shout out. I guess. Glkanter (talk) 21:05, 1 July 2010 (UTC)

It could only exist if there is some situation (particular initial door chosen by player, particular door opened by host) for which it would be (conditionally!!!) advantageous to stay!!!

Take the specific case mentioned by vos Savant (in an aside, and merely to help you visualize the problem - Whitaker didn't name any doors and vos Savant considered the simple solution complete).

Given player chose 1, the event {car is behind 1 and host opens 3} is contained in {car is behind 1} so has probability at most 1/3. The event {car is behind 2 and host opens 3} has probability exactly 1/3. The odds on car behind 1 to car behind 2 given host opens 3 (and player chose 1) are therefore at best 1:1. You can't strictly improve the overall success-rate by staying instead of switching - which is exactly equivalent to saying that all the conditional probabilities for car behind door 2 are at least 1/2. You should switch, whatever. You'll win unconditionally on average 2/3 of the time and you'll be at least as likely to win as a stayer, in each case separately. Stayers are dumb.

All I assumed was that initially all doors are equally likely.

OMG, what an ado about nothing. Gill110951 (talk) 17:39, 1 July 2010 (UTC)

What you seem to be missing is that everyone agrees that the simple solutions can be completed in various ways. Providing one more way is not really that interesting, unless it is sourced. The main problem with the completions is the lack of sourcing. Naturally, finding such a completion in some source would not change the fact that e.g. vos Savant's solution is incomplete as it stands, though.

It is not really essential to your argument above to compute the conditional probability, and since you have agreed before that answering Whitaker's question does not need a probability answer (it's a yes/no/doesn't matter question), I'll ignore that part. The essence is these two propositions:

1. Always switching is a strictly better strategy than never switching.
2. The new information the host provides by opening a door is not enough to help you to construct a better strategy than "always switch".

Your completion is proposition 2. Fair enough. This kind of argument is fine. Unfortunately, vos Savant's solution does not make this argument.

Vos Savant's incomplete argument goes: "Because proposition 1 is true, you should switch." Why is this incomplete? Well, because she hasn't shown proposition 2, so a better strategy might exist, where you sometimes switch and sometimes don't. Because proposition 2 is considerably more difficult to prove than proposition 1, this means that vos Savant's solution is only slightly closer to complete than a solution stating just "You should switch." Note that "considerably more difficult" here is relative - after all, this whole problem is elementary! But that doesn't mean that a solution that silently skips the hard part - such as it is - is somehow "complete". -- Coffee2theorems (talk) 22:03, 1 July 2010 (UTC)

@Gill - your conclusion here is essentially identical to that of Morgan et al. - if the car is uniformily distributed and the host must offer the opportunity to switch the player should switch. They show the conditional probability the car is behind door 2 is at least 1/2, specifically 1/(1+q) where q is the host's preference between two goats, and that you'll win unconditionally on average 2/3 of the time (i.e. a switcher will be at least as likely to win as a stayer). Sounds to me like you're a Morgan kind of guy. -- Rick Block (talk) 03:25, 2 July 2010 (UTC)

Rick, I should point out that even Morgan, Chaganty, Dahiya, and Doviak now agree that the answer is simply 2/3. Martin Hogbin (talk) 16:11, 2 July 2010 (UTC)

I essentially agree with all of you on many of your points, @Coffee2theorems and @Rick Block and @Nijdam, and I'm not ashamed to say that I have often changed my position on MHP in response to new points of view and new arguments which I learnt here from you all.

But I'm trying to say a little bit more. I'm saying that I don't believe that anyone could conceive of there being a strategy giving a *better* than 2/3 unconditional chance of winning. Therefore I understand why many quick and deep thinkers like vos Savant and @Glkanter "know" that vos Savant's solution is essentially completely complete. But OK, if you think that proof is needed that you can't improve on unconditional 2/3, then proof can be given. I think that mine is the shortest and most compelling yet. My point here is you don't have to compute the conditional probability and you don't have to make unwarranted assumptions, you just have to show the conditional probability is bounded. I have a strong feeling there should be an even shorter argument. Maybe it will come to me tomorrow during my morning cycle ride to work thinking about the latest news on MHP.

I'm also saying that my proof is so elementary that IMHO you don't need to source it; but in order to help you guys out, in case you're scared you'll get into trouble with other wikipedia editors or think I might be fooling you, I will revise the paper I wrote which had already been accepted but not yet published in a peer reviewed international mathematics journal. I was asked to add a bunch more references to the game theoretic/economics solution (some referees who clearly know the economics literature much better than me told me about them). Also I have to refer to the wonderful @Martin Hogbin and @Nijdam letter to The Am. Stat., and Morgan et al's fantastic resposne, including the original Whitaker letter. I'll get it published as soon as possible and ask all my friends to cite it in their next paper by writing "On a completely unrelated topic, Gill (2011) is a totally superfluous but not incorrect paper on the MHP". I will then quickly have the required citation statistics to be considered a reliable source by wikipedia rules. I've already made a point of adding this request on the last slide of all my conference talks (which are not on the MHP, by the way, but on forensic statistics or on quantum probability). At last then you guys will be able to cut all the crap out of the main page on the problem and keep it short and sweet. By the way I do acknowledge all wikipedia editors on this page for the stimulus and the fun which this has been.

But IMHO elementary mathematics comes with its own "reliable source" certification, namely the universal human facility for basic logic. We expect wikipedia editors to have basic numeracy and basic literacy. We don't ask them for a reliable source when they rewrite somebody else's claims in their own words. Why do we need that when they cut a corner in someone else's belaboured and ugly mathematics by noticing that it is all rather easy if only you notice that a number greater than or equal to a number x can never be strictly less than x ? Gill110951 (talk) 08:16, 2 July 2010 (UTC)

The GUT of the MHP

Silently assumed: quizmaster always reveals goat and always offers opportunity to switch. I assume the three doors are visibly numbered 1, 2, 3 in advance of the show.

Level 0. Popular solution:

If you have chosen Door 1 and if the car is initially equally likely behind every door, you should switch (unconditional odds of winning car converted from 1:2 to 2:1)

Level 1. Completion of popular solution:

Under the same conditions (you have chosen Door 1 and the car is initially equally likely behind every door) switching is the best strategy. i.e. You can't do better than unconditional odds 2:1 of winning car, or equivalently, conditionally on door opened by quizmaster, odds on car behind other door are always at least even.

Level 2. Game theoretic completion of previous solutions:

If you don't know for sure that the car is equally likely behind every door, ensure this yourself by randomization: choose your own door initially, 1 or 2 or 3, uniformly at random, and thereafter always switch.

Going from Level 0 to Level 1 there is a refinement or improvement in the conclusion. The conclusion is strengthened. It's not just better to switch, it's optimal to switch.

Going from Level 1 to Level 2 there is a weakening of the assumptions. No more assumptions are made at all, instead the player himself neutralizes all possible game playing by the host by randomizing his strategy. He doesn't have to assume the doors equally likely, he makes the doors equally likely.

BTW on probability interpretations: for a subjectivist understanding of probability, we are just interested in the player's initial feelings. If he initially has no preferences for any door, the three doors are equally likely, he may as well pick Door 1. Since he hasn't got any idea about the devious mind of the host, for him the host is equally likely to open Door 2 or Door 3 if the player's first choice is actually correct.

Ignorance is bliss. It guarantees all probability distributions uniform, it creates symmetry, the Door Numbers of specifically chosen doors become (subjectively) statistically independent from the two opponents' actions on those doors. Conditional and unconditional probabilities are the same and door numbers are irrelevant. Probabilities a priori and a posteriori truly are truly the same since the numbering of the doors is ignorable, superfluous.

However a clever player will not trust to his intuition. Maybe the host knows that subjectivists always pick Door 1 because foolish subjectivists think it doesn't make any difference. The clever player neutralizes this move of the host by choosing his initial door uniformly at random. The game theoretic approach is a true and valuable refinement to the previous levels' solutions. It also shows that the wiley host will in fact *hide* the car uniformly at random and open a door *uniformly* at random, so as to be ahead of all players, devious or not, subjectivists or objectivists, whether they have studied probabilty or economics or psychology. Gill110951 (talk) 08:53, 2 July 2010 (UTC)

B.S

This is complete B.S. If you want to make the case that one door out of three can magically have other probabilistic properties than the others, make it empirically. That is science, and that is what counts in Wikipedia. So, have anyone actually tried to always switch and see if they got better winnings? If so, why is it not in the article instead of all the paranormal crap? If not, why is there even an article about it? --62.199.168.98 (talk) 03:42, 7 July 2010 (UTC)

Carlton's Decision Tree

Please offer comments, criticisms, and questions.

Tree Showing Results and Probabilities As Per Carlton's (2005) Simple Solution to Whitaker's Question in vos Savant's Column.

Glkanter (talk) 19:21, 23 June 2010 (UTC)

I'm fixing to add this to Carlton's solution in the simple section. Any comments, etc. before I do so? Glkanter (talk) 10:53, 26 June 2010 (UTC)

This diagram does not match the diagram (Figure 5) in Carlton's paper. AFAICS this is original research. -- Coffee2theorems (talk) 14:16, 26 June 2010 (UTC)

That's correct, this tree does NOT support that solution. Rather, this tree supports the 'intuitive solution' he gives at the very beginning of section five, which is included in the Wikipedia article's Simple solution section. It is the only solution without a visual aid. The branching on car/goat rather than door's is Carlton's. The initial 1/3 and 2/3 come from Cartlton. The calculations of 1/3 * 100% = 1/3, and 2/3 * 100% are simple math. Everything else in the tree comes from Whitaker's letter, which Carlton includes in full earlier in the paper.

I don't agree, therefore, that there is any OR in this tree. Glkanter (talk) 14:26, 26 June 2010 (UTC)

For the solutilon you're talking about, Carlton doesn't say anything about "initial probability" or "probability after revealing a goat". What you're doing is caled synthesis, which is a form of OR - please see WP:SYNTHESIS. -- Rick Block (talk) 15:58, 26 June 2010 (UTC)

The words "say Door 1" and "say Door 3" should be removed from the diagram. Then it is is a perfectly good graphical representation of a rigorous proof of a perfectly reasonable mathematization of vos Savant's statement of the MHP and it was moreover authorized by herself. Both the proof and the formalization. If you want to call this OR then this means you are playing wikipedia in order to ban someone else's point of view which you don't like. (Or "whom" you don't like). Silly. I think we should sort out the facts first, and then figure out what we want to write in the article and how we are going to reference things later. There are a number of guiding principles for wikipedia and I believe in following them in the spirit. People who edit articles on basic probability questions ought to be at home enough in elementary probability calculus that they can use it to advantage to make a better article. Banning a simple correct understandable probability derivation because you can't copy it verbatim from an existing source seems to me quite nuts. Gill110951 (talk) 21:54, 4 July 2010 (UTC)

Carlton repeats Whitaker's letter in it's entirety in his paper.

Combining the premises of a puzzle with the solution is not WP:Synthesis. It's called 'visual presentation'.

Carlton certainly mentions the starting 1/3 and 2/3 probabilities as part of his 'intuitive solution'.

Wikipedia does not regard simple math as OR.

I find your objection unsupported by Wikipedia policies. Glkanter

Your diagram refers to a conditional probability ("Probability After Revealing a Goat"), even though Carlton's informal explanation, which the diagram is intended to illustrate, refers to no conditional probabilities at all. Further, the way this conditional probability is obtained is very odd. You said that it comes from your own calculation "2/3 * 100% = 2/3". Huh? The only conditional probability currently covered in the Wikipedia article is P(C = 2 | H = 3, S = 1), and this calculation is certainly not a justified way of computing that! I'm guessing that it is some other conditional probability instead. If so, it certainly should not be abruptly introduced in a diagram, but be covered in the article text before the diagram occurs. Of course you would also need a source mentioning that this new conditional probability is of interest in the context of the MHP. -- Coffee2theorems (talk) 22:55, 26 June 2010 (UTC)

Carlton's solution, like every other solution, relies on this, from the article: "The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it."

That's 100%. Glkanter (talk) 23:04, 26 June 2010 (UTC)

And it SURE DOESN'T say MUST BE DOOR #3, does it? Glkanter (talk) 23:05, 26 June 2010 (UTC)

The informal solution you're referring to here doesn't name which door is opened, or which door remains closed. Your figure does. This is a problem. We've been through this before in table format. The following table would be OK:

Probability	Player pick	Door the host opens	Remaining door
1/3	Car	Goat	Goat
2/3	Goat	Goat	Car

but a table or figure that is trying to imply this simple solution says the probability of door 2 being the car after the host opens door 3 is misrepresenting what this solution says. -- Rick Block (talk) 04:04, 27 June 2010 (UTC)

Any illustration that includes probabilities or frequencies seems (almost?) counterproductive, because the bulk of Carlton's argument is non-probabilistic, and that's its best feature because people have problems with probabilities! First Carlton assumes that you always switch, and then goes on to show using only classical logic that the proposition "you win" is the same as the proposition "you initially pick a goat". Then he notes the obvious(!!) consequence that P(you win) = P(you initially pick a goat) = 2/3. Sure, maybe you could illustrate the classical logic part using a truth table or something, but I don't think you can actually beat vos Savant's table that way. Vos Savant's table actually doubles as an illustration of Carlton's argument, and seems like par for the course. -- Coffee2theorems (talk) 11:40, 27 June 2010 (UTC)

Regarding the math, and Wikipedia policies, is there anything wrong with that tree? Glkanter (talk) 12:08, 27 June 2010 (UTC)

This has already been explained to you. The tree is WP:OR. The columns labeled "probability after revealing a goat" and "remaining: door 2" are not from anything in the source. -- Rick Block (talk) 14:04, 27 June 2010 (UTC)

Yes, I've already explained to you, that as always, you are misinterpreting Wikipedia policy, and ignoring a substantial part of Carlton's writings in his article. Glkanter (talk) 14:47, 27 June 2010 (UTC)

Regarding the math, "1/3 * 100% = 1/3" isn't really a correct way to compute a conditional probability (i.e. a "probability after"). It's an expression for a joint probability (i.e. a probability of "A and B"), where one of the events is certain.

The whole idea of conditioning on a certain event to get "Probabilities After Revealing a Goat" is ludicrous, because conditioning on a certain event does not change any probabilities whatsoever. In particular, it does not change the probability that the car is behind the door the host just opened, so even that is still 1/3. So you are staring at a door the host has opened, see plain as a day that there's a goat there, and think "the probability that that thing is really a car is 1/3". That's not right. -- Coffee2theorems (talk) 14:38, 27 June 2010 (UTC)

Congratulations! You finally understand the MHP. All it took was Carlton's tree for you to get it. Monty's action doesn't change the initial probability that the contestant chose a car or goat. Glkanter (talk) 14:47, 27 June 2010 (UTC)

You're not comprehending the point here. What Coffee2theorems is saying is that the "condition" that the host opens a door (which happens with 100% certainty) changes none of the original probabilities. Each door, even the one the host opens, still has a probability of 1/3. If you say the "probability" of the opened door is 0 you're talking about conditional probabilities given that the host has opened that specific door. Rick Block (talk) 15:26, 27 June 2010 (UTC)

It seems that you misread me. Rick got it right. When I said "the probability that that thing is really a car is 1/3", the phrase "that thing" referred to the goat that is standing before your very eyes, freshly revealed by the host. When you are staring at that goat, and think "that goat is in reality a car with probability 1/3", you're not thinking straight. Of course the probability that that goat is a car is zero. The probability of anything that is impossible is always zero. But if you think that it makes sense to compute "probabilities after" by conditioning on a certain event, you will necessarily conclude that the goat you are staring at is actually a car with probability 1/3. That's nonsense, and that's why the whole idea of conditioning on a certain event to compute "probabilities after" is nonsense. -- Coffee2theorems (talk) 15:59, 27 June 2010 (UTC)

Discuss it with Carlton. Or Selvin. Or Devlin. Or Adams. Or vos Savant. Or Rosenthal. They're the ones who published simple solutions. I'm just trying to give a visual aid to Carlton's solution, based on Whitaker's letter and Carlton's paper. No different than your [redundant] 'extended figure' for the conditional solution. Glkanter (talk) 15:37, 27 June 2010 (UTC)

Unreferenced Figure In The Conditional Solution section

Rick Block, it looks like you uploaded the various car/goat/door images in the 'expanded figure below', so I'll presume that you are the originator of that overall figure.

Neither the narrative or the figure are referenced. Can you please provide a source for this text and figure? Otherwise, I'm sure you would agree that it must be considered OR. Please advise.

"The conditional probability of winning by switching given which door the host opens can be determined referring to the expanded figure below, or to an equivalent decision tree..."

Car hidden behind Door 3	Car hidden behind Door 1		Car hidden behind Door 2
Player initially picks Door 1
Host must open Door 2	Host randomly opens either goat door		Host must open Door 3
Probability 1/3	Probability 1/6	Probability 1/6	Probability 1/3
Switching wins	Switching loses	Switching loses	Switching wins
If the host has opened Door 2, switching wins twice as often as staying		If the host has opened Door 3, switching wins twice as often as staying

Glkanter (talk) 19:12, 26 June 2010 (UTC)

The source is Chun and Grinstead and Snell (as it says in the article). The figure is in all respects identical to Chun's diagram, which is a simplified version of Grinstead and Snell's diagram. The text is paraphrased from Grinstead and Snell (and innumerable other sources). -- Rick Block (talk) 04:04, 27 June 2010 (UTC)

Here's the Grinstaed and Snell pdf: www.math.dartmouth.edu/~prob/prob/prob.pdf. Which diagram and page, please. I didn't see anything like your figure. Glkanter (talk) 05:16, 27 June 2010 (UTC)

Figure 4.3, page 138. The first row of the figure corresponds to the initial 3-way branch (car hidden behine each door with probability 1/3). The next row corresponds to the three branches "contestant picks door 1" branches in shown in the Grinstead and Snell figure (like I said, the figure is a simplified version). The next row (with the images) shows theis situation visually. The next row lists the possibilities from this point, corresponding to the available forks from the previous nodes in G&S's diagram. The next row (with the images) shows this visually. The next row shows the total probabilities (from Chun's version of the same diagram, which is simplified in exactly the same way as the figure you're challenging). -- Rick Block (talk) 14:04, 27 June 2010 (UTC)