Talk:Monty Hall problem

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[edit] Even Rick losing his patience

I suggest we:

  1. Archive this talk page.
  2. Stop squabbling.
  3. All admit the problem, however it's actually phrased, is not meant to have different conditional and unconditional answers.
  4. All admit that the current statement of the problem indeed satisfies the previous point.
  5. Tweak the wording of the solution section so that it appeases probability purists while avoiding explicitly rubbing the nose of the reader in conditional vs. unconditional probability considerations, i.e. change it so it has the form of a conditional solution but without making a big deal out of it.
  6. Add a discussion of the conditional and unconditional consideration, referencing Morgan et al, to a later section of the article.

A conditional approach to the solution is certainly not wrong, and if the solution shows the probability of winning by switching is 2/3 for each player regardless of which door they pick and which door the host opens then the unconditional probability surely must be 2/3 as well.

Note the solution section needs to be accessible to a reasonably intelligent non-specialist, e.g. your grandmother (I realize there's a small chance your grandmother is a mathematician) or your 15 year old kid (no, not the goat kind).

Is this an acceptable approach? -- Rick Block (talk) 19:49, 29 February 2008 (UTC)

  • I don't support this proposal. I think that the existing solution is fine, and that what we need to do is not change the solution, but change the problem statement. See my draft suggestion at Talk:Monty Hall problem/Matt. Matt 20:40, 29 February 2008 (UTC). —Preceding unsigned comment added by 81.156.127.87 (talk)
...I don't disagree with point 6 though... Matt 20:46, 29 February 2008 (UTC) —Preceding unsigned comment added by 81.156.127.87 (talk)
On point 5, unfortunately, there is no way to appease probability purists, at least not all of them, because there is more than one school of thought. Indeed, there is an unresolved (unresolvable?) problem in trying to define the meaning of probability and randomness in a fundamental and rigorous way: they tend to be circular. But that is a story for another topic. Seriously: it is a whole other article; it does not belong here.
The only way to achieve point 2 is to agree on point 3, i.e. to adopt a notably conventional meaning for what is notably agreed to be unfortunate wording of the original problem, and give poor Marilyn a rest. 67.130.129.135 (talk) 23:41, 29 February 2008 (UTC)
OK, yet another suggestion. How about explicitly identifying the existing solution as the unconditional solution with wording like "Over numerous trials of this game, there are three equally likely scenarios depending on the player's initial choice ...". This sort of begs the question of how we know these are equally likely (although it is intuitively obvious!), but it clearly turns the solution into the unconditional solution and makes it (I think) unassailably true. For this point in the article (the initial "Solution" section), I think this may be good enough so long as we include a more formal solution later (perhaps as the very next section, or even as subsequent text within this section). Comments? -- Rick Block (talk) 02:26, 2 March 2008 (UTC)

Wiki cream (written by Jgm):

"... I'm beginning to see Wikipedia articles (and, by extension, the Wikipedia itself) as analogous to whipping cream. At first there is great anticipation as the raw material is revealed and the process of whipping it into shape begins. When conditions are right, at some point, the churning and friction result in something approaching ideal -- well structured, tasty, substantive.

The problem is that the whipping never stops. What was near-perfect turns into homogenous goo; the main ingredient hasn't changed and it's still capable of meeting basic dietary needs, but far less enjoyable to consume." ...humbly offered by hydnjo talk 03:08, 2 March 2008 (UTC)

Goose notes that whipped cream turns into butter if you keep whipping it.--Father Goose (talk) 06:14, 2 March 2008 (UTC)

What about this two-part approach:

  1. Present the simple solution to the "main version" of the problem, with the least amount of fussing over the problem statement we can get away with to make it into a solution. Also present some other analysis methods of the problem, leading to the same solution. Following these presentations of the solution, briefly discuss unintended interpretations of the problem statement, and the difficulty in ruling them out unambiguously, while referring to a later section on variations.
  2. Have (like now) a section with variations on the problem, such as those in which the player may know more than in the main version, depending on the host's behaviour.

The terminology "unconditional solution" is unfortunate. The notion of a "conditional" solution has no meaning for the main problem version. It is only in variations of the main version that conditional issues arise.  --Lambiam 08:28, 2 March 2008 (UTC)

Seems like we're converging on an approach where we first present the intuitive (unconditional, but not using that terminology) solution, backed up with a more rigorous discussion - eerily similar to the approach our anonymous friend of many IP addresses was originally attempting several weeks ago [1]. Since this is a featured article I think it would be best if we backed up some of these changes with references. I think we end up with a structure like this:

  • Lead section as is, quoting Parade magazine's version
  • Problem statement, nearly as is, quoting Mueser & Granberg with the "equal goat door" addition. IMO we should include a source for the addition - Mueser & Granberg mention it but unfortunately don't include it in their unambiguous problem statement. [Matt - I think your suggested problem statement is similar enough that we might as well stay with a quoted version]
  • Expanded Solution section including:
    • Wording very similar to the current explanation, clearly labeled as the "intuitive" solution (is this OK?) per Matt's (and Goose's and Lambian's?) preference. This solution statement should have a reference - possibly Parade (don't laugh). I think we should add some sort of disclaimer (also referenced, possibly Morgan et al) about this solution being true "on average" while not necessarily applying to any particular player.
    • A more detailed (more rigorous if you will) conditional analysis showing the same 2/3 result, also with a reference (possibly Morgan et al), covering anon-many-ips's original criticism.
    • A discussion about unintended interpretations as Lambian suggests (also referenced, not sure where).

Then, we also add a variant for the biased host (or host preference with probability P) with an analysis showing the Morgan et al result that switching is better regardless of the host's strategy for picking between two goat doors.

One little detail (well, I think it's little) is that I suspect the more rigorous solution will be primarily based on where the car is, not whether the player picked the car or a specific goat. To keep the intuitive and rigorous solutions as parallel as possible I think we should adjust the intuitive solution (and Goose's diagram, which perhaps moves to the "rigorous" solution) so the primary selection is the location of the car. This means the three "equal probability" cases would be (wording quibbles aside, and assuming we can find a reference that presents it this way):

  • The player originally picked the door hiding the car. The game host has shown one of the two goats.
  • The car is behind one of the two doors the player did not originally pick. The player originally picked a door hiding a goat and the game host has shown the other goat.
  • The car is behind the other door the player did not originally pick. The player originally picked a door hiding a goat and the game host has shown the other goat.

Is this a plan everyone can live with? -- Rick Block (talk) 18:19, 2 March 2008 (UTC)

Thanks Rick, this satisfies my main objection, which is that we were proposing to replace a simple and, as you say, "intuitive" solution with a more complicated one. I know this has already gone on a for long time, and I'm going to shut up fairly soon I hope, but I have a few specific comments on your latest proposal:
  • In addition to a full-on mathematical treatment (if that's what you're proposing), I would very much like to see, after the intuitive solution, a plain-English explanation of the "host chooses randomly" and "specific probability versus average probability" issues, written at a similar or even lower technical level to my "version 4" wording, and covering the same sort of points. (Though I'm not especially wedded to that exact form of wording, and it may not be technically entirely correct as it stands.) I think that these issues are likely to be of interest to a general audience. It's certainly something that I'd never appreciated before, and after everyone's hard work it would be a shame to see it buried in a mass of symbols that only a few hard-core readers will be able to penetrate.
  • Referring to specific doors in the problem adds no value that I can see -- certainly as far as it concerns the intuitive solution or the layman's explanation of the technicalities that I just mentioned. In particular, the existing wording raises uncertainty about whether someone just forgot to insert the words "say" or "for example" and it's actually describing an example outcome (which is what I suspect most people intuitively want), or whether it really is describing a specific realisation of the game where that particular door outcome arose. (I don't want to reopen the debate about why it matters/doesn't matter/shouldn't matter, because we've done that to death.) I also think that the phrase "thoroughly honest" is unnecessary and raises doubts that wouldn't exist if we simply said "the host does this ... the host does that ..." If we say he does it then he does it; there's no need to complicate the issue by commenting on his honesty. However, I do understand your desire to use a quoted version.
  • I'm not convinced of the need to move to the slightly more complicated "intuitive solution" wording that you're proposing, especially if we avoid mentioning specific doors in the problem statement. Maybe it's just me, but just saying "player picks a door ... 1/3 chance of picking the car ... 1/3 chance of picking the first goat ... 1/3 chance of picking second goat" seems like so much less hassle.
  • Query: you say that you want a disclaimer about the intuitive solution being true "on average" while not necessarily applying to any particular player, but with the "host chooses randomly" constraint, which we are including, it does apply to any particular player, doesn't it? Please correct me if I'm wrong... Jeez, I've looked at this so many times now that I don't even know what my own name is any more.
... just to amplify ... groan, more verbiage ... although one early objection was that "the problem is that the reasoning in the existing solution exactly applies to a different problem and produces the wrong result" and "doesn't differ depending on whether or not we include the assumption that the host choose equally between two goat doors", I thought that somewhere along the line we concluded that the solution did implicitly rely on this assumption because in the specific-player-biased-host problem the statement that the three cases are equally likely is no longer true... Maybe I'm wrong. —Preceding unsigned comment added by 86.137.136.122 (talk) 22:05, 2 March 2008 (UTC)
The solution description pertains to the average solution and not the solution for a particular player pick and host door. This is true whether the equal goat constraint is there or not. If we include the constraint, the average solution and the particular player solution are both numerically 2/3. We've been presenting this solution as if it shows a particular player's odds of winning by switching are 2/3. Each player's odds are 2/3, but this reasoning doesn't show this result (it shows the average player's odds of winning by switching are 2/3 which is true even if the host always opens the rightmost door and any given player's odds are 50% or 100%). I think it will take some doing to adequately explain this point in the article. -- Rick Block (talk) 05:16, 3 March 2008 (UTC)
I'm not sure that we are quite on the same page here. Let's look at that wording again: "When the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with probability 1/3." Let's assume that we are taking into account which specific door the host has opened (a not wildly unnatural assumption, given the wording when the player is asked whether to switch). If the host chooses randomly then this statement is true whichever door the host opens. If the host does not choose randomly then the statement is false: there is not necessarily a 1/3 chance that the player originally chose the car, it depends on which door the host opened. So, the solution works either in the unconditional case with any host behaviour, or in the conditional case under the assumption that the host chooses randomly. In the latter case the assumption that the host chooses randomly is implicit in the statement that the three cases are equally likely. As far as adequately explaining this point in the article is concerned... although I originally thought it was irrelevant in the "general reader's" explanation I have now changed my mind, and that's precisely what I'm suggesting we should do -- not cluttering up and complicating the simple solution wording that we have, but immediately following it in a subsection. This is what I was attempting to do in my version 4 wording at Talk:Monty Hall problem/Matt#Version 4. I have slightly reworked this now; take another look if you feel so inclined. Matt 14:18, 3 March 2008 (UTC). —Preceding unsigned comment added by 86.146.47.184 (talk)
I agree with pretty much everything you've said here, but the equal probability phrasing is (I think) pretty clearly meant to reflect the equal probability of the player's initial pick rather than the resultant situation after the host has opened a door. The point I'm making is that this phrasing (reflecting the player's initial pick, before we know which door the host opens) shows us the "average" probability of winning regardless of the equal goat constraint. With the equal goat constraint we do indeed end up with the enumerated cases being equally likely after the host opens a door but we haven't said why they're equally likely (and given the length of this discussion over the last several weeks I don't think it's trivially obvious). -- Rick Block (talk) 15:19, 3 March 2008 (UTC)
Essentially I agree with you. However, if we are all agreed that the solution strictly is correct in the conditional interpretation, given that the host chooses randomly (which we say he does), then we don't need any disclaimers that imply it may somehow be faulty, incomplete or simplified. The implication of the equal probability statement as it applies to the conditional interpretation is, as you say, far from trivially obvious, and it may not become apparent until the reader reads the later section explaining this issue (if he ever does; if not then never mind). I have changed my position dramatically since the start of this -- largely because of my discussions with you -- and I now feel that the conditional interpretation of the question is a very natural one. Although we don't need to spoil the main solution wording by labouring this, I do feel that we need to explain it somewhere. To this end I am continuing to pick away at the wording. Unless we're going to sweep this completely under the carpet, which I think would be a mistake, at some point we're going to have to come up with a way of explaining it. Matt 17:51, 3 March 2008 (UTC). —Preceding unsigned comment added by 86.137.136.25 (talk)


Matt 21:13, 2 March 2008 (UTC).
This might be okay. I'd have to actually see it to know if it's what I'd want.
My overall desire is that the primary focus of the article should be on explaining the paradox (2/3 odds, not 1/2) to a general audience. To that end, I want to see the simplest possible presentation of both the problem and the solution (short of stating them ambiguously) at the beginning of the article, followed by a note about whatever simplifications were made, and a full explanation of the intricacies of unconditional vs. conditional probabilities addressed in later portions of the article.
As for labeling the unconditional solution "intuitive", that is a bad idea. Intuition is what makes most people believe the odds of switching will be 50-50. I'd go with "simplified solution", personally.--Father Goose (talk) 02:03, 3 March 2008 (UTC)
I agree that the label "intuitive" must not be used. But I also do not think it should be called a "simplified" solution, at least not if we maintain (as I believe to be the intention of the original problem) that the opening of the door by the host conveys no information whatsoever to the player about whether the initial choice was correct. (Note that this formulation differs subtly but tangibly from that used in the article now; it is also meant to cover the manner of opening.) Under that assumption it is a full solution — and the solution given as the main one now is unnecessarily complicated. Problem versions for which the assumption is not correct should be delegated to the variants.  --Lambiam 13:09, 3 March 2008 (UTC)
I agree with you that it is a full solution under that assumption; this is essentially the point I was trying to make, from a different angle, in my recent response to Rick, above. However I think that the phrasing the opening of the door by the host conveys no information whatsoever to the player about whether the initial choice was correct is inherently confusing and should not be used without further explanation. It strongly implies that the player might be using this information to decide whether to switch, and that the question is about optimal player strategies, whereas in fact, as I understand it, we're only considering the "always switch" and "never switch" strategies. In fact, we are not interested in the host door conveying information to the player per se, we are interested in whether it conveys information to the person who is calculating the probability. That might be the player, or it might be "us". Matt 14:24, 3 March 2008 (UTC)~. —Preceding unsigned comment added by 86.146.47.184 (talk)
I think we are very close, and hope we can achieve consensus with other editors who may not think alike. As to your specific points: in the "two-part approach" I proposed above (08:28, 2 March 2008), a clarification like the conveys no information phrase would only appear following the presentations of the solution, in a brief discussion of "unintended interpretations". Whichever the interpretation, if the problem is phrased like: Is it to your advantage to switch your choice?, we may equate, for the purpose of presenting an analysis, the player with whoever does the maths. An answer like Yes it is, but you have absolutely no way of knowing that would be thoroughly unsatisfactory.  --Lambiam 18:27, 3 March 2008 (UTC)
Right. I'm hoping that we might now all be able to agree that the solution currently presented in the article can be left fundamentally unchanged, and that the issues surrounding the host behaving randomly or non-randomly, and the conditional vs unconditional interpretations, can be covered in a subsequent section. Are we at the stage where we can proceed to settle exactly what wording changes, if any, we want to make to the opening sections (lead, problem statement, solution)? Then we can move on to the next part. After all this effort it would be nice to knock this on the head. Matt 01:08, 4 March 2008 (UTC). —Preceding unsigned comment added by 86.133.55.27 (talk)
I have tried to work the "conveying information" theme into the explanation at Talk:Monty Hall problem/Matt#Version 4. Matt 14:46, 3 March 2008 (UTC). —Preceding unsigned comment added by 86.146.47.184 (talk)
I've looked around for references we might add for
a) the equal goat constraint
Mueser and Granberg mention the constraint but (curiously) don't use it in their unambiguous problem statement. Our claim that "The problem as generally intended ..." pretty much begs for a reference. Perhaps the best we can do is the Mueser and Granberg reference (which says "the answer that the contestant has a 2/3 chance of winning if she switches follows if we assume ...", not quite the same as "as generally intended").
b) what to call the "intuitive" (unconditional) solution that we present
Morgan et al call this a "false solution" (to what they interpret as the conditional problem, and without the equal goat constraint). They later call this a solution to the unconditional problem, which they phrase as "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?" (note this is a general question, not about any specific player given what door they've initially picked and what door the host has opened). I haven't found any other reference (yet) that attempts to distinguish the unconditional and conditional solutions.
Update: Gillman's American Mathematical Monthly article [2] published 2 months after Morgan et all looks to be very similar to Morgan et al (based on the first page). Anyone have a copy of this? -- Rick Block (talk) 14:10, 4 March 2008 (UTC)
-- Rick Block (talk) 05:02, 4 March 2008 (UTC)
Rick, could you try to avoid using the terminology "unconditional and conditional solutions"? It is unnecessarily confusing. The notion of a "conditional" solution has no meaning for the main problem version. It is only in variations of the main version that conditional issues arise. I don't have access to the original Parade column in which the solution is explained, but the solution given there (and discussions in subsequent columns) may make it clear that the problem is not meant to be interpreted in a conditional way ("If the host did not wink, and you hear a bleating sound coming from behind door No. 2, then ..."). The title of the article by Bohl et al., to which I also do not have access, might be helpful, since it sounds like it will give the relevant tacit assumptions.  --Lambiam 14:20, 4 March 2008 (UTC)
I'm just using those terms here on the talk page, and I'm not sure if I want to get into this again but they both definitely apply. The original Parade column is quoted at http://www.marilynvossavant.com/articles/gameshow.html and the original question, which we quote in the lead, has the player picking door #1, the host opening door #3, and the player being asked "do you want to switch todoor #2?". This sure sounds like a conditional question. Call it what you will, but a solution that produces the "average" probability of winning by switching, ignoring the specific door the host opens, is different than a solution that does consider the door the host opens. For example Marilyn's table, at the link above, enumerates all the "equal probability" alternatives, including the case where the car is behind door 3. Clearly this is not an equal probability case where we're considering the probability of winning given the player originally picked door 1 and the host opened door 3. "Conditional" doesn't mean given any conceivable conditions, but simply given the conditions explicit in the problem statement.
I don't know if you've read my whole dialog with Matt about this, but my agenda here is NOT to introduce the distinction between conditional and unconditional analysis in the article in the initial solution section. However, I do want our solution to not be subject to the same criticism Marilyn's solution has been subject to (by Morgan et al, Gillman, and perhaps others). I think doing this requires that we understand the criticisms. I suggested above that we introduce our solution with wording like "Over numerous trials of this game, there are three equally likely scenarios depending on the player's initial choice ...". IMO this clarifies that we're talking about an "unconditional" analysis (without making a big deal of it, and without using the terms "conditional" or "unconditional"). -- Rick Block (talk) 15:55, 4 March 2008 (UTC)
This column from The New York Times of July 21, 1991, contains, next to other possibly usable material, discussions of variants of the problem using words like violating the spirit of Ms. vos Savant's problem, [but] not violating its letter, and loophole, which are quite explicit ways of saying we are dealing with unintended interpretations. It also gives the simple solution as offered by vos Savant, which is substantially simpler than what we have now in the article, and should be included as the first solution.  --Lambiam 14:40, 4 March 2008 (UTC)
[Written before digesting Lambiam's response above.] Regarding Rick's point (b), I think the issue is not what we call the solution, it's what we call the problem. The solution is just the solution to that problem, no more, no less. The problem that we want to be presenting at this point in the article is, presumably, the one we consider to be the "usual" or "most natural" interpretation (based on the evidence that we can find, or, failing that, our own best idea). The solution itself is valid for both the unconditional problem with no host behaviour constraints, or the conditional problem with the equal goats constraint, so in principle we can present either. The current M & G wording, in which the doors are numbered and we are placed at the point that a specific player is switching, implies the conditional interpretation. So, if we use the M & G wording then what you're calling the "unconditional" solution is, in the context we're initially using it, actually applying to the conditional problem. Whether we want to actually use the words "conditional" and "unconditional" at this initial stage I'm not sure.
I think we should first decide whether we want to present the conditional conditional-with-equal-goats or unconditional problem as the "usual interpretation". I suppose another option would be to bite the bullet and make the conditional/unconditional distinction very visible up front, and split the "usual interpretation" into two parts. Not sure about that... Matt 14:46, 4 March 2008 (UTC).
Another reference: Mathematical Fallacies, Flaws and Flimflam by Edward J. Barbeau (ISBN 978-0883855294), states: The standard analysis [of the Monty Hall problem; my emphasis] is based on the assumption that after the contestant makes the first choice, the host will always open an unselected door and reveal a goat (choosing the door randomly if both conceal goats) and then always offer the contestant the opportunity to switch.p.87  --Lambiam 16:11, 4 March 2008 (UTC)
Here is how I'd like to see this addressed. Here is the TOC:
1 Problem
2 Solutions
    2.1 Solution A
    2.2 Solution B
    ...
    2.Y Bayes' theorem
    2.Z Simulation
3 Unstated assumptions
4 Variants
    4.1 Other host behaviors
    4.2 Two players
    4.3 Sequential doors
    4.4 Quantum version
    4.5 Similar problems
5 History of the problem
6 References
7 External links
The only thing that is essentially different from the present article is a new section between Solutions and Variants. Other differences I'd like to see:
  • Split Problem and solution into two sections, and merge (large parts of) Aids to understanding into Solutions.
  • Use a condensed formulation of the problem statement in the lede, and the full one from Parade under Problem. Postpone the M&G formulation to the section on unstated assumptions.
  • Move the "simple solution" from the intro to Solution A. Leave the answer ("switch") in the lede, but remove the analysis.
  • Until we reach the section on Unstated assumptions, avoid mentioning them explicitly as additional assumptions (also in the lede!), but instead introduce them en passant in the analysis; for example thus: "Each door is initially equally likely to conceal the car, therefore ...", without drawing attention to the fact that this assumption of equal probabilities is actually unstated.
 --Lambiam 16:40, 4 March 2008 (UTC)

Some further comments on the recent points being debated. There is all the difference in the world between hosts winking, bleating sounds, and a million other things that we all agree no-one intends, and the knowledge of the door that the player chooses and the door that the host opens, which is a perfectly reasonable thing to assume that we know. Knowledge of the doors automatically makes it a conditional problem, and even if we don't want to highlight this in the initial presentation, and even though with the equal goats constraint the answer and the reason for the answer are the same, we still need to be clear in our own minds which question we're asking.

The problem as I see it is that most versions of the problem wording place us at the point at which the player switches, where we might reasonably be assumed to know the doors. In statements such as the M & G one currently in the article, and also the original Marylin quote, where the doors are identified by numbers, letters or positions, it is very explicit that we know the doors. Like it or not, and whether we call it by that name, we have a conditional question.

Originally I felt that if we used a form of wording such as at Talk:Monty Hall problem/Matt#Version 4, where we we didn't identify the doors, then we'd be fine, but I now think that this is a fudge. We're still at the point of switching, and it's still perfectly reasonable to assume we actually do know the doors. A proper statement of the unconditional problem requires some rather different wording that I'm not entirely sure is the "usual" one. So, I'm worried we might be in the position of presenting as "standard" a form of wording that isn't actually the one most often used. Hopefully this concern can be disposed of.

I rather like Lambiam's suggestion that we move the original quote out of the lead into the problem section. Then, rather than trying to find a quotable version that says exactly what we want to say, we can quote the original formulation, say that it needs clarifying (which we already do), and then list in our own words the specific clarifications, into which we slip the assumption, suitably worded and without labouring the point, that we're using the unconditional interpretation*. Then, as Lambiam says, the M & G quote and the explicit discussion of conditional vs unconditional analysis can follow in a later section. Matt 18:33, 4 March 2008 (UTC).

* assuming, of course, that this is the interpetation we want to lead with, which is the question I asked before: we need to decide, in our own minds at least. Matt 20:09, 4 March 2008 (UTC).
I am generally in concurrence with Lambiam's suggestion. My only reservation is that it entails quite a bit of surgery, albeit mere cut & paste for the most part. My only suggestion regarding structure is to offer first an "explanation" followed by a series of "proofs." This avoids the problem of calling the (putatively) simplest solution an "informal" one, and makes explicit that the various solutions are different ways of deriving the same result, not different results. 67.130.129.135 (talk) 18:39, 4 March 2008 (UTC)

[edit] Suggestion

I feel we are at the stage where we would make faster progress if we actually started looking at concrete versions of the wording, rather than talking in abstract terms. To this end I have knocked up something at Talk:Monty Hall problem/Matt#Version 5 based loosely on some of the suggestions, and removing quite a lot of what seemed like unnecessary repetition in the current version. At the moment it stops at the point where we think we have covered the standard interpretation and its solution. This is strictly a "straw man" -- it doesn't matter if this version is completely rejected, it doesn't matter if it is utterly changed, it doesn't matter if it eventually gets put back to how it is in the article now. I think the important thing is just that we progressively move towards the actual version of the wording that we are going to put in the article. Otherwise we're in danger of going round in circles forever. That's my opinion anyway. Matt 23:35, 4 March 2008 (UTC).

I think it helps if we have consensus on the major issues before we start sculpting the butter. Since the changes can't be just local, I also think it is easier to work from a copy of the full article, and transform it somewhat more gradually, with regular synchronization with the article. Otherwise, assuming that the "new" version will take some time to tale shape, merging it with the article will be a nightmare.
However, if everyone agrees that (eventually) we'll postpone formulating all these unstated assumptions (4 bulleted clauses in Talk:Monty Hall problem/Matt#Version 5, but I believe more assumptions are needed, such as that the player knows the host always behaves like that, and also that the way the host opens the door conveys no signal) until after the presentation of the solutions, we could already start work on the proposed new section "Unstated assumptions".  --Lambiam 00:17, 5 March 2008 (UTC)
There should be no need for "merging" or piecemeal edits to the article. We eventually end up with a complete new version of the article (or those sections we want to change anyway), and then copy and paste. I think it will be easier if we begin at the start of the article and try to get some stuff fixed (we can always go back and change it later), rather than planning everything out in theory... then when we come to implement it forgetting what we decided, trying to figure out how to actually word what we decided and then debating the exact form of wording all over again, and changing our minds about what we thought we decided, and having all the fiddle of trying to merge it into the article. Matt 01:43, 5 March 2008 (UTC).
What will you do with the edits to the article itself that were made in the meantime? Discard them?  --Lambiam 11:18, 5 March 2008 (UTC)
If they overlap the wording that we've so laboriously agreed (or will have so laboriously agreed), then yes. We're not going to agree it all once and then try to re-agree how to merge it with a bunch of independent edits that may have happened externally to this discussion. But to be kind to editors, significant edits that overlap what we are discussing now should probably be undone with a polite note directing the editor to the talk page discussion. But you're right, of course, there may be a few non-overlapping edits that we would have to reapply. Matt 12:20, 5 March 2008 (UTC). —Preceding unsigned comment added by 81.156.127.240 (talk)

[edit] Does knowledge of the doors automatically make it a conditional problem?

Does the player's knowledge of the door they initially chose make it a conditional problem? No, because of the symmetry in the problem. That symmetry would be broken if the player somehow had information about the initial position of the car. But the (unstated) assumption is they have no such information.

Likewise, knowledge of the choice of door opened by the host does not make the problem conditional. Why not? Also simply by the symmetry of the problem. Any reasoning that tells us it is worse|indifferent|better to switch if the initial choice is X and the host opens door Y, equally applies if the initial choice is X' and the host opens door Y'. The only way it might not is if there is something that breaks the symmetry, like a known preference of the host for opening doors to the left – which then conveys some information about the position of the car. This form of symmetry breaking is also ruled out by an (unstated) assumption.

The formulation in Parade, with "say No. 1" and "say No. 3", is obviously intended to indicate a symmetrical interpretation; "say" means so much as: it-does-not-matter-which-but-for-the-sake-of-concreteness-let's-name-something. The problem could have been phrased completely equivalently as:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, "Do you want to pick the other unopened door?" Is it to your advantage to switch your choice?

Giving names to the door makes it easier to grasp for the reader which door is "the other unopened door"; that's all there is to it.  --Lambiam 00:44, 5 March 2008 (UTC)

I broadly agree with what you say, but strictly speaking not quite with the statement: "Knowledge of the choice of door opened by the host does not make the problem conditional". Technically I think that it probably does; the thing is that with the symmetrical problem it doesn't matter. It's kind of like the difference between these two problems:
  • You toss a coin twice. What is the probability that you get the same result both times?
  • You toss a coin twice. What is the probability that you get the same result both times, given that the first toss was heads?
The answer's the same, but the second question is (IMO) technically a conditional one. Second point: I would prefer to see the usual-interpretation assumptions stated along with the problem, rather than offer a solution to what we acknowledge is an ambiguous question and then backtrack. Sorry, response is a bit rushed. Matt 02:01, 5 March 2008 (UTC).
If the host tells you that the goat's name is Billy, does that make the problem conditional?
  • Is it to your advantage to switch your choice?
  • Is it to your advantage to switch your choice, given that the goat's name is Billy?
 --Lambiam 11:25, 5 March 2008 (UTC)
The form of the wording indicates, strictly, that it is. But probably we are getting into terminological niceties here. Matt 12:23, 5 March 2008 (UTC). —Preceding unsigned comment added by 81.156.127.240 (talk)
This use of language such as "a door, say No. 1" and "another door, say No. 3" is so completely conventional that symmetry really is not an "unstated" assumption; this bit is completely explicit and unambiguous in the statement of the problem. I tried to make this point on day two of this discussion, some three weeks and 30,000 words ago, but it went over like a lead balloon. 67.130.129.135 (talk) 02:37, 5 March 2008 (UTC)
If the constraints in the problem statement do not make the problem symmetric, using "let's say" in the solution doesn't make it so - it makes it an unfounded assumption. Without the equal goat constraint the problem is not necessarily symmetric, so whether we say "let's say" or refuse to name the doors doesn't matter - the conditional probability of winning by switching given which door the host opens might be different from the unconditional probability (the average across all players who switch). This is the point of at least two different papers published about this problem (Morgan et al and Gillman). Like I've said elsewhere, what we're doing by assuming the problem is symmetric is assuming the host picks between two goat doors with equal probability. If the host doesn't do this the problem is not necessarily symmetric; that's all there is to it.
I'm fine with using the unconditional analysis because it's easier to understand (so long as we introduce it as the unconditional solution, using whatever understandable words we'd like), but let's please not delude ourselves into thinking there is no difference between an unconditional and conditional solution.
I've said this before, too, but I suggest we keep the M&G formulation, with the added equal goat constraint, and then present the unconditional solution prefaced with something like "on average, across many players ...." to make it clear we know what we're talking about. Then we can follow this up with: "There's a subtle difference between the solution above which applies to all players on average and a solution that applies to a specific player given the initially picked door and which door the host opens. This sort of solution, called a conditional solution goes like this ..." Given the constraints we've imposed (including the equal goat constraint) both will end up with a 2/3 chance of winning by switching. I think we should end the solution section with something like "See other variants for variations where the average probability across all players and the conditional probability given which specific door the host opens are different." -- Rick Block (talk) 04:29, 5 March 2008 (UTC)
You agree to conditionally present things in an unconditional simplified way provided that we make sure we introduce all kinds of conditional complications first. I'm going to drop out of the discussion; my contributions have had zilch effect.  --Lambiam 11:34, 5 March 2008 (UTC)
That's a shame. Your suggestions certainly have not had zilch effect on me. I think, for example, your idea of taking the analysis and quoted problem out of the lead section, and using the problem statement from Parade under Problem, is an excellent one, that I have tried to follow at Talk:Monty Hall problem/Matt#Version 5.
I am going to make one more push that we consider that wording as a starting point. Rick: can you fundamentally live with this sort of structure, or is it a total non-starter as far as you're concerned? Matt 12:32, 5 March 2008 (UTC).
Regarding 'using "let's say" in the solution,' (emphasis added) note that Marilyn used it in the original statement of the problem. I will grant that it is informal, and leaves a loophole for host partiality. To express original intent rigorously, some form of equal goat stipulation is useful, as is a stipulation of equal probabilities for the original placement of the car. (When presenting the problem as a parlor game, I always include both stipulations.)
My position is that the article should focus, in the lead and in the initial exposition, on original intent. When original intent and strict construction are at odds, one needs to make a judgment call. My post of 29 February explains the reason for my position on this call. Is the consensus is that the article should be principally about Marilyn's folly rather than Marilyn's interesting puzzle? (The "version 5" draft statement of the problem is, by rough word count, 70% about ambiguities.) 67.130.129.135 (talk) 18:56, 5 March 2008 (UTC)
Postscript: Sorry about the tone above. Just feeling a little cranky. I do appreciate the good faith efforts of everybody here. 67.130.129.135 (talk) 19:15, 5 March 2008 (UTC)

[edit] Working draft

I've made a copy of the article at Monty Hall problem/draft starting with the current version. This edit is my suggestion for how to respond to the majority of the comments on this thread.
Points about this version:
  1. It's not that different than what we have now (decide for yourself whether this is a good thing or a bad thing)
  2. The featured article criteria includes a requirement that the lead summarize the important points of the article (this is why some of the redundancy is there that Matt's version deletes).
  3. This version uses several of Lambian's suggestions, including the addition of Barbeau as a reference, as well as several of Matt's suggestions from his version 5.
  4. It is my hope that this version would be acceptable to our anonymous friend of many IP addresses (who hasn't commented in quite some time).
I hope Lambian continues to comment - I agree with Matt that his input here has been constructive and quite useful. -- Rick Block (talk) 19:38, 5 March 2008 (UTC)
Great stuff Rick... I think the most important thing is that we start making visible progress with the actual wording that we are going to use, rather than debating it in the abstract, which could go on forever... and at the end of all that we'd still need to actually write it up, with the possibility of that exercise reopening the debates all over again. I don't much mind what wording we start with. As soon as I get a chance I'll read through the draft and let you have my comments. Matt 01:24, 6 March 2008 (UTC). —Preceding unsigned comment added by 86.134.13.28 (talk)

(Undo indent) Here are my comments on the relevant sections of Monty Hall problem/draft:

  • The lead section looks much cleaner! (I don't understand what you mean by our being obliged to retain some of the stuff I deleted. Apart from the wording of the problem statement, your version looks virtually identical to mine. But never mind...)
    This was just in response to your comment about having removed unnecessary repetition. Repetition between the lead and the article proper is required. -- Rick Block (talk) 20:18, 6 March 2008 (UTC)
    At the level we currently have, yes, absolutely. Matt 01:30, 7 March 2008 (UTC).
  • As you know, I don't like the M & G wording, for the various reasons I've explained before. But if everyone else is happy then I'll drop it.
    We might be the only two left talking ( :( ), in which case it's a tie. I think you know my reasons for preferring it as well. -- Rick Block (talk) 20:18, 6 March 2008 (UTC)
    If it's a tie then I'm happy to leave it as it is. Matt 01:30, 7 March 2008 (UTC).
  • In a more precise statement of the problem ... the host is constrained to always open a door revealing a goat and to always make the offer to switch. This seems to be only a partial list of the reasons why the statement is more precise. For example, the new wording also makes it clear that the player has a 1/3 chance of picking the car, and the reported-speech cicumlocutions are, as I understand it, supposed to indicate that the player knows the rules in advance. I have no idea why that matters, but if it does we ought to mention it in the list.
    These two (always open a goat door and always make the offer) are typical knocks on the Parade version. They're specifically mentioned merely to highlight them. Would a reference to a criticism of the Parade version make it more obvious why these and not others are mentioned? -- Rick Block (talk) 20:18, 6 March 2008 (UTC)
    Yes, I think that would be a good idea, but I think it would be good to mention the others too if we can. After all, we're presenting this, in effect, as the completely unambiguous version, and ideally we should list all the ambiguities in the original that we're clearing up. Matt 01:30, 7 March 2008 (UTC)

So far so good. To be quite honest, though, I'm afraid I'm not very keen on the solution section. Sorry! These are my main criticisms:

  • The "intuitive" solution looks lost in the text. Even though it takes up a couple more lines, I think we should retain the existing format with the bulleted list. It's punchier, clearer and more eye-catching -- giving this common analysis its due prominence.
    I noticed this is as well, although I thought it would be confusing to present two bullet lists in the same section. I've included a very slightly modified version of your list (with the bullets), introduced as "A typical explanation" (and a note that we should add a reference). I'm not sure if "typical" is quite the right word, but the intent is to present this solution attributed to at least one source that uses it. Nothing here should be ours - we're simply presenting what others say (is this a potential way out of our disagreements?). -- Rick Block (talk) 20:18, 6 March 2008 (UTC)
    Great, I think that's an improvement. I'm not sure about "A typical explanation..." either. It somehow sounds a bit disparaging, or dismissive, as if we're going to follow it with "... but, this is wrong...". I do take your point, though, that ideally we need to somehow indicate that we didn't just "make it up one day at school". I can't think of the exact form of words just now but perhaps something will come to me. Regarding the multiple bulleted lists, I don't think they should be in the same section. I think we really do need a new sub-heading after the "simple" solution (still within the main "solution" heading would be fine I think).
  • I think we should stop the initial solution section once we've explained the "intuitive" solution, and treat the conditional vs. unconditional issues in a separate section. Then, in either the problem statement or "intuitive" solution section we need a forward reference to the conditional vs unconditional section, somehow tying it in to the equal goats constraint. Something simple to make sure that the issue is noted, but that the reader can just zip over ("I don't exactly see that now, but I understand this simple solution, and I know that nicety is explained later if I want to read on").
    Per the suggestion just above that we present what others say, I think this section is simply not complete if we don't also present something about the conditional/unconditional issues. Again, would it help to explicitly introduce this with a reference? E.g. The solution above applies to all players without regard to which specific door the host opens and is criticized by some (Morgan et al) for ignoring this information. A more formal conditional .... Note that "criticized by some" would need to be carefully worded in a WP:FA sort of way. -- Rick Block (talk) 20:18, 6 March 2008 (UTC)
    Absolutely we need to say something. In fact, we need to say rather more than we currently do -- not necessarily in terms of column inches, but in terms of information content. But the way I see it is this: the simple solution is accurate and complete, given the problem that we've stated, in both the conditional and unconditional interpretations. We present this first without troubling the reader about the distinction, beyond a brief forward reference to hint that there is more to come. Then we pause for breath. Then, under a new sub-heading we describe the two intepretations, and all the ramifications. I think your reference about the criticism could well go at the start of this new section, as a lead-in. But we need to be careful that we're not quoting criticism of a different problem (one without the equal goats) in such a way as to suggest that the solution for our formulation (with the equal goats) is incorrect.
  • In the new section about conditional/unconditional analysis we need a more coherent motivation. The statement The solution above applies to all players without regard to which specific door the host opens. A more formal conditional solution analyzes the problem at the point the player is asked whether to switch explicitly considering which door the host opens. is misleading at best. In the symmetrical problem, with the equal goats constraint (which we've explicitly included), the "intuitive" solution applies in all cases, whether or not we have regard to which specific door the host opens. We need to explain why and under what circumstances/interpretations the "intuitive" solution fails, how this is tied in with the host-behaving-randomly issue, and how and why the conditional analysis might give a different answer. At the moment it kind of looks like the conditional solution is just a complicated way of getting exactly the same answer, and we're not quite sure why we need the equal goats constraint but we thought we'd better include it just in case.
    Does my response above address this? I thought at one point you agreed that although the intuitive solution is a true statement when applied to the conditional question (given the equal goat constraint) the reason the cases are equally probable is far from obvious (includes reasoning like "if you pick goat A, since the car is equally distributed behind the doors the host opens either one half the time").
    I agree, the reason is not obvious. But if we're going to do this at all then we need to bite the bullet and do it properly, in my view. We need to actually explain this, and the other things I mentioned. Otherwise there's no point to the exercise that I can see, and all the effort that's been put into bottoming out these issues is wasted. Matt 01:30, 7 March 2008 (UTC).

In summary, I feel that the issues we've been discussing fail to come across at all clearly in the text. Sorry to be so negative. Matt 18:39, 6 March 2008 (UTC).

Yes, but I think we're making progress. However much we seem to agree or disagree, I very much appreciate your willingness to continue to engage in this conversation. -- Rick Block (talk) 20:18, 6 March 2008 (UTC)
I suggest that the first paragraph under Solution in the current version of the draft could be moved to Sources of Confusion. It describes the naive 50/50 error, not the correct solution. The fallacy is already mentioned in the lead. Its elaboration can be deferred to that later section so the Solution section can get right to the solution. We might also consider putting Sources of Confusion ahead of Aids to Understanding. 67.130.129.135 (talk) 22:51, 6 March 2008 (UTC)
Sounds good to me. Do you want to edit the draft, or wait for Matt (or anyone else who may still be lurking) to comment? -- Rick Block (talk) 01:04, 7 March 2008 (UTC)
I agree too. We say this in the lead, and we have another section about erroneous solutions. There's no need to say it a third time when we're presenting the correct solution. Matt 01:36, 7 March 2008 (UTC). —Preceding unsigned comment added by 86.136.195.87 (talk)

I've moved the paragraph under Solution to Sources of Confusion and moved the Sources of Confusion section so it follows Solution. I've also tried to address Matt's concerns about motivating the conditional analysis. I don't know if I'm the only one who thinks this (and if so I suppose I need to back off) but I really think we need to include both explanations in the primary Solution section. -- Rick Block (talk) 13:28, 7 March 2008 (UTC)

I made the changes above in two separate edits, so they can be viewed independently. To reuse the same bullet list between the two solution descriptions I changed it to "car location" based rather than "user pick" based. and added door numbers. This destroys the property of the previous version that it's a true statement (but not obvious!) even for the conditional question. I think door numbers actually help make it more concrete (less abstract) and therefore more accessible. The whole thing fits in one screenful of text and doesn't seem to me to be particularly difficult to follow. Is this getting closer to something others can live with? -- Rick Block (talk) 17:12, 7 March 2008 (UTC)
I've spent some more time tweaking the wording - it now says pretty much what I think it should say, in pretty much the way I think it should be said. -- Rick Block (talk) 00:49, 8 March 2008 (UTC)
I do apologise for once again knocking your hard work, but I have to give an honest opinion, and I really don't like it at all. I've taken one last shot at this at Talk:Monty Hall problem/Matt#Version 6. Do you think there's any way we can meet in the middle, or are our approaches irreconcilable? If the latter then I will bow out gracefully! Matt 01:02, 8 March 2008 (UTC). —Preceding unsigned comment added by 86.142.109.205 (talk)
That's funny - I was thinking I'd bow out about now as well. Reading the two versions side by side they don't really seem that different (I assume you read the most recent version of mine - post tweaks). Your version presents a solution that's effectively the same as the popularly presented version that's criticized in the literature, introduces a new header, and goes on at some length regarding the difference between conditional and unconditional. My version segues from one to the other (obliquely referring to the criticism) in an integrated solution section without making a big deal out of it (what I'm trying to do is present the two different approaches as being more like two sides of the same coin than actually "different" in any significant way). I think it would really be good if there were more people still willing to talk about this. I'm not sure, but I'd guess Lambian would prefer your version. Not at all sure about 67.130.129.135. I suspect the original instigator of all this (anon-many-ips) wouldn't like either one (although I'd assert he should be happier with mine). Don't know about Father Goose or Glopk or Wikiscient (has anybody else commented?). I've already opened an RfC on this, and solicited comments at Wikipedia talk:WikiProject Mathematics. I guess directly soliciting comments from these folks might be worth a try.
In any event, no it doesn't look like we're going to end up agreeing and given how this discussion has gone (and how many users have stomped off in huffs) perhaps it's time for somebody else to be the main shepherd for this article. I'll be around, and I'll be happy to help if asked, but it seems like it might be best to let somebody else try to herd the cats for while. -- Rick Block (talk) 03:18, 8 March 2008 (UTC)
I fatigued a while ago, and haven't been paying close attention to how the discussion was going. If it's back in a logjam, I can't add anything.--Father Goose (talk) 06:20, 8 March 2008 (UTC)

[edit] Possible wording of the problem

This problem confused me for years, but I recently got a good grasp on it which I think makes it a lot easier to understand. Please read below and tell me what you think.

Assume you have two players in a game. You have three doors. Behind one door is a car, behind the other two are goats. You choose one door in the hopes that it is the door behind which is the car. The other player chooses a door also hoping to get the car. Again, there is one door out of three that has the car. You have a one in three chance to pick the car. There is a 2 in three chance that the car is behind another door. The other player now has two doors to choose from (because the door you picked is not available to him). Because there is a 2 in 3 chance that the car is behind one of these two doors and he can only select 1 of the 2 doors, he has a 2/3 * 1/2 chance of picking the right door. 2/3 * 1/3 = 1/3. He has a 1 in 3 chance of picking the right door.

Now, let's change the rules of the game. The other player is not trying to win the car, he is, instead, trying to confuse you as to where the car is (and he already knows where the car is). You pick the first door and have a 1 in 3 chance of being right. If the other player picks the door with the car, then he doesn't confuse you as to where the car is (he, instead, shows you exactly where the car is). So, his goal is to pick a door which does not have the car behind it. Out of the 2 doors available to him, there is a 2 in 3 chance that the car is behind one of them. He deliberately selects the door without the car. Again, there is a 2 in 3 chance that the car is behind one of the doors which were closed at the start of his turn and he's just reduced that number of doors to one. So, there is now a 66% chance the car is behind the one door neither of you selected.-198.97.67.59 (talk) 17:00, 7 March 2008 (UTC)

[edit] No new information

Good lord this talk page has gotten long. Anyway-- has any consideration been given to pointing out that when the host opens the door with the goat behind it, no new information has been revealed to the contestant? The player already knows that there are two goats and one car, so no matter which door they initially pick, it's a given that one of the other two doors will conceal a goat. Thus, the host revealing a goat and then asking the contestant if he'd like to switch could be replaced by the host simply asking, "Do you want me to open the ONE door you picked, or BOTH the doors you didn't pick?". The probabilities of getting a car would be identical either way. Clayhalliwell (talk) 23:45, 11 March 2008 (UTC)

On the contrary, the player now knows where one of the goats is, and knows not to pick that door. That alone changes the odds of winning the car from 33% (could be behind one of three doors) to 50% (could be behind one of two doors). Other factors explained in the article further raise the odds (when switching) to 67%.--Father Goose (talk) 04:08, 12 March 2008 (UTC)
Ummm... you can't pick a door that's already been opened... Clayhalliwell (talk) 05:32, 12 March 2008 (UTC)
Exactly. One of the "goat doors" has been removed as a choice. Take the 100-door variation: your chance of picking the door with the car behind it initially is 1 in 100. Open 98 goat doors, and your average odds of winning have risen to 1 in 2: the car is either behind your door or the other unopened door.
However, if the host deliberately avoided opening a car door when opening the 98, the car will be behind the "switch to" door 99% of the time, for the reasons explained in the article.--Father Goose (talk) 05:52, 12 March 2008 (UTC)
Okay, somehow your version of these explanations is actually making me understand the problem less. Clayhalliwell (talk) 15:43, 12 March 2008 (UTC)

Let me try a different tack then. If the host were to reveal both of the other two doors -- and still let the player switch -- the player would win 100% of the time (he'd either see where the car is and switch to it, or see two goats and stick with his original choice).

If the host reveals the player's door but still allows the player to switch, then the player wins right away 1/3 of the time (oh, look, I picked the car!) If the host reveals that the player initially picked a goat (2/3 probability), the player still has a 50-50 chance of picking the car behind one of the two remaining doors. (His total odds of winning over the course of this game are 2/3: 1/3 initial win plus 50% chance of winning in the 2/3s of cases where he switches: 1/3 + 50% * 2/3.)

If the player were not allowed to switch after the host reveals any door(s), the probability of winning is 1/3 straight up: the player has to be lucky enough to pick the "car door" right off the bat.

These are, however, very different ways of playing the game than specified in the "problem" section.--Father Goose (talk) 20:19, 12 March 2008 (UTC)

I was also mislead by the "no information" viewpoint and propose the following addition: "A source of confusion is the notion that Monty reveals no information. It is given that Monty will always reveal the booby prize, a goat in the example (because it prolongs the suspense). Since we know Monty will reveal a goat and then proceeds to reveal a goat, we might conclude that no information was provided. But clearly some information was provided - that a particular door has a goat. Whether this information is useful may not be clear initially. But indeed it is. Here's why. Since it is a given that Monty never reveals the selected door immediately (as it would resolve the suspense) and always reveals a goat, he never reveals anything about the probability that the selected door has the car - and so the probability that the car is at the selected door remains at 1/3. But since it is given that a valuable prize is behind one door - and only one door (to minimize show expenses), the sum of the probabilities for all the doors must be one. That is, the probability that the car is at door one plus the probability that the car is at door two plus the probability that the car is at door three must total to 1. Since the probability that the car is behind door one (before or after the goat is revealed) is 1/3 and the probability that the car is at the open goat door is 0, the probability that the car is at the unselected closed door must be 2/3 since the total must be 1. Hence the probability that the car is at the unselected closed door is twice as likely as the probability that the car is at the selected closed door. Therefore, it would be very foolish not to change one's selection. If this informal "proof" still does not satisfy one's intuition, here is a fuzzier simpler one-sentence way to look at the situation: Since Monty never reveals anything about the selected door and does reveal something about the unselected doors (where the car isn't - and hence where it is more likely to be), aren't you better off playing among the doors that you have some information about? Indeed you are." Miclog (talk) 10:04, 4 June 2008 (UTC)

I kinda like your explanation, Miclog, but I don't think you should add it as is. The troubles are that it's somewhat long and wordy, contains opinion and interpretation, and doesn't reference others, making it close to original research. I think what you do here would be a great way to teach a class about the problem, but isn't quite right for the encyclopedia. Also, the essence of what you say is already in Combining doors. I like parts of your phrasing better, so if you can combine them without messing up the references, that would be good.
Things to leave out: "very foolish", "because it prolongs the suspense", restating stuff from the problem, "It is given that Monty will always reveal the booby prize, a goat in the example", "to minimize show expenses", the whole fuzzier way. Cretog8 (talk) 14:38, 4 June 2008 (UTC)
How is this different from the "Combining doors" section? Also note that the sentence Since it is a given that Monty never reveals the selected door immediately (as it would resolve the suspense) and always reveals a goat, he never reveals anything about the probability that the selected door has the car - and so the probability that the car is at the selected door remains at 1/3. is not strictly true. Consider a variant where Monty opens the rightmost door that he can. In this case he's not revealing the selected door and always revealing a goat, but is either revealing nothing at all about the selected door (for example, in the case the player picks door 1 and Monty opens door 3 - in this case the player is left with a 50/50 choice), or revealing that the player's selected door is guaranteed to be a goat (for example, in the case the picks door 1 and Monty opens door 2 - in this case Monty's action has revealed the car must be behind door 3). The point is that we can't just say "opening a losing door doesn't affect the player's 1/3 initial pick" without going into more details about how Monty picks which door to open, or specifying that we're only interested in the overall chances of winning without considering which door the host opens (this is NOT how the problem is usually phrased). -- Rick Block (talk) 14:51, 4 June 2008 (UTC)

Cretog8 - Below I've taken some of your suggestions. But it seems to me that convincing the reader of the surprising result is the prime objective and overrides other rules. For example restating some facts creates local continuity. And I'm all for concise writing but in technical writing clarity overrides conciseness IMO. Rick - In a sense all the sections are the same - the trick is getting people to believe the result and in my opinion, the new text is clearer - even my wife understood it. And as for variants, it seems to me that the focus is on one particular variant. Here's the tighter edit:

"A source of confusion is the notion that Monty reveals no information. It is given that Monty will always reveal the booby prize, a goat in the example. Since we know Monty will reveal a goat and then proceeds to reveal a goat, we might conclude that no information was provided. But clearly some information was provided - that a particular door has a goat. Whether this information is useful may not be clear initially. But indeed it is. Here's why. Since it is a given that Monty never reveals the selected door immediately and always reveals a goat, he never reveals anything about the probability that the selected door has the car - and so the probability that the car is at the selected door remains at 1/3. But since it is given that a valuable prize is behind one door, and only one door, the sum of the probabilities for all the doors must be one. Since the probability that the car is behind door one (before or after the goat is revealed) is 1/3 and the probability that the car is at the open goat door is 0, the probability that the car is at the unselected closed door must be 2/3 since the total must be 1. Hence the probability that the car is at the unselected closed door is twice as likely as the probability that the car is at the selected closed door. In summary, since Monty never reveals anything about the selected door and does reveal something about the unselected doors - where the car isn't - and hence where it is more likely to be - you are better off playing among the doors that you have some information about."Miclog (talk) 13:26, 8 June 2008 (UTC)

Again, this argument is already in the "Combining doors" section (specifically the third paragraph in that section). The diagrams in that section show this argument visually as well. The disclaimer that this argument depends on the host choosing between two losing doors with equal probability (provided as the 2nd paragraph in "Combining doors") is not provided in this version, but is necessary. The point above of mentioning the variant is to show why this disclaimer is necessary. The sentence Since it is a given that Monty never reveals the selected door immediately and always reveals a goat, ... is not strictly correct since this same argument applies to the variant but gives the wrong answer! This "no information" confusion (i.e. the notion that opening a different door cannot affect the player's initial 1/3 chance) is mentioned in the 3rd paragraph of the "Sources of confusion" section. If you want to expand the "Combining doors" section that would be OK, but I strongly object to introducing a new section that repeats content from another section - and even more strongly object to introducing any new content that is not sourced and is not mathematically correct. -- Rick Block (talk) 17:08, 8 June 2008 (UTC)

[edit] A Priori?

Is this a term used by Gill in his paper on a Bayesian approach to the problem? If not I think that phrase should be replace with something more neutral. A Priori has pretty big philosophical connotations and should probably be avoided in other contexts if possible. Perhaps just talking about "before the experiment starts" rather than "a priori" would be better. 137.222.230.13 (talk) 17:47, 13 March 2008 (UTC)

It is accepted and notable usage in the field of statistics. From A priori (statistics): "It is common in Bayesian inference to make inferences conditional upon this knowledge, and the integration of a priori knowledge is the central difference between the Bayesian and Frequentist approach to statistics." 67.130.129.135 (talk) 19:21, 14 March 2008 (UTC)

[edit] Addition to address conditional vs. unconditional solution description

I've added a paragraph (with this edit) to the Solution section meant to address the issue extensively belabored above. The point is to address the Morgan et al sort of criticism with the minimum amount of change to the article. Even if you don't think this addition is necessary, please don't delete it unless you address the Morgan et al criticism in some other way. -- Rick Block (talk) 21:44, 15 March 2008 (UTC)

After all this discussion, do you think that a lay person can understand this new paragraph ?!! 70.137.136.97 (talk) 00:02, 17 March 2008 (UTC)
I'd prefer Monty Hall problem/draft#Solution, but there was persistent resistance to that version (indeed, to any version that even hints of a conditional analysis). Under the conditions specified in the problem statement, the analysis that is presented is actually true regardless of which door the host opens. The added paragraph is an attempt to explain why this is so. Please feel free to try to clarify it. -- Rick Block (talk) 01:52, 17 March 2008 (UTC)

[edit] The Venn Diagram section is utterly wrong

First of all, the argument in this section is purely "original research" and stated without a reference. More importantly, it is utterly wrong. Probability does not move around as the author suggests by writing "...so the entire 2/3 probability of the two unchosen doors is now carried only by Door 2." The correct concept for such a situation is conditional probability and its calculation is different. Furthermore, the argument does not use the assumption that the host chooses each possible door with equal probability. The conditional probability of 2/3 depends on this assumption. The fact that this section has been there for a long time is irrelevant. It should be removed because it is wrong. -- 70.137.136.97 (talk) 23:46, 16 March 2008 (UTC) -- 70.137.136.97 (talk) 23:47, 16 March 2008 (UTC)

Certainly the following statement is incorrect:
Under the conditions of the problem statement the host is equally likely to open either unchosen door
This was not the statement. "Possible" means permitted by the rules of the game. 70.137.136.97 (talk) 00:26, 17 March 2008 (UTC)
This is clearly wrong, as whether the host is "equally likely" to open either unchosen door depends on where the prize is. If it is behind the chosen door, then this is correct. But if it is behind one of the unchosen doors, then the host is forced to open the other door. So the section is confusing at least.
Nonetheless, pedagogically the Venn diagram section is useful, and that's the main point of the article in my opinion. So rather than just cutting it out (as 70.137.136.97 wishes to do), there needs to be some judicious editing to make it both mathematically correct and pedagogically useful. Bill Jefferys (talk) 00:03, 17 March 2008 (UTC)
The Venn Diagram is WRONG, so it cannot be pedagogically useful. It causes the lay person to think he understands something, but the argument is fundamnetally false because conditional probability is not calculated by shifting probability the way it does. Some solutions consist of true statements but do not solve the problem, however this one contains false statements, and the whole approach is false. The article contains enough explanations and there is no need to add more confusion. 70.137.136.97 (talk) 00:26, 17 March 2008 (UTC)
Calm down. Take enough time to spell your words correctly, and then explain exactly why it is wrong and how you would correct it. Shouting WRONG doesn't get us anywhere.
In my opinion, having taught this concept for many years to unsophisticated students, the section could be improved, but pedagogically it is not WRONG. Bill Jefferys (talk) 00:50, 17 March 2008 (UTC)
Once again, the statement "...so the entire 2/3 probability of the two unchosen doors is now carried only by Door 2" is false because probability does not move around like this. If the host chooses which door to open (when he has a choice) with UNEQUAL probabilities, then the conditional probability is not 2/3 any more, and the chosen door has probability different from 1/3 to have to car. Therefore, the argument must use the assumption of equal probabilities; however, this assumption is not expressed in the Venn Diagram at all. So, it is bad service to students to show them such a diagram and have them believe they understood the probability calculation. Too bad for your students of many years. However, you are NOT the first one to teach this false solution. 70.137.136.97 (talk) 01:20, 17 March 2008 (UTC)70.137.136.97 (talk) 01:31, 17 March 2008 (UTC)
I didn't say that I used the Venn diagram analogy with my students (I don't), only that it could be pedagogically useful.
You continue to avoid the actual issue. Conditional probability can be used correctly here. How do you propose to do it? Bill Jefferys (talk) 01:53, 17 March 2008 (UTC)
Please see the long discussion above. Conditional probabilities have been written about extensively here and in the article. See the section Rigorous Solution in this discussion. Some lay persons refuse to mention conditional probabilities in the Solution section though. Lay persons have veto power in this encyclopedia. The Venn diagram section is worthless. The pedagogical value of this problem is exactly that it can help people understand conditional probability better, but this article (the solution section) is bad in this respect.70.137.136.97 (talk) 02:08, 17 March 2008 (UTC)
The host only has a choice if the player chose the door with a car. In this case, the actual choice of the host doesn't matter, because the player will lose either way if he switches, and will only win if he sticks—regardless of the host's choice. Whether the host picks either of the doors with equal probability or always picks the leftmost door doesn't change the fact that the "switching" strategy yields the car 2/3 of the time. I therefore believe that your following statement is incorrect: "If the host chooses which door to open (when he has a choice) with UNEQUAL probabilities, then the conditional probability is not 2/3 any more, and the chosen door has probability different from 1/3 to have to car." Phaunt (talk) 02:01, 17 March 2008 (UTC)
Please read Morgan et al. or the discussion above. The conditional probability can vary between 1/2 and 1. 70.137.136.97 (talk) 02:10, 17 March 2008 (UTC)
70.137.136.97 is certainly correct on this point, and Phaunt is not right. I believe that the standard presentation of the problem (a la Martin Gardner) specifies that if the contestant chooses the door with the car, the host flips a fair coin to decide which door to open. In such a case the probabilities shift from 1/3 prior to 2/3 posterior on the unopened, unchosen door. Bill Jefferys (talk) 22:59, 17 March 2008 (UTC)

Please, let's not go over all this again, but let's stay on topic (Venn diagrams is the topic). The Venn diagram is an illustration, directly corresponding to the initial solution. The sentence Under the conditions of the problem statement the host is equally likely to open either unchosen door is true (as the problem is posed here, the host opens either door 50% of the time), which means the only information content conveyed to the player is which of the two unpicked doors must hide the car if either does. This is shown by a case analysis as well. It's certainly an original illustration, but hardly "original research". This section isn't explaining why the player's chances of winning by staying remain 1/3 or, alternatively, why the host opening one of the doors has no impact on this 1/3 chance - but these are both nonetheless true (you're not arguing this, right?). The point of the section is to show visually that if the host's actions do not change the initial probability (and, with the constraints in the problem statement they don't) then the other door must have a 2/3 chance. -- Rick Block (talk) 03:32, 17 March 2008 (UTC)

It's Phaunt who reopened this. Anyway, the Venn Diagram section misleads people to believe that they understand the solution using a simple argument. But the argument is false. So, in my opinion this section is worthless unless you are interested in just comforting people. 70.137.136.97 (talk) 06:13, 17 March 2008 (UTC)
Ahem, I'm sorry to have inadventently brought up an apparent side matter. I'll read up on this matter, because apparently I don't quite understand what you're talking about. I'll leave this debate for now. Phaunt (talk) 11:42, 17 March 2008 (UTC)

Given the normal interpretation of this problem, the host's actions do not change the initial chance of having selected the door hiding the car. Can you suggest a wording that would make this clear? Would it help to refer to the table that enumerates the cases in the Solution section? In my opinion the understanding we're looking for here is that two unopened doors does not imply equal probability. I suspect at least 80% (probably far higher) of the population of the planet does not understand this and that this is the main intuitive disconnect. BTW - I agree the Morgan et al criticism is not yet adequately addressed in the article. On the other hand, tendentious editing is not helpful and won't fix it. -- Rick Block (talk) 14:21, 17 March 2008 (UTC)

To Rick Block: We are discussing here the Venn Diagram. The argument of the diagram must use the assumption of equal probabilities but I do not see a way to incorporate this assumption in a proof based only on such a diagram. Many people wrongly believe in this "proof" but it cannot be a valid proof because it does not use the assumption; note that if it were valid, then it would have "proven" the 2/3 conditional probability even without the assumption, hence a contradiction! This is why I think it would be better to leave that section out altogether. I repeat my opinion that the best way to educate people here is simply to explain what conditional probability is (it takes only a couple of lines) and then do the calculation with, and without, the assumption. This would, in particular, educate all the wikipedia editors who repeatedly demonstrate their misunderstanding. This would be much more valuable than all the "intuitive" explanations, some of which are false anyway. Morgan et al. give the correct solution and explain the falsehood of the other "solutions." The calculation does not require more than junior-high-school education. 70.137.136.97 (talk) 02:49, 18 March 2008 (UTC)

OK, well I understand this and agree with it. So (as I have been asking all along) how can we explain this intuitive but not properly set out example so as to make it clear to mathematically unsophisticated people what's going on. I think that that is the intention of this section.

Now, my view of this section, pedagogically, is that it is trying to show that both before, and after, the evidence

E = Monty (obligatorily) opens a door that was not chosen by the contestant and shows a goat (with probability 1/2 if Monty has a choice of door to open)

is learned, this information does not change the probability that the prize lies behind the door that the contestant has chosen, and that that probability is 1/3. Therefore, (the contestant should infer), the probability that the prize lies behind one of the doors that he has not chosen (of which only one viable candidate remains) is 2/3.

Now, if we assume that the Martin Gardner standard assumptions are valid, then we can compute, given

C=The prize is behind the door chosen by the contestant

that P(E|C)=1 (since Monty is obligated to open a door that shows a goat, if C is true), and even P(E)=1 (since Monty is obligated to open a door that shows a goat, regardless of whether C is true or false).

But then P(E|C)/P(E)=1 and thus P(C|E)=P(C)=1/3 (by Bayes theorem).

[Yes, we can refine this to talk about the particular door that is opened, but that just complicates the example and the results are the same, again assuming the Martin Gardner rules. So let's not go there.]

The section in question is trying to explain this concept without using the full concept of conditional probability. The problem is that 70.137.136.97 insists on mathematical rigor, even for people who are reading this article who probably don't have any chance of understanding the concept of conditional probability.

Even if you explain Bayes' theorem in a mathematically rigorous way, it doesn't mean that you have explained it to the people we are trying to reach.

The perfect is the enemy of the good. Bill Jefferys (talk) 03:52, 18 March 2008 (UTC)

It isn't just that some people don't have any chance of understanding conditional probability; it's that there are ways of using simple logic to explain the basic paradox correctly without delving into the conditional analysis. For some reason there are people who are incapable of understanding that. They have learned how to use a hammer, and insist that everything must be treated like a nail.--Father Goose (talk) 05:38, 18 March 2008 (UTC)
Father Goose - The probem description in this article says:
"You begin by pointing to door number 1. The host shows you that door number 3 has a goat. Do the player's chances of getting the car increase by switching to Door 2?"
Therefore, the question is precisely about conditional probability because the player learns information that was not available before (the car is not behind door 3) and that fact changes the probability that the car is behind door 2 (and possibly also the probability that the car is behind door 1). You cannot avoid the issue of conditional probability here because this is precisely what it means -- a revised probability in view of newly acquired information. Simple logic can explain conditional probability, but the exact meaning of the question remains the following:
"Given that the host has opened door 3, is the chance of winning by switching greater than the chance of wininng by sticking?"
So, the conditional probability is not one ("hammer") of several tools to approach this question. It is the very concept underlying the question. I suggest you read Morgan et al. I would say that a person who does not understand what is conditional probability also does not understand the question that is phrased in this article. Some people convince themselves that their simple explanations prove the result, but it turns out that by the same arguments they could prove a falsehood. If so, then the arguments do not actually prove the correct result. 70.137.136.97 (talk) 04:24, 19 March 2008 (UTC)
Bill Jefferys - is the above analysis published anywhere that you know of? If so, at the beginning of the Venn diagram section we could legitimately include a statement like "Under the standard interpretation of this problem, the host's opening of a door does not affect the player's probability of having selected the door hiding the car (with a reference)." This would be a very useful fact to be able to reference other places as well. -- Rick Block (talk) 13:34, 18 March 2008 (UTC)


The "solution" of Bill Jeffreys is what Morgan et al. call false solution 5. Correct is the enemy of Wrong, Bill. Here is how Morgan et al. phrase it:
"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."
Morgan et al. explain:
"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."
Prof. Jeffreys' "proof" does not use the assumption of equal probability, which is necessary for concluding the 2/3 conditional probability. Therefore, this is not a valid proof because the same proof would give the 2/3 conditional probability even when the host does not follow the equal probability rule, and that of course is not true.
The discussion here is consistent with the experience of Morgan et al.:
"Refusing to believe their arguments wrong, the whirlwind would begin again. Posting the problem on the math club bulletin board increased the clamor as undergraduate students grappled with their arguments. So it was that we repeatedly found what vos Savant's article and the quotes above demonstrate: that this apparently innocuous little problem can be erroneously "solved" in a variety of ways and that the nature of these errors can be quite subtle, at least to those not accustomed to thinking in terms of conditional probability (and sometimes even to those of us who are!). That this problem so entertainingly demonstrates the pitfalls of conditional probability calculations and interpretations alone warrants its wider dissemination for instructional purposes."
Let me also add that if one assumes that the choice of the host does not change the probability that the car is behind the chosen door (i.e., p(1)=1/3), then the question becomes trivial, and this is the proof: p(2) = 1 - 1/3 = 2/3. This certainly is not the correct solution to the problem at hand. 70.137.136.97 (talk) 04:24, 19 March 2008 (UTC)
You seem to have missed Bill's parenthetical Yes, we can refine this to talk about the particular door that is opened, but that just complicates the example and the results are the same, again assuming the Martin Gardner rules. So let's not go there. You're clearly not disputing the truth of the statement that given the normal interpretation of the problem the host's actions don't affect the player's 1/3 chance of having selected the car, or that this can be proven with a Bayesian analysis (right?). Your objection is purely that this assumption is tantamount to assuming the solution and its proof is as difficult as solving the (conditional) problem in the first place (do I have this right?). I'm not sure where to go from here. Would you be willing to work on a draft rewrite (I think you're basically suggesting the article needs to be rewritten) to offer up as a replacement article? I'm at a loss for how else to constructively proceed. -- Rick Block (talk) 05:13, 19 March 2008 (UTC)
NO, I have not missed that. The "solution" he offered shows that he did miss the point. Offering his solution to the readers is bad service. My objection to the assumption that the host actions do not affect the player's 1/3 chance of having selected the car is that it implies immediately that there is a probability of 2/3 that it is behind the other door ( 1- 1/3 = 2/3 ), and there is nothing more to say. The whole question here is to prove this particular no change assuming equal probabilities when the player chooses the door with the car.
I have good reasons to believe that anything I might write, would be deleted by the people who still maintain that conditional probabilities should not be used in the solution section. Of course I would like to be constructive -- this is how I started here if you remember. I offered simple explanations more than once here. In particular, the most important point is that switching is better under whatever allowable protocol the host may use (yet must open a goat door). Morgan et al. is a reference for that. 70.137.136.97 (talk) 05:57, 19 March 2008 (UTC)
70.137.136.97 is right and my "proof" is incorrect. I tried to avoid the additional information about which door was opened (and which then requires knowledge of the probability of the host's strategy for opening a door when the contestant has chosen the door with the prize). But I now recognize that you can't do that. (In my course I always do this with the proper assumptions and using full conditional probability). I found my copy of Morgan et. al. (which was packed away in a box), and reread their analysis.
It would be interesting to learn how Martin Gardner originally discussed this in his Scientific American articles; but I wouldn't be able to find this for at least a week (that is, assuming that the University of Vermont library has it).
So, I have come around to 70.137.136.97's point of view, while still hoping that a simpler way to present it might be found in the literature. I've been looking for my copy of Mosteller's 50 Problems but I've misplaced it. He probably has a pretty decent discussion. But until something defensible comes along, the Venn diagram section should remain out. Bill Jefferys (talk) 16:45, 19 March 2008 (UTC)
What would both of you think about changing the solution section to something more like the version at Monty Hall problem/draft#Solution, and recasting the Venn diagram section as purely an illustration to help visualize how the probability might not be evenly distributed between two unknowns (not a proof, simply an illustration of the trivial fact that 1 - 1/3 = 2/3)?
When I wrote my "proof", I had in mind (and stated) that the "standard" (Martin Gardner) assumptions were operative, namely that if the contestant had chosen the prize door, then the host would choose one of the two unopened doors with probability 1/2 ("at random", but that's ambiguous since there are many assignments that would be "at random", including rules that depend on previous behavior of this or other contestants). In this scenario, the fact that door 2 (for example) was chosen instead of door 3 transmits no information to the guest. This is because the likelihood is the same, regardless of the state of nature (prize behind the chosen door, prize not behind the chosen door, regardless of the door). Thus, in this scenario, all you need to know is that the host has (obligatorily) opened an unchosen door that does not have a goat. I admit that I didn't stress this particular point well, as I assumed that everyone would understand this, particularly our mathematically sophisticated friend 70.137.136.97, and that assumption has been proven wrong. But then, this is supposed to be a pedagogical section, and such failings cannot be allowed in the article.
So, to 70.137.136.97, I ask, if it is clear that no information is available from the fact that a particular door has been opened, is there still a problem? Bill Jefferys (talk) 00:56, 20 March 2008 (UTC)
As I pointed out before, you cannot assume that "no information is available from the fact that a particular door has been opened," because obviously there is new information, namely, the car is behind another unopened door. If you wish to assume that after door 3 has been opened the conditional probability that the car is behind door 1 remains the same 1/3 as the unconditional probability, then trivially the conditional probability that it is behind door 2 is equal to 1 - 1/3 = 2/3. So, essentially, you want to assume what actually has to be proven. 70.137.136.97 (talk) 01:40, 20 March 2008 (UTC)
(after edit conflict) Bill - Can you please explain in more detail why it is true that the host's chosen door transmits no information to the player ("the likelihood is the same" - what likelihood?)? This same point is discussed below as well. I know that this is in fact true, but other than by showing the conditional probability that the player's selected door remains 1/3 given either door the host might open I don't have an entirely satisfactory reason. And, what do you each think about the proposed Solution section? -- Rick Block (talk) 01:46, 20 March 2008 (UTC)

If you insist on using the unconditional probability calculation, you have to change the statement of the problem. Quoting Morgan et al.:

F1 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."

I know many of the editors here will say that this is the same as the original statement, but it is not. This is not my own Original Research. I would include the two statements of the problem, conditional and unconditional, and provide a solution to each -- all in the Solution section. The Venn Diagram could be offered as a solution to the uncondtional problem; until then, it should stay out. It should stay out on the grounds that it is a wrong original research that is contradicted by a refereed published authority (Morgan et al.) I think it goes without saying that people who do not fully understand a certain topic should not act as editors of the encyclopedia article on that very topic, let alone interject into the article their original research which is contradicted by the published literature. 70.137.136.97 (talk) 02:01, 20 March 2008 (UTC)

Are you saying you've read Monty Hall problem/draft#Solution and find it unacceptable? In this version of the solution the unconditional calculation is clearly identified as unconditional (isn't it?). I am sympathetic to the view that both unconditional and conditional answers should be provided, however it is FAR from a mere coincidence that both answers are 2/3. I think there's a reasonable argument that most people interpret the problem statement as the unconditional problem, even though to someone mathematically sophisticated it is obviously a conditional question. I'd venture a guess that 99% of the readers of this article will NEVER understand the difference, don't care to, and don't want to. I suspect the very reason the standard version of the problem includes the "equal goat constraint" is to force the unconditional and conditional answers to be the same in an attempt to avoid this very discussion. -- Rick Block (talk) 02:16, 20 March 2008 (UTC)
Rick - Sorry, I have not read it. I guess it is acceptable but have not checked it. I am not going to continue this hopeless discussion with the others though. I am glad that Prof. Jefferys agrees with me and I salute your patience. However, I find it impossible to continue dealing with more and more new editors reverting stuff and insisting on including in the article inappropriate material. By the way, I think the "goats" should also stay out -- it is simpler to deal with "car" and "no car" rather than "goat A" and "goat B." You have one good authoritative reference (Morgan et al.). Follow it and you will have a good article. Write a section on false solutions and it will be very instructive. Bye for now. 70.137.136.97 (talk) 03:39, 20 March 2008 (UTC)

People are editing so fast and furiously that I can't get a word in edgewise. I will wait until you guys have gotten your licks in. Bill Jefferys (talk) 02:21, 20 March 2008 (UTC)

Rick Block asked what I meant about the likelihoods. Here's what I meant. Instead of considering the three states of nature defined by the particular door the prize is behind, consider instead the two states of nature A: The prize is behind the chosen door and B: The prize is not behind the chosen door. Clearly, if the host follows the rule that when A is true, he reveals one of the unchosen doors with probability 1/2 each, then it is equally likely, if A is true, that either door will be revealed. But it is also the case that if B is true and A is false, the host will be forced to open each of the unchosen doors, again with (from the point of view of the contestant) equal probability. Thus, from the contestant's point of view, the probability that the host opens the leftmost unchosen door equals the probability that he opens the rightmost unchosen door, independent of whether A is true or false. Thus, the contestant gets no information about whether his chosen door has the prize or not...under these circumstances, the probability that the prize is behind the door he chose stays equal to 1/3, regardless of which door the host opens. The likelihoods are equal, and that's always the case when observing a datum tells you nothing new. Bill Jefferys (talk) 17:14, 20 March 2008 (UTC)

This is the same reasoning I used replying to Hydnjo, two sections down, about the sentence I added to the Venn diagram section that says Under the conditions of the problem statement the host is equally likely to open either unchosen door, so opening this door does not affect the average probability of winning the car by staying with the original choice which remains 1/3. The only bit about your explanation that is not utterly convincing is the last sentence ("Thus, ..."), which is the same point in my explanation below where I say "this means X=Y". My intuition says this is true. Your words claim it is true. Hell, my words claim it is true. But I'm not actually entirely convinced.
I'll repeat my question from above - what would you think about changing the solution section to something more like the version at Monty Hall problem/draft#Solution (it's MUCH shorter and probably easier to follow than this one talk page section), and recasting the Venn diagram section as purely an illustration to help visualize how the probability might not be evenly distributed between two unknowns (not a proof, simply an illustration of the trivial fact that 1 - 1/3 = 2/3)? We could introduce the Venn diagram section with words like The solution above shows that under the conditions of the problem statement the host opening either unchosen door does not affect the probability of winning the car by staying with the original choice. What this means visually is .... -- Rick Block (talk) 01:12, 21 March 2008 (UTC)

The Venn diagram problem in the draft proposal has the same problem that our anonymous friend was pointing out, since it does not specify that the host opens the unchosen doors with equal probability when the contestant happens to pick the prize door. Or am I missing something? Bill Jefferys (talk) 16:05, 21 March 2008 (UTC)

I apologize for not having been clear about this - the draft is not a full draft yet, it's more or less complete through only the initial couple of sections (including, specifically the "Solution" section). The entire rest of the draft including the Venn diagram section is still as it was in the article before the Venn diagram section was originally deleted by our anonymous friend. The question about the Solution section is only peripherally related to the Venn diagram section, so I probably should not have brought it up in this specific thread. This question is related to the more general objection raised by our anonymous friend about the article presenting an unconditional solution (in the Solution section as well as in the Venn diagram section, and probably several other sections as well). At this point I'm asking for your opinion about only the Solution section in the draft (not the entire draft - which is not complete), and the (not yet implemented in the draft) idea of changing the Venn diagram section as suggested above. -- Rick Block (talk) 16:26, 21 March 2008 (UTC)

My bad,I looked at the wrong section. The one you point to (down to where it says "propose solution section ends here") seems fine. Bill Jefferys (talk) 17:04, 21 March 2008 (UTC)

Thanks. And the idea of recasting the Venn diagram section? -- Rick Block (talk) 03:30, 22 March 2008 (UTC)

I think the Venn diagram section needs a clear statement of the conditions under which observing the host's goat door does not change the probability that the chosen door has the prize, and a convincing argument as to why this is the case. Bill Jefferys (talk) 17:13, 22 March 2008 (UTC)

I've edited the wording to try to address the original issue raised by 70.137.136.97, which was objection to the phrasing "the entire probability ... is now carried". I don't think a discussion of why opening a door doesn't change the probability needs to be repeated here, since that's covered in other sections, or should be. Warren Dew (talk) 01:40, 15 April 2008 (UTC)

I removed the dispute tag since no one has objected to my fix to 70.137.136.97's issue, and added the reference Rick Block found. Note that I wouldn't mind someone merging this section with the next - but please keep the illustrations, as they are the part that really make it clear what's going on, at least for me and presumably others like me. Warren Dew (talk) 08:40, 29 April 2008 (UTC)

[edit] Simple approach from game theory

My simple two cents from game theory. We have two seperate strategies (change and don't change the doors) and two cases (you chose the door with the car in your first try or you didn't).

Case Strategy 1: change Strategy 2: dont change
1. chose the car-door (1/3 prob) 0 % win 100% win
2. didn't chose the car-door (2/3 prob) 100 % win 0% win

Easy to see that its better to use strategy 1 all the time as it gets you 100% win in 2/3 of all cases.

The mindtwister is to think of the cases first and then think about how your strategy works in that cases. When Monty opens a second door in case two, he gives you a 100% chance to win if you change - that's the trick. --Unify (talk) 03:45, 19 March 2008 (UTC)

Lets bring this in one sentence: Changing the doors will guarantee you the winnings if you didn't chose the door with the car in the first try, which is more likely. --Unify (talk) 14:37, 19 March 2008 (UTC)

Are you suggesting we add this to the article as another explanation? The main topic that's generated so much activity on the talk page is the difference between a solution that applies on average to all players (which this one does) vs. a solution that applies when given which door the host opens (which this one doesn't). It's the same issue being discussed above (and below, now) with regard to the Venn diagram section. The subtle point is that the overall chances of winning by switching can be 2/3 while at the same time an individual player's chances of winning by switching might be any where from 50% to 100% depending on which door the host opens (say, left or right). The problem statement presented in the article precludes this quirk by including the constraint that the host choose between two goat doors randomly (if the player initially picked the car). However, if the solution does not use this constraint the same solution would apply to a problem statement without the constraint, meaning any such solution is not quite complete. -- Rick Block (talk) 23:25, 19 March 2008 (UTC)
For a game, game theory is the natural way to go. You don't need to distinguish between which goat-door the host opens if the player chose the right door at the start - both other doors would be wrong so the outcome is identical (you lose) and you don't need to seperate identical cases. I prefer this 'general' view on the topic as you don't need conditional prob or 'wandering prob' or anything that the average user doesn't understand. In this simple game with just one real decision (change oder don't) you don't even need to understand all the math that comes along with game theory. It is actually very similar to the first solution in this article, but it needs only 2 cases (picked the right door at the start or not) instead of 3 (right door, wrong door, other wrong door). The Extensive form game is also similar to the decision tree in this article - again you need only two cases to start with because the identical two cases (wrong door 1/3, other wrong door 1/3) can be treated as a single case with 2/3 prob. I'm working on a small decision tree for the game. --Unify (talk) 16:08, 20 March 2008 (UTC)
There is the Extensive form game, after all its just a decission tree with two cases. I'm not a native english speaker, if something is confusing or wrong, please say. --Unify (talk) 17:48, 20 March 2008 (UTC)
I think you're not getting my point. The literal question posed in the article is not "what are the average chances of winning by switching, across all players and ignoring which door the host opens" but "if a given player chooses one door and the host opens some specific other door, is that individual player better off switching". Your approach answers the first question, but not the second, and the answers can be different (even for a given version of the "game"). With the version presented in the article, these answers turn out to be the same, but that doesn't mean they aren't still different questions. This is explained above as well, but the simplest example is a version of the problem where the host opens the leftmost remaining door whenever possible (as long as doing so doesn't reveal the car). With this version the average chances of winning by switching, across all players and ignoring the door the host opens are 2/3. But, the chances for a specific player are different depending on whether the host opens the rightmost or leftmost door. If the host opens the leftmost door the player's chances of winning by switching are 50%. If the host opens the rightmost door the player's chances of winning by switching are 100%. If the question is "is an individual player better off switching", in this version the answer isn't a simple "switching wins 2/3 of the time" but "switching wins 50% or 100% depending on which door the host opens". The point really is that if you ignore which door the host opens, you're only answering the first question (average across all players) not the question about an individual player knowing which door the host opened. -- Rick Block (talk) 18:54, 20 March 2008 (UTC)
It's hard to understand how you seperate the individual player from the average player - I never heard of that concept before. It doesn't matter which door the hosts opens because it's always a non usable goat door (and why do you care if it is standing on the left or right?)
We should start do decompose the game in all of its 24 different possible combinations: We have 3 doors A,B,C. Behind one of them is the car and you start to choose one door. That makes 9 possible combinations, 3 of them where you got the car-door (A-A B-B C-C) and 6 where you got a goat door instead (A-B A-C B-A B-C C-A C-B). In the first case the host can open one of the two remaining goat-doors and you can change or not, that makes 3*2*2=12 combinations. In the second case the host can open only the one remaining goat-door and again you can change or not, that makes 6*1*2=12 combinations. Thats total 12+12=24 different outcomes of the game. In 12 outcomes you changed the door (and in the other 12 not), changing led you to win in 8 combinations and lose in 4 comb. (2/3 vs. 1/3). One might bring a complete table up to clear that once and for all. --Unify (talk) 23:30, 20 March 2008 (UTC)
The point is that it might matter which door the host opens. The distinction between the average player and the individual player is the same as the distinction in game theory between the chance of winning at the beginning of a game and the chance of winning some number of moves into a game - things have happened that might affect the average outcome. If I ask you what is the chance of winning a tic-tac-toe game, you'll probably say it should be a tie 100% of the time. However, if I ask what is the chance of winning a tic-tac-toe game after two moves you'll probably say it depends on what two moves were made. It's the same thing with the host and doors. With the usual interpretation (where if the host has a choice he opens either door with equal probability) the probabilities are as in the table in the next section (assuming the player picked door 1). In particular, on the line where the player has initially (unknowingly) picked the car the probabilities of the host opening each remaining door are the same (1/6). With other versions of the problem this doesn't have to be the case. In the (hypothetical) version where the host opens the leftmost door whenever he can, the probabilities on this line are 1/3 and 0. This means if you pick door 1 and host opens door 2 you have a 50% chance of winning by switching, but if you pick door 1 and the host opens door 3 you have a 100% chance of winning by switching. The simple game theory analysis doesn't distinguish this version from the regular version, so it's not answering the question about the specific player given which door the host opens (it IS answering the question about the overall odds as seen by all players). -- Rick Block (talk) 00:43, 21 March 2008 (UTC)
This means if you pick door 1 and host opens door 2 you have a 50% chance of winning by switching, but if you pick door 1 and the host opens door 3 you have a 100% chance of winning by switching.
-> That seems to be an impossible game situation. If you had door 1 at the start an Monty could open both other doors, that means the car is behind door 1 and switching will let you lose 100%. If you had door 1 and Monty opens door 3 and you have 100% by switching, that would mean the car is behind door 2. The car can't be behind door 1 and 2 in 'one' game. It seems to me that you are combining several 'distinct' settings into one game and so you create 50% prob. or other side effects. --Unify (talk) 15:29, 21 March 2008 (UTC)
I'm saying if you follow the path in the decision tree where you've picked door 1 and the host has opened door 2 (in the variant where the host opens the leftmost door if possible) you now at this point have a 50% chance of winning by switching. And similarly, if you follow the path to where the host has opened door 3 you have (at that point) a 100% chance of winning by switching. At the beginning you have a 2/3 chance of winning by switching. This number can change depending on what path you follow (depending on how the host selects from two goat doors). The decision tree in this variant following the point where you've picked door 1 splits unevenly based on which door the host opens (2/3 door 2, and 1/3 door 3). If the host picks evenly between two goat doors then at any point following your initial pick and whatever door the host opens, the chance of winning by switching is still 2/3. -- Rick Block (talk) 16:09, 21 March 2008 (UTC)
I think I know what you mean. I you chose door 1 and Monty opens door 2 the car might be in door 1 or 3, so you think of 50:50 possibility? But thats misleading, the car is not with 50:50 behind door 1 and 3 - the possibilities depend on your first choice. You are trying to seperate the second step from the first step and look at the second step isolated. That doesn't work if you want to describe the game because the first step affects the possibilities of the second step. You have to concider that Monty knows and reacts different if you had the car-door or a goat-door at the start - Monty might chose between 2 doors or the single remaining door and that biases the probabilities to land in a different kind of second step. It's more likely you missed the car in the first try and so it's more likely you landed in a second step (scenario) where changing helps. The car is with 2/3 behind door 3 and with 1/3 behind door 1, so change. --Unify (talk) 17:14, 21 March 2008 (UTC)
Again, the point is that there may be a difference between the overall odds (which are 2/3 win by switching, as you say) and the odds given which door the host opens. The odds at the beginning (before the host has opened a door) are from your table above:
Case Strategy 1: change Strategy 2: dont change
1. chose the car-door (1/3 prob) 0 % win 100% win
2. didn't choose the car-door (2/3 prob) 100 % win 0% win
If we expand this to also include in the cases which (left or right) door the host opens, and assume the host chooses between two goat doors equally we get
Case Strategy 1: change Strategy 2: dont change
1. chose the car-door and host opens left door (1/6 prob) 0 % win 100% win
2. chose the car-door and host opens right door (1/6 prob) 0 % win 100% win
3. didn't choose the car-door and host opens left door (1/3 prob) 100 % win 0% win
4. didn't choose the car-door and host opens right door (1/3 prob) 100 % win 0% win
which is still 2/3 win by switching (overall) and if you look only at cases where the host has opened the right door it is still a 2/3 win by switching (and also 2/3 if looking only at cases where the host has opened the left door).
However if the host opens the leftmost door whenever possible, we get
Case Strategy 1: change Strategy 2: dont change
1. chose the car-door and host opens left door (1/3 prob) 0 % win 100% win
2. chose the car-door and host opens right door (0 prob) n/a (can't happen) n/a (can't happen)
3. didn't choose the car-door and host opens left door (1/3 prob) 100 % win 0% win
4. didn't choose the car-door and host opens right door (1/3 prob) 100 % win 0% win
which is still 2/3 win by switching overall. But, if you look only at the cases where the host opens the left door it's a 50% win by switching, and if you look only at the cases where the host opens the right door it's a 100% win (!) by switching. If you're in this version, it doesn't hurt to always switch but your probability of winning depends on which door the host opens. -- Rick Block (talk) 17:46, 21 March 2008 (UTC)
if you look only at the cases where the host opens the left door it's a 50% win by switching, and if you look only at the cases where the host opens the right door it's a 100% win (!) by switching. -> Are you comparing the 1st and the 3rd case in your last table (1/3 vs 1/3 is like 50:50)? And the 2nd and 4th case (0 vs 100%)? Is that how you calculate the 50% and 100%? --Unify (talk) 19:44, 21 March 2008 (UTC)
No, I'm looking at the combination of the 1st and 3rd cases (which the player can't distinguish) or the combination of the 2nd and 4th cases (also which the player can't distinguish). You pick a door. The host then opens a door. At this point, after the host has opened a door, you're in case 1&3 or case 2&4 depending on whether the host opened the right or left door. In case 1&3, you now have a 50% chance of winning by switching because you're actually in case 1 with a 1/3 chance and you lose with certainty if you switch or case 3 with a 1/3 chance and you win with certainty (but you don't know which). In case 2&4 you now have a 100% chance of winning by switching because the only possibility is case 4. -- Rick Block (talk) 21:26, 21 March 2008 (UTC)
The 3rd and 4th case can't exist in that game. If you dind't choose the car-door at the start, the host has only one remaining door to open, so there is no left or right door choice for Monty to do. The single remaining door might be left, middle or right - that is predetermined by the position of the car and the players first choice. That squeezes the table:
Case Strategy 1: change Strategy 2: dont change
1. chose the car-door and host opens left door (1/3 prob) 0 % win 100% win
2. chose the car-door and host opens right door (0 prob) n/a (can't happen) n/a (can't happen)
3. didn't choose the car-door and host opens single remaining door (2/3 prob) 100 % win 0% win
One might delete the "impossible" event in case 2 and you end at the beginning:
Case Strategy 1: change Strategy 2: dont change
1. chose the car-door and host opens left door (1/3 prob) 0 % win 100% win
3. didn't choose the car-door and host opens single remaining door (2/3 prob) 100 % win 0% win
--Unify (talk) 22:55, 21 March 2008 (UTC)
3rd and 4th case can't exist in that game? Why not? If you don't choose the car at the start, the car is behind one of the other 2 doors. Assuming it was distributed with equal probability, then 1/2 the time it's under the left door and 1/2 the time it's under the right door. The host has to open the door that doesn't reveal the car, so if you don't pick the car the host opens the left door 1/2 the time (total of 1/3 since you don't pick the car 2/3 of the time) and the host opens the right door 1/2 the time. You can delete the impossible case 2 if you want, but can't combine 3&4 since in these the host is opening different doors (which is the point we're trying to examine). Again, the net over all players is 2/3 (as you've shown) but if we ask "what is the probability of winning if the host opens the left door?" your analysis is not responsive. -- Rick Block (talk) 23:15, 21 March 2008 (UTC)
You are slightly changing the rules from the game. In your first case the player chooses the car (and it's there and no more random) an Monty has a 50:50 choice to open one of the two remaining doors - thats how the game works. In your 3. an 4. case you assume that the player didn't choose the car and now you start again to randomize the car with even possibility behind the doors. But the car has its fixed position already from the start. Remember that Monty knows where the car is from the start (thats part of the game), so you can't re-randmomize the car's position again during a running game (after the player had its choice). That biases the possibilities because that happens only in case 3 and 4. --Unify (talk) 19:39, 22 March 2008 (UTC)
I understand what you mean: from the perspective of an unknowning (individual) player it might look as Monty opens the remaining doors with 50:50 in case 3 and 4. But from the perspective of Monty he opens the last remaining goat door with 100% without any further choice and if that door is left, middle or right is predetermined by the car's start postition and the players choice. --217.83.64.140 (talk) 20:04, 22 March 2008 (UTC)
We're clearly talking past each other. Let's start at the very beginning. The car is placed randomly behind one of the 3 doors (1/3 chance for each door). The player picks a door (say door 1). If we call the cases C1,C2,C3 (corresponding to the door the car is behind) we're in case C1 with 1/3 chance, or case C2 with 1/3 chance, or case C3 with 1/3 chance. Case C1 has two subcases - let's call them C1-L and C1-R (for host opens the remaining left or right door). In case C2 the host has to open door 3 (the right door) and in case C3 the host has to open door 2 (the left door).
Assuming we're still on the same page, in the standard problem (where the host chooses between two goat doors equally, i.e. C1-L and C1-R are equally likely) the analysis is as follows
Case Strategy 1: change Strategy 2: dont change
1. C1-L (1/6 prob) 0 % win 100% win
2. C1-R (1/6 prob) 0 % win 100% win
3. C2 (1/3 prob) 100 % win 0% win
4. C3 (1/3 prob) 100 % win 0% win
If we're still on the same page, this table means you win by switching 2/3 of the time. It also means in all games where the host opens the leftmost door (cases C1-L and C3), you win by switching 2/3 of the time. This is significant because when you initially pick a door you don't know what case you're in (could be any of them), but when the host opens a door if it's the leftmost door you know you must be in one of cases C1-L or C3 and not in cases C1-R or C2. Alternatively, if the host has opened the rightmost door you're in one of C1-R or C2 (but again don't know which). The door the host opens tells you which 2 of the 4 possible cases you might be in (but not specifically the single case), and even knowing this information you still have a 2/3 chance of winning by switching. Still with me?
In the variant where the host always opens the leftmost door if possible, the table changes to this
Case Strategy 1: change Strategy 2: dont change
1. C1-L (1/3 prob) 0 % win 100% win
2. C1-R (0 prob) n/a n/a
3. C2 (1/3 prob) 100 % win 0% win
4. C3 (1/3 prob) 100 % win 0% win
(i.e. if you pick the car first the host never opens the rightmost remaining door). Overall you still have a 2/3 chance of winning by switching, and which door the host opens still tells you which case(s) you might be in. But if the host has opened the leftmost door telling you you're in C1-L or C3 (but not which), switching now only gives you a 50% chance of winning. Similarly, if the host opens the rightmost door the only possibility is you're in case C2, and if you switch you win with 100% certainty.
There's no "re-randomizing". The only difference between these is the rule specifying that the host choose between two goat doors with equal probability vs. the host always open the leftmost door if possible. The overall odds are the same, but the odds given which door the host opens are not. -- Rick Block (talk) 20:28, 22 March 2008 (UTC)
Got a source?
Got a source?
We should come to a conclusion. Do you really want to add to the article that there is some kind of 50% chance? I wouldn't agree on that and don't know if there is any more or less important source to be found? --Unify (talk) 01:07, 27 March 2008 (UTC)
The source is Morgan et al (from the references section of the article). They show that the probability a given player wins the car by switching (given only that the host must open a door and knows how to avoid opening a door showing the car) is 1/(p+1) where p is the probability of the host opening the door that has been opened if he has a choice (i.e. if the player initially picked the car door). p ranges from 0 to 1 so the probability of winning by switching ranges from 1/2 to 1 (in the "usual" problem, p = 1/2, so the player's chances of winning by switching are 1/(1+1/2) which is, .. .. .. .. .. wait for it .. .. .. .. .. .. 2/3!).
I would like to add the Morgan et al analysis to the article, but not in the initial "Solution" section (which I would like to look more like Monty Hall problem/draft#Solution). The point of this discussion (and the point the anonymous editor who raised it is making) is that the current Solution section, and many of the analyses in the article, ignore the "equal goat constraint" and present an analysis based on the overall average chance of winning. It's somewhat pedantic, but this isn't what's asked by the usual problem statement so technically an answer based on an "overall average" approach is not strictly correct. Moreover, any reasoning that says "2/3 chance of winning by switching" that applies equally well with or without the equal goat constraint is manifestly not answering the question about a specific player - whether or not in the problem as posed the numerical answer is 2/3 (this is at best arriving at the right answer with faulty reasoning - sort of like a geometry proof where the answer is derived from measuring an angle in the example drawing with a protractor rather than using what's given in the problem statement).
If you read through the thousands of words above, you'll see many people argue "but just because the problem statement says You begin by pointing to door number 1. The host shows you that door number 3 has a goat. Do the player's chances of getting the car increase by switching to Door 2? doesn't mean we're asking about a conditional probability, i.e. the probability given what door the host opens, and even if we are asking about a conditional probability our problem statement forces the host to open one of two goat doors equally randomly so 2/3 is the right answer - who cares how we got there?". I'll admit I initially thought there was a pretty high likelihood the anonymous editor who first brought this up was a troll, but at this point I'm completely on his side. -- Rick Block (talk) 03:20, 27 March 2008 (UTC)

Let me see if I've got this right. I think the disagreement between the two of you is between the specific statements of the problem. Unify's version of the problem is the one I've always seen, where, given a "choice", the host picks at random which door to open; that is, the player has chooses the car first, then the host randomly chooses which goat to show him. Rick's version has further conditions on Unify's version: if the host has a "choice", then he'll choose a specific door, which without loss of generality, Rick has defined as the leftmost door. The way I see it, Unify's version is the one presented in the article as the "typical" version. As such, his original argument about the game theory approach is logically valid. Rick's version is an interesting statement of the problem, but I think it goes beyond the version stated in the article. It could be put in the "Variants" section, though. But tell me if I've completely missed something. 74.178.44.162 (talk) 05:15, 6 April 2008 (UTC)

No, the disagreement (if you want to call it that) is explained in my very first response in this thread and relates to the question the problem asks and then whether the simplistic game theory analysis is responsive to that question. If the question we're asking is "what is a player's chance of winning by switching given the host has opened a door", my assertion is that the proposed analysis is not fully responsive since it ignores the constraint on the host that he choose between two goats randomly. The only point of introducing another version of the problem (like the one where the host opens the leftmost door if possible) is to show that this analysis still applies but gives the wrong answer. The reason for this is that the analysis is not answering "what is a player's chance of winning by switching given the host has opened a door", but the subtly different question "what is the probability of winning by deciding to switch knowing the rules of the game (but before the host opens a door)". The usual question is the first one. This analysis (and many others, including most currently presented in this article) answer the second one. -- Rick Block (talk) 15:45, 6 April 2008 (UTC)

[edit] This part of the Venn sction:

The host now opens Door 3. Under the conditions of the problem statement the host is equally likely to open either unchosen door, so opening this door does not affect the average probability of winning the car by staying with the original choice which remains 1/3. The car is not behind Door 3, so the entire 2/3 probability of the two unchosen doors is now carried only by Door 2, as shown below. Another way to state this is that if the car is behind either door 2 or 3, by opening Door 3 the host has revealed it must be behind Door 2.

... is confusing (to me).
The host is equally likely to open either unchosen door if and only if the player has chosen the winning door. Otherwise the host is constrained as to which door he opens. That point is doesn't punch out clearly enough and may be a source of confusion and argument. --hydnjo talk 18:26, 19 March 2008 (UTC)

I added those words to try to address the objection 70.137.136.97 has raised. If the player picks door 1 and the car is behind it, we've constrained the host to open either remaining door with equal probability by the problem statement. If the player picks door 1 and the car is not behind it, the car is equally likely to be behind door 2 or door 3, forcing the host to open the other door. This means the overall odds of the host opening door 2 are 50% (and the same for door 3). My reasoning (which I admit is WP:OR and I'm not overly confident that it's true) that this means the host's choice doesn't affect the player's initial 1/3 probability is the following. Assume the player picked door 1. Let the probability of the car being behind this door if the host opens door 2 be X and let it be Y if the host opens door 3. (X*50%)+(Y*50%) must be 1/3. On the other hand, since these two scenarios are equally likely I think (this is the part I'm not overly sure of) this means X=Y. If X=Y (I'm entirely confident of the rest), (X*50%)+(X*50%) = 1/3, and X is therefore 1/3 (and similarly Y is 1/3).
An alternative, assuredly true, way to get to this same result is from the following probability table (assuming the player has picked door 1):
Car location Host opens door 2 Host opens door 3
Door 1 1/6 1/6
Door 2 0 1/3
Door 3 1/3 0
we see that the overall odds of the host opening Door 2 (or Door 3) are 50% (1/6 plus 1/3, for each door) and given either door the host opens the probability the car is behind Door 1 is (1/6) / (1/2) which is 1/3. This is the full conditional analysis, but it certainly shows the truth of the statement Under the conditions of the problem statement the host opening either unchosen door does not affect the average probability of winning the car by staying with the original choice which remains 1/3 (not the original phrasing from above). Perhaps we should discuss whether the host opening either door with equal probability is what forces this to be true (again, I'm not certain), but in any event I think we can say something as a lead in to this section justifying the illustration. IMO, the illustration is not meant to be a "proof" but a visual aid helping the reader understand how the probability can end up unevenly distributed between the two doors. Numbers, tables, and formulas help some people. Pictures help other people. -- Rick Block (talk) 22:41, 19 March 2008 (UTC)
Well, let's put it this way then: "The host now opens Door 3. Under the conditions of the problem statement the host is equally likely to open either unchosen door, so opening this door does not affect the average probability of winning the car by staying with the original choice which remains 1/3" is self-contradictory, so however you want to fix it, it's got to go. Every statement of probability fixes a sub-universe. "equally likely" implies that we're in the sub-case where the car has been correctly chosen, in which case the odds that the car was chosen are 100%, not 1/3. Regardless of whether there is sound reasoning lurking around here, the sentence has to be fixed - Dan Dank55 (talk) 13:50, 28 March 2008 (UTC)
It is true that (1) given that the prize is behind the chosen door, the host is equally likely to open either of the unchosen doors, and (2) given that the prize is not behind the chosen door, the host is equally likely to open either of the unchosen doors. (2) is true even though the host has no choice, because it is equally likely that the prize is behind unchosen door A as it is behind unchosen door B, again given only that the prize is not behind the chosen door. Thus, from the point of view of the contestant, the behavior of the host can't be distinguished between situations (1) and (2). Even if the contestant knew that he had/had not chosen the correct door, she would be unable to predict which door the host would open with better probability than a coin flip. This is really the reason why there is no probability shift regarding the chosen door, in the statement of the problem that dictates that the host opens unchosen doors with equal probability if the chosen door has the prize. Bill Jefferys (talk) 14:38, 28 March 2008 (UTC)
I'm not sure what you're saying, but let's be clear: probability is a state of mind of an agent (human, computer, animal, whatever) at a specific time (and you can say it has other conditions, such as "given an understanding of the problem", but leave that aside for now). There is no specific agent at a specific time (or specific point in a flowchart, if you diagram it that way) who believes everything in the self-contradictory statement I pulled out; it confuses the mind of the gameshow host with the mind of the contestant. - Dan Dank55 (talk) 01:56, 29 March 2008 (UTC)
I think you're not seeing the point. If the player picks the car the host is equally likely to open either remaining door (I think you agree with this). If the player doesn't pick the car, the car's location is evenly distributed between the two remaining doors, so even in this case the host is equally likely to open either remaining door (in the sense that if we run this 1000 times roughly half the time the host will open one and roughly half the time the host will open the other). These are the only two cases, and in each, the statement is true. So, we don't need the case qualifier. The host is equally likely to open either remaining door (period). The perspective is that of a third party observer (neither the contestant or the host), with knowledge of the game rules. -- Rick Block (talk) 04:00, 29 March 2008 (UTC)
I'll withdraw the objection, because you're right, it can be read the way you're describing. - Dan Dank55 (talk) 04:18, 29 March 2008 (UTC)

[edit] Proposed solution section

I've been working on the Solution section at Monty Hall problem/draft#Solution. Does anyone have any objections to replacing the Solution section currently in the article with this version? The major point is to address the conditional vs. unconditional concern extensively discussed in several of the threads above. -- Rick Block (talk) 18:13, 25 March 2008 (UTC)

I think it's more difficult to understand than what's in the current version. However, helping to fix it would require more attention than I am willing to give at this time.
May I ask you to show both versions to a few laymen and see which they respond to better?--Father Goose (talk) 19:26, 25 March 2008 (UTC)
I'll try. Just curious what you find more difficult to understand in the draft version - is it the words or the different images, or both? -- Rick Block (talk) 01:04, 26 March 2008 (UTC)
I've solicited comments from anyone watching my talk page. Please do the same if you're so inclined. -- Rick Block (talk) 02:10, 26 March 2008 (UTC)
Okay. As for the draft version, I think the replacement illustration table has too much text, and is less easy to follow since it doesn't have all the purely visual cues that are present in the illustration I created. However, considering I created the other illustration, it may just be that it's easier for me to understand.
Additionally, some of the new language is ponderous ("only two of the cases enumerated above are possible") and the table and final paragraph are a bombardment of numbers without a whole lot of explanation that would give the reader a more intrinsic sense of the solution. I can understand it, but only if I make the effort to try to figure out what you're trying to say.
I'm fine with noting that all the different unconditional solutions only work "on average", and I'm also okay with adding a layperson-accessible explanation of the unconditional probability. However, such an explanation should focus on why the conditional and unconditional problems are different, instead of simply ticking off what the probabilities are in the unconditional solution.--Father Goose (talk) 07:21, 26 March 2008 (UTC)
The extra text disrupts the intuitive benefit of the graphic since stopping to read it disrupts the visual flow. Also, having 9 table cells loses the message of "three equally likely cases"; this is worsened by the two equal sized "switching loses" options, which may confuse the viewer into considering this as possibly "four equally likely cases" grouped artificially into 3.
This impulse to over-explain breaks the graphic. Struggle with detailing the meaning of everything in the article text.
One advantage of the draft version is the toothy face is gone. A face probably helps to represent the contestant's choice, but that exaggerated grin is creepy. Having just the pointing hand probably loses some clarity because it can be either the host or the contestant; at some screen resolutions it may not be quickly apparent it is a pointing hand. / edg ☺ ☭ 08:10, 26 March 2008 (UTC)
The "creepy grin" can be toned down. 8-D --Father Goose (talk) 03:05, 27 March 2008 (UTC)
Another alternative (sans graphics) is this one. I've asked for graphics help at Wikipedia:Graphic Lab/Images to improve#Monty Hall problem images which I expect would address the pointing hand issue. We can arrange this set of images in a variety of table formats (I'm not too particular about this) - the salient point is that the initial "3 equal choices" reflect the position of the car rather than the "car-goatA-goatB" choice. I've taken some effort to make the current table arrangement reflect both the "unconditional" (3 equal choices) reasoning and the conditional host has opened door 3 reasoning (the "outcome" column shows the "host opens door 2" outcomes vertically aligned and the "host opens door 3" outcomes vertically aligned - I don't think it's much of a stretch to see the relationship between this figure and the table that follows with the 1/6, 1/6, 0, 1/3 ... numbers). -- Rick Block (talk) 04:20, 27 March 2008 (UTC)
Suppose I might get anyone to common on Monty Hall problem/draft#Solution? This version reuses Goose's table (rearranged so the primary case selection is by car location), and then reuses the graphics in the following probability table. -- Rick Block (talk) 22:17, 29 March 2008 (UTC)

[edit] Very simple, yet complete (both conditional and unconditional), solution

The following can replace the solution section, as well as the Venn Diagram section and more. Note that there is no need to mention a probability of 1/6, and there are symbols. Also, there is no need for pictures of silly goats. Only rudimentary probability.

Unconditional:

Before the host has opened any door, the probability that the car is behind the chosen Door 1 is 1/3. Therefore, probability that it is behind one of the two unchosen doors is 2/3. It follows, the probability that the car will be behind the unchosen door that will not be opened by the host is 2/3.

Conditional. The following probabilistic analysis reflects the state of knowledge of the player at various points of time:

  • After the player has chosen Door 1, and before the host has opened any door, the probabilities that the car is behind any of the three doors are the same: 1/3.
  • In the event the car is behind Door 1, the probability that the host opens Door 3 is 1/2. (This is assumed to be the host’s protocol).
  • In the event the car is not behind Door 1, the probability that it is behind Door 2 is 1/2. Therefore, in the event the car is not behind Door 1, the probability that the host opens Door 3 is 1/2.
  • Therefore, after the player has chosen Door 1, the probability that the host opens Door 3 is 1/2. This event contains the event that the car is behind Door 2, whose probability is 1/3.
  • Therefore, after the host has opened Door 3, the probability of the event that the car is behind Door 2 is equal to the relative size of the fraction (i.e., proportion) of the latter event (whose probability is 1/3) within the event that the host opens Door 3 (whose probability is 1/2). So, this fraction (conditional probability) is the ratio of 1/3 to 1/2, which is equal to 2/3.
  • In other words, given that the host has opened Door 3, the probability that the car is behind Door 2 is 2/3, and therefore the probability that it is behind Door 1 is 1 – 2/3 = 1/3.
70.137.136.97 (talk) 05:07, 26 March 2008 (UTC)
I might be shooting myself in the foot with this comment, but the probability that the car is behind Door 1 is not 1/3, it's unknown.--Father Goose (talk) 06:33, 26 March 2008 (UTC)
Goose - Probability is a subjective expression of the state of knowledge of an individual. The host knows where the car is, so his probability for any door to hide the car is either 1 or 0. The player, however, knows nothing in the beginning. He has three equally likely, mutually exclusive, possible situations. Therefore he has probability 1/3 for each door even though the car has been already placed behind one of the doors. The entire analysis is from the subjective point of view of the player. I hope this helps. If you do not accept this view of probability, then I do not see how you can proceed at all. 70.137.136.97 (talk) 04:25, 27 March 2008 (UTC)
Yes Father Goose, not to rub salt on your injured foot, that is exactly where all this over-analysis is inconsistent. The probability for an uninformed and unbiased initial guess is 1/3, regardless of the distribution of goats and cars. But the probability that the car is behind door No. 1 in particular depends entirely on that unknown distribution. It seems fairly natural to read the phrase "pick a door, say No. 1" to refer to an uninformed and unbiased choice rather than specifically to door No. 1. Yet, it is evidently somewhat less natural to read "another door, say No. 3" to refer to an informed (constrained by the car) but unbiased (among the goats) choice, for some read this as a specific reference to a particular door. If one reads it as unbiased and symmetric then it is "obviously" sufficient to treat it as a question of combinatorial probabilities, but if one reads it as specific to particular doors then it is "obviously" necessary to treat it as conditional on the a priori particulars. What is evidently not so evident is that, for the latter reading is consistent, there is no reason to suppose any particular probability that the car is behind door No. 1, as you note.
Of course this may be resolved by adding yet another stipulation to the problem statement, that the car is initially placed behind one of three doors with equal probability. Although, to be perfectly snarky about it, there is still the problem that we know there is a goat behind door No. 3. Read the problem again: every time door No. 3 has a goat, always and invariably, no matter how many times one reads it. The probability that the car is behind particular door No. 3 is exactly zero.
Silliness aside, my point is that for the best pedagogical value in the problem I prefer a simpler interpretation of its meaning. Rather than layer on more and more stipulations and constraints, simply interpreting "say No. 1" and "say No. 3" as indicating no a priori labeling of the doors would subsume multiple constraints and make clear that a simple combinatorial analysis is sufficient for this symmetrical situation. It seems clear to me, although clearly not to some others, that this was Marilyn's original intent. As she notoriously remarked about another missing constraint, "Anything else is a different question." 67.130.129.135 (talk) 23:44, 26 March 2008 (UTC)
Yes, yes, yes.
Sigh.--Father Goose (talk) 03:09, 27 March 2008 (UTC)
What is the harm in admitting there are two different questions that can be asked, with two different forms of appropriate solutions? I (for one) am not arguing about what Marilyn might or might not have meant and entirely agree the main interest in this problem stems from the "is it 1/2 or something else" question about the overall odds of winning. On the other hand, I also completely see the point that many (most? all?) statements of the problem specifically phrase it in a conditional manner. With "proper" constraints, the answer is still 2/3 - so what's the problem with presenting both an unconditional and conditional solution? Both Morgan et al and Gillman (American Mathematical Monthly, January 1992) distinguish between the unconditional and conditional solutions. I don't think there's any excuse for us to ignore this. The conditional question is a subtly different question, but depending on the assumptions can have a different result. -- Rick Block (talk) 03:51, 27 March 2008 (UTC)
Presenting both is fine; in fact, I do want to see an accessible explanation of the conditional probability in the article in addition to the unconditional explanation. But I also would like for us to offer the simplest expression of both the problem (short of ambiguities) and the solution (in terms of the overall odds of winning).
What I'd love to pull out of my magic hat is a statement of the problem that does not have ambiguities as far as the overall odds are concerned, and that does mention specific doors chosen and opened, but treats those specific doors as examples, not specific conditions requiring a conditional solution. Could such a statement of the problem be written, leaving aside issues of sourcing for the time being?--Father Goose (talk) 05:21, 27 March 2008 (UTC)
(Gillman 1992) suggests the unconditional problem is equivalent to a variant where the sequence of events is 1) initial pick, 2) player decides whether to switch or not before the host opens a door (but with the assurance a door will be opened, and that it will show a goat), 3) host opens a door. I'd say this variant is not as intuitively disturbing. If the sequence of events is 1) initial pick, 2) host opens a door, 3) player decides whether to switch, then to make a solution more like the unconditional one clearly apply I think we'd need to (highly artificially) keep the player (and the audience) from being able to know which door the host opens - like in Talk:Monty Hall problem/Matt#Variant 3.
I think the exquisitely disturbing characteristic of the problem relies on it being presented in a conditional form, which may well be why so many people apparently find the unconditional solution unconvincing. The unconditional solution clearly ignores which door the host has opened, and although I highly doubt anyone internalizes this as "well, that's an unconditional solution to a conditional problem - it clearly doesn't apply", I do think people's intuition tells them there's something fishy about it. It is certainly a valid way to arrive at the overall odds of winning by switching, but I think people intuitively pick up on the fact that it doesn't necessarily apply to them. -- Rick Block (talk) 13:41, 27 March 2008 (UTC)
It is exquisite indeed, but I am skeptical of this psychological analysis. Some people, such as myself, are conditioned by training to look for and exploit symmetries, while others are conditioned by training to look for and exploit a prioris. (It is not unlike the difference between algebraic and analytic temperaments among mathematicians.) But people's intuition leads to the naive 50/50 error, which I think has more to do with failure to differentiate than with unease about insufficient differentiation. 67.130.129.135 (talk) 01:05, 28 March 2008 (UTC)
I don't think we're arguing here. BTW - having seriously looked for references lately, I gather there are numerous papers about the psychology of the problem, none of which we reference (hmmm). In any event, I think the conditional presentation is certainly the usual form, the unconditional solution is certainly the usual explanation, at least some mathematical purists know the difference and scoff at those who don't, and the conditional solution is not terribly difficult to follow. I think the only real question is whether we want to better integrate the unconditional and conditional solutions. I'm not happy with the current solution section. 70.137.136.97 is not happy with the Venn diagram section (and, I'm sorry my friend, your prose above is much too technical for a general encyclopedia - I think we'll get it right eventually and you have my sincere thanks for bringing it up and sticking with it, if you had a login I'd give you a barnstar). Goose (and "anonymous who signs as Matt") are not happy with my Solution rewrite. I hope we don't get stuck again with people stomping off in huffs. I think we're sort of basically agreeing, but translating that into action we can all agree on seems inordinately difficult. -- Rick Block (talk) 02:13, 28 March 2008 (UTC)
Rick -- I am still amazed that after hundreds of pages of endless disucssion you still believe that there may be some "nontechnical" explanation. The "prose" is the simplest complete explanation. If you drop anything from it, it becomes incomplete. All that the reader has to understand is the meaning of probability. Some of the people who post here do not understand it though, and they prevent completing the article. Wikipedia has much more articles than this one. The reader has to read one bullet at a time and think how it follows from the previous bullets. This is how math should be read - there is really no other way. It requires thinking between the sentences. Some people may convince themselves without really understanding the correct proof, and then they object to the correct proof. 70.137.136.97 (talk) 04:58, 29 March 2008 (UTC)


I might ponder some rewriting myself (to see if I can address the conditional vs. unconditional aspects), but probably not any time soon. There are many other things on Wikipedia I'd rather be doing.--Father Goose (talk) 02:51, 28 March 2008 (UTC)

[edit] Monty Hall solution

I quite like this graphic:

or a more expanded version one more atempt:

Step 1 * 6 contestants pick a door. Each person represents one of the 6 outcomes. * The blue people pick door 1 * The green people pick door 2 * The purple people pick door 3 Step 2 * The host (as an honest broker) always reveals one goat at random. Step 3 * One of the blue/green/purple people swaps doors. * The other of the blue/green/purple people doesn't swap doors. Result *Swapper group: 2 Winners : 1 looser *Keeper group: 1 Winner : 2 loosers

Step 1
* 6 contestants pick a door. Each person represents one of the 6 outcomes.

* The blue people pick door 1
* The green people pick door 2
* The purple people pick door 3

Step 2
* The host (as an honest broker) always reveals one goat at random.

Step 3
* One of the blue/green/purple people swaps doors.
* The other of the blue/green/purple people doesn't swap doors.

Result
*Swapper group: 2 Winners : 1 looser
*Keeper group: 1 Winner : 2 loosers
  • 6 possible outcomes:
  • --> 'Swapping' gives two wins to one loss.
  • --> 'Keeping' gives one win to two losses.

Quantockgoblin (talk) 01:31, 29 March 2008 (UTC)

No ill will intended but, as an avid follower of this article, I find this graphic to be accurate and at the same time difficult to follow. I have difficulty imagining anyone but the most scholarly intelligent finding this graphic "helpful". But, this article is if nothing else, has become a sophisticated constellation of approaches to what I have always thought to be a fairly easily understood explanation of the anti-intuitive. I wonder if a separate article addressing the many branches (conditional vs. unconditional, host is 67% sure, player flips a coin, etc.)... would leave this article alone to address the nonintellectual initial MH problem in all of its glorious simplicity - nah, never happen. --hydnjo talk 02:37, 29 March 2008 (UTC)
Hi hydnjo, thanks for the comment. I just want to add that these diagrams are meant to be a simple pictorial flow chart which maps out the six possible outcomes. Maybe it only me that finds this diagram helpful!!! I've tried not to use words so that it would be useful in other wikis. PS I tried a third diagram!! -- Quantockgoblin (talk) 03:39, 29 March 2008 (UTC)
Are you suggesting replacing the current image in the Solution section with this image? I find it rather difficult to comprehend, and it does not address what I think is the central issue still on the table (the conditional vs. unconditional issue). -- Rick Block (talk) 16:56, 29 March 2008 (UTC)
Rick Block The above diagram assumes the host is a "fair broker", i.e. always removes one goat at random after the first pick. Personally I think the motives of the host is a bit of a red herring, if not an interesting aside to the problem. I've drawn the picture, so naturally I think it is clear. However, if others don't think so then it simply can't replace the current page image. If however, others liked the diagram, then I wouldn't object if they used it in the article. Thanks for the input, constructive criticism is better than silence. -- Quantockgoblin (talk) 17:19, 29 March 2008 (UTC)
Added a legend to graphic to explain graphic! - Quantockgoblin (talk) 17:44, 29 March 2008 (UTC)
A graphic should be easy to comprehend. This one is complexly detailed at the expense of comprehension. If one stares at it long enough, a solution may become apparent by which the car can reliably be selected. / edg ☺ ☭ 22:33, 29 March 2008 (UTC)
edg to be fair, I'm not sure that your criticism stands up. This is what the current solution looks like without text:





.



I submit that the current images (without text) are just three rather meaningless pictures that seems to have no inter-relationship at all, and centre around a random person pointing a doors with arrows coming out of his head and with the occasional elimination of a goat by means unknown. I submit that my image is more self-contained than the present set of images. I also believe incorporates the relative probabilities without extra explanation. However that all said, I think explanation text with the image is valid.


I suspect once a person understand the apparent paradox via one branch of reasoning, it less easy to latch onto another persons logic (I hope it is logic anyway). My solution is to explore the six possible choices available to the contestant - that is the six (and only six) possible “fates” of the contestant. Then by simply comparing these six “fates”, one can see that those 3 fates that involve "swapping doors" result in more wins than those 3 fates that involve sticking with the originally picked door. However, it seems that with 3 negative and no positives reply, I'm swimming against the tide here!! -- Quantockgoblin (talk) 04:30, 30 March 2008 (UTC)
I was not referring to the text, but the graphic. However, the fact that so much text is needed to explain the graphic itself is part of the problem. The graphic you are proposing seems like an unnecessarily complicated way to say the same thing as the current graphic. / edg ☺ ☭ 04:50, 30 March 2008 (UTC)


edg- I'm not sure that I want to beat my head against this door too much more ... (whether it has a car behind it or not!)
but … I believe that the present graphic is only comprehensible due to the meaty paragraphs leading and following it, which explains the line of reasoning used. I believe that your view of the comprehensibility of the present graphic relies on the fact that you already know what in effect these leading/following paragraphs say, and are reading that this information is implicitly known, and thus implicit in the current graphics themselves.
Wipe your brain of that information (impossible, but pretend) and see if the graphics makes any sense by themselves. My graphic does not have the benefit of these reasonably meaty introduction/concluding paragraphs. So far my graphic just has the short legend to explain it. If my graphic were couched in these leading/concluding paragraph, explaining about the "six fates", it would be just a comprehensible, and more (in my view anyway) comprehensive.
After all isn't this a comparison between two choices – that is, should the contestant "stick" or "twist". To me the present graphic doesn't really get to the heart of that. Currently it just shows six results, and not the sum of the three “sticking” results and the sum of the three “twisting” results. Perhaps it would be helpful (in my view) to have a fourth box which sums the LHS results (i.e. “sticking" results) vs the RHS results (i.e. twisting results). Sadly, I suspect I don’t have a convert on my hands … - Quantockgoblin (talk) 05:58, 30 March 2008 (UTC)
I just came due to a link from Talk:0.999..., and I don't recall ever being here before. On my first look at the current image, without reading the surrounding text, I was immediately able to understand why switching is a better option. To understand the images currently proposed I would have to spend some time figuring them out, which is not good for the reader. –Pomte 14:46, 31 March 2008 (UTC)
Pomte et al., ok, I officially give up,,, my ‘6-fates theorem’ will have to rest here on the talk page forever or until the server finally packs-up. I'll just have to be content myself with the idea that I’m a misunderstood genius :-) !! Thanks for the comments! -- Quantockgoblin (talk) 23:09, 31 March 2008 (UTC)
I love this graphic. Once it was made clear to me that the determining factor is the initial guess, not the door opening, all the pieces fell into place. It's rather like a magician's trick; the misdirection is placing importance onto the second event, and not the first where it truly belongs. It's the first time this has made intuitive sense to me. --75.173.4.172 (talk) 02:37, 3 April 2008 (UTC)
Which graphic, the one that's presently in the article, or the one that Quantockgoblin suggested at the beginning of this thread?--Father Goose (talk) 04:41, 3 April 2008 (UTC)
Sorry; I jumped right into a section without reading the whole thing. I meant the existing article and graphic. I didn't even bother to try to decipher the graphic about which this section was written. I'm sure that with an explanation, I'd get that one too. (Green guys? Purple guys?) But while a picture can replace a thousand words of explanation, it shouldn't need a hundred-word explanation. Looking at it now, I see an attempt to visually represent a combinatoric explication of all six possible outcomes. I'm not saying the existing graphic is objectively better at describing the full scope of the problem; I'm saying it ended all questions in my own mind within a few seconds of looking at it. I'm sorry, Quantockgoblin, but I do prefer the existing graphic. Perhaps there is room for your graphic in the later section of the article with the other combinatoric tables. --75.173.4.172 (talk) 01:13, 7 April 2008 (UTC)

[edit] Featured article review

Just so nobody misses it, see Wikipedia:Featured article review/Monty Hall problem. -- Rick Block (talk) 22:56, 28 March 2008 (UTC)

The concerns that have been raised are basically citations and length (as in too long). There is a specific list of sections for which citations are requested. I've looked around a bit and have found references to some citations that I think would be useful (but haven't found the actual article[s] that are referred to yet). If anyone can help, that'd be great.

  • "Increasing the number of doors"
    One of the original vos Savant Parade articles has a "1000000 doors" example. This is recapped at http://www.marilynvossavant.com/articles/gameshow.html, but I think the original Parade reference would be better. There's also apparently a German paper by Hell, W. & Heinrichs, C. from 2000 titled "Das Monty Hall Dilemma (Ziegenproblem) mit 30 Türen" from a conference (?) "TeaP 2000 - Experimentelle Psychologie" about an experiment where by increasing the number of doors to 30 the number of subjects choosing to switch increases from the typical 10-15% to 65% (supporting the claim that increasing the number of doors does indeed make it easier to understand). The reference to this paper is from "The Psychology of the Monty Hall Problem: Discovering Psychological Mechanisms for Solving a Tenacious Brain Teaser" by Kraus and Wang which was (apparently) published in Journal of Experimental Psychology: General in 2003 (vol 132, no 1, pp 3-22). I haven't seen this reference either, so can't legitimately cite it.
  • "Venn diagram"
    I haven't found anything (yet) that has this sort of analysis.
  • "Decision tree"
    I think the original reference should be Chun 1991 "Game Show Problem" OR/MS Today Vol 18 no 3, but I haven't found this (doesn't seem to be online). This is cited in a 1999 paper by Chun ("On the Information Economics Approach to the Generalized Game Show Problem" in The American Statistician, Vol. 53, No. 1, Feb., 1999). This later paper is a reference for other host behaviors. Another citation of Chun's 1991 paper is from "A Genuine Decision Tree for the Monty Hall Problem" by Andreas Hammer which is a 2004 (unpublished?) working paper from the International University of Germany.
  • "Combining doors"
    Perhaps Stibel et al, 2008, The Collapsing Choice Theory: Dissociating Choice and Judgment in Decision Making, http://www.springerlink.com/content/v65v2841q3820622/, from the journal "Theory and Decision" (published online 5 January 2008), or http://nd.edu/~rwilliam/xsoc592/appendices/xappxd.pdf (but this appears to be an appendix to an unpublished set of course notes)
  • "Simulation"
    Using playing cards as a simulation is mentioned lots of places including one of the original vos Savant Parade articles. I'm not sure what the first reference might be. I'll add something, but if anyone can find an earlier reference feel free to update it.
  • "Other host behaviors"
    Again, lots of references. We'll need to pick. Morgan et al (1991) have some. Mueser & Granberg (1999) have some. Others are in Chun (1999). We already reference the "forgetful Monty" variant (in a fairly recent vos Savant Parade column).
  • "Two players"
    I haven't found a reference for this variant yet.

-- Rick Block (talk) 01:10, 30 March 2008 (UTC)

Thanks for leading this work, Rick. A couple years ago I would have offered to help directly, but right now I have to cut back on major efforts. I hope you get whatever support you need!
At a glance, the above citations look like they go in good directions. I know there's been a lot of discussion about the "Venn diagram" section, but if it's really so outside the norm that it can't be referenced, I say kill it. This would also help with the length issue, although I wouldn't really worry too much about length. Melchoir (talk) 01:59, 30 March 2008 (UTC)
I would kind of like to see this quote from the "Ask Marilyn" column (and perhaps the next sentence too) in the article, because it's the response to the problem (both the response in 1990 and in general) that makes the problem notable, much more than the math, I think: "I'm receiving thousands of letters, nearly all insisting that I'm wrong, including the Deputy Director of the Center for Defense Information and a Research Mathematical Statistician from the National Institutes of Health! Of the letters from the general public, 92% are against my answer, and and of the letters from universities, 65% are against my answer. Overall, nine out of ten readers completely disagree with my reply." - Dan Dank55 (talk) 02:51, 30 March 2008 (UTC)

Though a little trimming may be in order, personally I like the way it reads now, and the pictures with the happy brown guy pointing (and the following 3 clarification statements in text) were VERY clear to me. The pictures suggested above, with the myriad arrows, are unnecessarily intricate to explain the results.

I agree that the first part of the Venn Diagram section (where there is a box around two of the doors representing the 2/3 chance) is confusing and not really accurate, since it does not specify taking into account that the host knows where the prize is and always reveals a goat after the player's initial selection--thus it would be incorrect if the host does not have knowledge, but doesn't state this. The chart just below that with the arrows works ok, though...it is pretty much just a probability-multiplying chart of the brown guy pics above.

I initially took a look at this article because I saw the movie 21 the other day and they discussed it in an MIT math class, and I recalled learning the problem and its result at some point in the past, but I couldn't explain it clearly to my mother on the spur of the moment. I sent her a link to this article and am confident that the way it reads now from the beginning, with the quote of the problem specifying the parameters of the host's knowledge & thus his conditional behavior, will allow her to understand the problem.

Thanks!! I bet you get a ton of hits the next few weeks, with that movie having just come out! PuppyAddict (talk) 17:34, 1 April 2008 (UTC)

[edit] Compare this article to the one on the "Three cards problem"

Wikipedia has a nice, concise and to-the-point arctile titled Three cards problem. The problem is with three cards:

  • a black card that is black on both sides,
  • a white card that is white on both sides, and
  • a mixed card that is black on one side and white on the other.

You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black. What is the probability that the other side is also black? This is essentially the same paradox as the Monty Hall Problem. I leave to the readers to verify the equivalence. The answer is that the other side is black with probability 2/3, even though some people think it is 1/2. Evidently, that article was not subject to attacks by many editors. I now raise the question: Why not rewrite the Monty Hall article in the same style. If not, would it be justified to edit the "Three cards problem" article and add to it pictures of goats, charts, etc.? —Preceding unsigned comment added by 70.137.136.97 (talk) 23:49, 30 March 2008 (UTC)

I suspect you mean something like this version of the "Three cards problem" (it has now been merged into Bertrand's box paradox). Articles about popular topics tend to increase in size (and gain illustrations) over time. For example, roughly three years ago the Monty Hall problem article looked like this (the images have been deleted so don't show up, but there were far fewer than there are now). In addition, "Three cards problem" doesn't meet Wikipedia:Manual of Style (mathematics) (specifically the "Writing style in mathematics" section). I'd imagine there are other issues with regard to Wikipedia:Featured article criteria as well. Other mathematics FAs might be more comparable. The ones that are not biographies are 0.999..., 1 − 2 + 3 − 4 + · · ·, Infinite monkey theorem, Polar coordinate system, Prisoner's dilemma, and Trigonometric functions. -- Rick Block (talk) 01:36, 31 March 2008 (UTC)
Yeah, Bertrand's box paradox needs references, a history section, better prose style, better unification of variants, and it probably has some parts that will ultimately have to be removed. Not the best example to follow just yet. And the topic is not, as far as I can tell, equivalent to Monty Hall, despite their similarities. Some differences:
  • The analysis of Monty Hall is complicated by the fact that the player gains information from the actions of the host, who knows where the car is. In Bertrand's box, the dealer does not have privileged information and cannot cheat the player by adopting an unexpected strategy.
  • Bertrand's box is older. Monty Hall is better known in recent times, and therefore has a greater variety of analyses. This is both a blessing and a curse.
Melchoir (talk) 01:55, 31 March 2008 (UTC)

Both the "Three cards problem" and the "Bertrand's box paradox" are equivalent to the Monty Hall Problem as presented in this article. Because the host in this article must randomize 50-50 when the player picks the door hiding the car, and the other constraints, the host does not have any discretion whatsoever, and that makes the problem equivalent to Three cards problem. So, why is this article so long and controversial? 70.137.136.97 (talk) 02:13, 1 April 2008 (UTC)

"and that makes the problem equivalent"? Does it really? Melchoir (talk) 02:44, 1 April 2008 (UTC)
This article is long and controversial because "no other statistical puzzle comes so close to fooling all the people all the time" (attributed to Piattelli-Palmarini by vos Savant in her "Power of Logical Thinking" book). Greater than 80% of the population gets the answer wrong, and is extremely hard to convince. This article presents numerous solutions (currently most are to the unconditional problem) in an attempt to be convincing to more than 20%. There are still people who read this and conclude it's simply bunk. -- Rick Block (talk) 03:37, 1 April 2008 (UTC)
Response to 70.137.136.97 : The Monty Hall problem is not exactly equivalent to the Three cards problem#Card version, because in the latter, the person who picks a card has no further decision to make after his original choice. In Monty Hall, he makes one choice, gets a response to what he did, and then is offered a further choice between two options. You seem to be arguing that the sample spaces of the two problems are isomorphic even though the number of choices is different. I'd need to see the equivalence shown before I'd believe it. EdJohnston (talk) 04:25, 1 April 2008 (UTC)

Here is a version of the Three Cards Problem that is equivalent to Monty Hall. Suppose the player wins the game if he picks the mixed-color card. Obviously, the player has a probability of 1/3 to initially pick that card. In this game, the player initially picks a card and then the host reveals one side of the card. The host picks the side randomly. (50:50). Suppose the host reveals a black side. This means that the chosen card is not the white card. Suppose the host also removes the white card at this point. The host now offers the player to switch to the other remaining card. As we know, the conditional probability that the other remaining card is the winner is 2/3 and if he does not switch, the probability of winning is 1/3. 70.137.136.97 (talk) 06:13, 1 April 2008 (UTC)

Very interesting! Yes, this variant of Three Cards is awfully close. Certainly it reaches the same final state as Monty Hall, once all choices and labels have been fixed. It still leaves a few things to be desired, though, if you want to compare explanations.
In Monty Hall, the usual fallacy comes from abusing the observation that the location of the car has been narrowed down to two doors from three. But in Three Cards, the fallacy doesn't come from narrowing down the position of the mixed card. Unless you insert the step where the dealer reveals the white card, the player has no idea where the mixed card is; its probability of occurring in each position remains 1/3. Instead, the fallacy in Three Cards comes from abusing the observation that the identity of the chosen card has been narrowed down to two possibilities from three.
So, in order to draw a parallel between the paradoxes, one would really want to identify doors with color (or metal) combinations rather than with cards (or boxes). This possibly explains the differences in intuitive difficulties between the two problems: it is more tempting to imagine a ghostly car evenly redistributing itself behind closed doors than to imagine a fixed card repositioning itself in some abstract color space. Melchoir (talk) 03:55, 2 April 2008 (UTC)
Yet, even if you do not insert the step where the dealer reveals the white card, the probability that the chosen card is a loser remains 2/3 even though it can only be either the black or the mixed (leading to a mistaken belief that it is 1/2). This is the same as in Monty Hall -- sticking with the choice is losing with probability 2/3. 70.137.136.97 (talk) 06:34, 2 April 2008 (UTC)
Right, you've established a dictionary between elements of the final situations of the two problems, including a restatement of their final questions, such that the answer is 2/3 in either language. However, this dictionary doesn't seem to translate between the larger structures of the problems, such as their setups. Compare, for example, the dictionary between Monty Hall and Three Prisoners. Melchoir (talk) 08:34, 2 April 2008 (UTC)

[edit] "Variable Change?"

I have heard the term "Variable Change" referenced in the solution to this problem. I assume that it means that the host's behavior changes based upon the player's initial choice, but I'm not sure about that. I'd LOVE to see you guys mention the term early on in the page as part of the definition, probably linking it to a separate page with a brief explanation of this term. I think you will get more people wondering about this aspect of the problem since the movie 21_(2008_film) has come out, since they mention Variable Change in the film when discussing the problem.

Thank you! You guys have done a great job already, and I appreciate all the hard work. PuppyAddict (talk) 17:52, 1 April 2008 (UTC)

The "Why the probability is 2/3" section does detail why the host's behavior changes. As for the term "variable change", that may just be 21's way of phrasing the concept. Maybe they're referring to something known in bridge as the principle of restricted choice, which is already mentioned in our article, though possibly it could be worked into the 2/3 section as well.--Father Goose (talk) 03:42, 2 April 2008 (UTC)
Yeah, 21 just uses the phrase "variable change" to describe a plot point; it's a vague slogan that ultimately comes back to bite one of the characters. Unless I'm remembering wrong, the movie doesn't suggest any particular meaning for the phrase, so I'd rather not attempt to insert it in the article. Possibly if someone finds an interview with the film's writers, something meaningful could be said? Melchoir (talk) 04:08, 2 April 2008 (UTC)
Interestingly, the page has had a spike in views the past couple of days... related to the film's release on Friday?... Mebbe. :-) --Father Goose (talk) 03:50, 2 April 2008 (UTC)
Sure enough, daily page views have been more than triple that of our prior average: [3].--Father Goose (talk) 09:43, 6 April 2008 (UTC)

[edit] "If the host is clueless"...

Correct me if I'm wrong, but doesn't vos Savant make a mistake on the following statement: "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch". Now, once you pick a door, the host has to show you the contents of one of the other two. Obviously, if the host (unwittingly) shows you the door with the car, you know where the car is and this whole thing doesn't apply. But, if the host shows you the other goat door, you end up with the original "Monty Hall problem" by default - you've originally picked either a goat, a goat, or a car, thus you have the same percentage if you switch, 66%. So the host being clueless doesn't matter unless the host's cluelessness reveals the car, right? All Hallow's Wraith (talk) 07:42, 6 April 2008 (UTC)

No, because you're disregarding some information that's been provided to you -- the very fact that the host didn't screw up. A clueless host is more likely to show you a goat if you picked the car in the first place (100% vs 50%). So if he does show you a goat, that should make you suspicious that it's because you've already picked the car. This new information is just enough to raise the expected value of your door from 1/3 to 1/2.
This kind of explanation would probably be clearer if it accompanied a Bayesian analysis, complete with numbers. The section "Other host behaviors" touches on the variant, but it isn't explained at length. Perhaps we need better coverage there, maybe in a sub-article to avoid stealing focus? Melchoir (talk) 08:09, 6 April 2008 (UTC)
I thought this whole situation was under the assumption that you wouldn't be suspicious when dealing with this clueless host. Maybe I should have said it clearler - if the host is clueless but we don't know that he is clueless, then the percentage is the same? (and this is still presuming the doesn't show us the door with the car by accident) All Hallow's Wraith (talk) 08:52, 6 April 2008 (UTC)
Switching when a goat is revealed will win 2/3 of the time given a standard host and 1/2 of the time given a clueless host. These percentages, interpreted as frequencies, are real, and they don't depend on what the player thinks about the host's knowledge. Melchoir (talk) 11:16, 6 April 2008 (UTC)
The "Why the probability is 2/3" section explains in full detail why the two versions of the problem are different. I believe our coverage of that is adequate. As for a Bayesian analysis making it clearer... it would to statisticians, but not to the bulk of our readers. The Bayesian section is pretty much in a foreign language to me, although I understand the problem quite well in all other regards.--Father Goose (talk) 09:33, 6 April 2008 (UTC)
Hmm, so it does! Okay, I'm embarrassed, but also frustrated. The structure of the article could be a lot more helpful for a reader who is trying to locate a specific bit of information. The entire section "Aids to understanding" should theoretically be about why the probability is 2/3, so it doesn't make sense to have a subsection titled "Why the probability is 2/3". The subsection currently with that title is mostly about why the probability isn't 1/2, and it doesn't indicate that it covers a variation until you get to the second paragraph of the body. Meanwhile there exists a "Variants" section which isn't actually the place to read about variants, and a reader who clicks to it from the Table of Contents is going to miss things.
Besides being more readable, I suspect that a hierarchical, modular article structure would be also easier to maintain. Section X and its subsections are for the "conditional" version, Y is for "unconditional" analysis, Z for host behavior variants, W for structural changes like additional players and doors. Imagine, no more endless arguments about what style belongs where: just look at the header of the section you're editing! And since the natural logical flow isn't so rigid, one has an Introduction section at the very top, maintaining a high-level overview providing intuitive summaries of the analysis below. Here variants may be contrasted against each other to make a point, but details follow later. Melchoir (talk) 10:55, 6 April 2008 (UTC)
Yeah... the article is kind of disjointed. Long ago, I suggested the possibility of regrouping most the "aids to understanding" subsections as "alternative solutions" (or maybe "alternative explanations"). As for the 2/3 section, I think that's different from the other solutions in that most of them show that the odds are 2/3, but don't explain what specifically it is that causes it to not be the instinctive answer (1/2) -- that the host is playing with a restricted choice. The more I think about it, the more I want to add that specific term to the "2/3" section, though I'd need to read up on "restricted choice" theory first. Our article on the subject isn't quite applicable to the Monty Hall problem.
Or maybe I can sucker Rick into doing the research... ;-) --Father Goose (talk) 11:32, 6 April 2008 (UTC)
Agreed that that's why the "2/3" subsection is different. It's also why it should be given a better title to differentiate its main thrust from the other subsections, barring a complete restructuring of course.
I am proud to officially support suckering other people into doing research! This kind of headache is why I restrict myself to articles on narrow topics. Thank heavens for braver editors, eh? Melchoir (talk) 11:54, 6 April 2008 (UTC)
Given the request for references at the FAR, without a better reference I'm not sure the current "2/3" section isn't original research (the Parade reference that's cited has no analysis at all - Marilyn's entire response is quoted). Certainly the current restricted choice article isn't applicable to the Monty Hall problem - hard to believe there isn't a more general article about this already (although quickly looking through Category:Probability theory, I don't see anything obvious). -- Rick Block (talk) 16:25, 6 April 2008 (UTC)
I'm not too surprised to hear that Parade isn't helpful here. This search appears to get a hit. Melchoir (talk) 21:19, 6 April 2008 (UTC)
...And that article suggests that the original publication is Barbeau's 1993 article The Problem of the Car and Goats. Melchoir (talk) 21:35, 6 April 2008 (UTC)
The article in Bridge Today already referenced in the article might be of use; a portion of it (along with additional analysis) is in this book. This site, while not having all the cachet of a perfect "reliable source", does give an analysis similar to mine (using a "roulette wheel" diagram instead of the tables). It's from UC San Diego's math department, apparently as supplemental material to their "Introduction to Cryptography" (Math 187) course. Reliable enough? I think so, especially given that the analysis in the "2/3" section that it would back up documents something not too difficult to understand (the consequence of the restricted choice).--Father Goose (talk) 00:39, 7 April 2008 (UTC)
I don't know... a how-to book from "Master Point Press"; and an unsigned webpage starting with the words "I can only assume..." that could have been written by some bored TA? Lacking cachet is putting it mildly. I feel a little bad for not tracking down Barbeau myself, but I know we can do a lot better. Melchoir (talk) 01:44, 7 April 2008 (UTC)
I haven't read the Bridge Today article, so I don't know how much of the analysis in that book is from that article. I hope the Bridge article is not regarded with suspicion as well.
My general view of these things is that when information is useful and uncontroversial (I feel the 2/3 section verges on self-evident, the way it's explained), then sourcing can be relaxed some if needed. Sourcing is always preferable, but not at the expense of losing important parts of an article. The 2/3 section explains how the "host must reveal a goat" condition produces a restricted choice, which is what makes the probability distribution non-random, and is a really fundamental aspect of the problem. Tossing it just because we couldn't find perfect sourcing for it would be... unfortunate. It would be great if we could find a pedigreed source that does delve into this exact aspect of the problem. Frankly, it's very surprising that more sources don't discuss it. But I don't take that as an indication that it's wrong, just that it's not an angle that statisticians or psychologists dwell upon. But bridge players have apparently dwelt upon it, as a group of people needing to employ probability in "the real world", not just on paper.--Father Goose (talk) 04:48, 7 April 2008 (UTC)
I don't conclude that the 2/3 section as presently written is at all wrong; it makes perfect sense to me. But if it's an angle that academics don't dwell on, I suspect they have better reasons than a cynical disregard for the real world. Published sources include educators and, yes, psychologists; these of all people should know which explanations best zero in on the interesting issues.
Also, I think a couple of different things are being conflated here. Analyzing the "clueless host" variant is not quite the same as analyzing the effect that restricted choice has in producing 2/3. There isn't a canonical way to remove restricted choice from the problem; that would be like trying to do a partial derivative without saying what's held constant. With "clueless host" one chooses to alter the possible outcomes of the game, adding the possibility that the host reveals a car. By contrast, you could also choose to preserve the set of outcomes but alter the rules: let the host open a random door and move the car behind the scenes, if necessary, to show a goat. Then there are no possibly-restricted choices visible to the player, and the probabilities hinge on whether the host is allowed to mess with the player's door.
Now, I don't know which of these variants is more psychologically relevant, or even which is better attested in the literature. These aren't questions we can answer by insight alone. Melchoir (talk) 08:48, 7 April 2008 (UTC)
Well, each have their areas of interest. Statisticians aren't concerned with helping people understand the paradox as much as providing mathematical analyses of it. Psychologists focus on people's thought processes and perhaps not so much on the underlying puzzle. Bridge players focus on having a good practical understanding of probability. If what's in the section is reasonable on its face and backed up by outside sources which are not optimally reliable but not necessarily unreliable either, I'm not going to tear my hair out over it.--Father Goose (talk) 09:48, 7 April 2008 (UTC)
Marilyn's explanation about the forgetful host variant is on parade.com here. The Barbeau article doesn't seem to be online. I'm actually accumulating a list of stuff to look up at a university library sometime. I'll add this Barbeau paper to the list. -- Rick Block (talk) 03:05, 7 April 2008 (UTC)
Interesting, she does explain it after all... Melchoir (talk) 08:54, 7 April 2008 (UTC)
Not in a way that backs up most of what's in the 2/3s section, unfortunately. She doesn't get into how the host's behavior is altered by the "must reveal goat" constraint. Have to look to other sources for that.--Father Goose (talk) 09:48, 7 April 2008 (UTC)

[edit] Unsourced additions

In a series of edits, a new variant was added (this is the aggregate diff). I've reverted these changes since no source was provided. As a featured article, content must be appropriately sourced - the article is currently undergoing a featured article review in part because of unsourced content. Adding more is not helpful. -- Rick Block (talk) 01:04, 7 April 2008 (UTC)

[edit] NY Times Source

The April 8th, 2008 issue of the Science Times section of The New York Times featured the Monty Hall problem in an article on cognitive dissonance. The article goes into detail about the Monty Hall problem and provides an online version to play. From the article by John Tierney:

The Monty Hall Problem has struck again, and this time it’s not merely embarrassing mathematicians. If the calculations of a Yale economist are correct, there’s a sneaky logical fallacy in some of the most famous experiments in psychology.

The economist, M. Keith Chen, has challenged research into cognitive dissonance, including the 1956 experiment that first identified a remarkable ability of people to rationalize their choices. Dr. Chen says that choice rationalization could still turn out to be a real phenomenon, but he maintains that there’s a fatal flaw in the classic 1956 experiment and hundreds of similar ones. He says researchers have fallen for a version of what mathematicians call the Monty Hall Problem, in honor of the host of the old television show, “Let’s Make a Deal.”

(I posted this source in both article talk pages) — Becksguy (talk) 17:58, 9 April 2008 (UTC)

Here's Chen's paper: [4]. I'd say it's much more relevant to cognitive dissonance than to this article, unless the connection between them gains prominence in the future.--Father Goose (talk) 06:32, 10 April 2008 (UTC)

The NYT article is actually about cognitive dissonance, but about 28% of the articles' 1298 words are on the Monty Hall problem (plus a playable game and a link), which is considerably more than a passing mention. I was offering that portion of the NYT coverage in case someone wanted to use it here. Chen's paper has only one brief mention of the Monty Hall problem, so I agree that it's definitely not applicable to this article, unless a connection becomes notable in the future. Thanks. — Becksguy (talk) 07:48, 10 April 2008 (UTC)

I did read the article, and pondered how it might be worked into our article, but ultimately it was just a rehash of Tierney's earlier article about the MHP and an attempt to relate it to the statistical error outlined by Chen.
If Chen's analysis proves to be correct and the field of cognitive dissonance is turned upside-down by it, we'll probably be able to include a sentence or two about it here, demonstrating how a comparable problem caused problems in that field. One to watch.--Father Goose (talk) 08:29, 10 April 2008 (UTC)
Pages 3 and 4 of the pdf are pretty compelling. Rather than pointing out some generic gotcha in statistics that's vaguely similar to the MHP, and stretching out a comparison for rhetorical purposes, the form of the experiment he describes is actually equivalent.
If and when the paper actually gets published, I'd certainly appreciate a couple sentences here, probably in the History section. I don't think we need to demand a revolution in the field. In any case, though, it shouldn't be its own top-level section. Melchoir (talk) 02:36, 12 April 2008 (UTC)

[edit] Analysis about different prize levels

I've deleted the following analysis from the "Other host behaviors" section since it was added without a reference (and lack of references is perhaps the primary WP:FAR complaint). This analysis was added in this sequence of edits by an anonymous user, who signed a different edit as RWFarley. -- Rick Block (talk) 03:36, 15 April 2008 (UTC)

Another analysis considers three types of hosts and three prize levels. The Benevolent Host always shows the worst remaining prize after you choose, the Random Host randomly picks a remaining door to show, and the Malevolent Host always shows the best remaining prize. The prizes are bad, middle, best; e.g. Goat, Luggage, Car. The Player is unaware of which prize is which. He may expect to be choosing among Pigs, Goats, Blenders, Luggage, Cars, and Houses.

If you always switch, the results for each host are:

Results after switching, expanded behaviors
Host Bad Middle Good
Benevolent Host 0% 33% 67%
Random Host 33% 33% 33%
Malevolent Host 67% 33% 0%

If each host were equally likely, the total probability for each prize would be 33%—the same as not switching. Without knowing the type of Host and the prize mix, you can make no meaningful statement about the success of a switching strategy.

The only way for a Player to "improve the odds” is if he or she can get some meaningful information from the prize shown. I.e., if you know what the three prizes are and what type of Host you have, then you can develop a winning strategy. If you were wrong about either the Host or the prize mix, that strategy may be harmful.

I actually rather like the look of this section. It is certainly correct, and it seems to me even to be a rather useful extension of the problem. I agree that it should have at least one citation to be replaced, though, but I wish it would make it into the article. 76.190.157.141 (talk) 18:42, 17 May 2008 (UTC)

[edit] Three Prisoners Problem

In regards to this:

History of the problem An essentially identical problem appeared as the Three Prisoners Problem in Martin Gardner's Mathematical Games column in Scientific American in 1959 (Gardner 1959). Gardner's version makes the selection procedure explicit, avoiding the unstated assumptions in the Parade version. This puzzle in probability theory involves three prisoners, a random one of whom has been secretly chosen to be executed in the morning. The first prisoner begs the guard to tell him which of the other two will go free, arguing that this reveals no information about whether the prisoner will be the victim; the guard responds by claiming that if the prisoner knows that a specific one of the other two prisoners will go free it will raise the first prisoner's subjective chance of being executed from 1/3 to 1/2. The question is whether the analysis of the prisoner or the guard is correct. In the version given by Martin Gardner, the guard then performs a particular randomizing procedure for selecting which name to give the prisoner; this gives the equivalent of the Monty Hall problem without the usual ambiguities in its presentation.

I highlighted the "which of the other two" as I came to this interesting brainstorm. I'm sure I must have missed some, but I like to hear what others think. If he asks which OTHER of the two will go free, he rules himself out of the question. Otherwise he must ask: name one of the three of us that will go free. If the guard names either two of the others, leaves him and one of the others a 50% chance of being executed. If the guard would name him, of course he will have a 100% change of surviving and both others will have a 50% chance.

But now he ask which of the OTHER two will go free. Now the guard can only name one of the two others. Which one he names doesn't matter, because the one he doesn't name, will be executed. Leaving the asker a 0% chance of being executed. Because if the person asking was the one to be executed, the guard could not name either of the other two and could only say something like "what do you mean which other two..." which would reveal the person asking is the one to be executed.

So this means that this DOES reveal information about whether the prisoner will be the victim, if he askes "which of the other two" will go free.

The wording here is not a precise quote of the problem. The request is for the name of one of the others who will go free (there are two, so at least one must be going free). -- Rick Block (talk) 14:05, 16 April 2008 (UTC)

[edit] Venn diagram (again)

Shall we just cut the "Venn diagram" section and archive it in the Talk page? It looks like Rick has "Combining doors" under control, so I think this is the last remaining issue. Melchoir (talk) 02:56, 26 April 2008 (UTC)
I'm thinking about merging it into "Combining doors" (it's fundamentally just a visual for the same argument). Perhaps we should move this thread to the talk page? -- Rick Block (talk) 04:24, 26 April 2008 (UTC)
(moved) That works too. As long as the phrase "Venn diagram" goes away, and content doesn't get repeated. Melchoir (talk) 06:35, 26 April 2008 (UTC)
Actually, the initial explanation offered by Devlin [5] matches the text in the Venn diagram section. How about if we include both "Combining doors" and "Venn diagram" in a new section discussing the difference between unconditional approaches (which these are) and conditional approaches (which are more rare, but mathematically more sound)? -- Rick Block (talk) 14:26, 27 April 2008 (UTC)
Well either way you're doing the actual work <grin>, and you have a better idea of what will succeed. All I can see is a mess to clean up, not the ideal result. Go for it! Melchoir (talk) 20:06, 27 April 2008 (UTC)

[edit] Remaining FAR issues

I believe the remaining FAR issues are:

  • "Solution"
The first solution and large figure use a distinguishable goat analysis that is unsourced and (as far as I can tell) uniquely presented here. In the sources I've found, a solution of this form varies the car location and keeps the player's choice as Door 1 (matching the problem statement). There is a draft of a solution in this form at Monty Hall problem/draft#Solution I've suggested including before that could be sourced to numerous references, but there has not been what I'd call consensus to switch to this version.
  • "Why the probability is 2/3"
Comparing the standard version and the "forgetful Monty" version is essentially not sourced.
  • "Venn diagram"
Still tagged as disputed since it presents an unsourced solution. Per comments above, Devlin could be a source and the section could be part of a new section about the difference between conditional and unconditional solutions.
  • "Two players"
I have been unable to find a source for this, other than http://www.daviddarling.info/encyclopedia/M/Monty_Hall_problem.html (and I have a suspicion that this is based on some version of the Wikipedia article rather than the other way around).
This section was apparently originally added by user:AxelBoldt and he doesn't remember where it came from (see user talk:AxelBoldt#Monty Hall problem). I've deleted it with this edit (it is recoverable from the history). -- Rick Block (talk) 14:23, 28 April 2008 (UTC)
  • Length
The original list of concerns cites WP:FACR #4 (appropriate length). I'm not sure what to do about this.

-- Rick Block (talk) 18:38, 27 April 2008 (UTC)

I for one don't think the article is too long in terms of scope and detail of ideas, or in terms of the bare amount of text. It probably feels long due to the space taken by tables, formulas, and diagrams, and due to repetition of very similar explanations. Melchoir (talk) 20:11, 27 April 2008 (UTC)
Is there a way to prevent the math tags from using a ginormous font? 67.130.129.135 (talk) 18:34, 28 April 2008 (UTC)
Not that I know of, see Help:Displaying a formula. We could think about not using math tags. -- Rick Block (talk) 00:37, 29 April 2008 (UTC)
The "Bayesian analysis" section could be shortened by deleting all the |I\,s. It would be easier to read, and a couple of the displays could be rewrapped. Melchoir (talk) 01:28, 29 April 2008 (UTC)
Yikes! Please don't. IMHO the main value of this article (other than as a large display of trivia :-) is pedagogical. As someone who time and again has had to purge naive/outdated/wrong intuitions about probability out of young minds, I found that having to always write the conditioning symbols is a quite useful exercise: a reminder that there's _always_ something conditioning your P's that, unless you account properly for it, is coming back to bite you in the end. Besides, the reduction in length is likely to be minimal, and the gain in readability is debatable.The Glopk (talk) 15:51, 29 April 2008 (UTC)

[edit] Bayesian analysis in an appendix?

Well, it looks like the section with a Bayesian analysis has been pushed into what is visually an appendix. Perhaps we should make that explicit, rename the section "Appendix: Bayesian Analysis" and refer to it as such. Or just make it into another article and reference it from here. Any thoughts/suggestions? When I re-wrote that section last year I did it because - it seemed to me - this kind of analysis was the only one that did not require any special jumps of intuitive imagination, visual or otherwise, but just a willingness to write down carefully the hypotheses and follow a couple of general rules. In fact, I find it remarkable that, in the recent flurry of activity and debate about different formulations of the problem, no commenter has questioned the validity of that section, or offered suggestions to improve the clarity of its presentation.The Glopk (talk) 16:09, 29 April 2008 (UTC)

I boldly did this thinking it would not be overly controversial (note there is a link to it from the "Solution" section - which I think is where it logically belongs). I thought about renaming it "appendix" or moving it to a separate article, but ended up just moving it. I am not aware of a standard way for articles to include appendices (neither WP:MOS or WP:LAYOUT mention this possibility), or of any articles that have an appendix. The point is not that it is of lesser importance than anything else in the article, but it is considerably more technical and therefore less accessible to general readers. I suspect the observation about it not being edited much indicates most people don't read it - and, for those who do, it inherently reflects a conditional approach so the whole conditional/unconditional argument doesn't apply. -- Rick Block (talk) 17:00, 29 April 2008 (UTC)

[edit] #4 Appropriate length

I have a few suggestions but am not so bold as to commit triage without consensus:

  • Quantum version - This digression is notable and worthy of mention, but could it be trimmed to 50 words?
  • Sequential doors - This digression is notable and worthy of mention, but could it be trimmed to 50 words? It is surely not worth 150 words to say Deal or No Deal is different.
  • Venn diagram - Delete! Yes, a picture is worth a thousand words, but we didn’t need another thousand words. As an elaboration on "In the 2/3 case where the player initially chooses a goat, the host must reveal the other goat making the only remaining door the one with the car," I think Combining doors covers the point better, and is better sourced.
  • Solution - This runs on a bit. Could the final paragraph and image dealing with decision trees be moved (perhaps to Aides to understanding), or at least sub-headed?

By the way, I really like the work that has been done on Sources of confusion. This is very much better than before the FAR. 67.130.129.135 (talk) 23:16, 29 April 2008 (UTC)

Briefer is better, but 50 words? Hmmm. I'd like keep the images from the "Venn diagram" section, maybe in "Combining doors". The decision tree bit in the "Solution" section is (according to sources like Morgan et al.) the only real (i.e. conditional) type of solution. I'd prefer to keep this in the "Solution" section. -- Rick Block (talk) 01:39, 30 April 2008 (UTC)
Yeah, put the Venn diagrams in the "combining doors" section, as it is a diagram of that concept anyhow.--Father Goose (talk) 06:08, 30 April 2008 (UTC)
Sounds good to me 67.130.129.135 (talk) 00:34, 1 May 2008 (UTC)
Combined, preserving the diagrams. -- Rick Block (talk) 03:17, 1 May 2008 (UTC)

Does it seem to anyone else that the "Sequential doors" variant discusses the same issue as the "Increasing the number of doors" section? I'd suggest deleting one or the other of these. Warren Dew (talk) 05:29, 1 May 2008 (UTC)

These are certainly related, but they are actually two different ways to generalize to N doors. The "Increasing ..." section is a 3-step game - choose 1 of N, host opens N-2, switch or not - while "Sequential doors" is a 2(N-2)+1 step game - choose 1, host opens 1, optionally switch to any other, host opens 1, <repeat until only two left>, switch or not. They could perhaps be merged into a single section but they serve two different purposes. Exploding the number of doors from 3 to 1000000 lets some people grok the problem better. The sequential variant is truly an N-door variant. -- Rick Block (talk) 14:04, 1 May 2008 (UTC)

[edit] Remaining FAR issues, again

Of the issues from above I think we're down to "Solution" and "Why the probability is 2/3". My inclination is to delete "Why the probability is 2/3" and replace the first section of "Solution" with the version at Monty Hall problem/draft, but I know both of these would be controversial. Other opinions, specifically other suggestions for how to proceed? -- Rick Block (talk) 04:13, 1 May 2008 (UTC)

I'd be in favor of replacing the solution section, especially as I prefer the aesthetics of the modified diagram. I think that for some people, the "why the probability is 2/3" section would be useful, though. I note that the alternative it presents is the same as the third example under "other host behaviors" - does anyone have a copy of Vos Savant's book, and could it serve as a source? Warren Dew (talk) 05:35, 1 May 2008 (UTC)

I have vos Savant's book (and nearly all of the other references). The closest I've seen to what we have in the "Why ..." section is [6] (which is not currently referenced). This site was proposed above, but rejected by Melchoir as not being a reliable source. I don't have the 1993 Barbeau article from College Math Journal v24 (I can chase this down this weekend). -- Rick Block (talk) 14:31, 1 May 2008 (UTC)

Perhaps it would help to explain why I think the "Why the probability is 2/3" section is helpful. In my opinion, it is helpful because the table is a succinct visual representation of the six equal probability paths, and illustrates why switching is better in four of them, in a way that can be seen at a glance. The other diagrams don't do this at present - the main diagram because it doesn't label probabilities in the second column, and the decision tree diagram because the visual representation shows four outcomes with about the same visual size, which tends to obfuscate their weights. Thinking about this, I think that revising the main diagram to add probabilities in the final column - "probability 1/6" by each of the two small pictures, and "probability 1/3" by each of the big pictures - would obviate the need for the extra "Why the probability is 2/3" section. In the proposed version of the graphic, the "probability 1/3" text would have to be added back to the three pictures in the first column, too. Does that make any sense? Warren Dew (talk) 19:49, 2 May 2008 (UTC)

Perhaps I should add that my primary objection to the "Why the probability is 2/3" section is the title - the whole article is about why the probability is 2/3! Warren Dew (talk) 19:49, 2 May 2008 (UTC)
We could change the section title to "Why the probability is 2/3 and not 1/2". Most of the other sections explain, in various ways, why it's 2/3, but don't detail the specific constraint of the game, and its effect, that causes the odds of winning by switching to be other than a toss-up.
I'd like to suggest that where this section is concerned, it's a good candidate for ignoring all rules -- in this case, FAR's rules. The information in it is uncontroversial; I've seen no one, neither laymen nor mathematicians, contend that it is wrong. It provides a fundamental explanation of the source of the underlying paradox that none of the other sections do. To remove it in service of FAR's criteria would produce a poorer article -- clearly not the goal of FAR or Wikipedia in general. Even if we can only manage to source it with the UCSD page, that source is not necessarily an unreliable source; it doesn't have the pedigree of many of the other sources in the article, but as part of the course material of a mathematics course (Math 187 - Intro to Cryptology) from a major university, I consider it sufficient affirmation of information not in dispute anyway. I note that we are already using other course materials (Williams 2004) as a reference for other parts of the article.--Father Goose (talk) 21:11, 2 May 2008 (UTC)

[edit] Car=Win, Goat=Loss?

Sorry if this has been brought up before, but the problem automatically assumes that the player wants to select the car rather than a goat. What if the player prefers a goat? Applejuicefool (talk) 19:34, 1 May 2008 (UTC)

Yes, it is brought up once or twice a year. It is generally accepted that, even for readers who are unfamiliar with the original game show, this goes without saying. After all, if you really prefer a goat you would swith to the door Monty has opened. Not much of a riddle. 67.130.129.135 (talk) 20:42, 1 May 2008 (UTC)
If that's allowed. It's not intuitive that the opened door would necessarily still be an allowable choice. 69.94.176.215 (talk) 16:17, 2 May 2008 (UTC)
Ok, I was being silly about choosing the open door. But is it being seriously suggested that this needs to be stipulated explicitly? The metaphor is an American game show. If anyone is unsure of the values implicit therein they can refer to Let's Make a Deal (referenced in the first sentence of the lead) for an explanation. Use and treatment of animals as booby-prizes ("zonks" in the parlance of the game) is discussed at length in that article. 67.130.129.135 (talk) 19:14, 2 May 2008 (UTC)
True, I agree that LMAD certainly promoted valuation of the "prizes" over the "zonks". But a bunch of flashing lights, a bit of applause, and Monty's approval are not, to my way of thinking, the greatest inducements in the world. True, a car does have a significantly greater monetary value in western society than does a goat. And given the nature of game show contestants, it might seem that monetary value is paramount in their decision-making processes. But it's not inconceivable that a contestant might appear (or might have appeared!) on the program, perhaps with a political agenda, who might have preferred the goat. For such a player, it would be to his/her advantage not to switch. Applejuicefool (talk) 15:00, 7 May 2008 (UTC)
This was clarified in a previous version of the article. I've re-added a clarification in the "Problem" section. -- Rick Block (talk) 01:36, 8 May 2008 (UTC)

[edit] Mosteller's "Prisoner's dilemma" problem

I finally located my copy of Frederick Mosteller's Fifty Challenging Problems in Probability with Solutions. I had thought that it discussed the Monty Hall problem, but in fact it discusses a related (and mathematically equivalent) problem, which he calls "The Prisoner's Dilemma" (Problem #13) [but not to be confused with the game-theoretic Prisoner's dilemma, which is quite different].

The statement of the problem is simple (p. 4).

Three prisoners, A, B, and C, with apparently equally good records have applied for parole. The parole board has decided to release two of the three, and the prisoners know this but not which two. A warder friend of prisoner A knows which are to be released. Prisoner A realizes that it would be unethical to ask the warder if he, A, is to be released, but thinks of asking the name of one prisoner other than himself who is to be released. He thinks that before he asks, his chances of release are 2/3. He thinks that if the warder says, "B will be released," his own chances have now gone down to 1/2, because either A and B or B and C are to be released. And so A decides not to reduce his chances by asking. However, A is mistaken in his calculations. Explain.

This is mathematically equivalent to Monty Hall: Instead of choosing a door, it just happens that the prisoner knows the warder; and instead of opening a door without the prize, the warder would be asked to reveal the name of someone who will be released. But the math is the same.

Mosteller remarks in his solution (p. 28) that this is the one problem that most people have difficulty with (no surprise here!) His solution points out that A hasn't listed all the possible events properly, i.e., in statistical jargon he does not have "the correct sample space." Mosteller's version of the problem doesn't have the problem of host (warder) behavior that the Monty Hall problem has, since the prisoner has no reason to suppose that the warder friend would preferentially choose to reveal the name of B or C; and, the question asked is not "whether to switch doors," but rather "whether to ask my warder friend to reveal someone else's name."

I'm not sure how or whether we should use this in the article. One thing that could be done is to more clearly identify that the source of the confusion is basically because if misidentifying the sample space. It may be useful to list this in the references as an alternative version (or maybe create a section devoted to this version?) Any comments would be welcome. Bill Jefferys (talk) 23:26, 2 May 2008 (UTC)

I believe this is the "Three Prisoners Problem" to which there is a link in the "Similar Problems" section, though the description is slightly different. It might be worth listing as a variant on that page. —Preceding unsigned comment added by Warren Dew (talk • contribs) 00:59, 3 May 2008 (UTC)
There's also a link in the "Problem" section, and a description of Gardner's version in the "History" section. I think at most we'd want to add a mention of this version in the "History" section. -- Rick Block (talk) 16:40, 3 May 2008 (UTC)

[edit] Yet another suggestion for the solution section

In my perpetual quest for the perfect solution description, I've written up yet another version at User:Rick Block/MH solution. This version essentially collapses the decision tree into the large figure with the doors. Visual queues indicate both the relative probabilities of the various alternatives (column width) and the conditional probabilities given which door the host opens (different rows for each door the host opens). Comments? -- Rick Block (talk) 23:23, 4 May 2008 (UTC)

I like it. It's very clear, with a slight exception for the paragraph starting 'The reasoning above applies' - but I understand why that paragraph needs to be there. It doesn't need to be perfect to put it in, and I think it's much better than the present section. Warren Dew (talk) 01:37, 5 May 2008 (UTC)
That's not bad, though I don't see why you put "switch to middle" and "switch to right" results on different rows. Seems less clear than just putting it all on one row.--Father Goose (talk) 02:39, 7 May 2008 (UTC)
The point is to make the conditional solution visually accessible. Rotating the graphic 90 degrees so the rows are the equiprobable car location and the columns are the door the host opens doesn't work too well with an HTML table (it's way harder to constrain row height than column width - and I suspect column width is intuitively more meaningful than row height anyway). Note that this version keeps its columnar proportionality across browser window widths and largely independent of font size (forces horizontal scrolling if the window gets too narrow and makes the text multiline if the font gets very large - although at least in Firefox the table column widths break down as the font gets insanely large because "Switching" ends up taking more than 16% of the window width). I've tried other ways to show the conditional choices without making them separate rows and haven't found anything more satisfactory (background color is not good for blind users - although the whole thing is so visual I'm not sure a blind user can "grok" this from any image presented in a table).
BTW - feel free to edit this version. I put it in my user space only to avoid cluttering up article space with yet another version. -- Rick Block (talk) 03:43, 7 May 2008 (UTC)
I have slightly edited it - mostly trying to explain that the "unconditional" analysis is an answer to a question of statistics (since its conclusion applies to a population), whereas the "conditional" analysis answer the related question of decision (it applies to a particular player) . Of course, this makes sense if you believe that "unconditional" questions of probability may legitimately be asked, i.e. that "statistics" is something fundamentally different than probability theory. As a Bayesian, I don't see any such distinction, but am not going to start another flamewar :-) The Glopk (talk) 03:36, 8 May 2008 (UTC)
Hmmm. You've reordered things so that the "conditional" paragraph ("In the figure above, ...") comes before the explanation about the difference between "conditional" and "unconditional". This paragraph is meant to explain things in a conditional sort of way (given that the host has opened Door 3 [or Door 2], the probability of winning by switching is still 2/3). I'm not sure if this was simply an oversight or deliberate.
It was an oversight: I thought to reorder so that all comments about the figure would be closer to the figure itself, but I can see now that you meant that sentence to be exactly the "slightly different analysis" to apply to the "conditional" (decision) case. I was confused by the "regardless" clause, which is often used to suggests (conversationally, when discussing an inference) an intent to marginalize w.r.t. a prior. I have re-reordered - please let me know what you think of the current version.
Of course, "unconditional" is simply conditioned on fewer events (in this case omitting whatever we may learn from the host opening a door). I don't think anyone is arguing that "unconditional" in this case means anything other than "given the problem statement at the point the user has picked a door but before the host has opened a door."
Actually, and that was the main intent of my edit, I take "unconditional analysis" to mean "averaged over a population of random players and games" (in the language of orthodox statistics) or, equivalently, "marginalized w.r.t. a flat prior on the player's strategy, the host's strategy and the location of the car" (if one prefers to speak Bayesianese). In Bayesian terms, the "unconditional" problem is: "Compute the *sum* over S and D of P(W,S,D | H) = P(W | S,D,H) P(S | H) P(D | H) P(H)", where W == "Player wins", S == "Player switches", D == "Car is behind initially chosen door" and H == "Host's nehavior". It isn't that we condition on less things. Rather, we average over some of the conditions, using a uniform probability distribution for both P(S|H) and P(D|H). So it is a fundamentally different question than the "conditional" (i.e. decision) one, which is: "Compute the maximum over S of P(W | S,D,H) P(S | H) P(D | H) P(H), given P(S|H), P(D|H) and P(H)" (i.e. one particular summand in the previous summation).
It would certainly be lovely if this could be phrased in a way that was both mathematically precise and comprehensible to normal humans (not that mathematicians are abnormal - oh, who am I kidding, of course they are). In any event, or perhaps I should say unconditionally, there are some stylistic issues with your revisions. I'll try to address these and move the "conditional" paragraph where I think it belongs. -- Rick Block (talk) 04:38, 8 May 2008 (UTC)
I'll be happy to help. My $0.02 suggestion is that a lay person is likely to understand intuitively the difference between "what's better for the people" and "what's better for me" , i.e. the difference between between a statistics and a decision, at that maps rather well to what we have called "unconditional" and "conditional" analyses.The Glopk (talk) 16:53, 8 May 2008 (UTC)
Father Goose, the extra row was puzzling to me too at first, but reading through the text explained it - as Rick mentioned, it has to do with the issue of conditional probability. Warren Dew (talk) 16:51, 8 May 2008 (UTC)
Does everyone think the alternative is better than what's there now, even if it's not perfect yet? If so, do we want to go ahead and put it on the main page, and continue to clean it up there? Warren Dew (talk) 16:51, 8 May 2008 (UTC)
I've replaced the Solution section with this draft. A couple of notes about this:
  • The images now have alternate text, which makes the figure accessible if using a screen reader like JAWS (although the rows don't have row headers, which would help).
  • This version drops the decision tree that at one point was in a separate section (at this point, it's simply gone). Dropping the tree also orphans the Grinstead and Snell reference.
I suggest we find a place to put the decision tree (possibly integrated into the last paragraph in "Solution"). -- Rick Block (talk) 20:25, 11 May 2008 (UTC)
I'd combine the two "switching" rows together. The explanation for why they are on separate rows comes three paragraphs later and even after reading it, what each row signifies still isn't clear. Unless it can be made immediately, visually clear what their being on separate rows signifies (possibly with a caption), I wouldn't separate them: it adds a confusing, unexplained element to what is otherwise a clear diagram.--Father Goose (talk) 20:52, 11 May 2008 (UTC)

[edit] 21 (2008 film)

I know this may be controversial, but it might be noted that this problem, called the "Game host problem" played a prominent role in the movie 21 (2008 film). Young mathematican hunk Sturgess(?) comes up with the solution to MIT prof Kevin Spacey's statement of the problem, and thus goes on to get the adventure, girl, and money (not just a car!).

I wouldn't include this, however, unless it was stated that Spacey mis-states the problem. He says that the game show host is trying to fool the contestant (see the section on alternative host behaviors), so Sturgess's solution was actually wrong. This brings up a potential problem of WP:OR, unless a mathematician-movie reviewer has actually written something on this in a reliable source. Has anybody reviewed that literature?

Smallbones (talk) 13:52, 5 May 2008 (UTC)

The sentence The Monty Hall problem (though it was called "the game show host problem" in the film) appears in the film 21, in which the main character, Ben, correctly answers the question in his MIT college math course. already appears in the "History" section. Are you questioning the word correctly in this sentence? How about if we just delete the details? I don't think details of how it's stated in the film are relevant for this article (although it might be an interesting tidbit for the article about the film). -- Rick Block (talk) 14:14, 5 May 2008 (UTC)
I've deleted the details and added a reference to Andy Bloch's (an MIT blackjack team emeritus) review of the film which includes a brief discussion that stops just short of saying they blew it. -- Rick Block (talk) 18:11, 6 May 2008 (UTC)

[edit] Alternative Explanation

what's about the following explanation: it requires two small modification:

  1. the game is not played only once but say 9 times (this is to illustrate probability as frequencies).
  2. the player is not alone but accompanied by his brother (this is to illustrate the benefit of the change).

Now the player sticks to his decision independent of what the host says and therefore he will win 3 of 9 cars. The game is played 9 times so there are 6 cars left and The players brother will receive them. So instead of a 3 of 9 (1/3) he had a 6 of 9 (2/3) change. Oub (talk) 19:59, 12 May 2008 (UTC):

It's a nice explanation, but it shouldn't be included in the article unless you can find a source for it, in accordance with Wikipedia:verifiability.Warren Dew (talk) 15:24, 13 May 2008 (UTC)

[edit] Passed FAR

Warm thanks to Rick Block and congratulations to all who contributed to improving this article during featured article review. 67.130.129.135 (talk) 18:28, 19 May 2008 (UTC)

[edit] An Excel spreadsheet to demonstrate the problem

Hi,

I found this article really interesting. Thanks to everyone who contributed.

I love maths, and read a lot of the "popular" books, but to be honest a lot of it passes over my head!

But I so enjoyed this article, I wrote an Excel spreadsheet to prove it to doubters like me :)

I'm new to all this, so apologies in advance, but I don't think I'm allowed to upload it.

Is someone able to upload it for me or tell me how to (without making 10 (unneccessary) edits?).

I can email a copy if anyone wants to check it out.

Regards,

Mathalogical (talk) 20:45, 2 June 2008 (UTC)

For security reasons Wikipedia doesn't support very many file types, see Help:Images and other uploaded files#Uploading non-image files - there simply isn't any way to upload a spreadsheet here. You may be interested in the page at the Wikibooks project with source code for simulators in various languages, see wikibooks:Algorithm implementation/Simulation/Monty Hall problem, but you won't be able to upload a spreadsheet there either. A while ago a user added Excel formulas for a simulation on this talk page, now archived at Talk:Monty Hall problem/Archive4#MS Excel Experiment proves this true KeyStroke 17:48, 17 April 2006 (UTC) (note his initial formulas had a bug). -- Rick Block (talk) 00:51, 3 June 2008 (UTC)

Thanks for clarifying that for me - and pointing me to the archives. Mathalogical (talk) 05:45, 3 June 2008 (UTC)