Template:Editnotices/Group/Wikipedia talk:Requests for mediation

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Conditional probability solution

Groundrules

In most of the successful mediations I've been involved with, there is a transition from dispute to collaboration. The best mediations involve learning. One thing we can all learn more about is how to navigate through the maze of WP policies. It is usually beneficial to agree on some "first principles" that will promote a shift to collaborative editing. In my opinion it will improve our chances of a successful outcome if all participants sign their agreement to the following principles:

Focus on content rather than the contributor. Note: This is to be interpreted literally, as worded.
Be guided by WP content policies, particularly WP:V and WP:NPOV
Commit to being as economical as possible in posts to this discussion page.
Work towards consensus in editorial decisions.

Easy right? If you have any questions or comments, by all means start a new section below the signatures. Please signify your agreement below:

Agreements

Agree Rick Block (talk) 23:09, 10 August 2010 (UTC)[reply]
Agree Glkanter (talk) 23:47, 10 August 2010 (UTC)[reply]
Agree--Kmhkmh (talk) 07:26, 11 August 2010 (UTC)[reply]
Agree Martin Hogbin (talk) 08:10, 11 August 2010 (UTC)[reply]
Agree Colincbn (talk) 09:56, 11 August 2010 (UTC)[reply]
Agree Nijdam (talk) 17:37, 11 August 2010 (UTC)[reply]
Agree glopk (talk) 21:37, 11 August 2010 (UTC)[reply]
Agree Gerhardvalentin (talk) 23:41, 14 August 2010 (UTC)[reply]
Agree Gill110951 (talk) 07:47, 15 August 2010 (UTC)[reply]

General discussions

Wikipedia talk:Requests for mediation/Monty Hall problem/General discussion

Proposed compromises

Wikipedia talk:Requests for mediation/Monty Hall problem/Compromises

Merry Christmas

I'll be away for some days, only to return just before Christmas. I take this opportunity to wish you all, AGK, Gerhardvalentin, Glkanter, glopk, Kmhkmh, Martin Hogbin, Richard Gill, Rick Block, Sunray, a Merry Christmas and the best wishes for the coming year. Nijdam (talk) 22:43, 15 December 2010 (UTC)[reply]

I would also like to wish everyone a happy Christmas and best wishes for the coming year. Here is my Christmas present, it's for Nijdam and Rick. Rick said that he thought that the reason you must compute a conditional probability is because of the moment when the decision has to be made to change. Nijdam also has this point of view: the conditional viewpoint corresponds to the player who makes his decision after he has seen Door 3 opened. However, probability theory is not normative, it does not prescribe what you *must* do. However, there are some very good reasons why it is *rational* to compute the conditional probability "at that moment of time". Here they are: I will give two reasons, one for each of the main streams of thought:

Subjectivist: for the subjectivist, information comes in and beliefs about the state of the world are continually updated. If we have modelled the MHP as a four stage process going on in time (car hidden, door chosen, door opened, reconsideration of choice) and if we are a rational subjectivist then our beliefs are updated in each phase: car hidden -> 1/3,1/3,1/3 -> door 1 chosen -> 1/3,1/3,1/3 -> door 3 opened -> 2/3,1/3,0. Subjectivist probability theory is a theory of how a "rational person" *must* adjust his degrees of belief about unknowns as new information is received. Our rational degree of belief at any stage is *the* conditional probability distrbution given *all* that we know at each stage.

Objectivist: for the objectivist, there are no normative rules coming from probability theory. But in a decision problem like MHP it is agreed that the problem is solved by finding the *optimal* strategy, and by *proving* that it is optimal. The overall probability of success is the target quantity which we want to optimize. The simple arguments of the simple solutions tell you that always switching has overall (unconditional) success probability 2/3. We have not finished "solving" MHP if we haven't mathematically proven that 2/3 can't be beaten. The only way it could be beaten, would be if there was some situation recognisable to the player (door chosen, door opened) in which the conditional probability that switching gave the car was strictly less than 1/2. This follows easily by consideration of the law of total probability: for any strategy whatsoever, the overall success chance equals the sum over the six possible situations of the probability of each situation multiplied by the conditional probability of success with that strategy, given you are in that situation.

I would suggest to the conditionalists that they search for reliable sources which give these arguments, in order to make their case stronger. The reliable sources from probability theory and statistics which write on MHP don't give these reasons explicitly, because their intended audience already knows them, often instinctively rather than explicitly; and also because sources usually try to express themselves "probability interpretation free", in order not to get into heated controversy. Richard Gill (talk) 04:36, 18 December 2010 (UTC)[reply]

Thanks Richard, for your good wishes and Xmas present. I especially invite Martin, Gerhard and G. to study it. I also want to emphasize that the solutions presented in the sources, and referred by us (me cs.) as "simple solutions", are NOT the solution using the symmetry to calculate the conditional probability. See for instance what Devlin writes as solution, or the solution known as the combined door solution. These solutions are mathematically wrong, and hence have to be criticized. Nijdam (talk) 16:14, 30 December 2010 (UTC)[reply]

I hope you had a good Xmas Nijdam. All solutions, including the simple solutions and the conditional solutions in the article, can be subject to criticism. The point is, when and how do we criticise them? My suggestion is the we present the solutions (in particular the simple solution) first, complete with explanation, and later on we criticise them in a sober and scholarly manner. The reason for this is simple. Most people who come to the article will find the hardest thing to understand and believe is that the answer is 2/3 and not 1/2, regardless of whether the problem is conditional or not. First we should explain why this is so.

I have used the example of good text books (and encyclopedia articles) before to explain this approach. It is how almost all such sources do things. Simple, maybe glossing over some details, first, then a more in-depth discussion. You have never explained what you have against this approach, Nijdam. Martin Hogbin (talk) 17:20, 30 December 2010 (UTC)[reply]

I also like a "simple explanation", as long as it mentions something about the probability before and after the opening of a door by the host. I.e. it could mention that the initial probability for door No. 1 to hide the car is not influenced and hence the probability after door No. 3 is opened also is 1/3. That would be correct reasoning, other than the simple solution does. Nijdam (talk) 17:42, 30 December 2010 (UTC)[reply]

Why not? So long as we can do this in a way that does not draw unnecessary attention to it in a way that might put off non-expert readers. Why do you not propose something that could be added to the simple solution. Not to make or prove a point but just to make it more correct. Martin Hogbin (talk) 20:13, 30 December 2010 (UTC)[reply]

Marilyn vos Savant's original simple solution was, as she explicitly confirmed later, from the outset based on a host who, in opening of a door, would certainly not give away any closer hint on the actual location of the car. So, by the host's opening of a door, we've learned absolutely nothing to let us revise the odds on the door originally selected by the guest (1/3). In this version they remain unchanged. As much to the odds on the car originally selected by the guest, before and after the opening of a door by the host. Independent of the door No. Gerhardvalentin (talk) 23:08, 30 December 2010 (UTC)[reply]

There are several sources which make this same point. Martin Hogbin (talk) 23:20, 30 December 2010 (UTC)[reply]

Well, how can opening of a door with a goat, not tell anything about the location of the car? I hope I do not have to explain this further.Nijdam (talk) 08:52, 31 December 2010 (UTC)[reply]

Nijdam, you are just being awkward now. It is quite obvious that showing a goat behind door 3 tells us the car is not behind door 3. The point, that you make yourself above, is that (in the symmetrical case) this does not alter the probability that the car is behind door 1. How would you wish to make his point (as you propose above) in the article? Martin Hogbin (talk) 10:15, 31 December 2010 (UTC)[reply]

Who is awkward here? Yes, the car is not behind door 3, yet the probability the car is there is, due to the random placement, 1/3!! Got it?Nijdam (talk) 11:00, 31 December 2010 (UTC)[reply]

You continue to quibble about a simple mistake but make no attempt to engage in cooperative editing. You yourself say that, in the symmetrical case, 'initial probability for door No. 1 to hide the car is not influenced' yet you will not discuss how we might add something to that effect to the article in a way that might be acceptable to everyone. Martin Hogbin (talk) 11:06, 31 December 2010 (UTC)[reply]

Sorry Martin, I'm definitely willing to formulate a correct "simple explanation". But all in its proper time. For the moment I do not understand what you refer to as "simple mistake". My efforts here still are mainly to make you, and I hope Gerard too, understand why the simple solution is wrong, because of the lack of the right arguments.Nijdam (talk) 00:59, 1 January 2011 (UTC)[reply]

The simple solution gives a clear argument for the correct answer to the MHP-question "is it to your advantage to switch?", i.e. for the decision asked for: "YES or NO". – The answer clearly has to be YES, and the decision is SWITCH, for staying never can be better. – Any correct conditional approach gives a "correct conditional result", but no correct conditional result has ever been able to proof the contrary, to proof that staying could ever be a better decision, in any specific game-show, if there should have been more than one.

So it is more important to criticize the illusion that, under the reasonable assumptions, a "conditional probability" is needed as a "better basis" for the decision asked for. And that a conditional approach really is indispensably necessary as a "better" basis for that decision. Such illusions are missing the point and therefore fail to heed. And so I somehow agree to Martin and to W.Nijdam. Regards, Gerhardvalentin (talk) 00:23, 31 December 2010 (UTC)[reply]

I don't understand what you mean to say here. The simple solution says "switch", but based on the wrong argument. And it does not say: "for staying never can be better", whatever that should mean. And then, why should a conditional approach prove that staying could ever be a better decision? Where is your logic? Nijdam (talk) 09:05, 31 December 2010 (UTC)[reply]

W. Nijdam: The argument of the simple solution never is that it is guaranteed to win by switching. Never guaranteed.

>>No one said so, so why do you mention this?

The simple solution just says: Pws is (only) 2/3, on the long run.

>>No such thing as "on the long run". You confuse probability with its interpretation.

That's enough.

>>Enough for what? Certainly not enough to base your decision on.

Therefore it cannot ever be of any advantage to stay.

>>In the specific situation you do not yet know this.

Based on this argument you never should stay, but always should switch.

>>The word never is not appropriate here. And the argument fails.

And this argument is confirmed to be correct by any correct conditional approach.

>>No, the conditional approach gives the correct argumentation, the simple explanation does not.Nijdam (talk) 15:19, 31 December 2010 (UTC)[reply]

Even any tattletale host is out of position to make staying more advantageous, even if you should (but you cannot!) "know" the exact conditional probability in any single act. Even any tattletale host cannot avoid that Pws will be 2/3, on the long run. Never. That proves the argument of the simple solution to be correct, and no conditional approach can ever contradict this implication of the simple solution. You always should switch, and staying never can be "of any advantage". Never. Absolutely never. Gerhardvalentin (talk) 11:13, 31 December 2010 (UTC)[reply]

Gerhard, I have to agree with Wietze here. Indeed, the simple solution tells us that "always switching" has an overall success chance of 2/3. inspection of the conditional probabilities shows us that this couldn't be improved by sometimes switching, sometimes not, depending on the configuration numbervof door chose, number of door closed. The simple solution only compares "always switching" with "always staying". But there are other possible strategies. BTW, by "simple sokution" I mean any correct argument that always switching has success probability 2/3. I think that by "simple solution", Wietze usually refers to some incorrect argument. Fortunately he does now agree that there exist mathematically correct proofs that "always switching" has overall success probability 2/3. Richard Gill (talk) 16:29, 31 December 2010 (UTC)[reply]

I have to correct Richard here: I never was of a different opinion. There are several proofs concerning the overall probability, being 1/3 under the right assumptions. Like Morgan, I consider this a correct solution to a slightly different problem, but not appropriate for the usual MHP. Nijdam (talk) 00:41, 1 January 2011 (UTC)[reply]

Let's Get This Party Started!

I propose the editors supporting the extensive discussions of the differences between the solutions and the limitations of the simple solutions so prominently featured in the Wikipedia MHP article present the reliable sources that comprise the majority viewpoint of the literature in the field. Along with some text and links that everyone can read and interpret for themselves.

Should they fail in this simple requirement for maintaining the prominence of this POV in the article, I suggest that stuff get reduced in prominence to the tenuous significant minority viewpoint status, or less, that it deserves in the article, as per long standing Wikipedia policies. Glkanter (talk) 19:16, 18 January 2011 (UTC)[reply]

The Wikipedia Editors' Remaining Criticisms Of The Simple Solutions Are?

So, we know from reliable sources that the simple solutions can solve the conditional problem where door 3 has been opened.

So, what's the problem that *still* requires?:

• Health warnings

• Extensive discussions of problems different than the MHP, that have host bias premises other than 50/50, in the Conditional solution section before the conditional solution has even been presented

• Morgan's 'always =>1/2 and averages 2/3' 'correct resolution' in the conditional solution section right after the decision tree solution, which is also a variant without the 50/50 host premise

It seems to me we are addressing just 3 issues now:

One or more editors' insistence that the verbiage supporting the simple solutions from the reliable sources are incomplete
One or more editor's insistence that the article must take the POV that the simple solutions are wrong
One or more editor's insistence that the article must take the POV that the simple solutions solve the wrong problem (are false)

This seems simple enough. Those editors that are arguing the above should present their reliable sources. But they don't have any.

This is OR. There are no reliable sources that take issue with the manner that simple solutions are presented. I'm not even sure this POV is in the current article, but it serves to stalemate the mediation. This pseudo issue should be eliminated from the mediation, and kept out of the the article.
There is one editor in the mediation who is using OR to say the simple solutions are wrong. Period. There are no reliable sources that say this. Again, this pseudo issue should be eliminated from the mediation, and kept out of the the article.
From my readings, only one source ever called the simple solutions false or wrong. That was Morgan, and they based their claim entirely on vos Savant's failure to include the 50/50 host premise in her column. The first problem is that Morgan calls the conditional solutions false for the same reason.

So, if the article *is* going to emphasize this POV based on the writings of 1 source, as it currently does, the article needs to make it clear that both the simple and conditional solutions, including the Bayes Theorem solution are false. The article currently only diminishes the simple solutions for this. Here's how that is accomplished: Instead of omitting the 50/50 premise, the article contrives new host bias premises, creating problems that are different than the MHP, that require the conditional solutions, and that the simple solutions cannot handle.

But, of course, this POV is not a significant minority viewpoint of the reliable sources, and the article currently suffers from substantial UNDUE WEIGHT, NPOV and OR violations. These violations are first seen in the 4 paragraphs in the Conditional solution section, 3 of which are entirely about premises that are different than the 50/50 host. This is of no value to the reader, and these same points are made many more times throughout the article.

As far as I can tell, this is what the stalemate is all about. It's time to eliminate:

• The OR argument of 'the simple solutions are not described properly by the reliable sources' from this mediation

• The OR argument of the 'simple solutions are wrong' from the mediation and the article

• The fallacy argument that 'here are variants where the simple solutions don't work' from the mediation and as a criticism of the simple solutions

Posted by Glkanter (talk) 12:39, 19 January 2011 (UTC)[reply]

The sources that say the simple solutions are incomplete or solve a different problem include at least

Morgan (full citation in the article)
Gillman (full citation in the article)
Rosenthal Monty Hall, Monty Fall, Monty Crawl
Grinstead and Snell (full citation in the article)
Eisenhauer The Monty Hall Matrix
Lucas, Rosenhouse, and Schepler The Monty Hall Problem, Reconsidered
Falk (full citation in the article)

What reliable sources say "the simple solutions can solve the conditional problem where door 3 has been opened"? Please name them (be specific). -- Rick Block (talk) 13:37, 19 January 2011 (UTC)[reply]

Rick, so that we don't argue in circles, could you please include the actual text from those sources? And could you please group your sources by criticism, in a manner similar to the way I listed the 3 remaining criticisms? Thanks. Otherwise, these sources support my statement:

• Morgan's rejoinder specifically states that due to symmetry

• G & S specifically states that due to symmetry

• Carlton provides both solutions

• Selvin provides both solutions

• vos Savant

• Every source that describes the problem with door 3 open and provide a simple solution, or a simple solution and a conditional solution, is using a simple solution to solve the conditional problem

• Fundamental principals of probability as per Boris Tsirel and Richard Gill

• Wikipedia editor Boris Tsirel states that due to symmetry

• Wikipedia editor Richard Gill states that due to symmetry

Posted by Glkanter (talk) 13:50, 19 January 2011 (UTC)[reply]

Further, I challenge at a first glance, your inclusion of the following as those who say the simple solution does not solve the door 3 opened problem:

• Morgan

• Rosenthal

• Eisenhauer

• Lucas, Rosenhouse, and Schepler

And I will research the others promptly. Glkanter (talk) 14:08, 19 January 2011 (UTC)[reply]

Postings here by Wikipedia editors Boris Tsirel and Richard Gill are not reliable sources, moreover I think at least Richard does not agree with you (per extensive discussions not on this page). I'm not sure how you want these grouped, but here are direct quotes:

Morgan et al.: Ms. vos Savant went on to defend her original claim with a false proof and also suggested a false simulation ...

Morgan et al.: Solution F1: If, regardless of the host's action, the player's strategy is to never switch, she will obviously will the car 1/3 of the time. Hence, the probability that she wins if she does switch is 2/3. ... F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand.

Morgan et al.: Solution F2: The sample space is {AGG, GAG, GGA}, each point having probability 1/3, where the triple AGG, for instance, means the auto behind door 1, goat behind door 2, and goat behind door 3. The player choosing door 1 will win in two of these cases if she switches, hence the probability that she wins by switching is 2/3. ... That it [F2] is not a solution to the stated conditional problem is apparent in that the outcome GGA is not in the conditional sample space, since door 3 has been revealed as hiding a goat.

Gillman: Marilyn's solution goes like this. The chance is 1/3 that the car is actually at #1, and in that case you lose when you switch. The chance is 2/3 that the car is either at #2 (in which case the host perforce opens #3) or at #3 (in which case he perforce opens #2)-and in these cases, the host's revelation of a goat shows you how to switch and win. This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3.

Rosenthal Monty Hall, Monty Fall, Monty Crawl: This solution [what he calls "Shaky Solution" which basically says your original chance of selecting the car is 1/3 and this doesn't change since you knew the host would open a door revealing a goat] is actually correct, but I consider it "shaky" because it fails for slight variants of the problem.

Grinstead and Snell: This very simple analysis [as a preselected strategy, staying wins with probability 1/3 while switching wins with probability 2/3], though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3. To solve this problem, we set up the problem before getting this information and then compute the conditional probability given this information.

Eisenhauer: ::When asked the probabilities of winning by switching and not switching, the vast majority of respondents answer that with two doors unopened and a prize behind one of them, the doors are equally attractive: the chance of winning is 1/2 in either case. On the other hand, it is sometimes argued that since one knows in advance that Monty will reveal a zonk, the revelation provides no relevant information to the contestant (see, for example, Nalebuff 1987). In this view, which Falk (1992) calls the "no news" argument, the probability that the prize is behind the chosen door is 1/3, just as it was a priori, whereas switching doors gives a 2/3 probability of winning.

The clash of these two perspectives led to something of a public uproar when columnist Marilyn vos Savant (1990a,b,1991) discussed the problem in a series of magazine articles, and asserted that the probabilities of winning are 2/3 for switching and 1/3 for not switching. Some 10,000 readers, including many irate teachers and college faculty, responded with letters, 90% of them vehemently disagreeing with vos Savant (and some personally denouncing her) in favour of the 50-50 odds (Tierney 1991). Adding to the confusion, vos Savant then gave a 'dubious analogy . . . [and] went on to defend her original claim with a false proof and also suggested a false simulation as a method of empirical verification (Morgan et al. 1991, p. 284). Consequently, what could and should have been a correct and enlightening answer to the problem was made unconvincing and misleading. Subsequent work by Gillman (1992) and Falk (1992) applied the correct Bayesian mathematics to derive the general solution, but several other authors continued to perpetuate the "no news" argument, which at best relies on an unstated assumption (see, for example, Engel and Ventoulias 1991; Gilovich et al. 1995).

Lucas, Rosenhouse, and Schepler: This [the "high numbered Monty" variant - Monty opens the highest numbered door without revealing the car] shows that any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete.

Falk (this is actually from her book [1], but the paper says effectively the same thing): Truly, Monty can always open one of the two other doors to show a goat, and [emphasis in original] the probability of door No. 1 remains unchanged subsequent to observing that goat, still, it is not because [emphasis in original] of the former that the latter is true. The probability of winning the car by sticking with door No. 1 remains unchanged due to a specific combination of priors and likelihoods characterizing this problem.

-- Rick Block (talk) 14:28, 19 January 2011 (UTC)[reply]

Rick, is the 50/50 host premise a part of the 'standard' MHP? Are there sources, in addition to Morgan, who give an answer other than 2/3 & 1/3 with no host bias premise whatsoever? Would you agree that the host bias premise must be either 50/50 (Selvin,K & W) or unknown (Morgan)? Glkanter (talk) 14:42, 19 January 2011 (UTC)[reply]

Of what relevance are these questions? You asked for sources supporting specific claims that you want removed from the article. These sources are clearly talking about what they consider to be the Monty Hall problem, they are all extremely reliable sources (predominantly peer reviewed academic articles), and they say almost to the word precisely what you want removed from the article (i.e. that the simple solutions do not quite address the door1/door3 problem).

What is your suggestion for how to "fairly, proportionately, and as far as possible without bias" (per WP:NPOV) represent the views expressed by these sources in the article? -- Rick Block (talk) 15:17, 19 January 2011 (UTC)[reply]

First, we have to know in what way they say the solution is flawed. Then we can determine how prominent each criticism is, and how relevant each criticism is to the MHP as described in the article. This would help us determine whether a criticism should be addressed in the solution section, or elsewhere in the article. Glkanter (talk) 15:26, 19 January 2011 (UTC)[reply]

I guess the criticism categories would be:

Solves wrong problem

Omits 50/50 premise

Inadequate description of why symmetry allows the simple solution to solve the conditional problem

Fails for variants

They're just wrong

Please add any other you think relevant. Thanks. Glkanter (talk) 14:52, 19 January 2011 (UTC)[reply]

Morgan, Gillman, G&S all directly say the simple solutions solve the wrong problem (i.e. does not address the conditional probability where we know the player has picked door 1 and the host has opened door 3). Rosenthal, Eisenhauer, and Lucas et al are saying the same thing, but somewhat less directly. There are scores of others (mostly academic mathematical sources) that present only conditional solutions - these are implicitly agreeing that the conditional approach is the "right" approach and can be construed to be indirectly agreeing with the criticism that the simple solutions solve the wrong problem. You disagree with this, but I consider Carlton to be in this camp because he explicitly distinguishes what he considers to be his "solution" (which is conditional) from his "intuitive explanation" (which is one of the simple solutions).

Rick, you wrote:

"There are scores of others (mostly academic mathematical sources) that present only conditional solutions - these are implicitly agreeing that the conditional approach is the "right" approach and can be construed to be indirectly agreeing with the criticism that the simple solutions solve the wrong problem."

I think you are putting words into people's mouths. Maybe it's a chapter on conditional probabilities, and the source didn't want to interrupt the chapter with a comment on symmetry? Maybe he's not aware of the simple solutions. Which category of criticism would you attribute such a 'silent criticism' to? Is that really a valid method of determining that a significant minority of sources (or more) hold the viewpoint that the simple solutions are flawed, when they made no comment whatsoever? I don't buy it. At all. Is the converse true, then? Are the sources that give simple solutions implicitly criticizing the conditional solutions? Why not? Glkanter (talk) 16:25, 19 January 2011 (UTC)[reply]

Rick, your explanation and conclusion are flawed from a logic and Wikipedia standpoint.

What they actually write: Here is my conditional solution to the MHP

What you conclude they mean: All simple solutions to the MHP are flawed.

No. That's not right. Glkanter (talk) 17:51, 19 January 2011 (UTC)[reply]

Morgan criticizes even conditional solutions that assume without it being given in the problem statement that the host's choice between goats is 50/50. I don't recall any other source doing this.

Other than Falk (and the discussions on Wikipedia talk pages), I'm not aware of any source that makes a point about symmetry.

Many sources (Morgan, G&S, Rosenthal, Eisenhauer, Lucas et al, Falk) specifically use variants to show that the simple solutions indeed solve a different problem than the door 1/door 3 problem (because, for these variants, the simple solutions still say the answer is 2/3 chance of winning by switching even though for these variants this is not the correct answer for the door 1/door 3 case). These sources are not criticizing the simple solutions because they fail to address these variants, but rather using these variants to demonstrate (prove) that the simple solutions are not addressing the conditional problem. -- Rick Block (talk) 16:15, 19 January 2011 (UTC)[reply]

With symmetry, the simple solutions, which I will call 'shortcuts' *do* solve the conditional problem. But, of course, they don't generalize well, and in many other cases the conditional solution would be needed. But that does not make their application to the MHP wrong. Countless sources use these shortcuts, and make no comment about their suitability for other problems. I think this is where the article veers from being about the MHP into being a course on probability. Glkanter (talk) 16:43, 19 January 2011 (UTC)[reply]

No, they don't. Richard Gill has repeatedly tried to explain this to you (on other pages). Since a professor of statistics is apparently unable to convince you I don't see much point in anyone else trying. The bottom line here is it doesn't matter whether you agree. Numerous reliable sources say the simple solutions do not address the conditional problem. This is not a fringe or minority viewpoint, but the mainstream academic opinion published in numerous sources (and contradicted by none, at least none that I'm aware of). If you can find a reliable source (not Wikipedia talk pages) that says otherwise, please cite it. -- Rick Block (talk) 17:49, 19 January 2011 (UTC)[reply]

The Simple Solutions Solve The Conditional Problem

Morgan

Here's what Morgan said in their rejoinder to vos Savant in the same issue as their letter:

"We do find it interesting that, in the simulation described in the February article, an explicit randomization technique is described for every stage of the process except the choice of which door to open when the player's initial choice contains the car. From this and her previous solutions, one is tempted to conclude that vos Savant does not understand that the conditional problem (of interest to the player) and the unconditional problem (of interest to the host) are not the same, and that 2/3 is the answer to the relevant conditional problem only if p = q = 1/2. Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one. But then, it may be that only an academician, and no one connected with a game show, would ever consider p # q."

This is consistent with all the sources that offer simple solutions to the symmetrical conditional problem. Glkanter (talk) 17:01, 19 January 2011 (UTC)[reply]

This by no means contradicts their claim that the simple solutions address the unconditional problem (for which the answer is 2/3 regardless of p and q) rather than the conditional problem. Indeed, this is the very point they're making here, i.e. the simple solutions say the answer is 2/3 without considering p and q when the answer is 2/3 only if p = q = 1/2. A simple solution that says it is premised on p = q = 1/2 (as Selvin clarified in his 2nd letter) is correct (or, more properly, results in the correct numeric answer), but a simple solution that blithely ignores how the host chooses (e.g. vos Savant's - despite repeated revisits to the problem and extended discussions in response to thousands of letters) is not. This is precisely the same point all of the other sources I've quoted make. -- Rick Block (talk) 17:37, 19 January 2011 (UTC)[reply]

No, Morgan didn't criticize any simple solution's presentation. They criticized them all, and the conditional solutions, for solving the wrong problem. Then, in the bold type, above, explain that with the 50/50 host premise, they *are* solving the conditional problem. Glkanter (talk) 18:00, 19 January 2011 (UTC)[reply]

No, they are again saying that symmetry allows you to easily deduce the answer to the conditional probability question from the answer to the unconditional probability question in the unbiased host case. They never leave their standpoint that MHP must be solved by finding the conditional probability and that vos Savant (just as Glkanter, it seems) is apparently incapable of being able to distinguish between the two concepts. Richard Gill (talk) 15:29, 20 January 2011 (UTC)[reply]

And, Richard, do you agree with them? Nijdam (talk) 00:53, 21 January 2011 (UTC)[reply]

Boris' Final MHP Revision In His Sandbox

I include Boris' work here in order to show, from a qualified subject matter expert, that the symmetry argument is a recognized mathematical concept for simplifying a conditional problem. While the proof may be Boris' OR, the underlying principle is not.

This is an old revision of this page, as edited by Tsirel (talk | contribs) at 10:19, 7 December 2009. It may differ significantly from the current revision.

Notation:

Three doors with numbers 1,2 and 3
C is the random variable indicating the number of the door with the car
X is the random variable indicating the number of the door chosen by the player
H is the random variable indicating the number of the door opened by the host

Assumptions:

C is uniform on the set {1,2,3}
C and X are independent
H is conditionally uniform on the complement of the set {C,X} in the set {1,2,3} (be the complement a 1-point or 2-point set)

The "unconditional" solution:

A pure strategy of the player is a function that maps every possible pair (X,H) to either X or the "third" element of {1,2,3} (different from X and H).

A mixed strategy of the player is a function that maps every possible pair (X,H) to a probability measure on {1,2,3} that vanishes on {H}.

The winning probability is a function of a strategy. It is invariant under the rearrangement group of {1,2,3} (by the symmetry, of course). Therefore it is sufficient to consider only invariant mixed strategies. Such a strategy is nothing but a probability of switching. It is sufficient to consider only extremal values (0 and 1) of the probability. Thus, the question boils down to: to switch or not to switch. The simple unconditional calculation completes the analysis.

The "semi-conditional" solution:

The conditional winning probability without switching, P(X=C|X,H), is a function of X and H. It is invariant under the rearrangement group of {1,2,3} (by the symmetry, of course). Therefore it is the constant function. Taking into account that its expectation is equal to the unconditional probability (the total probability formula) we see that P(X=C|X,H) = P(X=C); that is, the conditional probability is equal to the unconditional probability. The simple unconditional calculation (the same as above) completes the analysis.

P(X=1,H=2) + P(X=1,H=3) + P(X=2,H=1) + P(X=2,H=3) + P(X=3,H=1) + P(X=3,H=2) = 1;

therefore

That is, the conditional probability is equal to the unconditional probability (in the symmetric case, of course).

Posted by Glkanter (talk) 17:12, 19 January 2011 (UTC)[reply]

What is the relevance to this discussion? -- Rick Block (talk) 17:39, 19 January 2011 (UTC)[reply]

Boris is showing here how to solve the conditional problem by using the answer (2/3) to the simple problem (what is the unconditional probability?) together with symmetry and the law of total probability. He is not saying that the simple solution solves the conditional problem. He is saying that you can get a cute solution to the more tricky and subtle conditional problem by combining the simple solution with symmetry. BTW he also thinks that MHP ought to be solved by finding the conditional probability, at least, that is what he once said to me. Richard Gill (talk) 15:26, 20 January 2011 (UTC)[reply]

Lemme see if I have it right, now:

A simple solution is just one part of a tool kit that gives me the ability to correctly calculate the 2/3 & 1/3 probabilities of the conditional MHP with the 50/50 host bias premise, without directly calculating a conditional probability.

The required tools in this tool kit are:

• A simple solution

• Symmetry

• The law of total probability

Am I right so far? Glkanter (talk) 22:15, 20 January 2011 (UTC)[reply]

YES! It's a smart trick to calculate the conditional probability without any sweat. (And it's reliably sourced in my two papers.) Richard Gill (talk) 19:21, 26 January 2011 (UTC)[reply]

I am delighted to read your response, Richard. So, tell me, what has the argument been about for over 2+ years?

Much ado about almost nothing, I am afraid (and have always thought so, and always said so). Richard Gill (talk) 19:39, 26 January 2011 (UTC)[reply]

What do you think of the "Conditional solution" section in the current article, then? Do you agree with me that paragraphs 1, 2 & 4, and the 2nd diagram add little or no value, and actually serve to confuse the solution sections badly? Glkanter (talk) 19:48, 26 January 2011 (UTC)[reply]

I agree. I once got set upon by a pack of dogs because I deleted one of the more excruciating paragraphs since in view of the symmetry argument it seemed to me totally superfluous, worse than superfluous. Richard Gill (talk) 20:02, 26 January 2011 (UTC)[reply]

Question: Is this correct?

The simple solution gives the unconditonal probability to win by switching of 2/3.
Unconditional in the sense of irrespective of which one of his two doors the host will open / has opened, and therefore able to give that answer, both knowing as well as without knowing which door the host opens. Irrespective. The unconditional answer will be 2/3 in any case.

Conditional probability calculation however is a tool, able indeed to exactly consider that very special door actually having been opened, and in addition able indeed to exactly consider any possible "assumed influence" on the odds, that might have been evoked by the opening of that very special door. And even the greatest influence has to define Pws to be always within the range of 1/2 (but never less) to 1, with an average of 2/3. – So nothing can be better than always to switch.

But note: If, in 1/3, the host has got two goats to choose from, then his procedural method in choosing of his door, as well as its direction, will forever be completely unknown. So, irrespective of which one he opens, by that an outsider will have learned absolutely nothing to enable him to revise the odds on the door first selected by the guest. So a correct conditional solution has to give exactly the same answer as the simple solution: Probability to win by switching 2/3, just as well independent of the door opened by the host. Gerhardvalentin (talk) 23:36, 20 January 2011 (UTC)[reply]

No. Irrespective here specifically means without knowing which door the host opens. If you do know which door the host opens then you're talking about a conditional probability, not an unconditional probability. The difference is whether you're talking about (say) all 300 plays of the game show where the player has picked door 1 and you don't know whether the host has opened (or will open) door 2 or door 3 vs. only half of these where you do know which door the host opened. If you don't know which door the host opened (or will open), we have no choice but to be talking about all 300 cases (and it includes both cases where the host opens door 2 and cases where the host opens door 3). As soon as you do know which door the host opened, we're not talking about all 300 cases anymore but only the subset where the host opens one of door 2 or door 3.

By watching the shows, and keeping track of how many times a player switches and wins in each of the 6 possible combinations of initial player pick and door the host opens, an outsider sees the result of the host's bias. It is this history that lets the player revise the odds on the door first selected upon seeing which door the host opens. The fact that this history is accessible is why some sources say the conditional probability is the only acceptable answer. Avoiding this "case dependency" is why urn problems talk about "indistinguishable" balls. If the doors in the MHP were indistinguishable (but, since we're talking about a gameshow they are inherently distinguishable) neither the player or the host could tell the difference between door 2 and door 3. In this (unrealistic) version of the problem, the host CANNOT have any bias between these doors, the problem is forced to be symmetrical (actually, the individual cases no longer exist since we can't tell the difference between the host opening "door 2" vs. "door 3"), and the conditional and unconditional solutions must be the same. As soon as anyone can tell the difference between door 2 and door 3 (and, the next tine we run the show the same doors remain door 2 and door 3), the shows where the player picks door 1 and sees the host open door 2 can be tallied separately from the shows where the player picks door 1 and sees the host open door 3 - and these tallies may show different odds of winning by switching (depending on which door the host opens). The combination will still show a 2/3 overall chance of winning by switching - but the individual chance (by door) may be different (depending on the host bias). -- Rick Block (talk) 06:24, 21 January 2011 (UTC)[reply]

Thank you Rick, for your answer. But I fear you are not correct. Did you read my question?

You say "irrespective here ... means without knowing which door the host opens".

And you say "If you do know ... you're talking about conditional probability, not an unconditional probability." That's not the fact, I'm not forced to talk about conditional probability only. Please understand that I was in fact talking about the unconditional probability. You can do that, even if you do know which door has been opened, and then I talked about conditional prob. later on. Please read once more my question above.

I tried to say: No matter whether you already have seen the door opened, showing a goat, or not: With "Irrespective here" I wanted to say "no matter which door", does mean "whatever door", does mean "regardless of which door", does mean "never mind which one of the two doors", "not caring which one, even if you should know which one", does mean "irrelevant which one of his two doors". Because the "unconditional", the simple solution does not care about which one of his two doors the host just has opened. And it does not need to make further assumptions.

You said By watching the shows, and keeping track of how many times ..." – But the question is about "a game show" the contestant is in. You just do not know whether it's the first one or whether it's the mphth one, or whether this show will be repeated at all. You can assume that it could be repeated. But do you know how often it will be repeated in the future? Three times or even 5 times? Of course you can "assume" it could be repeated 600 times, in the future. But that's just an assumption. And who holds the records of all identical shows that already have taken place? Anyone can "assume" 5 shows or 100 million shows. Just unproven assumptions. Who holds the logfile?

And, knowing that "future repetitions" is just an "assumption", it's not necessary to focus on the individual chance (by door) that may be different (depending on the host bias). No prerequisite. And who says that anyone can read any host's bias out of repetitions? In reading the records of a 100 million repetitions-logfile, you can bet that you and I and any other will guess every now and then, in parts, to have traced some clear host's bias here and there, even if there is absolutely none. That is the proof that you risk to be absolutely wrong in "reading" such host's bias out of records. So you can just base on mere assumptions that never are given to having anything to do with the real world or with the answer to the MHP-question asked for. Really nothing in effect, just flat assumptions.

So, knowing and considering which door the host has opened in that one MHP game, is of no benefit whatsoever. Any host's bias will forever be totally unknown to any outsider. With no exception. Check my logfile! So all of that, in addressing the MHP-question, and in addressing the answer asked for, is just a mere gimmick.

But, on the other hand, all of that is really extraordinary helpful as an example in teaching and learning conditional probability theory. Without any effect to the "MHP" indeed, but with good effect and benefit for students of mathematics. And that is the only reason why it is necessary to mention conditional probability in the MHP lemma. And it has to be shown without deception and beguilement that this could be of any help for the contestant in this one game show that is reported of, in that famous PARADOX question — that many are out of position to answer correctly, and, as the "sources" show, some were even out of position to read and understand correctly that little simple question, and therefore, presuming all the world of "assumptions", were reasoning about "existing records" on past and future repetitions. Gerhardvalentin (talk) 16:34, 21 January 2011 (UTC)[reply]

Are you talking about your personal opinion about the MHP here, or a position advanced by reliable sources? If sources, which ones? -- Rick Block (talk) 20:30, 21 January 2011 (UTC)[reply]

Sources? Thank you for your question, Rick. Did you ever read those recent comments of Marilyn vos Savant, (not) regarding that whole issue? For years it has just steadily been her identical comment: "Too paltry at all to comment about." And you can understand that. Just have a look on the bossiness of those pageant "sources", steadily having invented dazzling imagination of hundreds of game shows (see your own words above), and deliberately inventing folly "readable obvious imbalance" (see your own statements above), and further "improvements" and and and, and compare this funfair with the decent and presuppositionless, but famous question "Supposed you are in a game show, ...". Please comment yourself that discrepancy to what actually is asked for: "Is it to your advantage to switch?".

Please give me your personal opinion on those facts, just to show me whether (where) I am wrong to refuse tampering without naming it tampering, that really does address "quite other questions".

Conditional probability is clear and is helpful in many ways, and is really necessary in many problems. So you just should show its ability to give the one and only correct answer just as well, without desperately seeking any (forever unknown) "special disposition", and without flimsy plea that Pws could have "changed" by opening of one door. Would be great if you could help the article to be what it should be: clearly laid out and never puzzling. Kind regards, Gerhardvalentin (talk) 22:19, 21 January 2011 (UTC)[reply]

Wikipedia

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

Symmetry occurs not only in geometry, but also in other branches of mathematics. It is actually the same as invariance: the property that something does not change under a set of transformations.

Two objects are symmetric to each other with respect to the invariant transformations if one object is obtained from the other by one of the transformations.

Posted by Glkanter (talk) 17:25, 19 January 2011 (UTC)[reply]

Richard Gill

Conclusion to his paper:

"This little article contains nothing new, and only almost trivial mathematics. It is a plea for future generations to preserve the life of The True Monty Hall paradox, and not let themselves be misled by probability purists who say “you must compute a conditional probability”. - The Three Doors Problem...-s Richard D. Gill Mathematical Institute, University Leiden, Netherlands http://www.math.leidenuniv.nl/!gill February 15, 2010 Posted by Glkanter (talk) 18:18, 19 January 2011 (UTC)[reply]

Are you claiming this quote is saying the simple solutions address the conditional probability? If so, you are completely misinterpreting it. Richard's point (which in no way contradicts anything we're talking about here) is that it is not necessary to compute the conditional probability in order to decide whether to switch or not. His advice is to pick a random door and decide to switch before seeing which door the host opens and if you do this you'll have a 2/3 chance of winning the car (assuming only that the host always opens a door showing a goat and makes the offer to switch). This is an entirely different question than "should you switch if you've picked door 1 and have seen the host open door 3" (the conditional question). Richard does not and has not ever claimed that the simple solutions solve the conditional probability. If you don't believe me, ask him yourself. -- Rick Block (talk) 19:36, 19 January 2011 (UTC)[reply]

Rick's interpretation of my words is pretty much OK. I make a point in my two papers of showing that the simple solutions and the conditional solutions answer different questions. I say that it is a matter of opinion which question you might want to answer. I also point out that a game-theoretic point of view leads one to try to answer yet another question. I make a plea for diversity. I show how every approach has advantages and disadvantages. I show how they are mathematically closely related and I argue for synthesis, not for thesis and antithesis. I argue that the mathematician's job is to present the spectrum of interpretations, questions and answers as a kind of menu, and indeed, like a menu, there is a price list. You want a Mercedes rather than a Volkswagen, you'll have to pay more. Do you need a Mercedes or a Volkswagen? I don't know. My job (and I think wikipedia's job too) to give you the information on the basis of which you can make your own choice.

There's a problem that some (many?) editors and readers and sources are actually not capable or willing to understand the subtle differences between conditional and unconditional probability (in the probability context, these are technical concepts, with precisely defined meanings; but the common language of the man in the street doesn't make these distinctions and uses the same words in different ways.

I just don't know how to get all this across to the man in the street. My experience as a statistical expert in legal and medical and physics issues is that it can be completely hopeless. Probably only the young and as yet unformed have the mental flexibility to appreciate the points. Richard Gill (talk) 15:39, 20 January 2011 (UTC)[reply]

I think talking about it in frequency terminology is the most approachable method. Imagine 900 tapings of the show. The simplest solution (you have a 1/3 chance of initially picking the car, so a 2/3 chance of winning the car if you switch) is talking about all 900 shows. If all 900 players pick randomly or if the car is randomly distributed and they all pick door 1, 300 will initially pick the car. I don't think anyone fails to get this. Moreover, given that the host must open a door showing a goat (so the other remaining door is a goat if you've picked the car and vice versa) if 300 will initially pick the car if they switch they'll get a goat, so 600 will initially pick a goat and if they switch they'll get the car. Again, I don't think anyone fails to understand this either (it is the "simple" solution, after all). Where things get tricky is when you think about having picked door 1 and seeing the host open door 3. Back to our 900 tapings - how many is this? If the players' first picks are random, about 300 will pick door 1 (nobody fails to get this). How many of these will see the host open door 3? Well, if everything is symmetrical about 150 (half see the host open door 2 and half see the host open door 3). So, let's think about the 150 who see the host open door 3. First, of the 300 players who've picked door 1, we'd expect the car to be behind each of the doors 100 times. We know the host must open door 3 if the car is behind door 2, so of the 150 players who pick door 1 and see the host open door 3 we'd expect for 100 of them the car will be behind door 2. This leaves only 50 for whom the car is behind door 1. What happened to the other 50? They see the host open door 2. I think talking about concrete numbers in this way is understandable to nearly anyone. -- Rick Block (talk) 23:04, 20 January 2011 (UTC)[reply]

Why It's 2/3 & 1/3 And Not 50/50

If the section title is 'the MHP paradox', then the host bias is 50/50. Nothing else. Glkanter (talk) 21:26, 19 January 2011 (UTC)[reply]

Are you saying all reliable sources agree on this? Or is this simply your own opinion? It is a plain fact that numerous reliable sources talking about the Monty Hall problem (whether you call it the MHP paradox or not) bring the topic of host bias into the discussion in order to distinguish conditional vs. unconditional solutions. Your opinion that this topic is not worth talking about in the article is contradicted by these sources. The sources win. -- Rick Block (talk) 00:52, 20 January 2011 (UTC)[reply]

You don't agree with the 'if-then' statement I posted above? Which sources prove otherwise? Which sources conclude with something other than 2/3 & 1/3? Glkanter (talk) 01:53, 20 January 2011 (UTC)[reply]

Are you saying that any time a source discusses different premises in their paper, they are saying these premises, that are different than the 'standard' K & W formulation, and have a result other than 2/3 & 1/3, comprise a valid problem that can claim the name the MHP paradox"? If so, that is a false statement. Glkanter (talk) 02:16, 20 January 2011 (UTC)[reply]

False according to whom? Again, it is a plain fact that numerous reliable sources that say they are talking about the Monty Hall problem talk about host preferences other than 50/50. Generally they do this to distinguish conditional vs. unconditional solutions. For the article to talk about this is not in the least POV, but is reflecting the mainstream academic view that the simple solutions do not address the conditional probability. What is your suggestion for how to represent this view "fairly, proportionately, and as far as possible without bias"? If what you are suggesting is not that this view be eliminated, what exactly are you suggesting? -- Rick Block (talk) 03:15, 20 January 2011 (UTC)[reply]

I'm just trying to bring an end to the endless circular discussions exactly like this one.

♠ Selvin defined it as '2/3 & 1/3 vs 50/50'

♣ K & W calls it standard

♥ It's the only thing vos Savant & her mass audience, including 1,000s of math PhDs argued about

♦ 'Everyone' says '2/3 & 1/3 vs 50/50'.

The High Priests in their professional journals may agree with you, but the Wikipedia readers and the bar patrons are interested in 'why is it 2/3 & 1/3 and not 50/50'. You are demonstrating *perfectly* the issue I want to bring to a head. Are these simply my opinions? Yes. Do I think they are reasonable, correct and valid statements? Yes. Have I provided support from reliable sources for these statements? Yes. Thank you. Glkanter (talk) 03:33, 20 January 2011 (UTC)[reply]

If you're actually interested in why it is 2/3 & 1/3 and not 50/50 many sources say the answer lies in the proper use of conditional probability. This is the point you are refusing to acknowledge. How about if you just say it out loud - you don't give a damn about NPOV, you want to ignore these sources because you think what they say is just too darned complicated and you'll continue arguing about this forever. -- Rick Block (talk) 04:35, 20 January 2011 (UTC)[reply]

I resent your attempt to 'read my mind', tell me what I really mean, and then Aunt Sally/Straw Man me. As you have no valid response to the above explanation/justification, you have chosen to attack me personally. This is a perfect example of the GAMESMANSHIP you practice to the detriment of the article, the readers, the other editors, and this mediation. Glkanter (talk) 05:01, 20 January 2011 (UTC)[reply]

No valid response? "many sources say the answer lies in the proper use of conditional probability". I've asked at least twice (in this thread, today) what your suggestion is for how to address what these sources say. If you're not suggesting simply ignoring them, then what are you suggesting? -- Rick Block (talk) 05:21, 20 January 2011 (UTC)[reply]

Glkanter: I invite you to answer his question. AGK [•] 09:55, 20 January 2011 (UTC)[reply]

Are you done attacking me without provocation? I suggest we agree that the 50/50 premise is the only host bias that will be addressed in the solution sections. Other stuff can go other places. Limitations as to the generality of the simple solutions vs the conditional solutions could also be addressed in the solution section. The requirements for symmetry, and how symmetry enables the simple solutions to solve the conditional problem would also be addressed in the solution section. Glkanter (talk) 05:28, 20 January 2011 (UTC)[reply]

Who's attacking whom here. GAMESMANSHIP???? How many times have you accused me of this?

The 50/50 premise is the "usual" interpretation of the problem. The article ALREADY says this (and has for a long time). In the solution section, it is perfectly appropriate to mention that sources presenting conditional solutions distinguish conditional solutions that explicitly make use of this premise from the "simple" solutions (which do not), and that this is a difference between the conditional and simple solutions. The first mention of host preference in the article is currently in the "Conditional probability solution" section. What it says is that given the 50/50 host preference both the unconditional and conditional answers are 2/3, but without this premise the conditional and unconditional answers might be different. Later in this section it says the probability might vary from 1/2 to 1 depending on this preference. Are you suggesting deleting both of these statements, or perhaps only the second? Both of these seem relevant to me. -- Rick Block (talk) 06:01, 20 January 2011 (UTC)[reply]

You have repeatedly badgered me for my opinions. Then you relish in telling me why my opinions are wrong. Until you, and Nijdam, and the others agree that my proposals have merit and are reasonable, there is no point in discussing this further. Glkanter (talk) 06:07, 20 January 2011 (UTC) [Restored AGK's deletion, sans profanity - Glkanter (talk) 11:35, 21 January 2011 (UTC)][reply]

If you have any genuine interest in what I actually believe, and why, regarding the MHP, I suggest you look at Martin's talk page. Go back no more than a week. Glkanter (talk) 06:34, 20 January 2011 (UTC)[reply]

I have no interest whatsoever in what you actually believe, and why, regarding the MHP. What I care about is what the reliable sources say and whether the article represents what they say fairly, proportionately, and as far as possible without bias. What I'm asking here is what specific edit or edits you're suggesting regarding host bias. Again, nothing other than the usual 50/50 bias is mentioned in the article until the "Conditional probability solution" section, where the possibility that this bias might not be 50/50 is raised in order to distinguish conditional vs. unconditional solutions (which seems perfectly consistent with how the majority of sources that mention this topic treat it). To make this even more concrete, here's the current text from the "Conditional probability solution" section. Please mark up what you would change or delete. -- Rick Block (talk) 15:27, 20 January 2011 (UTC)[reply]

Immediately above, you begin with (in part):

"I have no interest whatsoever in what you actually believe, and why, regarding the MHP. (talk) 15:27, 20 January 2011 (UTC)"[reply]

In that case, keep your now-admittedly uninformed, biased opinions about what I care about, or think, or plan, to yourself.

Posts like this one are clearly not based on fact, and are nothing more than a personal attack intended to disrupt the mediation, and cast doubt on my contributions:

If you're actually interested in why it is 2/3 & 1/3 and not 50/50 many sources say the answer lies in the proper use of conditional probability. This is the point you are refusing to acknowledge. How about if you just say it out loud - you don't give a damn about NPOV, you want to ignore these sources because you think what they say is just too darned complicated and you'll continue arguing about this forever. -- Rick Block (talk) 04:35, 20 January 2011 (UTC)[reply]

Get it, Mr. Admin? Glkanter (talk) 20:18, 22 January 2011 (UTC)[reply]

Please explain to me how asking you what edits you'd make is being disruptive. And, I strongly suggest you stop attacking me. -- Rick Block (talk) 22:01, 22 January 2011 (UTC)[reply]

Conditional probability solution

The simple solutions show in various ways that a contestant who is going to switch will win the car with probability 2/3, and hence that switching is a winning strategy. Some sources, however, state that although the simple solutions give a correct numerical answer, they are incomplete or solve the wrong problem. These sources consider the question: given that the contestant has chosen Door 1 and given that the host has opened Door 3, revealing a goat, what is now the probability that the car is behind Door 2?

In particular, Morgan et al. (1991) state that many popular solutions are incomplete because they do not explicitly address their interpretation of vos Savant's rewording of Whitaker's original question (Seymann). The popular solutions correctly show that the probability of winning for a player who always switches is 2/3, but without additional reasoning this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens. That probability is a conditional probability (Selvin 1975b; Morgan et al. 1991; Gillman 1992; Grinstead and Snell 2006:137). The difference is whether the analysis is of the average probability over all possible combinations of initial player choice and door the host opens, or of only one specific case—to be specific, the case where the player picks Door 1 and the host opens Door 3. Another way to express the difference is whether the player must decide to switch before the host opens a door, or is allowed to decide after seeing which door the host opens (Gillman 1992); either way, the player is interested in the probability of winning at the time they make their decision. Although the conditional and unconditional probabilities are both 2/3 for the problem statement with all details completely specified - in particular a completely random choice by the host of which door to open when he has a choice - the conditional probability may differ from the overall probability and the latter is not determined without a complete specification of the problem (Gill 2010). However as long as the initial choice has probability 1/3 of being correct, it is never to the contestants' disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2.

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 as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138), or formally derived as in the mathematical formulation section below. For example, the player wins if the host opens Door 3 and the player switches and the car is behind Door 2, and this has probability 1/3. The player loses if the host opens Door 3 and the player switches and the car is behind Door 1, and this has probability 1/6. These are the only possibilities given host opens Door 3 and player switches. The overall probability that the host opens Door 3 is their sum, and we convert the two probabilities just found to conditional probabilities by dividing them by their sum. Therefore, the conditional probability of winning by switching given the player picks Door 1 and the host opens Door 3 is (1/3)/(1/3 + 1/6), which is 2/3.

This analysis depends on the constraint in the explicit problem statement that the host chooses uniformly at random which door to open after the player has initially selected the car (1/6 = 1/2 * 1/3). If the host's choice to open Door 3 was made with probability q instead of probability 1/2, then the conditional probability of winning by switching becomes (1/3)/(1/3 + q * 1/3)). The extreme cases q=0, q=1 give conditional probabilities of 1 and 1/2 respectively; q=1/2 gives 2/3. If q is unknown then the conditional probability is unknown too, but still it is always at least 1/2 and on average, over the possible conditions, equal to the unconditional probability 2/3.

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

Unchallenged Logical Fallacies Rule This Mediation

The following is a full quote of an editor's posting. The emphasis is mine:

"Morgan, Gillman, G&S all directly say the simple solutions solve the wrong problem (i.e. does not address the conditional probability where we know the player has picked door 1 and the host has opened door 3). Rosenthal, Eisenhauer, and Lucas et al are saying the same thing, but somewhat less directly. There are scores of others (mostly academic mathematical sources) that present only conditional solutions - these are implicitly agreeing that the conditional approach is the "right" approach and can be construed to be indirectly agreeing with the criticism that the simple solutions solve the wrong problem. You disagree with this, but I consider Carlton to be in this camp because he explicitly distinguishes what he considers to be his "solution" (which is conditional) from his "intuitive explanation" (which is one of the simple solutions).

"Morgan criticizes even conditional solutions that assume without it being given in the problem statement that the host's choice between goats is 50/50. I don't recall any other source doing this.

"Other than Falk (and the discussions on Wikipedia talk pages), I'm not aware of any source that makes a point about symmetry.

"Many sources (Morgan, G&S, Rosenthal, Eisenhauer, Lucas et al, Falk) specifically use variants to show that the simple solutions indeed solve a different problem than the door 1/door 3 problem (because, for these variants, the simple solutions still say the answer is 2/3 chance of winning by switching even though for these variants this is not the correct answer for the door 1/door 3 case). These sources are not criticizing the simple solutions because they fail to address these variants, but rather using these variants to demonstrate (prove) that the simple solutions are not addressing the conditional problem. -- Rick Block (talk) 16:15, 19 January 2011 (UTC)"

http://en.wikipedia.org/w/index.php?title=Wikipedia_talk:Requests_for_mediation/Monty_Hall_problem&diff=408802958&oldid=408795734

That first paragraph contains astounding logical errors *and* Wikipedia policy violations. He has repeated that particular argument countless times in the 2 years I've been discussing this. And he uses similar fallacies and opinions masquerading as facts in many (most?, all?) of his arguments and counter-arguments.

• Q: How come nobody but me calls him out on this, and I'm the only editor who perceives this as GAMESMANSHIP? Glkanter (talk) 09:57, 21 January 2011 (UTC)[reply]

I consider this a personal attack. -- Rick Block (talk) 15:34, 21 January 2011 (UTC)[reply]

Yeah? How so? Glkanter (talk) 16:09, 21 January 2011 (UTC) A second possible response:[reply]

Join the club, Rick Block! Your diff of 04:35, 20 January 2011 has the following summary, in full (emphasis mine):

"just admit it - you don't give a damn about NPOV"

And the actual edit says, in full:

"If you're actually interested in why it is 2/3 & 1/3 and not 50/50 many sources say the answer lies in the proper use of conditional probability. This is the point you are refusing to acknowledge. How about if you just say it out loud - you don't give a damn about NPOV, you want to ignore these sources because you think what they say is just too darned complicated and you'll continue arguing about this forever. -- Rick Block (talk) 04:35, 20 January 2011 (UTC)"[reply]

Wouldn't you also call your personal, biased, uninformed, demeaning opinions of my motivations and future plans 'a personal attack' had they been leveled at you, Mr. Admin?

The hypocrisy you exhibit with your crocodile tears is another unambiguous example of GAMESMANSHIP. Glkanter (talk) 19:42, 21 January 2011 (UTC)[reply]

Apart from the issue of whether this is the umpteenth case of Glkanter personally attacking Rick Block, I'd be curious to learn exactly what are these alleged (a) "WP policy violations" and (b) "astounding logical fallacies". Rick block's paragraph cited above reads like a fair summary of the sources he cites and he's read. The only mildly astounding thing I see is Glkanter's belief that usage of screaming bold sans-serif somehow makes one's point weightier. AKG, care commenting on this too? glopk (talk) 01:29, 22 January 2011 (UTC)[reply]

That's amazing glopk! You know exactly where to type & post your response, yet exhibit no evidence of any reading comprehension skills whatsoever. Glkanter (talk) 04:12, 22 January 2011 (UTC)[reply]

ps The emphasis isn't so that you'll believe me, glopk. It's so the relevant portion of the paragraph presents itself readily. Kapisch?

But that's exactly the point: it does not present itself readily, in bold or not. What is there in that comment of Rick Block's that is such an obvious (to you) "WP policy violation"? I just don't see it, care directing a mere mortal to the relevant policy pages? Specifically, what WP policy is Rick Block allegedly violating when he states his opinion that:

Morgan, Gillman, G&S all directly say the simple solutions solve the wrong problem (i.e. does not address the conditional probability where we know the player has picked door 1 and the host has opened door 3). Rosenthal, Eisenhauer, and Lucas et al are saying the same thing, but somewhat less directly. There are scores of others (mostly academic mathematical sources) that present only conditional solutions - these are implicitly agreeing that the conditional approach is the "right" approach and can be construed to be indirectly agreeing with the criticism that the simple solutions solve the wrong problem. You disagree with this, but I consider Carlton to be in this camp because he explicitly distinguishes what he considers to be his "solution" (which is conditional) from his "intuitive explanation" (which is one of the simple solutions).

He is plainly summarizing his reading of the sources and, politely, stating a position different from yours. What's so "astounding" in that? Scass'U'Kazz? glopk (talk) 06:01, 22 January 2011 (UTC)[reply]

Well, none of that debatable stuff was bolded and bigged for emphasis, was it, glopk? Which is the whole point, isn't it? Why don't you ask me about the "implict" and "indirectly" BS Rick makes up, that I *did* emphasize? Glkanter (talk) 06:57, 22 January 2011 (UTC)[reply]

I am asking you know. Please explain why it is BS, and how in your opinion it is a violation of specific WP policies (which ones?). You may do so in bold, in cursive, in cuneiform, whether in prose or rhyming sonnets. Just, please-pretty-please, put up or shut up: what exactly is being violated already? glopk (talk) 04:23, 23 January 2011 (UTC)[reply]

Summary

Thus far we have:

M0: (Conditional version) The version of the MHP with identified initially chosen door and identified opened door by the host. The general assumptions of randomness are made. As typical example door 1 is chosen and door 3 opened. Only then is the player offered to switch.

M1: (Unconditional version) The version of the MHP with just the rules and the assumption that the car is placed randomly. The player or the public is asked wheter to switch or not even before anything has happened.

M2: (General conditional version) The version of the MHP with just the rules and the assumption that the car is placed randomly. As typical example door 1 is chosen and door 3 opened. Only then is the player offered to switch.

S0: (Conditional solution using Bayes) The decision is switch, based on the conditional probability that the car is behind the remaining closed door, given the chosen and the opened door.

S1: (Unconditional solution) The decision is switch, based on the unconditional probability being 2/3 to get the car when switching

S2: (Conditional solution using symmetry) The decision is based on the conditional probability that the car is behind the remaining closed door, but the probability is not actually calculated, but due to the symmetry has to be equal to the unconditional one.

S3: (Wrong simple solution) The player initially hits then car with probability 1/3. As the opened door shows a goat, the remaining door must hide the car with probability 2/3.

S4: (Combined doors solution) The not chosen door together have probability 2/3 to hide the car. As the opened one shows a goat, the other one must have all the 2/3 probability on the car.

S5: (Many doors solution)

Sim: (Simulation) Simulations of the MHP are available on the internet and also suggested in some of the sources

Relations

S0 solves M0
S1 solves M1
S1 does not solve M0
S2 is not the same as S1
S2 may serve as a kind of simple solution to M0
S3 is logically wrong
S4 is equivalent to S3
S5 is similar to S4 and hence lacks the same logic
Most of the available simulations do not simulate M0, but simulate M1.
Any (probabilistic) solution to M0 has to calculate the conditional probability given the chosen door 1 and the opened door 3.

Nijdam (talk) 09:56, 29 January 2011 (UTC)[reply]

Table of what sources say

You're talking here (particularly in the "Relations" section) about your opinion of the The Truth, rather than what reliable sources say. I think it might help to categorize which sources say which of these. As a start, here's a table (incomplete) listing the references currently in the article, with which problem and solution they present (and some notes). -- Rick Block (talk) 19:45, 22 January 2011 (UTC)[reply]

Thank you Rick, you did a marvelous job. I would have suggested something like this, in order to get an overview of how the sources are in line with The Truth, not knowing you had already done so. Nijdam (talk) 09:59, 23 January 2011 (UTC)[reply]

Source	Type of source	Problem presented	Solution presented	Notes
Adams	popular, self published	M0	S4	Questions the assumption that the host would always make the offer to switch
Bapeswara Rao	academic journal, reviewed paper	M1	S1	Also computes the conditional probability for M0 obtaining (??) 1/2
Barbeau 1993	academic journal, column	M0 (and other variants)	many are mentioned	Literature overview
Barbeau 2000	book	M0 (and other variants)	many are mentioned	Literature overview
Behrends	book	M0	S0 (explicitly assumes p=1/2)	Also presents S1 as solution to M1
Bloch		na	na
Carlton	academic journal, reviewed paper	M0	S0 (uses p=1/2 without mention)	Also presents S1 as an "intuitive explanation" to M1
Chun	academic journal, letter	M0	S0 (uses p=1/2 without mention)
D'Ariano et al
Devlin	academic journal, column	M0	S3 and S4	Followup column presents S0 (using p=1/2)
Economist
Falk	academic journal, reviewed paper	M0	S0	Criticizes S1 and S3 as solutions to M0 (discusses the p=1/2 assumption)
Fitney and Abbott
Fox and Levav	academic journal, reviewed paper	na	na	Experimental psychology paper about how people judge probabilities
Gardner	popular journal (Scientific American), column	na	na	Presentation of the Three Prisoners problem
J.Gill	probability textbook	M0	S0
R.Gill 2010	academic journal, reviewed paper
R.Gill 2010b	academic journal, reviewed paper
Gillman	academic journal, column	M0 (but with unknown host preference)	S0 (1/(1+p) where p is the host preference, 2/3 only if p=1/2)	Criticizes S1 ("This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3")
Granberg	popular book (vos Savant's book)	M0	S1 (sort of)	States without proof that the player's initial 1/3 probability does not change given a precise set of conditions (including p=1/2)
Granberg and Brown	academic journal, reviewed paper	M1	S1 (sort of)	States without proof that the player's initial 1/3 probability does not change given a precise set of conditions (including p=1/2)
Grinstead and Snell	probability textbook	M0 (with p=1/2)	S0	Also presents S1 as a solution to M1 saying this "does not quite solve the problem that Craig posed" (i.e. M0)
Hall	letter	M0	S1
Henze
Herbranson and Schroeder
Krauss and Wang	academic journal, reviewed paper	M0	several, including S0 and S1	source of the fully explicit problem in the article (with p=1/2)
Mack	book	M1	S1
Magliozzi	popular, column
P.Martin	popular (bridge column)	M0	S3
R.Martin	book	M0	S3
Morgan et al	academic journal, reviewed paper	M0 (but with unknown host preference)	S0 (1/(1+p) where p is the host preference, 2/3 only if p=1/2)	Criticizes S1 and S3 as solving the wrong problem, and criticizes those presenting S0 who assume p=1/2 without it being given in the problem statement
Mueser and Granberg	academic paper (self published?)	M0	S0 (using p=1/2 with a comment about its importance)	Also presents S3 as a "common explanation"
Nalebuff	academic journal, column	M1	S1
Rosenhouse	book	many	many	entire book about the problem
Rosenthal	academic journal, reviewed paper	M0	S0	Criticizes S3: "actually correct, but I consider it 'shaky' because it fails for slight variants of the problem". Stresses importance of p=1/2 ("This assumption, callously ignored by the Shaky Solution, is in fact crucial to the conclusion")
Selvin 1975a	academic journal, letter	M0	S1
Selvin 1975b	academic journal, letter	M0 (including p=1/2)	S0
Seymann	academic journal, comment on a paper	na	na	Comment on Morgan et al., emphasizing the importance of clearly defining the problem
Stibel
Tierney 1991	popular, column
Tierney 2008	popular, column
vos Savant 1990	popular, column	M0	million door analogy
vos Savant 1990b	popular, column	M0	S1
vos Savant 1991	popular, column	M0	simulation of S1	Simulation randomizes car placement and player choice but not host selection between goats
vos Savant 1996	book	M0	S1	reprint of columns
vos Savant 2006	popular, column	host forgets variant	no reasoning offered
Schwager	book	M0	S3
Williams	academic, self published course notes	M0	S3 and S4	Discusses other problem variants, focusing on whether the host must make the offer to switch
Wheeler	book
Whitaker	letter	M0	no solution offered

Comments on the above

The design of a table can promote a bias. To avoid that, maybe you could add 4 columns:

Does the source give roughly the same problem statement as vos Savant - specifically, door #3 opened (for these purposes, at this point, we temporarily avoid the issue of the 50/50 host bias) ?

Does the source's solution return the result 2/3 & 1/3?

Is this source found in the 'popular' or the 'professional' literature?

Does this source specifically mention some problem with simple solutions, or is the criticism 'implicit' and/or 'indirect' by virtue of the source providing a conditional solution?

Thanks. Glkanter (talk) 20:05, 22 January 2011 (UTC)[reply]

Your column 'Solution presented' is problematic, as the data entered requires drawing a conclusion based on the very POV we are debating. Could you please add a column indicating that the source, right, wrong, or otherwise, has provided a simple solution as if it actually solves the MHP? And a 2nd column indicating if the source provided *only* simple solutions? Thank you. Glkanter (talk) 20:34, 22 January 2011 (UTC)[reply]

M0 is the problem statement presented by vos Savant, so any source listed as presenting M0 is one presenting roughly the same problem statement as vos Savant. There is no dispute about the 1/3:2/3 answer, so all sources present this solution (except as noted). Popular and professional is a good idea - I've added that (actually peer reviewed vs. "self published" is a reasonable distinction as well). Criticism is mentioned in the notes column already. -- Rick Block (talk) 21:53, 22 January 2011 (UTC)[reply]

Your approach requires the reader to master the 'M' & 'S' codes which are based on Morgan's POV. I find that intentionally biased, and unnecessarily complicated. My approach allows each column to be analyzed in a 'yes/no' or 'true/false' manner. And I still can't tell the difference between High Priest sources (and the intended audience of the publication) and popular sources the way you've done it.

There should also be a column for the '50/50 host bias' premise, with the following values:

•Expressly stated (Selvin)

•Not addressed (vos Savant)

•Expressly not a premise (Morgan)

Posted by Glkanter (talk) 22:12, 22 January 2011 (UTC)[reply]

The "readers" you're talking about here are the editors involved in this mediation. If anyone here does not understand the M&S codes by now (and, these are based on Nijdam's POV, not Morgan's) I suggest they haven't been paying attention. I don't understand what "bias" you think this introduces. I also don't know what you mean by "High Priest" sources, but the "type of source" column clearly indicates which are "popular" vs. "academic journal" (distinguishing reviewed papers from essentially self-published columns or letters) vs. "books" vs. "textbooks". I've made some host bias updates (not complete), although I'll not here that none of the popular sources mention anything about host bias (50/50 or otherwise). -- Rick Block (talk) 03:33, 23 January 2011 (UTC)[reply]

The 'S0' entry for Morgan is incorrect, because, as your table describes, there is no 50/50 host bias, as you point out in the table:

"M0 (but with unknown host preference)"

Morgan's F6 therefore says the conditional solutions are false:

"Solution F6 (cf. Mosteller 1965) attempts to correct the wrongly specified sample space of F4. One must ask, however, how the probabilities for this sample space are determined. It turns out that this is a correct specification only if one assumes a certain strategy on the part of the host. We will show that the problem can be solved without any assumptions of this type, which is to say the problem can be solved." Glkanter (talk) 20:51, 22 January 2011 (UTC)[reply]

ps I look forward to Rick Block's argument that Rosenthal's 'actually correct, but...' is a criticism of the simple solutions (as he included above), while Morgan's inclusion of the conditional solutions as one of 6 equally and vehemently criticized 'False' solutions, means they support the conditional solutions. vos Savant, as is the entire point of the Morgan paper, left out the 50/50 host bias. Glkanter (talk) 20:59, 22 January 2011 (UTC)[reply]

The Rosenthal paper is online. He presents the problem statement and says what the numeric answer is. Then he says "This fact is often justified as follows:" and presents a solution he labels "Shaky Solution" (his bold, not mine) and then says "This solutions is actually correct, but I consider it 'shaky' because it fails for slight variants of the problem. For example, ..." Then he goes into extensive detail about a conditional probability solutions. If you're not reading this as criticizing the "Shaky Solution" you're simply not understanding what he's saying. -- Rick Block (talk) 03:33, 23 January 2011 (UTC)[reply]

Hey, glopk! You starting to get a sense of the "(a) "WP policy violations" and (b) "astounding logical fallacies"" you claimed you couldn't see before:

"Apart from the issue of whether this is the umpteenth case of Glkanter personally attacking Rick Block, I'd be curious to learn exactly what are these alleged (a) "WP policy violations" and (b) "astounding logical fallacies". Rick block's paragraph cited above reads like a fair summary of the sources he cites and he's read. The only mildly astounding thing I see is Glkanter's belief that usage of screaming bold sans-serif somehow makes one's point weightier. AKG, care commenting on this too? glopk (talk) 01:29, 22 January 2011 (UTC)"

Looks pretty obvious to me. Glkanter (talk) 21:26, 22 January 2011 (UTC)[reply]

What are you talking about? There is no Wikipedia policy that I am aware of that forbids Rick Block's from stating what he thinks the source is saying, especially when he is so obviously right, and from calling you on your refusal to acknowledge the plain meaning of a source's text. A source's words have meaning, whether you like it or not. An author does not call a solution "shaky", and then spends pages and pages detailing alternative solutions, if he think the former isn't deserving of criticism. Please stop trying to play the wikilawyer, you are definitely ill equipped for the role. glopk (talk) 04:22, 23 January 2011 (UTC)[reply]

Rick, why don't you include the fact that Morgan called the conditional solutions, their F6, your S0, 'FALSE' in the comments section right alongside the simple solutions that you *do* include? I sense a POV bias there, and with all the vagueness in the other Morgan fields, but that can't be, can it? Glkanter (talk) 22:19, 22 January 2011 (UTC)[reply]

I've updated the Morgan entry to indicate their criticism of conditional solutions. I didn't originally include it since it's more nuanced than their criticism of the unconditional solutions. -- Rick Block (talk) 03:33, 23 January 2011 (UTC)[reply]

Y'all have already told me I don't understand what it means, but please be sure that Richard's conclusion & plea from his 2010 paper is reflected in the table, above (at least in the comments, OK?):

"This little article contains nothing new, and only almost trivial mathematics. It is a plea for future generations to preserve the life of The True Monty Hall paradox, and not let themselves be misled by probability purists who say “you must compute a conditional probability”. - The Three Doors Problem...-s Richard D. Gill Mathematical Institute, University Leiden, Netherlands http://www.math.leidenuniv.nl/!gill February 15, 2010"

Thanks. Glkanter (talk) 22:34, 22 January 2011 (UTC)[reply]

That paper and its sequel are both pleas for diversity and against dogmatism. They present mathematically sound versions of the simple solution, the conditional solution, and others, alongside one another, emphasizing the interrelations between them, and the advantages and disadvantages of all. They emphasize that no-one has a monopoly on the interpretation of Vos Savant's words. Richard Gill (talk) 10:07, 23 January 2011 (UTC)[reply]

Richard, if you'll take a look at the table above, you'll see that just about every source describes the same 'door 3 opened' problem with the 2/3 & 1/3 result. I disagree, again, that (1) there are various versions of the MHP Paradox problem statement, and (2) any non-K&W-style version has a worthy claim to be The MHP Paradox.

Whether or not the verbal description of the problem mentions "door 3 opened" is a completely different matter from whether the solution is focussed on determining a conditional probability or an unconditional probability. The word "conditional" has a rather precise technical meaning. It would help if every editor talking about these things would at least appreciate the intended meaning of any technical language. Richard Gill (talk) 14:15, 25 January 2011 (UTC)[reply]

Yes, I've acknowledge I'm a dumbass many times, not fit to discuss such topics with the High Priests. I will not engage or disturb you any further, Richard. Thank you. Glkanter (talk) 14:22, 25 January 2011 (UTC)[reply]

Isn't the question moot anyways, in the case of the 50/50 host bias? With symmetry & the law of total probability, the simple solutions do solve the conditional MHP Paradox, right?

No, the simple solutions present one stop on the way to solve the problem for those who insist on basing their decision whether to switch or stay on the conditional probability that switching will win. Richard Gill (talk) 14:15, 25 January 2011 (UTC)[reply]

Glkanter (talk) 10:30, 23 January 2011 (UTC)[reply]

Thank you for the excellent, thorough, time consuming work you have done above, Rick. Would you please:

• Include the code for any simple solutions from Nijdam's list that K & W offers

• Include the code for any simple solutions from Nijdam's list that Rosenhouse, Barbeau 1993 and Barbeau 2000 offer

• Indicate below whether you agree with Richard's reply to Nijdam from his own talk page (partial quotation only):

"You're right, please tell Martin from me. I don't believe I ever can have said the simple solution solved the conditional problem. Simple solution plus law of total probability and symmetry does." Richard Gill (talk) 03:36, 23 January 2011 (UTC)[reply]

• Indicate below whether you agree with this new conclusion from Glkanter:

Any source that gives a simple solution to the 50/50 host bias MHP that is 'incomplete' could have made it 'complete' via the inclusion of 'symmetry' and 'the law of total probability'.

• Indicate your approval or disapproval for me to highlight (maybe using 'bold' and/or 'big') the 'exception' items in each column, for example, the sources that explicitly offer 'M1' when describing the MHP?

Thank you. Glkanter (talk) 05:03, 23 January 2011 (UTC)[reply]

By 'High Priest' I mean PhD in Mathematics. I think a column indicating 'yes' or 'no' would be very valuable. Glkanter (talk) 05:06, 23 January 2011 (UTC)[reply]

Would you agree that Selvin also offers a simple solution, via Monty Hall, in his 2nd letter? Glkanter (talk) 05:23, 23 January 2011 (UTC)[reply]

Rick would you please:

Explain what "S0 (explicitly assumes p=1/2)" means for Behrands. This seems like a contradiction.

Indicate whether "S0 (uses p=1/2 without mention)" that you include for Carlton & Chun is the same thing that Morgan criticizes vos Savant for?

I count the following 5 sources as being explicitly critical of simple solutions as per your definition. Have I overlooked any?:

• Falk

• Gillman

• G & S

• Morgan

• Rosenthal

Thank you. Glkanter (talk) 06:06, 23 January 2011 (UTC)[reply]

K&W is online. They offer an unconditional solution (i.e. S1). I've updated the table.

Regarding Richard's reply to Nijdam - in his paper, Richard shows a solution (which he credits to Boris, which you've quoted elsewhere) to the conditional problem consisting of showing the unconditional probability is 2/3, and then (by symmetry) showing all conditional probabilities are the same, and then (by the law of total probability) showing this means the conditional probabilities must also be the same as the unconditional probability. What he's saying in his reply to Nijdam is that he agrees that the simple solution does not solve the conditional problem, but one form of a conditional solution uses a simple solution plus additional arguments based on symmetry and the law of total probability. Do I agree with this? I do, but why do you care?

Regarding your conclusion - if you're saying exactly the same thing Richard is, then I've already agreed (just above). If you're saying anything that's even slightly different, then you'll need to clarify what the difference is. But again, why do you care?

If you want to munge the table, feel free.

I'm not exactly sure this is complete, but I believe the following are all not just math PhDs but math professors: Rao, Barbeau, Behrends, Carlton, Chun, Devlin, J.Gill, R.Gill, Gillman, both Grinstead and Snell, Henze, Morgan (and every et al), Rosenhouse, Rosenthal.

Quite a few are indeed "maths professors" but one should take account of their specialism: quite a few are specialists in maths education, not specialists in statistics and/or probability. Richard Gill (talk) 14:22, 25 January 2011 (UTC)[reply]

Sevlin's solution in his 2nd letter is conditional. He mentions Monty Hall's letter, but does not present this as a solution - so no, I would not agree Selvin present a simple solution in his 2nd letter.

He praises Monty Hall's solution, saying, "I couldn't have said it better myself". Omitting this is BS. Glkanter (talk) 00:04, 24 January 2011 (UTC)[reply]

Hear, hear! And Monty Hall's solution sees randomness in the choice of the player, not in the actions of the host and his TV-show team! It's a beautiful little mathematical-economist's / game-theoretician's solution. Richard Gill (talk) 14:22, 25 January 2011 (UTC)[reply]

Behrends presents a conditional solution using p=1/2 (which is needed for the 2/3 solution) after saying he's explicitly assuming p=1/2.

Carlton and Chun use p=1/2 without saying anything about it, which is Morgan's criticism of F6 (not what he criticizes vos Savant for, which is that she uses an unconditional solution).

Morgan criticizes 4 simple solutions as false, attributing only 1 of them to anybody at all, that being vos Savant. What else besides the 50/50 host bias (which means there is no symmetry) being missing can Morgan be criticizing about these unattributed solutions, when they call them 'false'? Glkanter (talk) 10:06, 23 January 2011 (UTC)[reply]

5 explicitly critical of simple solutions in this list sounds right. -- Rick Block (talk) 07:27, 23 January 2011 (UTC)[reply]

Indeed, Gill (2010, 2011) uses Boris's symmetry argument to solve the conditional problem via a simple solution to the simple unconditional problem. He also shows that no 50-50 host no bias assumption is needed to find the simple solution. According to Gill, Selvin in his second letter approved of Monty Hall's simple solution, where randomness is seen in the player's choice, not in the host's choices (decision theory avant la letttre). Selvin is a "fusion" person, an eclectic person, a pragmatic person, he likes all solutions. Selvin and I are not dogmatic. We present various solutions using various assumptions and with various targets and we let the reader choose for themselves. Richard Gill (talk) 14:26, 25 January 2011 (UTC)[reply]

Well, I care, because by knowing your POV, then I can look for the information that I can bring forward that would address your objections to my POV. And I can then better state (correct?) my POV. Not unlike a salesperson trying to understand exactly why the customer won't say 'yes' to a closing question. Until the salesman can remedy whatever the (unstated) concerns are, he can't change the customer's decision.

But anyways, in my own words, are you saying (1) that the proper combination & explanation of a simple solution + symmetry + the law of total probability create a conditional solution? And (2) this new conditional solution is a proper method of solving the conditional 50/50 MHP? Glkanter (talk) 08:25, 23 January 2011 (UTC)[reply]

Here's an example of how you guys tell me I'm wrong, etc., but I'm trying very hard to get it right. In the paragraph above you describe Richard's response as:

"What he's saying in his reply to Nijdam is that he agrees that the simple solution does not solve the conditional problem, but one form of a conditional solution uses a simple solution plus additional arguments based on symmetry and the law of total probability.

And, as I've pointed out a few times, Richard wrote this:

"It is a plea for future generations to preserve the life of The True Monty Hall paradox, and not let themselves be misled by probability purists who say “you must compute a conditional probability”. - The Three Doors Problem...-s Richard D. Gill Mathematical Institute, University Leiden, Netherlands http://www.math.leidenuniv.nl/!gill

So, I find the above statements contradictory. And confusing. Glkanter (talk) 09:29, 23 January 2011 (UTC)[reply]

I wrote the following prior to an edit conflict, and did not modify it.

Another helpful column would have, for the sources that directly criticize the simple solutions, any comments they may have made about symmetry. Like Morgan's rejoinder. I'm pretty sure Gillman makes a comment as well. G & S mention that the 50/50 host (symmetry) is required for the simple solution to solve the conditional problem

Also, I think G & S's full statement would be more informative:

"This very simple analysis, though correct, does not quite solve the problem that Craig posed."

I think it should also be indicated that G & S purposefully made the problem simpler, which no other simple solution source does:

"We begin by describing a simpler, related question. We say that a contestant is using the "stay" strategy if he picks a door, and, if offered a chance to switch to another door, declines to do so (i.e., he stays with his original choice). Similarly, we say that the contestant is using the "switch" strategy if he picks a door, and, if offered a chance to switch to another door, takes the offer. Now suppose that a contestant decides in advance to play the "stay" strategy. His only action in this case is to pick a door (and decline an invitation to switch, if one is offered). What is the probability that he wins a car? The same question can be asked about the "switch" strategy.

I think it should also be indicated that G & S require changing the 50/50 host bias to some other values to make the simple solutions fail:

"At this point, the reader may think that the two problems above are the same, since they have the same answers. Recall that we assumed in the original problem if the contestant chooses the door with the car, so that Monty has a choice of two doors, he chooses each of them with probability 1/2. Now suppose instead that in the case that he has a choice, he chooses the door with the larger number with probability 3/4. In the \switch" vs. \stay" problem, the probability of winning with the \switch" strategy is still 2/3. However, in the original problem, if the contestant switches, he wins with probability 4/7. The reader can check this by noting that the same two paths as before are the only two possible paths in the tree. The path leading to a win, if the contestant switches, has probability 1/3, while the path which leads to a loss, if the contestant switches, has probability 1/4."

I think these items are important,as they more fully inform the reader of this table what each source said. It allows that reader to evaluate the criticisms for himself.

Maybe footnotes are the solution. Thank you. Glkanter (talk) 07:57, 23 January 2011 (UTC)[reply]

Morgan's criticism of the simple solutions is consistently the same for all of them, i.e. that they address the wrong probability (the unconditional probability rather than the conditional probability). Specifically, they mention F2 (which they attribute to vos Savant) includes in its analysis the case where the car is behind door 3 (which is impossible if we're thinking about the situation where the player has picked door 1 and the host has opened door 3 showing a goat)! Although much more obvious in this solution, this is true for all simple solutions (it's one way to tell the difference). Their criticism of F6 (a conditional solution, but based on a problem statement that does not say the host has no bias) is that this solution assumes the host has no bias without it being given in the problem statement. This criticism is not even remotely the same as the criticism levied against the simple solutions.

The fact is that assuming the 50/50 host is why Morgan calls the conditional solution false. The absence of this same 50/50 premise is all that Morgan would require to call the simple solutions 'false, because they can only solve the unconditional problem'. Put the 50/50 premise in, where it belongs, of course, and the conditional solution is valid, and at the same time the simple solutions, plus the symmetry from the 50/50 premise, plus the law of total probability, become valid solutions to the conditional problem. Highlighting in red, the unavailable 'door 2 opened' row in vos Savant's solution (which is essentially the same as Selvin's original simple solution) makes it just like the conditional tree in the article where the unavailable options are shown in red. Arguing against this, and failing to indicate this in the table is BS. Glkanter (talk) 00:04, 24 January 2011 (UTC)[reply]

Please stop caring about my POV. Care instead about the POV of the sources. I've said this before, but my desired POV is NPOV.

About the contradiction between Richard's response to Nijdam and the quote from his paper - this has already been explained to you multiple times. I see no point in my attempting to explain it again.

G&S say nothing at all like what you claim ("50/50 host (symmetry) is required for the simple solution to solve the conditional problem"). Perhaps you mean the 50/50 host bias is required for the answer to be the same, but even if the answer is the same the simple solution is not "solving the conditional problem".

No one other than the editors involved in this mediation is ever going to read this table (it's not intended for the article). Are you confused about this? -- Rick Block (talk) 16:49, 23 January 2011 (UTC)[reply]

New Synthesis

Let me suppose that the door chosen by the player X and the door hiding the car C are statistically independent and both uniformly distributed on the set of doors, which I name "L", "M" and "R" for left, middle and right (as seen by the audience). I'll fix the player's door later, so by independence it won't matter that for the time being I take it uniformly distributed. Let H de the door opened by the host and Y be the remaining closed door, i.e. the door different from both X and H. Assume no host bias. By symmetry, (X,H,Y) is a uniform random permutation of ("L","M","R").

We also know H is different from C and we may define G to be the remaing door with a goat. Thus (C,H,G) is also a uniform random permutation of ("L","M","R"). And obviously, either (C,H,G) equals (X,H,Y) or it equals (Y,H,X). Either C=X or C=Y.

We know from the simple solution that the probabilities of these two, complementary, events are 1/3 and 2/3 respectively. By symmetry, these events must be statistically independent of (X,H,Y), the identities of the three specific doors chosen by player, opened by host, and not opened by host, respectively.

So the player should switch, getting the car with probability 2/3, and this is independent of the identity of his initially chosen door and also of the identity of the door opened by the host. Since the door-identities are independent of whether or not X=C we cannot improve on this strategy by taking account of the door numbers. And the player might as well start by choosing door "L". Richard Gill (talk) 14:35, 23 January 2011 (UTC)[reply]

What this proof is saying is that as far as the joint probability distribution of C, X, H (and Y, and G) is concerned, we might just as well define them by choosing (X,H,Y) to be one of (1,2,3), (1,3,2), (2,1,3),(2,3,1),(3,1,2),(3,2,1) each with probability 1/6 and then fixing C=X or C=Y with probabilities 1/3 and 2/3. They have the same joint probability distribution as when we set them up in the usual way. Richard Gill (talk) 14:43, 23 January 2011 (UTC)[reply]

If you get a chance, Richard, could you please comment on the following, please?:

1. My 'tool kit' analogy on Martin's (new) talk page

I've not been able to locate this yet.Richard Gill (talk) 14:07, 25 January 2011 (UTC)[reply]

It's just a few sections above. Glkanter (talk) 14:13, 25 January 2011 (UTC)[reply]

2. Rick's clarification of your comments, just above on this page:

"Regarding Richard's reply to Nijdam - in his paper, Richard shows a solution (which he credits to Boris, which you've quoted elsewhere) to the conditional problem consisting of showing the unconditional probability is 2/3, and then (by symmetry) showing all conditional probabilities are the same, and then (by the law of total probability) showing this means the conditional probabilities must also be the same as the unconditional probability. What he's saying in his reply to Nijdam is that he agrees that the simple solution does not solve the conditional problem, but one form of a conditional solution uses a simple solution plus additional arguments based on symmetry and the law of total probability. Do I agree with this? I do, but why do you care?"

Rick's understanding of my comments seem to be correct. Richard Gill (talk) 14:07, 25 January 2011 (UTC)[reply]

3. The conflict that I perceive between #2 and the conclusion to your 2010 paper. That is, Rick calls it a 'conditional solution', while your conclusion says 'conditional solutions' must *not* be seen as the only solutions

There is no self-contradiction. I believe in the value of a multiplicity of mathematizations of MHP (ie model assumptions and solution desiderata). I am against dogmatism. I think that the consumer needs to be given neutral information about the input and output to any decent "solution", so that they can decide for themselves what they want to take home. By "decent" I mean that the argument to go from input to output must be mathematically/logically sound. Richard Gill (talk) 14:07, 25 January 2011 (UTC)[reply]

While this may make perfect sense to you and Rick, it makes zero sense to a dumbass like me. Glkanter (talk) 14:13, 25 January 2011 (UTC)[reply]

4. Respond to my response regarding your comment that [paraphrasing] 'no one interpretation can claim to define the MHP' just above on this page

Well, I just did that in my answer to your question 3.

Thank you. Glkanter (talk) 16:03, 23 January 2011 (UTC)[reply]

A Direct Appeal To The Mediation Committee

As the above section with the table clearly shows, there is no 'significant minority viewpoint' of the reliable sources to support the favored POV that currently causes violations of Wikipedia policy in the Monty Hall Problem article.

The article currently suffers from violations of NPOV, UNDUE WEIGHT, and OR in misguided support of the proposition, 'All simple solutions are flawed'.

The table above shows:

• There are no more than 5 reliably sourced critics of the simple solutions (this statement has been confirmed by Rick Block). The table above lists about 50 MHP sources, there are hundreds, maybe thousands more who do not criticize the simple solutions in any way

• Each of the 5 'critics' could (more) reasonably be construed as 'supporters' of the simple solutions

It takes Orwellian constructs to interpret phrases like Rosenthal's "...simple solution is actually correct, but..." into a criticism

• There are reliable sources who directly defend the simple solutions

• There are reliable sources that provide both types of solutions

• There are countless hundreds, maybe thousands of reliably sources that use only a simple solution

Please, members of the Mediation Committee, guide me to the resources necessary to end this unwarranted intrusion into the MHP article after nearly 3 years. Thank you. Glkanter (talk) 18:48, 23 January 2011 (UTC)[reply]

The table above clearly shows that in reliable publications it is mainly (M0,S0) or sometimes (M1,S1), and some of these sources criticize (M0,S1). Let's end this stupid years too long discussion, and present M0 as the (basic) MHP with solution S1 or S2 if that's easier to understand. Let us not forget to mention the criticism on the sources that present (M0,S1) and especially let us point to the logical error in S3 and S4. Nijdam (talk) 00:29, 24 January 2011 (UTC)[reply]

Just for the record, a couple of corrections to the above:

The article does not currently and (as far as I know) has never said "all simple solutions are flawed". The claim that it "suffers from violations of NPOV, UNDUE WEIGHT, and OR in support of [this] proposition" is completely unsubstantiated. Glkanter has been asked repeatedly to identify exactly where in the article these alleged violations occur [2] [3] [4] [5] [6] [7], responding perhaps most eloquently here.
There are indeed 5 of the existing sources referenced in the article which "criticize" simple solutions (actually, they clarify what probability the simple solutions address), however this does not mean that these are the only such sources that exist. There are indeed other such sources.
Suggesting these 5 sources "could (more) reasonably be construed as 'supporters' of the simple solutions" is simply ludicrous.

I agree. Richard Gill (talk) 16:13, 29 January 2011 (UTC)[reply]

As explained above, the Rosenthal paper is structured as a criticism of one simple solution. What Glkanter is doing here is taking one quote from this paper out of context and using this quote to invert the meaning of the entire paper.

From the table above it is quite obvious there are sources presenting simple solutions and sources presenting both simple solutions and conditional solutions. There are indeed far more that present simple solutions as well as far more that present conditional solutions. There are also sources that discuss the difference between simple and conditional solutions (5 in the table above, more exist as well), which certain editors here apparently consider to be "criticism" of simple solutions. The question all along has been how to deal with this plethora of sources, and in particular the discussion of the difference between simple and conditional solutions, in an NPOV manner. -- Rick Block (talk) 16:29, 24 January 2011 (UTC)[reply]

Martin Hogbin's opinion

As I have withdrawn completely from this mediation now I have given my complete thoughts on the article below. Further discussion seems pointless as there is no sign of anyone changing their opinion or of a consensus emerging. Please do not intersperse comments but feel free to collapse this section.

Proposal for the article

I believe that the article should give simple solutions to the MHP first, citing some of the many sources that give such solutions.

These solutions should not include any prominent 'health warnings' stating that they are incomplete or answer the 'wrong question'. I would not object strongly to careful wording of the simple solutions in order to reduce or remove what some see as technical errors in them, provided this is done in a non-obtrusive way that will not obstruct their understanding and acceptance by the average reader.

The simple solutions should be followed with a section explaining, with reference to the simple solutions (and maybe discreetly to the unconditional formulation) why the answer is 2/3 and why it matters that the host knows where the car is. This is what everybody has difficulty with and gets wrong.

Following this I would support a scholarly and comprehensive discussion of all aspects of the problem, including criticism of the simple solutions, responses to that criticism, and game theory, arranged in any logical manner.

Rationale

There are no 'bomb-proof' arguments, including mine, regarding the simple solutions to the MHP (despite the protestations of some). What I propose to give is a collection of reasons, from varying perspectives, why the article should be as I propose. I assume throughout this discussion the standard game rules, that the host always opens an unchosen door to reveal a goat and always offers the swap. I also use the word 'answer' to mean the value of the probability that the player will win the car by switching in the specified circumstances.

The MHP is a problem is an unconditional problem

By far the most notable and well known statement of the MHP is the question by Whitaker to vos Savants column. This was a question by a member of the general public to a regular column in a popular general-interest magazine. The question was changed by vos Savant to make it easier to explain by the addition of door numbers.

It is far from clear whether the intention was to ask a conditional or an unconditional question. What did Whitaker actually want to know? Although Whitaker actually asked if it is better to switch, as there is no dispute about this point, I have extended his question to ask by how much. Was it:

1) I am on a quiz show in which the standard rules apply. I have chosen a door and I have just seen another door opened to reveal a goat. Given this information, and taking into account the specific door opened by the host, what is the probability of winning if I switch. (Conditional formulation)

2) There is a quiz show in which the standard rules apply. Is it better for a contestant to switch or stick and by how much? (Unconditional formulation)

Is it really possible that Whitaker meant to ask question 1)? Would a member of the public be asking a conditional probability question in which the door number opened by the host is of significance, to a popular magazine. Hardly. It was some ten years after vos Savant's reply before anybody even proposed this interpretation. (Although Selvin, in response to comments to his problem even earlier, had noticed that possibility and quickly acted to nullify it. However, neither vos Savant nor the original critics, Morgan et al, seemed to be aware of the previous discussion).

There is no doubt in my mind that what Whitaker actually wanted to know was the answer to 2) In fact had he meant to ask 1) it is quite remarkable that he did not immediately respond to vos Savant's solution with an argument along the lines of Morgan. There is no sign that he did anything like this.

Even if given a condition we are entitled to ignore it

In Whitaker's statement we are told that the host says, 'Do you want to pick door No. 2?'. We are not told that he always uses these words. Consider for a moment what the situation would be if we were specifically told that sometimes the host says, Do you want to pick door No. 2?' and other times he says, 'Do you want to choose door No. 2?'. We would then need to take the host's words as a significant condition of the problem. We would need to consider introducing a parameter such as s, the probability that the host will say 'choose' given that the car is behind door 1. As it is, the situation is not so clear cut and people makes the decision to ignore what could be a vital clue as to the actual location of the car. They do so at their peril, however. In a show in which the host is permitted to influence the outcome, it is quite possible that he might give strong clues to the player as to the present location of the car.

Why then do we ignore this possibility. Because common sense tells us that it is not intended to be an important condition of the problem. Common sense also tells us that the door number opened by the host is not meant to be important. Even though we have been given that information, we should ignore it.

Suppose the problem is conditional

How does the host choose between goat-doors when he has a choice

Let us be a little perverse then. Let us assume that Whitaker actually meant to ask the conditional version of the problem. What is the correct solution.

We now have to ask the question of how the host chooses between the legal goat-doors. Does he choose uniformly or might he have some bias?

If we are subjectivists we may well decide to apply the principle of indifference and take the unknown distribution to be uniform. If, on the other hand, we are frequentists we might observe that a random choice by the host is a requirement of quiz shows and again take the distribution to be uniform.

Regardless of what our view of probability is we might conclude that, if we are consistent in our approach, and do not take the unknown distributions in Whitaker's statement to be uniform, then the problem is rather boringly insoluble. On any reasonable basis we must take the host's legal goat-door choice to be uniform; no source which addresses this issue does otherwise.

Solving the symmetrical conditional problem

Having decided, as we must, that the host is unbiased we now ask ourselves what the difference is between the two formulations, one in which a unknown door is opened by the host and the other in which a known, but irrelevant, door is opened. Sure, there are two subtly different concepts, an unconditional case in which no special condition is given, and a conditional one in which a specific door is opened. One requires our sample space to be conditioned and the other does not. However, even that distinction is not that clear cut. In what we generally call the unconditional case, there is a condition, that the host opens an unchosen door to reveal a goat. It is only by convention that, because that event is part of the game rules, and therefore happens with certainty, we do not use it as a condition.

So are the simple solutions valid?

The simple solutions solve the unconditional problem and, by symmetry, the conditional problem must have the same answer as the unconditional problem. Thus the simple solution solves the conditional formulation.

The common objection to this solution is that neither we nor any of the sources include this symmetry argument in their simple solutions thus they are incomplete. However, no solution is totally complete. In particular the conditional solution shown in the article is far from complete. The player is stated to have chosen door 1, but that is not a rule of the game, it does not happen with certainty, so our diagram should show the ample space including all the possible original choices that the player might have made and then, in some way, condition it based on the choice given. For most people there is no need to do this because it is quite obvious to them, because of the complete symmetry with respect to door number, that we need only show one case that the player initially chooses door 1. The situation is similar with the simple solution. We should state that the simple solution gives the answer to the unconditional formulation and that this formulation has exactly the same answer as the conditional case, but we have no need to say this because it is quite obvious.

How to present the solutions

Let us now be perverse and pedantic. Whitaker intended to ask the conditional question and the simple solution does not solve it. How should we present this in an encyclopedia intended to be accessible to the general public. Let me first give two approaches that I consider wrong:

1) Ignore the fact. This is only an encyclopedia for ordinary people, they will not notice a minor technical error.

2) Hit them with it. Give only complete and rigorous solutions right from the start or, at least, add prominent 'health warnings' to the simple solutions pointing out that they are, in fact, wrong. If the readers cannot understand the 'proper' solution that is just too bad, they do not deserve to understand the subject if they cannot do it properly.

I do not support either of the above approaches and neither do most good text books and encyclopedia articles. What they do is start with a simple version of the subject, sometimes with certain technical difficulties glossed over, and then, when the main issues have been understood (in our case that the answer is 2/3 and it does matter that the host knows where the car is) go through the details. It is very common to see in text books things like 'Equation 1 is not strictly correct but actually relies on the instated assumption... We will now look at that in more detail...'. To do things in this way is not structural bias but writing an article so people can understand it It is true that some authors do insist on getting things correct right from the start but they are usually writing for well defined technical audiences. We are writing for a wide range of readers, with the vast majority being non-expert.

In summary I do not think that a perverse and pedantic approach should be allowed to spoil the primary function of this article which is to inform our readers.

I have just noticed this rather rude and unnecessary edit summary by Glopk. 'Collapsed Martin's section (either you are in the mediation, or you are out. If out, please avoid spamming)'. I believe that I am entitled to leave a clear record of my opinion for the mediators to refer to if they wish. I stated at the top that I was quite happy for the section to be collapsed but it is not spam. Martin Hogbin (talk) 12:06, 29 January 2011 (UTC)[reply]

Richard Gill's opinion
I like the wise thoughts expressed in this article, Martin. Richard Gill (talk) 15:50, 29 January 2011 (UTC)[reply]

Let's Fix The Single Most Wikipedia Policy Violating Section In The Article

Would each editor please offer their response to the proposal that follows:

posted by Glkanter (talk) 20:46, 26 January 2011 (UTC)[reply]

Copied from above:

Proposal

What do you think of the "Conditional solution" section in the current article, then? Do you agree with me that paragraphs 1, 2 & 4, and the 2nd diagram add little or no value, and actually serve to confuse the solution sections badly? Glkanter (talk) 19:48, 26 January 2011 (UTC)

Richards's Response

I agree. I once got set upon by a pack of dogs because I deleted one of the more excruciating paragraphs since in view of the symmetry argument it seemed to me totally superfluous, worse than superfluous. Richard Gill (talk) 20:02, 26 January 2011 (UTC)

The response above was moved here by copy-paste. It's ambiguous, since two questions were asked. Actually I only responded to the first question.

So to be more precise and factual: I agree that the section on the conditional solution needs a thorough overhaul. I believe that it could and should be a whole lot shorter, and at the same time it could and should be a whole lot more accessible to readers without formal mathematical training. The ideas have to be brought across and the results have to be obtained in a transparent way. I have not answered, and am not answering, the second question, which makes specific proposals concerning particular paragraphs. Richard Gill (talk) 16:27, 29 January 2011 (UTC)[reply]

Glopk's Response

Disagree entirely. Paragraph 1 explains why it is worth considering a Conditional Probability Solution at all, as it is not trivially identical to the unconditional one (in the opinion of relevant sources and in fact). Paragraph 2 summarizes the sources themselves, and Paragraph 4 describes how the Cond. Prob. Solution is not trivially identical to the unconditional one. The points addressed by these paragraphs are all necessary for a section on the Cond. Prob. Solution is to be at all included in the article. If Glkanter thinks they are obscure, he should propose improvements. If he thinks there should not be such a section, he should simply say so, and then explain how the so-reduced article would present a NPOV when the POV of important sources is ignored. Last, the second figure is an alterntive pictorial illustration of the decision tree, and I think it is helpful to the reader. glopk (talk) 00:54, 27 January 2011 (UTC)[reply]

Thank you for the thoughtful response. Can you explain how these problems with differing premises help the reader understand the '2/3 & 1/3 vs 50/50' paradox? Could you associate your comments and conclusion to reliable sources? By the way, Glkanter thinks there should be a single solution section that appends the conditional decision tree paragraph with 1 image and the Bayes Theorem solution onto the existing Simple solution section. Glkanter (talk) 01:16, 27 January 2011 (UTC)[reply]

No, I don't have to explain anything - the sources already have, and the article provides the text/sources association already. The text itself may be improved, but to suggest that the Cond. Prob. Sol. section is not well sourced is plainly wrong. Further, I remind you that this is intended to be an encyclopedic article on the Monty Hall Problem, rather than a mere a tutorial for illiterate readers who need help to "understand the '2/3 & 1/3 vs 50/50' paradox": there are important sources speaking at a non-elementary level, and what they say isn't stuff for "High Priests" (your words) that needs be moved/hidden elsewhere, but rather valuable contributions that need be reported in a balanced and neutral manner. To do otherwise would be a disservice to the readers.

Responding to your suggestion that there should be only one section, I rather doubt, given the diversity of the sources and the conceptual difference between simple and conditional solutions, that one single section could be inclusive and readable without becoming a monstruosity. However, I won't pass judgment until I see what exactly you have in mind - care actually writing what you propose somewhere? glopk (talk) 16:35, 27 January 2011 (UTC)[reply]

Rick Block's Response

I don't think the current "Conditional solution" section is very well written, and have suggested improvements in clarity and writing style (such as here) - but this is an entirely different matter than what is being suggested here, which is clearly an attempt to make this section conform to Glkanter's POV by simply deleting the (very well sourced) content he disagrees with. It's not entirely obvious, but Richard's comments above were not written specifically in response to this request but rather copied from another thread on this page. -- Rick Block (talk) 01:51, 27 January 2011 (UTC)[reply]

I find your characterizations of this thread untrue, bogus and unwarranted. Bordering on an personal attack.

The same paragraphs that I suggest deleting occur numerous other times throughout the article. And let's be honest here, it's your favored POV that I'm trying to eliminate from the section due to UNDUE WEIGHT violations, NPOV violations, and OR violations. Very little of all that lengthy narrative comes from any sources, and there's no benefit to the reader by that stuff being in the solutions section. Glkanter (talk) 02:15, 27 January 2011 (UTC)[reply]

Other Editors' Responses

The 5 alleged critics that dominate the article

It's Go Time boys. I will demonstrate that the alleged critics of the various simple solutions are *not* saying what some editors are claiming. I will do this with nothing more than the English language and Logic.

One source at a time, please. Starting, of course, with Morgan. Who wants to go first?.

What is Morgan's criticisms, in their own words? Glkanter (talk) 00:40, 30 January 2011 (UTC)[reply]

Until AGK provides a further update, I suggest you give it a rest. -- Rick Block (talk) 01:41, 30 January 2011 (UTC)[reply]

I offered. Glkanter (talk) 02:55, 30 January 2011 (UTC)[reply]

Morgan's Criticisms Of The Simple Solutions

What I Will Demonstrate

Morgan calls vos Savant's simple solution, F2, 'False'. They also call 3 other unattributed simple solutions, F1, F3, and F5 'False'. They also call a conditional decision tree (in word form only), F6 'False'. None of the solutions are called 'false' because they are 'incomplete', or poorly phrased. The 4 simple solutions are called false because they do not address the specific situation in the problem statement that door 3 has been opened. Morgan concludes these simple solutions are only capable of solving a different, unconditional problem where 'another door' is opened, rather than specifically door 3. The conditional solution is called 'false' only because it relies on the 50/50 host bias premise, which Morgan claims is neither explicitly stated or implied by the Whitaker/vos Savant problem statement. Morgan's rejoinder includes Selvin's 2nd letter as a reference, but they make no mention of Selvin's explicit 50/50 host bias premise in that letter.

Their paper begins: "In a trio of recent columns, ...", i.e. they're talking about Whitaker's problem statement including clarifications vos Savant published in all 3 Parade columns, perhaps most tellingly the experimental procedure she suggests in her third column where the initial location and player's choice are both explicitly randomized but the host's choice is not. -- Rick Block (talk) 17:16, 30 January 2011 (UTC)[reply]

They conclude that vos Savant is solving an unconditional problem because one of the 3 equally like outcomes she shows includes doors 1 & 2 having goats and door 3 having the car (notated as 'GGA'). They point out that the problem statement has door 3 being opened, so GGA is not a possible outcome. (It's interesting they don't have this same concern for the conditional solution F6, which offers 4 outcomes, and, as Morgan notes, the same lettering scheme as vos Savant's F2. Perhaps if vos Savant had highlighted the 'door 3 being the car' outcome in red type, as is done for the 2 cases in the conditional decision tree in the Wikipedia article, this would be a moot point.)

The unconditional sample space for F6 is {AGG2, AGG3, GAG3, GGA2} and the solution then computes the conditional probability given a goat is behind door 3 (i.e. involves only AGG3 and GAG3). Your objection here is nonsensical. -- Rick Block (talk) 17:16, 30 January 2011 (UTC)[reply]

In their rejoinder, Morgan mentions somewhat casually of the 50/50 host premise:

"...and that 2/3 is the answer to the relevant conditional problem only if p = q = 1/2. Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one. But then, it may be that only an academician, and no one connected with a game show, would ever consider p <> q.

In that statement, Morgan is saying that every solution that relies on the symmetry from the 50/50 host bias premise solves the conditional problem. According to Morgan then, this 50/50 host bias premise changes the conditional solutions from false to valid and changes the simple solutions from false to valid. That is, the simple solutions are no longer solving the 'wrong problem'.

No, that's not what they're saying. This has been explained to you before [8]. -- Rick Block (talk) 17:16, 30 January 2011 (UTC)[reply]

Morgan seems to have (finally) recognized the universal interpretation that the 50/50 host bias premise is a part of the MHP, when in their 2010 response to Martin and Nijdam, they write:

"In any case, it should not distract from the essential fact that 1/(1+q) ≥ 1/2 regardless of q. Simply put, if the host must show a goat, the player should switch.

We take this opportunity to address another issue related to our article, one that arose in vos Savant’s (1991) reply and in Bell’s (1992) letter, and has come up many times since. To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period.

In their original paper, they say the result is not 2/3 & 1/3 because the 50/50 host bias premise is not stated by vos Savant.
In their rejoinder they say the result is only 2/3 & 1/3 with the 50/50 host bias premise.
In their response to Martin and Nijdam, they say, "To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period."

Those 'conditions' are the 50/50 host bias premise, which, contrary to their paper, they are now acknowledging is part of the MHP.

Therefore, Morgan has corrected themselves, in a very opaque, but provable way, that the simple solutions, with symmetry and the law of total probability solve the conditional problem where door 3 has been opened. Morgan is no longer calling the simple solutions, or the conditional solutions that return the result 2/3 & 1/3, false.

Your conclusion here is erroneous. From their rejoinder to vos Savant: "One of the ideas put forth in our article, and one of the few that directly concerns her responses, is that even if one accepts the restrictions that she places on the reader's question, it is still a conditional probability problem." They have never backed away from this stance. -- Rick Block (talk) 17:16, 30 January 2011 (UTC)[reply]

Unedited quotes follow:

Morgan's Original Paper

1. TO SWITCH OR NOT TO SWITCH

Solution Fl. If, regardless of the host's action, the player's strategy is to never switch, she will obviously win the car 1/3 of the time. Hence the probability that she wins if she does switch is 2/3.

Fl is immediately appealing, and we found its advocates quite reluctant to capitulate. Fl 's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand. Fl is a solution to the unconditional problem, which may be stated as follows:

"You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?" The distinction between the conditional and unconditional situations here seems to confound many, from whence much of the pedagogic and entertainment value is derived.

Solution F2. The sample space is {AGG, GAG, GGA}, each point having probability 1/3, where the triple AGG, for instance, means auto behind door 1, goat behind door 2, and goat behind door 3. The playerchoosing door 1 will win in two of these cases if she switches, hence the probability that she wins by switching is 2/3.

Solution F2 is offered by vos Savant in the December article, and may perhaps best be thought of as an attempted solution to the unconditional problem. That it is not a solution to the stated conditional problem is apparent in that the outcome GGA is not in the conditional sample space, since door 3 has been revealed as hiding a goat.

Solution F3. Play the game a few hundred times with the "host" using three cards: two jokers for the goats and an ace for the car. This will verify that the player who switches wins 2/3 of the time.

Several people, frustrated by contradictory arguments or failing to believe their arguments wrong, suggested schemes like F3 to settle the issue, which was proposed by vos Savant in the December article (compare, also, the classroom experiment proposed by vos Savant in the February column). It is so appealing because it models Fl: This is a correct simulation for the unconditional problem, but not for the conditional problem. The correct simulation for the conditional problem is of course to examine only those trials where door 3 is opened by the host. The modeling of conditional probabilities through repeated experimentation can be a difficult concept for the novice, for whom the careful thinking through of this situation can be of considerable benefit.

Solution F4. The original sample space and probabilities are as given in Solution F2. However, since door 3 has been shown to contain a goat, GGA is no longer possible. The remaining two outcomes form the conditional sample space, each having probability (1/3)/(1-(1/3)) = 1/2. Hence the probabilities of winning by switching and by not switching, given that door 3 has a goat, are both 1/2, and the player's choice is a moot one in so far as probability is concerned.

In F4 we at last have an attempt to solve the conditional problem. This attempt fails because the original sample space is incorrectly specified. We suspect this may be the solution used by the Ph.D. 's quoted in the December article. This answer is also obtained by the naive statement that, since there are two doors left, one of which contains a goat, each has probability 1/2. By whatever method of determination, 1/2 has in our experience been the most popular "wrong" answer.

Solution F5. The probability that a player is shown a goat is 1. So conditioning on this event cannot change the probability of 1/3 that door 1 is a winner before a goat is shown; that is, the probability of winning by not switching is 1/3, and by switching is 2/3.

Solution F5, like F1, is a true statement that answers a different problem. F5 is incorrect because it does not use the information in the number of the door shown.

Solution F6. The sample space is {AGG2, AGG3, GAG3, GGA2} where the letter triples have the same meaning as in Solution F2, and the number indicates the door opened by the host. The probabilities for the sample points are, in order, 1/6, 1/6, 1/3, and 1/3. Labeling events W, = "win by switching" and D3 = "goat shown behind door 3," Pr(W, I D3) = Pr(W, and D3)/Pr(D3) = Pr(GAG3)/Pr(AGG3,GAG3) = (1/3)/(1/6 + 1/3) = 2/3.

Solution F6 (cf. Mosteller 1965) attempts to correct the wrongly specified sample space of F4. One must ask, however, how the probabilities for this sample space are determined. It turns out that this is a correct specification only if one assumes a certain strategy on the part of the host. We will show that the problem can be solved without any assumptions of this type, which is to say the problem can be solved.

2. CONCLUSIONS In general, we cannot answer the question "What is the probability of winning if I switch, given that I have been shown a goat behind door 3?" unless we either know the host's strategy or are Bayesians with a specified prior. Nevertheless, in the vos Savant scenario we can state that it is always better to switch. The fact that Pr(W | D3) >= 1/2, regardless of the host's strategy, is the key to the solution.

Morgan's Rejoinder To vos Savant In The Same Issue

From this and her previous solutions, one is tempted to conclude that vos Savant does not understand that the conditional problem (of interest to the player) and the unconditional problem (of interest to the host) are not the same, and that 2/3 is the answer to the relevant conditional problem only if p = q = 1/2. Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one. But then, it may be that only an academician, and no one connected with a game show, would ever consider p # q. ADDITIONAL REFERENCES Rasmusen, E. (1989), Games and Information: An Introduction to Game Theory, Oxford: Basil Blackwell. Selvin, S. (1975), "Letter to the Editor," The American Statistician, 29, 134.

And how about: "One of the ideas put forth in our article, and one of the few that directly concerns her responses, is that even if one accepts the restrictions that she places on the reader's question, it is still a conditional probability problem. One may argue that the information necessary to use the conditional solution is not available to the player, or that given natural symmetry conditions, the unconditional approach necessarily leads to the same result, but this does not change the aforementioned fact." Richard Gill (talk) 13:19, 2 February 2011 (UTC)[reply]

Morgan's Responses To Martin And Nijdam In 2010

Response Our kind thanks to Mr. Hogbin and Dr. Nijdam for correcting our mistake. We will add that should the player have observed any previous plays of the game, those data, too, will modify the prior, and can produce posterior calculations other than 2/3 even with a symmetric prior. This, of course, is something else that we should have pursued. In any case, it should not distract from the essential fact that 1/(1+q) ≥ 1/2 regardless of q. Simply put, if the host must show a goat, the player should switch. We take this opportunity to address another issue related to our article, one that arose in vos Savant’s (1991) reply and in Bell’s (1992) letter, and has come up many times since. To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period.

Posted by Glkanter (talk) 06:48, 30 January 2011 (UTC)[reply]

Responses From The Other Editors To Morgan

In my opinion, Morgan et al. (2010) maintain their (1991) point of view that MHP has to be solved by computing a conditional probability. Their comment about Bell, 1992, refers to the fact that for q=1/2, the case many people - including Glkanter - take as implicit in the problem - the conditional probability is also 2/3. Morgan et al. also admitted during the 1991 round of comments and rejoinders that this conditional probability could have been got more easily by using the symmetry of the q=1/2 situation (a reference which Glkanter himself pointed out to us some time ago, and I think Rick also knew about).

This is what William Bell wrote in 1992: "The authors stress the difference between the conditional and unconditional problems, and I wish to point out something I feel is interesting in this regard for the case where, as vos Savant assumes, the host acts solely as an agent of chance. I would agree with her that this is a reasonable assumption, given her intended audience."

He goes on, in the symmetric case, to show that (when the player has chosen Door 1), whether or not a car is behind door 1 is statistically independent of whether the host opens Door 2 or Door 3. He says he'll leave it to readers to decide for themselves whether or not this statistical independence is immediately obvious, or whether it needs formal verification. He also says "If the independence [...] is judged intuitively obvious, then the use of a solution to the unconditional problem (e.g. the solution offered by vos Savant that the authors label F2) is valid."

Already in response to Seymann, Morgan et al wrote "... the condition p = q = 1/2 should have been put on ... . It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one". Richard Gill (talk) 13:37, 2 February 2011 (UTC)[reply]

By the way, when Morgan et al. (2010) write "to wit, had we adopted conditions implicit in the problem, the answer is 2/3, period" my understanding is that these words are intended to be a paraphrase of Bell's (1992) and Vos Savant's (1991)n point of view.

They continually reiterate that they want to solve MHP from the player's point of view and for them, this entails probabilistic conditioning on host opening Door 3. In 2010 they firmly reiterate the result of their (1991) paper that the conditional probability of winning by switching is 1/(1+q) which is at least 1/2, whatever q might be, hence the player should switch, anyway. They consider that Vos Savant solved the problem from the host's point of view. They merely agree in 1991 and in 2010 that q=1/2 seems to be the natural assumption and that one can get the conditional probability from the unconditional in this symmetric case. Richard Gill (talk) 13:52, 2 February 2011 (UTC)[reply]

Grinstead and Snell

It makes sense that a textbook on (conditional) probability would uncritically present a formal conditional solution to the MHP. And then, since it's such overkill for the problem, devise reasons & complexities (non-symmetric host bias, forgetful Monty, devious Monty) that engage the students , in order to show the capabilities of conditional probability.

I think this is the relevant passage from G & S. They are *not* describing the recognized MHP, nor are they claiming to. They intentionally restate the problem in order to make 'some' point more evident, and less ambiguous:

"We begin by describing a simpler, related question. We say that a contestant is using the "stay" strategy if he picks a door, and, if offered a chance to switch to another door, declines to do so (i.e., he stays with his original choice). Similarly, we say that the contestant is using the "switch" strategy if he picks a door, and, if offered a chance to switch to another door, takes the offer. Now suppose that a contestant decides in advance to play the "stay" strategy. His only action in this case is to pick a door (and decline an invitation to switch, if one is offered). What is the probability that he wins a car? The same question can be asked about the "switch" strategy.

"This very simple analysis, though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3."

"At this point, the reader may think that the two problems above are the same, since they have the same answers. Recall that we assumed in the original problem if the contestant chooses the door with the car, so that Monty has a choice of two doors, he chooses each of them with probability 1/2. Now suppose instead that in the case that he has a choice, he chooses the door with the larger number with probability 3/4. In the "switch" vs. "stay" problem, the probability of winning with the "switch" strategy is still 2/3."

G & S do *exactly* what I describe above. Which is *not* a criticism of any other solution, just an example of why you *might* need a conditional decision tree or Bayes, for some other problem. G & S are not 'critics' of simple solutions at all. Glkanter (talk) 11:48, 31 January 2011 (UTC)[reply]

The point they are making is precisely the point you keep refusing to acknowledge (e.g. [9] [10]), which is that asking whether always switching is preferable to always staying regardless of which door the host opens (this is their "simpler, related question") is different from asking whether it is better to switch given which door the player has picked and which door the host has opened (which is the conditional probability that they say, and you quote, "Craig asked for"). They introduce a host preference variant to demonstrate these are indeed two different questions. In this variant, the always switch answer is STILL 2/3, but the conditional probability given the door 1 and door 3 combination is not. This, of course, has already been explained to you numerous times. -- Rick Block (talk) 15:07, 31 January 2011 (UTC)[reply]

Rick, take my advise, stop responding to Glkanter's nonsense; it's pointless. Nijdam (talk) 20:56, 31 January 2011 (UTC)[reply]

Is it expecting too much of you Nijdam, to ask you, for once, finally, to explain or otherwise substantiate a post you make? Or will you continue to make veiled 'hit and run' disruptive troll attacks? And will Rick Block, The Admin, continue, after 2 years, to turn a blind eye to your gamesmanship, fake and perverse use of 'The Socratic Method', and other non-contributory tactics? I'm predicting 'yes' and 'yes'. Glkanter (talk) 21:14, 31 January 2011 (UTC)[reply]

These are among the logical fallacies in your argument, above, that have led me to consider, then reject your 'explanations':

You ignore that fact that Selvin and vos Savant each offered unconditional solutions to the problems that they *each* developed.
You incorrectly presume that the problem *must*, for some unknown reason, be solved with a conditional solution that makes specific notice of the door #1 and door #3 pairing.
You ignore the obvious symmetry of all 6 door pairings being equally likely as per the 2 'uniformly at random' premises that are explicit from Selvin, K & W, implicit from vos Savant, and consistent with "Suppose you're on a game show..."
G & S criticizes a different problem statement. Not the solutions.
You incorrectly assume that because G & S associates the simple conditional solutions with their '...simpler, related question.', that *all* sources who provide a simple conditional solution are solving that '...simpler, related question.'
G & S has to contrive a *third* problem that removes or replaces the 50/50 host bias premise in order to show the simple conditional solutions have limitations. You incorrectly interpret that as a valid 'proof' that the simple conditional solutions are being criticized.
You incorrectly conclude that the 5 so-called critics comprise a 'significant minority viewpoint' for Wikipedia purposes.
You incorrectly believe that your interpretation of the 5 so-called critics is the only possible, reasonable interpretation.
You incorrectly use repetition as a means of attempting to make your argument valid. That is known as 'propaganda'.
You incorrectly assume that anyone who does not agree with the above nonsense does not 'understand' your argument.

So, each time you repeat any of these flawed arguments, I will respond, as above. I will do so for another 2+ years, or longer, as necessary. Glkanter (talk) 15:48, 31 January 2011 (UTC)[reply]

Responses From The Other Editors To G & S

I think Rick is right and Glkanter is wrong concerning G & S. I recently re-read their contribution to MHP. It is a pity in an otherwise fine book that they take such a dogmatic position and just as Morgan et al, claim the right to know what Craig Whitaker was really asking. Which, let us not forget, was: "I’ve worked out two different situations (based on ... whether or not he knows what’s behind the doors) - in one situation it is to your advantage to switch, in the other there is no advantage to switch. What do you think?" Richard Gill (talk) 13:57, 2 February 2011 (UTC)[reply]

Rosenthal

Excerpted from his paper

2 The Monty Hall Problem and Variants

The original Monty Hall problem may be summarised as follows:

Monty Hall Problem: A car is equally likely to be behind any one of three doors. You select one of the three doors (say, Door #1). The host then reveals one non-selected door (say, Door #3) which does not contain the car. At this point, you choose whether to stick with your original choice (i.e. Door #1), or switch to the remaining door (i.e. Door #2). What are the probabilities that you will win the car if you stick, versus if you switch?

Most people believe, upon first hearing this problem, that the car is equally likely to be behind either of the two unopened doors, so the probability of winning is 1/2 regardless of whether you stick or switch. However, in fact the probabilities of winning are 1/3 if you stick, and 2/3 if you switch [3]. This fact is often justified as follows:

Shaky Solution: When you first selected a door, you had a 1/3 chance of being correct. You knew the host was going to open some other door which did not contain the car, so that doesn't change this probability. Hence, when all is said and done, there is a 1/3 chance that your original selection was correct, and hence a 1/3 chance that you will win by sticking. The remaining probability, 2/3, is the chance you will win by switching.

This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem. For example, consider the following:

Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?

[end of excerpt]

What's to argue about? He says,

"This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem."

OK. He gives a problem statement very similar to vos Savant's, where door 3 is opened, then the contestant makes his decision. He says a 100% conditional solution is 'actually correct' for this MHP problem statement. Then he starts changing premises and shows that the solution doesn't apply to these variants. So what? He's not saying it's flawed in any way for solving the MHP. This is not a paper critical of the 100% conditional solutions. This paper gives unambiguous support when he says, This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem.... Glkanter (talk) 13:24, 5 February 2011 (UTC)[reply]

Responses From The Other Editors To Rosenthal

Says Who?

Who decided that the MHP requires a solution where door 3 was specifically opened? Not Selvin. Not vos Savant. They wrote the darned problem statements that they solved. And they both gave unconditional tables of all possible outcomes. Countless reliable sources offer solutions that are either unconditional, or are conditional and indifferent as to which door is opened.

Morgan had to purposely misquote vos Savant to make their argument. Other than the 5 debatable critics, I think it's just Wikipedia editors with a POV saying this. Glkanter (talk) 13:14, 30 January 2011 (UTC)[reply]

Both Selvin and vos Savant found the 'symmetry' so obvious, even without expressly stated premises, that they used the simplest form of explanation, a table of all possible outcomes. These are the only unconditional simple solutions. All other simple solutions, which describe the original 1/3 not changing, are conditional solutions. Selvin even lumped [paraphrasing] 'opened either of the other doors' together in a single row in his original 9 row table. This ain't rocket science. No matter what any High Priests stamps his feet and says. Glkanter (talk) 19:09, 30 January 2011 (UTC)[reply]

What Glkanter calls "unconditional tables" of all possible outcomes are given by numerous authors in order to explicitly compute

Prob(Player chooses Door 1 and Host opens Door 3 and Car is behind Door 1)/Prob(Player chooses Door 1 and Host opens Door 3)

= Prob(Player chooses Door 1 and Host opens Door 3 and Car is behind Door 1)/

(Prob(Player chooses Door 1 and Host opens Door 3 and Car is behind Door 1)+Prob(Player chooses Door 1 and Host opens Door 3 and Car is behind Door 2)),

which has been the accepted definition of the conditional probability in question,

Prob(Car is behind Door 1 | Player chooses Door 1 and Host opens Door 3)

at least since Laplace (1814).

It looks to me as though Glkanter is not aware of the technical meaning of the word "conditional" when used in the phrase "conditional probability". That sure does explain a lot. Richard Gill (talk) 14:04, 2 February 2011 (UTC)[reply]

I'm not sure which authors you're talking about Richard, but Selvin shows (using his table of 9 equiprobable cases ignoring which door the host opens, i.e. one case is keys [car] in box A, player picks box A, host reveals box B or box C) that Prob(player switches and wins the car) = 6/9 and Prob(player doesn't switch and wins the car) = 3/9 and vos Savant shows (assuming the player has picked door 1) that Prob(player switches [to door 2 or door 3, whichever one the host does not open] and wins the car) = 2/3 and Prob(player stays with door 1 [whichever door the host opens] and wins the car) = 1/3. Selvin (in his first letter, where he describes this table) and vos Savant make no mention in these solutions of the conditional probability and make no attempt whatsoever to address it even though their respective problem statements clearly suggest the conditional probability is what is of interest. The "probability" they both talk about is the (unconditional) probability of winning by staying vs. the (unconditional) probability of winning by switching. Neither of their tables can even be used to compute the conditional probability since they both combine cases involving different doors the host opens. -- Rick Block (talk) 15:28, 2 February 2011 (UTC)[reply]

Sorry, I was very unclear. Let's agree that there are simple tables which don't distinguish which door is opened by the host (Vos Savant, Selvin), and there are more extensive tables which do. The tables themselves are not conditional or unconditional. The simple tables can only be used to give a simple solution. An extensive table can be used for either purpose. With the player's initial choice of Door 1 fixed, I don't see how we can possibly know whether Vos Savant and Selvin were aware of the statistical independence between whether or not the car is behind Door 1 and whether the host will open Door 2 or Door 3, which makes the door numbers irrelevant to the decision "switch or stay". In Vos Savant's discussion with Morgan et al. in Amer. Statist., she seems to refuse to take up the point which Morgan and his friends found so important. I don't know if her book gives any clarification.

Vos Savant and Selvin's problem statements might suggest to someone who knows about conditional probability that those authors are after the conditional probability, but such readers are disappointed. Richard Gill (talk) 09:32, 3 February 2011 (UTC)[reply]

Glkanter, Having seen further into the thinking of your ally Martin Hogbin at centrifugal force, I now see even more clearly what is going on here at Monty Hall. I recall Rick Block explaining to me that the simple solution is like using average speeds and times to calculate an acceleration when we really should be using instantaneous speeds and times. But that in the case of constant rates of change, the two happen to coincide. I think that Rick was making the point that Monty Hall just happens to suit the simple solution, but that the simple solution is not technically correct in general. I think that Martin on the other hand takes the view that since the simple solution works at Monty Hall, then it is adequate. I have noticed that Martin seems to be adopting a similar approach at centrifugal force. He points to water flying out of a spin drier. He points out that the water is not being constrained to the circular motion and hence it continues in its straight line motion, and hence it leaves the drum. He argues that this is an adequate explanation and that there is therefore no need to involve centrifugal force. So centrifugal force gets cast into the same pit as conditional probability. Rick would no doubt counter-argue that there are other problems in probability where conditional probability is absolutely necessary. Likewise, I can point to other case scenarios where the concept of centrifugal force is absolutely necessary. Best of all is in celestial mechanics, where in the radial gravitational field, centrifugal force comes face to face with gravity as an outward radial inverse cube law force. We then have to solve a very complicated radial differential equation which leads to orbital solutions that are either ellipses, parabolae, or hyperbolae. We cannot do without centrifugal force in this analysis.

So in that respect, I am now getting a clearer understanding of Rick Block's point of view in this dispute. He feels that a crude method is being used which only happens to work because it is one of the special cases in which it works. Rick feels that it is better to present the most general and technically correct solution to this probability problem, rather than to present a simple solution which only happens to work in the simple circumstances.

Nevertheless, I would still take the view that since the article is about Monty Hall, and not about conditional probablity, that the simple solution would be adequate to address the curiosity of all readers who had fallen into the initial trap of assuming wrongly that the answer is 50/50. And as such I think that the simple solution should take primacy in the Monty Hall article, but that the conditionalists should still have their full say further down the page. If however there happened to be an article on conditional probablity and it was decided to use the Monty Hall problem as a demonstration, I would oppose any attempts by simplists to get involved. David Tombe (talk) 11:42, 4 February 2011 (UTC)[reply]

David, as regards the MHP, it's so much simpler than all that. Simply put, there are *no* reliable sources that say all those things Rick Block says. It's OR. Richard Gill and Nijdam are also espousing OR. Glkanter (talk) 11:48, 4 February 2011 (UTC)[reply]

I do not agree with Nijdam at all. And what Glkanter calls OR is actually common knowledge and common sense among all mathematicians. Glkanter does not know what the word "conditional" means when used in a probability context, and his argument about the reliable sources who are mathematicians are based on the few snippets of their words which he happens to think he understands. Also, there are also plenty more "reliable sources" who say that MHP must be solved by conditional probability (personally, I do not agree with them, and I've have published OR explaining why I disagree). Just about every standard probability and statistics text includes MHP as an example and solves it, whether for good reasons or out of some kind of automatism I cannot say, by conditional probability.

I think that anyone who is smart enough to argue about MHP should be smart enough to understand what conditional probability is, and to decide for themselves whether they prefer a solution with or without conditional probability. The onus is on the conditionalists to present their section in as transparent and as appealing a way as possible. (So they have a lot of work to do!). Then let the people decide for themselves. Richard Gill (talk) 20:17, 4 February 2011 (UTC)[reply]

Glkanter, Yes, maybe there is an element of hobby horsism going on, in that the conditionalists are making an issue which is disproportionate to what is important for the purposes of the article. The article is all about the fact that most people intially think that the answer is 1/2, but then eventually realize that it is 2/3. The simple solution is adequate for the purpose of making that point. But I do have a certain degree of sympathy with the conditionalists, because they obviously feel that their expertise is not being taken seriously enough. And their point of view would be that the simple solution is rather crude and only happens to work, in the same sense that we can obtain an acceleration by using averages in the special cases where we have a constant rate of change. So you do need to give them their say, but lower down in the article. And also, they should be encouraged to make up a special new article on conditional probability where they can have a total field day with the Monty Hall problem exactly on their own terms. David Tombe (talk) 15:03, 4 February 2011 (UTC)[reply]

+1 to David. You are clearly referring to all those aspects here, and nevertheless you are able to see "the whole". You analyze and describe it vividly. Thank you, I agree fully with. Just with one (small) exception: The aspect of conditional probability threatens to overburden the MHP. Mathematically correct, yes. Regarding the MHP, it nevertheless is offering the false syllogism that, regarding any one very single game, you indeed "would know much better" – given the premise that you only just "knew better". Without being able to submit any log-list of "assumed previous or future repetitions". All of that is a curious side-aspect and should be presented as what it is. More important: To show a proof that probability to win by switching never can be below 1/2, and – otherwise – can likewise even be "1" under that approach. Gerhardvalentin (talk) 15:30, 4 February 2011 (UTC)[reply]

@Gerhard Valentin, The proof you refer to is (or rather: can be) very easy. The result was first given by Morgan et al., and it is given in a much easier form in my own recent publications.

The article is about what the reliable sources say. Period. Full stop. There are plenty of sources presenting simple solutions. There are plenty of sources presenting conditional solutions. Considering just these sources, per NPOV the article must not favor the simple solutions over conditional solutions (or the other way around). There are also a significant number of sources that explicitly compare simple solutions and conditional solutions, and make the (uncontested by any source) point that the simple solutions are not addressing the specific case where the player has chosen door 1 and the host has opened door 3. Also per NPOV, this point must be reflected in the article. -- Rick Block (talk) 15:35, 4 February 2011 (UTC)[reply]

Rick, you say "The article is about what the reliable sources say. Period. Full stop." – Accordingly, the lemma MHP has to disappoint and to discourage readers who are seeking for coverage? As in the past? In my opinion the importance of the necessary aspects has to be weighted, and especially the lemma must stop to be an unnecessary textbook in conditional probability only. Gerhardvalentin (talk) 16:05, 4 February 2011 (UTC)[reply]

David, the only thing that matters is the reliable sources:

Per the table above, there are only 5 so called 'critics' of simple solutions
I have already shown how calling 2 of these 'critics' is folly
I can do the same for the other 3
The critics are *not* a 'significant minority viewpoint'
There are 3 types of solutions

unconditional

conditional indifferent to the door #s

conditional concerned about the door #s

All reliably sources solutions that make up at least a 'significant minority viewpoint' belong in the article
Rick will claim showing simple solutions first violates NPOV. That is false.
The current article violates UNDUE WEIGHT, NPOV and OR, beginning with the 'Conditional solution' section

Posted by Glkanter (talk) 17:29, 4 February 2011 (UTC)[reply]

When discussing a mathematical problem (originated by Steve Selvin and by Martin Gardner, two mathematicians), are 50 articles by psychologists and science popularizers, which repeatedly present the same old news, to be weighted 10 times heavier than 5 articles by mathematicians and statisticians and maths educationalists, each of which presented something new? If Fox News puts out 50 articles on its web page about some new development in science, are those 50 articles to be weighted heavier than the handful of papers in Science and Nature which introduce and advance that new development?

Fortunately we are not talking about rocket science or quantum physics. The editors who aggressively oppose the conditional solutions have never taken the trouble to understand them. The editors who like the conditional solutions have never taken the trouble to simplify them (ah yes, that would be OR of course, none of that allowed here). Sad, sad, sad. Richard Gill (talk) 20:29, 4 February 2011 (UTC)[reply]

Rick, The problem is, that in disputes like this, both sides have reliable sources, and so we need to try and see the higher picture. The Monty Hall article is not an article about conditional probability. It is about correcting most peoples' misconception that the odds are 50/50. Once most people have seen the simple solution that it is in fact 2/3, they will be satisfied and will soon lose interest in the matter. I doubt if many will go on to read about the conditional solutions, but I have no objection to you having the conditional stuff there anyway. I do however object to the notion that conditional stuff should in anyway take primacy in the article because that only clouds the issue for most casual readers. I do however have sympathy with you if its a case of not having been able to convince these other editors of the correctness of your arguments. I am assuming that you are correct, but I have never taken sufficient interest in conditional probablity to bother checking it out. But I know how annoying it is to know about a topic and to find oneself arguing with people who clearly don't know about it, but who think that they know about it. In your case, what I would do would be to set up a new article on conditional probability and make Monty Hall into the flagship demonstration. That should keep all sides happy. David Tombe (talk) 18:29, 4 February 2011 (UTC)[reply]

David - Assuming you are talking to me ("Richard" here is generally Richard Gill - (Richard changed some "Richard"'s into "Rick"'s in David's writing above) ) you are missing the point. I am not insisting that the "conditional stuff should in anyway take primacy in the article" but instead that the article represent all significant views, fairly, proportionately, and without bias. This means not favoring the simple solutions and also not favoring the conditional solutions. The issue seems to me to be about how to describe the difference between the simple and conditional solutions. Glkanter apparently believes that clarifying the difference between the simple solutions (which a significant number of uncontested sources say technically address the unconditional probability) and conditional solutions (which address the conditional probability) is POV-laden criticism of simple solutions that should be eliminated from the article. -- Rick Block (talk) 19:46, 4 February 2011 (UTC)[reply]

Glkanter can speak for himself. Glkanter makes the case that criticisms of the simple solutions, and the over-emphasis of variants, etc. do not belong in the solution sections. They are found numerous times throughout the article. Glkanter (talk) 22:26, 4 February 2011 (UTC)[reply]

Glkanter wrote this less than 2 hours before you posted the above lies. Glkanter does not need anyone to post what he 'apparently believes'. He has posted this many times.

"* All reliably sources solutions that make up at least a 'significant minority viewpoint' belong in the article

"* Rick will claim showing simple solutions first violates NPOV. That is false.

"* The current article violates UNDUE WEIGHT, NPOV and OR, beginning with the 'Conditional solution' section

"Posted by Glkanter (talk) 17:29, 4 February 2011 (UTC)"[reply]

Posted by Glkanter (talk) 22:26, 4 February 2011 (UTC)[reply]

@David, thanks for your comments, to whomever they are addressed. Indeed I do try to see a higher picture, ind fact, a higher picture which reaches a synthesis of both the points of view usually argued here. My opinion is that both the simple and the conditional solutions are valid and that both have advantages and have disadvantages. (And personally, I prefer yet another point of view, namely the game-theoretic or economists' point of view). So I too have been arguing that the simple solutions should be presented first without health warnings of any kind. Also I have been pointing out that some of the conditional solutions to MHP do already exist on wikipedia, as examples, on the page on conditional probability, and do not need to be duplicated on the MHP page. It would be better to have the MHP page concentrate on the conditional solutions which avoid technicalities but instead use much concepts closer to ordinary intuition such as symmetry, or such as Bayes' rule, rather than going back to a mathematical definition of conditional probability, which won't mean anything to anybody except for students and teachers of probability. The mathematical definition was invented in order to match intuition, not the other way round. The mathematical formalists tend to forget these points. Read Laplace's philosophical essay on probability, 1814. I think that Laplace would have loved the full conditional solution as argued using symmetry, almost without calculation. As another famous mathematician Riemann once said, the aim of good mathematics is to replace computations by ideas. So please let's do the same in MHP. Richard Gill (talk) 19:55, 4 February 2011 (UTC)[reply]

In fact, the problem is that the "conditionalist" editors aren't confident enough in probability to follow informal arguments which can be easily formalized by any student of probability, while "simplist" editors don't know enough formal probability theory to understand what the conditionalists are going on about. As I have said elsewhere, it is a storm in a tea-cup, a babylonian confusion of languages of people who don't even realise that they actually agree - their troubles are that neither of their languages is rich enough to appreciate the subtleties carried in the other's language. That's why I have left the mediation, along with the other "moderate voices". Richard Gill (talk) 20:00, 4 February 2011 (UTC)[reply]

Rick, While you may have the higher picture as regards the maths, I think that Glkanter has the higher picture as regards the focus of the article. I do actually see your point of view more clearly now than I originally did. It has taken a while, but I am not an expert in conditional probability. As I understand it, you are adopting the general maths, and you are treating the opening of door 3 as a condition. Glkanter argues that this is irrelevant and that we can see ahead to the final result, irrespective of any such conditions. But you are arguing that Glkanter's line of reasoning is comparable to calculating accelerations using average speeds and times, when we should be using instantaneous speeds and times, but that in the special case of constant rates of change, the crude method produces the correct result. So maybe you are correct. I never liked the way those sort of approximations were used in A-level physics. I always preferred the rigorous calculus methods of applied mathematics, and maybe you see the simple solution at Monty Hall as being a kind of bastardization of the probability problem. In physics, I always felt that those crude approximations at an early stage led to misconceptions.

Glkanter can speak for himself. Glkanter makes the case that criticisms of the simple solutions, and the over-emphasis of variants, etc. do not belong in the solution sections. They are found numerous times throughout the article. Glkanter (talk) 22:26, 4 February 2011 (UTC)[reply]

But as regards the Monty Hall problem, I do think that the focus of the article is really only about showing that the answer is 2/3 and not 1/2, and it should be directed at a Friday night public house level audience, or supermarket check-out counter level audience. As such, as a neutral observer in this conflict, I would suggest that the simplists get 2/3 of the article and that the conditionalists get 1/3. But I would also support the idea that you start a new article on conditional probability and use the Monty hall example as a flagship demonstration, and link your new article to your 1/3 share at the lower end of the Monty Hall article.

You need to eventually come to an agreement to end this long running feud, and that doesn't mean a 50/50 compromise. As an outsider, I am looking in on this problem and I can see that the readers will by and large not be interested in the conditional solutions, even if the conditional solutions are more correct. David Tombe (talk) 20:05, 4 February 2011 (UTC)[reply]

@David, are you addressing "Rick" or "Richard" here? Richard (me) does not think either solution is "more correct". He thinks they are different. The "weaker solution" (the one which uses less assumptions) has more wide applicability. Makes less assumptions! Is easier to understand! And on its own terms it is completely correct. So he (me) (Richard) agrees with you that most readers need the simple solutions and need them without "health warnings". He would like to see the conditionalists make the effort to make their solutions simple and appealing to ordinary people. It is easy to do provided one dares to step away from the safety (but incomprehensibility) of mathematical formalism, and instead dares to use some mathematical insights, some mathematical ideas.

The "conditionalist" do know (or should know) that one can derive the conditional solution by adding a simple appeal to symmetry after the simple argument. Or alternatively one could say at the outset that the door numbers are irrelevant by symmetry, and that therefore the simple solution cannot be improved. This insight is not due to Glkanter but can be found in some of the earliest publications (e.g. Bell's response to Morgan et al, 1982), and it has been often been repeated here by other editors. Unfortunately it is not prominent in the "reliable sources" and hence I went to all the trouble to get two peer reviewed papers published which do explicitly use those arguments, appearing 2010 and 2011.

All the editors ought to quit quarelling and go home and read some more reliable sources. And if they don't know any standard probability calculus they ought to take the trouble to learn it. It isn't difficult. And a load of wikipedia readers are students and teachers of probability and statistics courses. They also need to see the problem treated in their language. Good encyclopeadia editors will minimalize the difference between the appropriate solutions for the two classes of readers, not maximize it. That requires taking a higher view, indeed. Richard Gill (talk) 20:42, 4 February 2011 (UTC)[reply]

I'm arguing that the whole article ought to be accessible to the Friday night public house level audience, or supermarket check-out counter level audience. Unfortunately that requires the editors leaving their entrenched positions (egos) and taking the trouble to learn the others' point of view. Richard Gill (talk) 20:46, 4 February 2011 (UTC)[reply]

Richard and Glkanter, I appreciate what both of you are saying. But it's time now to try and look at a formula which will end the quarrel once and for all. I suggest that the Monty Hall article itself should be given 2/3 to the simplists with them being allowed to corner the entire top part of the article, and without the inteference of conditionalist ideologies. The article would effectively become Monty Hall I (for Friday night drunks). The conditionalists on the other hand should be given total free reign on the bottom third of the article, and in addition to this they should be allowed to have their very own unofficial Monty Hall II (for Monday morning actuaries). This would consist of having a major section on Monty Hall in the existing conditional probability article, where they could have total freedom to explain the problem in their own terms. Rick does have a legitimate point of view, but it is unlikely to be of interest to most readers once they have realized that the answer is in fact 2/3. Can you imagine the situation in the pub on a Friday night? Somebody poses the question. They all boldly respond 50/50. They talk with bold self confidence and they talk down any attempts to explain that it is really 2/3. They stand shaking their heads and saying "it's 50/50, we've got two choices, both are unknown, so it has to be 50/50". The next week, one of them admits that they had thought about it during the week and played it out with a set of cards, and they now realize that it is in fact 2/3. The others look at him like he's a traitor. Then the questions start. A practical demo ensues, and soon they all get the point. It's 2/3 of a chance that the car is behind one of the other two doors, and so if the host eliminates one of those two doors, then it's 2/3 chance that the car is behind the remaining door. All now satisfied, sadder and wiser, suddenly a conditionalist arrives on the scene and starts to explain that there is a more accurate way of establishing the fact. As soon as he would mention about opening door 3, most of them would fall asleep. David Tombe (talk) 22:50, 4 February 2011 (UTC)[reply]

You're confusing what EDITORS have to say with what SOURCES have to say. EDITORS get no say whatsoever. Glkanter's view, your view, my view, Richard Gill's view are completely irrelevant. vos Savant's view, Selvin's view, Morgan's view, Gillman's view, etc. are what is relevant. Lots of sources (typified by vos Savant) present simple solutions. Lots of other sources (typified by, say, Jeff Gill) present conditional solutions without saying anything at all about simple solutions. A significant number of sources (starting with Morgan, but including Gillman, Grinstead&Snell, Falk, etc - more than the 5 that are currently referenced in the article) present conditional solutions AND simple solutions, and say how they're different. The article needs to reflect ALL of these sources in a neutral, unbiased way. Starting the article with an extensive section devoted to simple solutions without mentioning conditional solutions or how these two types of solutions are different is not representing all significant views, fairly, proportionately, and without bias.

Imagine an article on God. Would it satisfy NPOV to start such an article with an extensive section written by True Believers justifying the existence of God without mentioning that there are any other viewpoints until a much later section? -- Rick Block (talk) 00:55, 5 February 2011 (UTC)[reply]

Rick, There's nothing more that I can do here. Once the argument comes to a battle of sources, then the readers become the losers. An article ideally needs to be written with a focus and with comprehension, and not as a list of points of view of many different sources. In my view, the focus of this article is that it is intended to be a light hearted article with pictures of donkeys, and a riddle from a TV programme designed to shock people into realizing that their initial hunch that the answer was 50/50 is wrong. It's better that this point can be delivered without it being picketed by the complexities of conditional probability. I do actually fully appreciate your point of view. It's equivalent to wanting to derive v = u + at correctly using calculus and as a particular case where acceleration is constant, rather than using the straight line graphs that are used to teach this formula to form 4 pupils, leaving them wrongly thinking that this formula applies to all accelerations. Years later, the applied maths university lecturers then have to emphasize to them that that formula only applies for constant acceleration. Yes, I do see your point of view, but it is not suitable for the top part of the article. Having said that, I think that the time has come when we really need to hear the verdict of the mediation committee and get it all over with, once and for all. David Tombe (talk) 01:24, 5 February 2011 (UTC)[reply]

One last, very important point, David. ~~The *only* difference between the 'simple' conditional solutions and the 'traditional' conditional solutions is indifference to exactly *which* equally likely door has been opened.~~ Upon further review, I don't see how one tree differs from the other for solving the MHP. They both show door 1 has been selected, that door 3 has been opened, and rely on the various premises that doors 2 & 3 are equally likely to be opened. Nothing about 'averages'. Nothing about 'overall'. Nothing about 'before a door has been opened'. About the only difference is one tree is much less complex. This, then, is *not* similar to the centrifugal force comparison you have made above, and numerous times previously:

Posted by Glkanter (talk) 03:18, 5 February 2011 (UTC)[reply]

So, if a goal of the article is to show the reader the difference between 2 types of conditional solutions, the above trees accomplish that task very well. No need to clutter up the (misnamed) Conditional solution section with lengthy, repetitive tangents on premise-conflicting non-MHP 'changed host bias premises', and nonsense about 'before vs after' or 'averages' or 'overall'. Simply, the same math tool, the conditional decision tree, used in 2 ways:

~~Indifferent to which door is opened.~~ Identifies the condition as the 100% the host will reveal a goat from another door. Plugs in the specific, but equally likely door #s in the column headings
~~Aware of which door is opened.~~ Identifies the condition as one of two doors is equally likely to be opened to reveal a goat. Puts the specific, but equally likely door #s in each row.

And why are we indifferent? Because:

There is a premise that says the car has a 1/3 probability of being behind each of the 3 doors
The contestant selecting a door doesn't change those remaining equal 2/3 probabilities for each of the other 2 doors being opened to reveal a goat
There is no premise given that a host's bias will convey the location of the car to the contestant
"Suppose you're on a game show..." conveys to the reader "the processes and items in the problem are 'fair'"

As in, 'a fair die is thrown fairly.'

Being on a game show, the contestant, due to intentional ignorance, can only model his decision making by assuming all distributions are equally likely, absent any evidence to the contrary

This is, of course the expectation of the readers. Discussing a 'biased host' comes out of the blue, creating an unnecessary, nonsensical contrivance and likely confusion for the reader who is still in the solution sections, and may still think the 2 remaining doors are 1/2 & 1/2:

There is an explicit premise that the host selects between 2 goats equally
A 'host bias' that informs the contestant of the location of the grand prize is not a part of game shows
It is never explained how any bias becomes part of the contestant's SoK

There is a premise that says the host chooses between 2 goats equally
Opening either door gives the contestant no new information on the likelihood he has selected the car

The reader can easily see the single difference in these 2 conditional decision trees, and draw his own conclusions.

So yes, Glkanter *does* understand what 'conditional' and 'conditional probability' mean. He also understands the special case of a condition that has a distribution of 100% goat & 0% car.

Maybe it's Richard who doesn't understand the term 'conditional probability'? Glkanter (talk) 03:55, 5 February 2011 (UTC)[reply]

Possibly nobody on earth understands the term "probability". Glkanter is certainly willfully unaware that possibly one quarter of the readers of the MHP page think differently about it than he.

The first tree is superfluous (since the simple solution is more simple without it) and incorrect: the references to specific door numbers Door 2 and Door 3 needs to be erased. You can state, either before or after, that because of indifference you are well justified in ignoring the door numbers. If you don't mention the indifference explicitly you can be criticized for missing an essential part of the solution. After all, the reader can't know if you missed it deliberately because for you it is all so obvious, or you missed it accidentally because you rushed to conclusions without thinking. Richard Gill (talk) 07:05, 5 February 2011 (UTC)[reply]

I have always disagreed with your suggestion to remove the door #s from the headings. Why? The problem being solved talks about door 3 being opened, and 100% of the time that door 3 is opened, the host reveals a goat. This depiction of the various conditional solutions where the condition is the 100% host opens a door (say, #3) addresses every aspect of the MHP as thoroughly as the traditional conditional solutions that are door-centric and use the 1/2 & 1/2 premise in the calculations. It's just *way* simpler to understand. Glkanter (talk) 11:16, 5 February 2011 (UTC)[reply]

Why treat this tree differently than the 2nd tree? That tree also could have the door #s removed, and the 'other' doors could be 'leftmost unchosen door' and 'rightmost unchosen door'. But why? Why is that better, and why treat the trees differently? Glkanter (talk) 11:30, 5 February 2011 (UTC)[reply]

It would seem both trees need some explanation to accompany them. My only point is that other than Selvin's 9 row table and vos Savant's 3 row table, all the other 'simple' solutions *are* conditional. And are therefore capable of solving the conditional MHP. Despite what some Wikipedia editors insist via their unsupported OR. Glkanter (talk) 07:31, 5 February 2011 (UTC)[reply]

How Can Morgan Call The "Traditional" Conditional Solutions "False"?

In their paper they write:

"Solution F6. The sample space is {AGG2, AGG3, GAG3, GGA2} where the letter triples have the same meaning as in Solution F2, and the number indicates the door opened by the host. The probabilities for the sample points are, in order, 1/6, 1/6, 1/3, and 1/3. Labeling events W, = "win by switching" and D3 = "goat shown behind door 3," Pr(W, I D3) = Pr(W, and D3)/Pr(D3) = Pr(GAG3)/Pr(AGG3,GAG3) = (1/3)/(1/6 + 1/3) = 2/3."

"Solution F6 (cf. Mosteller 1965) attempts to correct the wrongly specified sample space of F4. One must ask, however, how the probabilities for this sample space are determined. It turns out that this is a correct specification only if one assumes a certain strategy on the part of the host. We will show that the problem can be solved without any assumptions of this type, which is to say the problem can be solved."

Well, here's how, from directly above:

"It turns out that this is a correct specification only if one assumes a certain strategy on the part of the host."

So, Morgan is criticizing vos Savant for failing to explicitly state that the host chooses equally between 2 goats.

Of course, Selvin's second letter makes it very clear that the 'certain strategy' *is* part of the problem:

"The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random." - Steve Selvin The American Statistician, August 1975, Vol. 29, No. 3

With Selvin's premise, Morgan is saying:

That Selvin's and vos Savant's unconditional tables of all equally likely possible outcomes are false because they don't solve the conditional question posed.

It would seem that Selvin and vos Savant knew what question they were asking, and answered that question. Morgan is stating an opinion that contradicts the very sources they criticize

The indifferent conditional solutions don't specifically answer the question posed, where door 3 has been opened

Morgan must be overlooking the obvious indifference as to which door is opened, as they are both equally likely. The indifferent conditional solutions, like the 1/2 & 1/2 conditional solutions (decision tree & Bayes), solves the puzzle for either door 2 being opened or door 3 being opened.

The 1/2 & 1/2 door conditional solutions (decision tree & Bayes) rely on the exact same Selvin premise, "...when he can open either of two boxes without exposing the keys, he chooses between them at random." as do the indifferent conditional solutions.

It's only *with* Selvin's premise that these are valid solutions

The editors of the Wikipedia article must be consistent in their treatment of Morgan's comments regarding the host bias:

Either the host bias premise is missing, so:

Morgan says all 3 styles of solutions are 'False', and the 2/3 & 1/3 paradox doesn't exist

Or Selvin's premise is included:

Morgan is calling 'False' the unconditional tables presented as solutions by the 2 originators of the problem, for solving their own problems
Morgan is calling 'False' the indifferent conditional solutions for not specifying door 3
Morgan accepts the 1/2 & 1/2 door conditional solutions (decision tree & Bayes) as correct

(1) is nothing more than Morgan's opinion of what Selvin and vos Savant wanted to know by their question. Obviously, Selvin and vos Savant disagree with Morgan. vos Savant told them directly. Selvin offers his unconditional table, Monty Hall's indifferent conditional solution, and the 1/2 & 1/2 conditional solution, without criticizing any of them. And he gives high praise to Monty Hall.

(2) is contradicted by Morgan themselves in their rejoinder to vos Savant:

"Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one. But then, it may be that only an academician, and no one connected with a game show, would ever consider p # q."

Morgan, but not any sources that offer either unconditional or indifferent conditional solutions, contrived this 'unconditional problem':

""You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?" The distinction between the conditional and unconditional situations here seems to confound many, from whence much of the pedagogic and entertainment value is derived."

It looks to me like Morgan is having problems with the term 'conditional problem', as they describe a door being opened. I think they mean 'indifferent'.

and from their response to Martin and Nijdam:

"To wit, had we adopted conditions implicit in the problem, the answer is 2/3, period."

Morgan's whole paper is predicated on the missing host bias premise. The Wikipedia article treats Morgan's criticisms in an inconsistent manner, leading to the mess in the Conditional solution section, UNDUE WEIGHT, NPOV and OR violations in the article. The Wikipedia article makes criticisms about the indifferent simple solutions that that are not supported by reliable sources, and treats as facts Morgan's opinion as to what problem Selvin and vos Savant should be solving.

No other source calls these solutions 'False', and Morgan no longer does so. The article needs to be corrected for this. Glkanter (talk) 06:12, 5 February 2011 (UTC)[reply]

Nice research, Glkanter!

You told us that you *do* know what conditional probability means. That's good. Maybe you can now answer my question to you of some weeks ago.

Do you agree that it is a different matter to explain why

1. when the contestant walks onto the stage and chooses Door 1, knowing he'll be shown a goat behind one of the other two doors, his odds are 2 to 1 that switching will give him the car,

from explaining why

2. when thereafter the host has actually opened Door 3, the contestant's odds are (still) 2 to 1 that switching will give him the car?

I'm still wondering.

I'd also be interested in David's answer. Richard Gill (talk) 06:54, 5 February 2011 (UTC)[reply]

Yes, they are different because #1 is incomplete. Please describe further what it is I am modeling. The MHP (#2) has the 4 steps you've previously described. Please list the steps, and the point at which the probabilities are to be determined for #1. Glkanter (talk) 07:38, 5 February 2011 (UTC)[reply]

Further, as the question is answered from the contestant's SoK, I need you to tell me when the contestant becomes aware that the opening of either door will not cause him to derive a value other than 1/3 for his door being the car. Does this happen before a door, say #3, is opened, or only after? Glkanter (talk) 07:51, 5 February 2011 (UTC)[reply]

Splendid! Here's my list of the four steps:

1. Initially the odds are 2 to 1 against that the car is behind Door 1

2. The contestant chooses Door 1. His action doesn't change the odds: so it is two times as likely for him that his choice is wrong as that it is right.

3. The host opens one of the other two doors, revealing a goat. Let's suppose that for the moment, the contestant doesn't take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind Door 1, the information that an unspecified door is opened revealing a goat does not change the contestant's odds that the car is indeed behind Door 1; they are still 2 to 1 against.

4. Finally, the contestant also gets informed which specific door was opened by the host - let's say it was Door 3. Does this piece of information influence his odds that the car is behind Door 1? No: from the contestant's point of view, the chance that the car is behind Door 1 can't depend on whether the host opens Door 2 or Door 3 - the door numbers are completely arbitrary, interchangeable.

Therefore, also knowing that the host opened specifically Door 3 to reveal a goat, the contestant's odds on the car being behind his initially chosen Door 1 still remain 2 to 1 against. He had better switch to Door 2.

This is the main content of the page on the Monty Hall problem at [11]. It's quite possible that this page gets approved in the near future by the mathematics editors of citiziendium (there is a distinction between "authors" and "editors" there). If that happens, it then becomes a reliable source in wikipedia terms. Richard Gill (talk) 07:54, 5 February 2011 (UTC)[reply]

Regarding Glkanter's second question: I have artificially separated the player's observation of the host opening a door into two steps for pedagogical purposes. I pretend he doesn't think, I pretend he just observes and that he uses all the observed information to logically update his state of knowledge regarding the question whether or not the car is behind Door 1. That's all. Richard Gill (talk) 08:02, 5 February 2011 (UTC)[reply]

I would suggest there is a 5th step, then. The contestant announces his decision. When does that happen in #1, and when does he realize steps 3 & 4 will happen? Because you and I know they will happen, and I haven't seen a door opened yet. Glkanter (talk) 08:06, 5 February 2011 (UTC)[reply]

Is this contestant the rare individual who recognizes he should switch, or does he think it's 1/2 & 1/2? Or are you asking me what he should do, based on what he's been told and has seen? But I know about steps 3 & 4 at the beginning, even if the contestant doesn't.Glkanter (talk) 08:14, 5 February 2011 (UTC)[reply]

I attempted to present a solution which unifies simple and conditional, verbal-logical and formal-probability-calculus. It's a simple (and useful) exercise for students of introductory probability calculus to convert each step into mathematical formalism. Since elementary probability calculus is based on elementary logic, see Laplace (1814), the "translation" works smoothly - there cannot be a logical contradiction between two logical arguments based on the same premisses. Of course, this is Own Research in wikipedia terms, though based on the tremendously stimulating discussions here, inlcuding much literature research. The solution is meant to be self-explanatory and neutral regarding the simple versus conditional controversy. Whether or not a solution is complete without *explicitly* doing step 4 is, I think, a matter of opinion, of context.

I have to quit internet now for the weekend. Everyone can think what they like about what I wrote. Richard Gill (talk) 08:11, 5 February 2011 (UTC)[reply]

Richard, For your two case scenarios listed above, I have to honestly state that I can't see the difference. If we know the probability in advance, the probability will not change just because we begin to act out the case scenario. But what I do accept is that there are two ways of calculating the probability, one of which is dependent on the condition of opening door 3. And I further recognize that there may be more complicated case scenarios in which the simple solution is not adequate. But in this case scenario, I believe that the simple solution is adequate. David Tombe (talk) 12:27, 5 February 2011 (UTC)[reply]

David, my question is not about different ways of computing "the probability" but about the logical argument necessary to prove that a probability is 2/3, in two different situations. Will the same logical argument do, or do you need a different argument? Please think about the question some more. We are talking about a logical puzzle, not about physics. A brainteaser is considered solved when not only is "the right answer" written down, but also when a sound argument is provided that that is the right answer. If you can't judge the logical correctness of some piece of reasoning, you really can't contribute much to discussions on MHP. All our sources are agreed that the answer is "2/3, switch" and all are prepared to take the same assumptions on board. So apparently the difference of opinion between sources, and editors, concerns the "reasoning" part of the solution, not the "answer" part of the solution. The main difference indeed at the moment is whether skipping one small step in the argument which is intuitively immediate and obvious, should be considered to make a solution "wrong". Richard Gill (talk) 16:30, 5 February 2011 (UTC)[reply]

The real question is 'is there a significant minority viewpoint amongst the reliable sources'? for what certain editors are insisting on. The fact is, there is not. Glkanter (talk) 16:40, 5 February 2011 (UTC)[reply]

Glkanter, In my view, the rule about sources gets taken too literally in these kinds of disputes. Sources are really only for the purpose of verification in relation to non-controversial topics. Ideally articles should be written coherently by somebody who has a full comprehension of the topic, and written based on their own accumulated knowledge. Sources should only become important if a fact is challenged. For example, some horse expert writes a big article on horses, which includes sections on zebras and donkeys. He then adds a section about the quagga. A reader who has never heard about the quagga might get a bit worried and ask if there is any additional confirmation that this animal ever existed. That's where a source would be needed. But in these long running affrays such as at Monty Hall or Centrifugal Force, it is obvious that the existing literature is diverse and full of contradictions, because the topic matter involves issues of logic which have been debated for centuries, and so it becomes totally pointless trying to replace socratic debate with a card game involving sources. Have you all agreed yet which source is the ace of spades? So I would like to see this dispute resolved by the mediation committee making a decision, based on what they feel would be of most interest to the readers. Now see my next response to Richard. David Tombe (talk) 17:21, 5 February 2011 (UTC)[reply]

Believe me, I *know* I am the least experienced Wikipedia guy around. But I think the reliable sources requirement serves more uses than you describe. For example, I've demonstrated that most of the arguments from the loyal opposition are not derived from a significant minority viewpoint of reliable sources. And without that, no editors' favored OR POV belongs in these discussions, or in the article. Now, If I could just get a mediator to take notice... Glkanter (talk) 19:21, 5 February 2011 (UTC)[reply]

Glkanter, Yes, but in a controversial topic like MHP, centrifugal force, East Jerusalem etc. it doesn't just end at producing a source. There will be further arguments such as that you haven't interpreted the contents of the source correctly, or that the source is unreliable, or that the source is a primary source etc. It doesn't end when the sources are produced. Sources become irrelevant in a controversial topic. All sides will have sources. And you should know that, because you claim that you have most sources on you side, yet you can see that nobody is paying any attention to that fact. You need to get the mediators to make a decision. David Tombe (talk) 20:07, 5 February 2011 (UTC)[reply]

I only claim the other guys have two or less reliable sources. Otherwise, I agree with you completely, and have done all I can to engage the mediators. Glkanter (talk) 20:20, 5 February 2011 (UTC)[reply]

Well, unless you count these 'critics':

"...There are scores of others (mostly academic mathematical sources) that present only conditional solutions - these are implicitly agreeing that the conditional approach is the "right" approach and can be construed to be indirectly agreeing with the criticism that the simple solutions solve the wrong problem... Rick Block (talk) 16:15, 19 January 2011 (UTC)"

But, really, ya gotta read the whole diff. It's a freakin' hoot! Glkanter (talk) 04:47, 6 February 2011 (UTC)[reply]

Richard, I did genuinely try to address your query. And I genuinely don't see the difference between the two situations. Maybe I am at fault because I have overlooked something, but I have tried. All I can see as regards the conditional scenario when door 3 is opened, is that we have already begun to act out the scenario, the answer regarding which we had already ascertained from the outset. The unconditional scenario is a 'would be' scenario. We can see ahead to the fact that the answer will be 2/3. So now we put it all into action, and the host opens door 3 and asks the question. I can't see the difference. Nothing has changed. We knew in advance he was going to open one of the doors, and we knew in advance that it wouldn't make any difference which door that would be. So the answer will still be 2/3 based on the original unconditional scenario. I don't see that we need to recalculate that fact using conditional probability. But what I do see is that we could recalculate it using conditional probability.David Tombe (talk) 17:21, 5 February 2011 (UTC)[reply]

I didn't say that you should recalculate using some tedious formulas. I talked about arguments, logic... not about calculations. The question was: is there a difference in establishing the answers to the two questions? And you gave it yourself: "we knew it wouldn't make any difference which door that would be". Indeed, it doesn't make a difference, and that is the remark which needs to be added, to convert an argument for the first result into an argument for the second.

The key words here should not be conditional probability but statistical independence. Which door is opened by the host is independent of whether or not the car is behind Door 1, so it doesn't make a difference. Unfortunately, the key reliable sources didn't seem to realise that the problem they want to solve can be solved so easily. It's hidden in Bell (1991), one of the responses to the infamous paper by Morgan et al. It's repeated in Gill (2010, 2011) and on [12], so fortunately we have reliable sources now from the side of the statisticians and mathematicians which acknowledge that mathematically it's not a big deal. Richard Gill (talk) 03:00, 6 February 2011 (UTC)[reply]

Richard, I'll answer you later today. But in the meantime, can you answer me the question as to which camp you are in? David Tombe (talk) 10:15, 6 February 2011 (UTC)[reply]

Again

The point I repeatedly seem to mention, and what is not apparently in much of what recently have been discussed with David Tombe is not whether the simple solution as such is correct, i.e. a correct reasoning, but what it is a solution to. At least, Rick and Richard agree with me that the simple solution is a correct solution to the unconditional formulation of the MHP, whereas it is not a solution to the conditional formulation, with the conditional solution. The discussion in fact is about which formulation should be considered the MHP. Also the conditional solution is not necessarily to be presented with the terminology of conditional probabilities, but may be explained in an easy understandable way for almost every interested reader. A whole lot of the recent discussion is pointless without reference to these points. Nijdam (talk) 08:55, 5 February 2011 (UTC)[reply]

Nijdam, I did get the point about difference between the conditional and the unconditional formulation of the problem. But I am looking in on this problem as an outsider, and assessing what is likely to be of interest to the average reader. The average reader will not be interested in that kind of distinction. Therefore a simple explanation is all that is necessary. There is a basic underlying logic behind the problem which runs that there is a 1/3 probability that the car will be behind the chosen door, and that hence when one of the other doors is eliminated, there will be a 2/3 benefit in swapping. This overview transcends any details about whether we are solving the unconditional or the conditional problem. David Tombe (talk) 12:44, 5 February 2011 (UTC)[reply]

Sorry David, you're really an outsider. I never said we should explain the distinction to the average reader. On the contrary, I only want to present the natural conditional formulation. Form this formulation, the average reader, seeing two closed doors, one hiding the car, will tend to think the odds are even. No need to trouble them with the unconditional formulation, which is rather unnatural. Alas, does the simple solution not solve the conditional formulation of the problem. No big deal, the conditional solution is not difficult to understand. It seems idiotic to me to change the problem formulation, in order to make the simple solution work. Most people will confuse the unconditional formulation with the obvious conditional formulation, and then accept the simple solution wrongly as a correct solution to this problem. And no, the simple solution, you correctly formulated it, does not trancend the differences. Nijdam (talk) 22:31, 5 February 2011 (UTC)[reply]

Nijdam, OK. You want to actually begin the entire story from the point after the first door is open. Is that the bit which I've missed? David Tombe (talk) 22:47, 5 February 2011 (UTC)[reply]

It's not a bit, it's the crux of the problem. Right then two closed doors are left, between which the player has to choose. Nijdam (talk) 23:56, 5 February 2011 (UTC)[reply]

Does the host not have to open one of those two remaining doors and then ask him if he wants to swap? David Tombe (talk) 00:40, 6 February 2011 (UTC)[reply]

Yes. but here lies a difficulty a non probabilist/statistician will hardly understand: the final answer about the required probability depends on the way the host acts. Nijdam (talk) 14:50, 6 February 2011 (UTC)[reply]

I see they have already grown tired of your act over at Citizendium, Nijdam. Did it take a full week? Glkanter (talk) 14:56, 6 February 2011 (UTC)[reply]

Nijdam, So the host then opens one of the other two doors. The probablity is 2/3 benefit for a swap. We knew that before we started. So what has changed? David Tombe (talk) 19:28, 6 February 2011 (UTC)[reply]

There is a difference between the probability before the host opens a door and the probability thereafter (the posterior or conditional) probability. We know that before the host opens a door the chance on the car by switching is 2/3. But that does not logically tell me what the probability is after the host opened a door, It may be argued, that with the right assumptions this probability is also 2/3, but it needs some reasoning, which fails in the simple solution. Nijdam (talk) 22:43, 6 February 2011 (UTC)[reply]

Nijdam, I've tried my best, but I just can't accept your argument. What could possibly change when the door is actually opened, putting aside irregularities like host bias etc.? David Tombe (talk) 23:07, 6 February 2011 (UTC)[reply]

Try harder, or do not take part in the discussion. Anyone with basic understanding of probability theory should know the difference, Nijdam (talk) 07:36, 7 February 2011 (UTC)[reply]

Well no Nijdam, that's not a satisfactory attitude. You are telling me to try harder to see the Emperor's new clothes or to depart from the debate. Well I don't see the Empereor's new clothes in this situation, and what you have said convinces me all the more that this is an clash between two camps, one of which wants to emphasize the Monty hall problem and the other which wants to emphasize conditional probablity. I agree with you that after the door is opened that we can indeed use a conditional probability based solution, but I don't agree with you that it is necessary to do so. David Tombe (talk) 11:57, 7 February 2011 (UTC)[reply]

Nijdam, On reflection I have retracted the above comments. I see your point now. You genuinely believe that the problem can only be solved using conditional probability once the door has been opened. I don't agree with you on that point. I do accept that it can be solved by conditional probability once the door has been opened, but I don't accept the imperative. But since this is only my opinion, I can see now that the problem that has been raging here for years is intractable. It is therefore imperative that the matter is swiftly brought to a conclusion by the mediation committee. Without the intervention of some kind of authority, the arguing will go on forever. David Tombe (talk) 15:44, 7 February 2011 (UTC)[reply]

"Unconditional" Problem Statement

Are there any sources:

That present a so-called "unconditional" problem statement of the MHP (like Morgan's), then offer their solution?
Do #1 above without calling the solution 'false'? (Morgan)
Do #1 above without using it to demonstrate the need for a tool that can calculate a condition that is not 100% with all outcomes symmetrical? (G & S)

In short, other than the so-called 'critics', does any reliable source restate the MHP without door 3 being opened?

If so, what solutions do they offer?

I think the Morgan-named, and incorrectly named "unconditional problem":

"You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?" The distinction between the conditional and unconditional situations here seems to confound many, from whence much of the pedagogic and entertainment value is derived."

Only exists amongst the so-called critics. The are no sources that describe a problem that omits door 3 being opened, then offer either Selvin's and vos Savant's unconditional tables, or the various conditional solutions that define the condition as the 100% the host will open another door to reveal a goat. Morgan's is an Aunt Sally or Straw Man argument. Morgan says any solution that doesn't address door 3 specifically must be answering the above "unconditional" problem statement. The problems with this argument are:

The problem statement *is* conditional. It's conditioned on the 100% that the host will open another door to reveal a goat
Their conclusion is flawed. They have no means mathematically or logically to redefine the problem statement that the reliable sources have given.
Morgan's conclusion is based on what they decided is the missing host bias premise. They're wrong about that, and have admitted it.
Morgan states that the false solutions *must* be solving this other problem. That's a flawed argument.
They are saying any solution that is not door 3 specific is flawed. That's just an uninformed opinion.
The solutions based on the 100% condition *are* conditional solutions. And they are able to solve the MHP where door 3 has been opened to reveal a goat.

Posted by Glkanter (talk) 09:42, 5 February 2011 (UTC)[reply]

For It To Be 2/3 & 1/3 From The Contestant's SoK...

In order for the 'why it's 2/3 & 1/3 rather than 1/2 & 1/2' paradox to exist from the contestant's SoK, the contestant must be informed of certain premises prior to selecting his door. Otherwise, the variants:

Random Monty (might reveal a car)
Deceitful Monty (doesn't always offer the switch)
Biased Host (doesn't choose between 2 goats equally)

are possible, thus rendering the puzzle unsolvable.

Therefore, the contestant is advised, prior to his door selection:

The car & goats have been distributed uniformly at random
The host will reveal a goat behind an unchosen door
The host will select between 2 goats in an equal manner
The host will offer the switch

So, before anything has actually happened on stage, the contestant's SoK, like our own, is sufficient, if he's clever enough, to calculate that the odds will still be no more and no less than 1/3 that his selection is the car, even after he has made a selection and a door has been opened.

In order to make the best decision, the contestant need not wait for the 100% condition to occur to know that with symmetry & indifference, he's going to make the switch.

The argument from the Wikipedia editors, not from the reliable sources, that finds fault with Selvin's and vos Savant's unconditional tables is not supported by the math, or by the reliable sources, and violates Wikipedia's NPOV and OR policies. Glkanter (talk) 16:57, 5 February 2011 (UTC)[reply]

1. The contestant is given an unconditional explanation of the rules before he makes his door selection

He is quite capable of determining his strategy at this point

2. The contestant experiences the playing of the game in which door #s are assigned

Of course, he has gained no knowledge about the likelihood that he has selected a car

It takes balls to say that the problem must be construed as 'Only conditional, because I say so.'

Posted by Glkanter (talk) 03:17, 6 February 2011 (UTC)[reply]