Talk:Monty Hall problem

Please note: The conclusions of this article have been confirmed by experiment

There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment. If you find the article's arguments unconvincing, then please feel free to use the space below to discuss improvements.

Archives
Archive 1: July 2004 Archive 2: July 2004 – July 2005 Archive 3: August 2005 – February 2006 Archive 4: March 2006 – November 2006 Archive 5: November 2006 – April 2007

Archiving notes

I've moved the existing talk page to Talk:Monty Hall problem/Archive2, so the edit history is now with the archive page. I've copied back a few recent threads. Older discussions are in Talk:Monty Hall problem/Archive1. Hope this helps, Wile E. Heresiarch 15:28, 28 July 2005 (UTC)

I've done similarly to produce Talk:Monty Hall problem/Archive3. In keeping with Wile E. Heresiarch I moved the page so the edit history is with the archive page, and copied back the current (March 2006) discussions. _└ ^UkPaolo/_talk^┐ 13:04, 10 March 2006 (UTC)

Likewise for Talk:Monty Hall problem/Archive4. Gzkn 06:15, 27 December 2006 (UTC)

Talk:Monty Hall problem/Archive5 archived via pagemove.--Father Goose 05:49, 22 May 2007 (UTC)

The new images

I'm all for racial harmony and all, but why do the images include a bald black man as the player? Could we be using a more abstract, symbolic face? (I know I know, I'm a racist because I wouldn't have noticed if it were a white guy. Granted, but I think the question is still valid.) --P3d0 02:00, 3 May 2007 (UTC)

Cause the player is Curly Neal. Perhaps I should've gone with Mr. T?--Father Goose 05:11, 3 May 2007 (UTC)

Deal or No Deal

The two mentions of Deal or No Deal contradict one another: under 'Sequential Doors' is an explanation of the critical difference between the NBC Prime Time SMASH HIT and the Monty Hall problem as stated, but under the 'History of the problem' heading, Deal or No Deal is essentially considered to be a minor variant with the same sorts of conclusions. The numbers directly contradict one another as well. Though I'm fairly positive the first description is accurate, I'll refrain from editing for now so people can argue and yell and scream and then somebody smarter than myself can fix it. I also kinda like the fact that both are in this article, as it seems the Monty Hall problem's best contribution to society is watching people flailing about entirely confused yet certain that they're right in the face of contradicting evidence. 66.188.124.133 17:32, 16 May 2007 (UTC)

I think there might be a valid place in the article for a section comparing and contrasting Deal or No Deal with the Monty Hall problem. They differ (and DOND was incorrectly analyzed in the section that was recently deleted) because the cases removed from play are chosen by the contestant, not the host of the show, and are chosen without knowledge of which case holds the big prize; thus, it is possible (and common) for the highest prize to be eliminated through the case-opening during the game. Thus, if it gets down to the last two cases, and one of them still holds the million dollars, then the odds are 50-50 for the contestant whether he/she keeps the original case or swaps it. *Dan T.* 19:49, 17 May 2007 (UTC)

DEBUNKED!!

The problem with this is that the 3 scenarios are actually 4.

For the first scenario where the person picks the car it is listed as host showing "either Goat A or B". Actually these are two different scenarios:

Scenario 1: Contestant picks car, Host shows Goat A, Contestant switches, Contestant LOSES

Scenario 2: Contestant picks car, Host shows Goat B, Contestant switches, Contestant LOSES

Scenario 3: Contestant picks Goat A, Host must show Goat B, Contestant switches, Contestant WINS

Scenario 4: Contestant picks Goat B, Host must show Goat A, contestant switches, Contestant WINS

You can see that switching yields the expected 50% success.

It is rather alarming to me that this is missed by experts. I believe Quantum computing falls into this same "smoke and mirror" science. —The preceding unsigned comment was added by 143.182.124.2 (talk • contribs).

Mindful of the note at the top of the page, the key is that scenarios 1 or 2 occur 1/3 of the time, while scenario 3 and 4 occur 2/3 of the time. The decision tree from the article lays out all four options. As an aside, however, the talk page of an article is not really the place to discuss the topic of the article, only to discuss improvements. A better place to ask questions is reference desk. Thanks, --TeaDrinker 19:33, 5 June 2007 (UTC)

This objection has been brought up twice recently, so I think it's worthwhile to specifically note and address it in the article. I've made an attempt to do so.--Father Goose 20:18, 5 June 2007 (UTC)

I've reverted this, tweaking a single word in the original that may help clarify. The detailed decision tree analysis is already presented not too far away. -- Rick Block (talk) 02:45, 6 June 2007 (UTC)

Now that I think about it, a different, possibly better way to address it is to change my diagram so that there's two panels in the upper right corner; switching to Goat A or switching to Goat B. I'll see if I can pull this off without making it ugly.--Father Goose 05:54, 6 June 2007 (UTC)

Perhaps we need something in the article to directly address this misconception that "there are n possibilities, therefore each has a probability of 1 / n ." Maelin (Talk | Contribs) 06:33, 6 June 2007 (UTC)

I was similarly confused when I first read this article, until I wrote a perl script to simulate the problem 1000 times, and was surprised to discover that switching really DOES win 2/3 of the time. Then I made the image I put up under "Not Switching" to help me wrap my head around it. Scenario 1 and scenario 2 ARE different scenarios, but they're the same choice, and the problem concerns the possible choices, not the possible scenarios; it's not important whether the host reveals goat A or goat B, what's important is that the host reveals "an option that is both a goat and not the player's choice" (as I hope is illustrated well in the image I added), thus scenarios 1 and 2 represent only one of the three possible choices. Luvcraft 01:18, 24 August 2007 (UTC)

Simulation will always produce the Monty Hall Fallacy because the simulation program adopts two sequential steps. The first step will give the fixed probability of 1/3 for the choice of each door. Then, in the second step, when you choose the car in the first step, the program says your changing fails. The probability is 1/3 as given in the first step. The program simply ignores two exclusive cases and considers them as one case because the first step already decides it. In the second step, when you choose the goat in the first step, the program says your changing succeeds. The probability is 2/3 because the program adds up two exclusive cases of wrong choices into one superficial event of failure (choosing goat) in the first step. Thus simulation simply gets the Monty Hall Fallacy. Simulation gets only what you want to see. However, four exclusive cases do not occur at the same time. You have to, therefore, treat four cases separately. The real probability of failure of changing is 1/2(2/4), and the probability of success is 1/2(2/4). Do not assert simulation is smarter than the program-writer. (Mingull Jeung) —Preceding unsigned comment added by 203.253.35.156 (talk) 07:13, 3 December 2007 (UTC)

Simulation works. The simulation doesn't care about scenarios. It simply plays out the Monty Hall game: It randomly places the prize, randomly picks a door, reveals a non-winning door and then switches. The problem with your four scenarios is that you assume all four are equally likely, which is not the case.

Scenario 1: Contestant picks car, Host shows Goat A: probability 1/6

Scenario 2: Contestant picks car, Host shows Goat B: probability 1/6

Scenario 3: Contestant picks Goat A, Host must show Goat B: probability 1/3

Scenario 4: Contestant picks Goat B, Host must show Goat A: probability 1/3 --RLent (talk) 16:36, 20 December 2007 (UTC)

Simulate it using a friend and a few coins in real life then. mattbuck 08:24, 3 December 2007 (UTC)

Mingull Jeung, you're wrong. You either haven't thought through the problem, haven't read the article, or are a troll. If either of the first two are the issue, please tell us how we can help. If the latter, begone. Jouster  (whisper) 12:49, 3 December 2007 (UTC)

Awww, now that's not nice. No reason to assume he's done anything but misunderstood the problem. But it is a good idea to actually read all the parts of the article that are intelligible before claiming "it's wrong".--Father Goose 17:03, 3 December 2007 (UTC)

Thanks! When I wrote the above, I did not read the entire article. Anyhow, will anybody make it clear for me? In Monty Hall setting, you have the chance of 2/3 to pick the door that has a goat. When you picked one of the goat doors, however, your real chance (fact) to pick the door was 1/3, not 2/3. (You never picked two doors.) Peoples think they picked two goat doors (2/3) even though they actually did only one (1/3). After you picked a goat door, you had only one chance (not two) to switch. The other chance was gone by the host. Mathematical expression is just a tool to represent your thinking like any other language. Mathematical expression itself does not think nor understand. Mingull. —Preceding unsigned comment added by 203.253.35.156 (talk) 12:27, 4 December 2007 (UTC)

Does the decision tree in in the "Decision tree" section help? Say you've picked Goat A (with, as you say, a 1/3 chance). The host now must reveal Goat B. You can't tell if you're in this position or any of the others, but if you assume you're in this position you now have a 0% chance of winning by staying and a 100% chance of winning by switching. You have a 1/3 chance of being n this position, which means through this fork your chances of winning (by switching) are 1/3 and your chances of winning by staying are 0. The other "goat" fork contributes similar chances to the overall picture, while the fork where you picked the car with a 1/3 chance contributes a 1/3 chance of winning by staying (and a 0% chance of winning by switching). You can't tell which fork you're on, so the total probability is the sum of these, i.e. 2/3 chance of winning by switching and 1/3 chance of winning by staying. As a check to make sure we've accounted for all the possibilities, these probabilities should add up to 1 (and they do) because you must either win by staying or win by switching. -- Rick Block (talk) 15:50, 4 December 2007 (UTC)

Thank all of you. I met the Monty Hall problem in a book. The author gives the answer without explanation in his book. At that time I thought my first explanation above, and asked his explanation. He said that switching means giving one more chance to open a door after the host's removal of a goat door, i.e., by switching we have chance to open two doors. I argued that we never open two doors until we open the first chosen door then switch. We just open a door in any situations. He introduced this article, and I was here. I assumed this article would give the same explanation. After reading the introductory part of the article I moved to this discussion section and wrote my first explanation. As Father Goose said that was not intelligible. After reading nice persons' (Mattbuck, Jouster and Father Goose) remarks, I started my defense without reading the article because I had to do a business and didn't have time to read. While doing the business I considered deeply. Finally I got the right answer. However, I would like to disclose how I was unintelligent and what the mistake in my thinking was. And also I wanted to show how people can give a bit awkward explanation after reading the Monty Hall article. So I wrote my question, the second words, intentionally. I am sorry for the bothering you. My version of explanation is "the host can remove the present, but the host cannot remove the past." In the first step choosing a goat door is twice as probable as choosing the car door. After the host's removal of the other goat door (B), the person who has chosen a goat door (A) loses a chance to pick the other goat door (B) by switching, and always wins a car by switching. If the host doesn't remove the other goat door (B), the person will have half a chance to win by switching. I don't think this explanation is mathematical, but it is easily understood. This explanation is similar with Rick Block's explanation, but less mathematical. I suggest using the term "Monty Hall Fallacy" to describe the fallacy that people think exclusive cases should have even chance although they do not. Mingull

Re: new B+ rating

Might I ask for more specificity as to what this article needs in order to become an FA? It's not clear how to act upon the suggestions left in the Mathematics rating box.--Father Goose 02:31, 23 June 2007 (UTC)

Well, the article is a Featured article, so I think pretty much by definition it has to have a FA rating. -- Rick Block (talk) 02:41, 23 June 2007 (UTC)

The comments are at Talk:Monty Hall problem/Comments. -- Rick Block (talk) 03:33, 23 June 2007 (UTC)

Yes, that's what I was referring to. It's not clear how to act upon them. "Might be too long" is extremely vague, and the other comment seems to be demanding redundancy.--Father Goose 06:19, 23 June 2007 (UTC)

Many thanks to Rick for his constructive response to my concerns, and for signing the current rating with a good comment. Geometry guy 21:14, 23 June 2007 (UTC)

From my talk page, but it makes more sense to comment here:

Hello and thanks for the curiosity and comments. Yes, I did see the star, but on the grading scheme, FA-Class is for articles which have "received featured article status after peer review, and meet the current criteria for featured articles". I came to this article because the maths rating was not signed and dated, so I looked at the article and read the recent FAR. In my view the latter got side-tracked by irrelevant inline citation arguments and I don't think the article should have passed: in fact I doubt it would survive a good article review at the moment. Now I can't sign a rating I don't agree with, so I changed it, and added my comments to the rating. I do actually quite like the article (it is great for the portal, for example), and the most obvious flaw is pretty easy to fix: see WP:LEAD.

However, this is just my opinion. If someone else believes that the article does currently meet the FA criteria, they are of course free to uprate the maths rating and replace my comments and signature by their own. Geometry guy 10:47, 23 June 2007 (UTC)

The current lead is a teaser, which is great for a magazine, but not for an encyclopedia. I'm sorry that my other concerns are vague, but it seems that other editors believe that this article meets FA standards, in which case please will someone replace my comment and signature by their own. Thank you! Geometry guy 10:56, 23 June 2007 (UTC)

The lead used to include the solution, but there was an objection to this (to the extent that for a while there was a "spoiler" warning). I've been viewing the current version as a exception to WP:LEAD based on the desire of some readers to not have the solution displayed in the lead. The recent FAR was apparently OK with this. We could of course change it back. Any other opinions on this? -- Rick Block (talk) 17:00, 23 June 2007 (UTC)

Yes I noticed that in the FAR the view was expressed that information should be taken out of the lead: certainly detail should not go in the lead, but the lead is supposed to be able to stand alone; indeed I believe some fixed editions of WP will contain only the lead for some articles. As regards including the solution, I guess one has to seek guidance from WP:SPOILER. My view is that at least the problem should be summarized briefly. Geometry guy 17:41, 23 June 2007 (UTC)

Anyone have any objections restoring the lead to approximately this version? -- Rick Block (talk) 20:33, 23 June 2007 (UTC)

My feeling is that this is the kind of precise formulation of the problem that SandyGeorgia wanted removed from the lead at the last FAR, and that it would be better to summarize (approximately) the problem rather than state it in full. Geometry guy 21:21, 23 June 2007 (UTC)

At one point there was a summary statement of the problem, but it was constantly wordsmithed to include this or that constraint on the host's behavior or to avoid this or that ambiguity. Including the problem statement as a quote was a device to end the wordsmithing. The other problem with summarizing the problem statement is that the "correct" solution depends on the detailed constraints assumed or placed on the host's behavior. To conclude switching results in a 2/3 chance requires the host to always offer the choice to switch, to have knowledge of what's behind each door, and to always open a "goat door". Without including these constraints, any solution offered in the lead can be argued to be incorrect (and people do argue this, believe me). If anyone has a suggestion for an unambiguous, but summarized, statement of the problem please speak up. Short of that, quoting the Parade version seems to me like a reasonable approach (it's only 72 words). -- Rick Block (talk) 22:29, 23 June 2007 (UTC)

I think moving the Parade version into the lead is a good approach. It doesn't look like the article would need to be rewritten much at all to move it in there and still present the information in an almost identical way.--Father Goose 02:08, 24 June 2007 (UTC)

Here's my suggestion. -- Rick Block (talk) 02:55, 24 June 2007 (UTC)

---

In search of a new car, the player picks door 1. The game host then opens door 3 to reveal a goat and offers to let the player pick door 2 instead of door 1.

The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. A widely known statement of the Monty Hall problem appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (vos Savant 1990):

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

Since there is no way for the player to know which of the two unopened doors is the winning door, many people assume that each door has an equal probability and conclude that switching does not matter. However, as long as the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door. As there is only one other unopened door, the probability that this door conceals the car must be 2/3.

The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive. For example, when Marilyn vos Savant offered the problem and the correct solution in her Ask Marilyn column in Parade, approximately 10,000 readers, including several hundred mathematics professors, wrote to tell her she was wrong. Some of the controversy was because the Parade statement of the problem fails to fully specify the host's behavior and is thus technically ambiguous. However, even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.

---

That looks very good, and it's got an excellent concise explanation of the solution. That might run into the "spoiler" objection again, but I'm not sure this article should ever be considered a suitable place to first hear (and try to solve) the riddle anyway.--Father Goose 05:03, 24 June 2007 (UTC)

I agree. I think if anyone complains, they can be pointed towards WP:SPOILER, which states "Concerns about spoilers should play no role in decisions about the structure or content of an article, including the article's lead section", and WP:LEAD, which states "The lead should not "tease" the reader by hinting at but not explaining important facts that will appear later in the article." Taken together, I would say that is pretty decisive! Geometry guy 12:02, 24 June 2007 (UTC)

Not to mention: "Spoilers and spoiler warnings should not be used in articles on non-fictional subjects" from WP:SPOILER. --P3d0 19:36, 11 July 2007 (UTC)

I'm sorry guys, (especially Rick Block), but the lead section contains a 'fundamental error' -- the following reasoning in the current lead is not true:

"However, as long as the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door. As there is only one other unopened door, the probability that this door conceals the car must be 2/3."

Here is the truth. The conditional probability that the car is behind the player's initially chosen door may be between 0 and 1/2, depending on the strategy of the host with regard to which door to open when the car is behind the initially chosen door. Therefore, the complementary conditional probability that the unopened door conceals the car is between 1 and 1/2. Therefore, it is better for the player to switch regardless of the host's strategy. —Preceding unsigned comment added by 70.137.163.193 (talk) 05:54, 12 February 2008 (UTC)

About the length of the article

I found another exchange in the FAR which resonates a little bit with my reaction to the article and may help clarify the vaguer part of my comment. I quote:

• Comment: I am of the opinion that the article could be cut in half, and its quality would dramatically improve. I am aware that not everyone shares this opinion. But in its current state, I find it overwhelming, directionless, and confusing. - Abscissa 02:49, 10 January 2007 (UTC)
  • Abscissa has a point. The article contains several explanations of the same result, and many of them could be struck. I suspect, however, that one will make sense to one reader, and another to another reader, so I'm not sure which ones, if any, to remove. This should be taken to the article talk page. Septentrionalis PMAnderson 03:07, 10 January 2007 (UTC)

I wouldn't go as far as Abscissa, but it is not dissimilar from my reaction. Anyway, this exchange seems to have got lost in a sandwich between the arguments about inline citation that opened the review, and the impressive copyediting drive that ended it. I couldn't find it in the talk archive. Did anything come of it? Geometry guy 17:41, 23 June 2007 (UTC)

Would it be completely unreasonable to determine the pedagogy at work in each of the explanations, and make subpages for each one? Or, better yet, to select the consensus-best explanation of the problem, and relegate the others to an "alternate explanations"-type page? Jouster  (whisper) 18:27, 23 June 2007 (UTC)

I think the biggest problem is not article length but lack of structure. The "Aids to understanding" section lumps together too many things: fundamental explanations of the problem (Bayes, "Why the problem is 2/3", and the decision tree); explanations which aren't so much fundamental as alternative ways of viewing the problem; and an orphaned "Sources of confusion" section. Can we split it into "Fundamental explanations" and "Alternative ways of viewing the problem"? (Something like that.) "Sources of confusion" repeats a few things found elsewhere in the article; most of it can probably be merged with that stuff.

At an absolute minimum, "Combined doors" and "Venn diagram" can be merged together: they say nearly the same thing.

--Father Goose 02:40, 24 June 2007 (UTC)

I wrote that comment that was quoted above. I still hold that opinion. My brother and I were laughing at the "quantum version" of the problem. Who is the target audience for that? At least people have given up trying to convince us that this is a problem of game theory... we (editors of this article) hold a place amongst the lamest edit wars on Wikipedia. - Abscissa 19:30, 4 July 2007 (UTC)

I agree that there is a lot of redundancy/rehashing in this article. Maybe the problem statement should be changed, so that instead of "Goat A" and "Goat B" two of the doors lead to a "Stick" and a "Dead Horse". It would be more fun to read that way. But seriously, I think the most compelling and helpful explanations are (1) switching gives you the "best" (or maybe your pick) of what is behind the other two doors, and (2) assuming 100 doors instead of 3 makes the true relationship obvious. If it were up to me I would archive all the rest.--Canistota (talk) 06:37, 15 January 2008 (UTC)

A flaw in the variations?

I am just wondering if it is a flaw in one of the Variations  Other host behaviors  “The host does not know what lies behind the doors, and the player loses if the host reveals the car.”

The answer is allegedly “The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time.” Shouldn’t switching here also result in a 2/3 probability of victory (since it is purportedly the same result (so far) as in the original Monty Hall problem)? --85.164.95.143 14:15, 9 July 2007 (UTC)

We may safely discount what the host does here, since he's choosing a door at random. (Though I think that description might need some work, or be outright misleading. But I just woke up, so I could be out of my mind.) The host choosing a door at random is the key to the whole thing, really—if the host is running off of the same set of information we have (namely, none), then we have a straight 50/50 shot at picking the right door amongst the two that remain. We haven't "narrowed the probability field", if you will, at all through our (and the host's) earlier actions. Jouster  (whisper) 18:28, 9 July 2007 (UTC)

I believe as well there is a flaw here. Let's extend the problem to 100 doors the same way we reason the standard problem. The host luckily opens 98 goats in a row. But once this has happened, certainly you would switch, because even without the knowledge of the host, you only had a 1/100 chance of landing on the car in the first place, which has not changed. Your odds of WINNING this sort of game are of course lower because of the probability of the host opening the door to the car causing a loss. However, after he opens a goat, even if it is random, it does not affect the probability of you having picked the car in the first place, and thus the advantage of switching. Again, the 100 doors intuition here seems to support this argument. — Preceding unsigned comment added by 151.190.254.108 (talk • contribs)

The difference is whether the host acs knowingly or by random chance. In the 100 door version if the host knows then any player is still in the game after 98 doors have been opened and still has a 1/100 chance that the originally picked door hides the car. If the host doesn't know, 98/100 times the player loses (because the host randomly opens the door with the car). The player picked a random door. The host opened 98 random doors. Whether the player picked the car door is now a 50/50 chance, made so by the host's random choices. In the "host knows" case the only random event affecting whether the player wins the car is the player's initial choice. In the "random host" case, the host's actions affect the player's chance of winning as well. -- Rick Block (talk) 13:54, 17 July 2007 (UTC)

No, because there are only two base circumstances involved in the 100-door variation: you have a goat, or you have a car. All other chances are randomly distributed.

Let's try another way--suppose we ran this test as, "Contestant picks door #1, hosts opens doors #2-#99, switch or not?", 100 times, one time with a car in each of the 100 positions. 98 times out of 100, the game would fail because the host would reveal a car. One of the remaining two times, switching would lose, as the car is in position #1. The remaining time, switching would win, as the car is in the last position.

If the host has no knowledge of the doors' contents, it's all random. Jouster  (whisper) 14:00, 17 July 2007 (UTC)

Perhaps I'm confused as to what probability we are talking about. The probability of winning the game when the host doesn't know anything, with the rules making you lose if he reveals a car, is in fact 0.5 for the 3-door experiment. I was talking about the conditional probability of winning by switching given that you already have not lost from the host opening doors.

Let's say we run your example. What I am talking about is only that situation when the 98 doors the host opened revealed goats. It is in this case that I feel it is advantageous to switch doors, because your original probability of picking a goat was 0.99. The method by which the other possible goats were eliminated does not change the experiment at this point. It is only the overall probability of winning that is affected because you can lose before this situation even arises.

I started doing a little bit of testing with a random number generator. The numbers agree with what I felt, which leads me to think we are in fact talking about different points. You win the game about half the time, like you said. However, if we discount the games in which you lose because the host opens the door with the car, of the remaining times you win twice as much as you lose by switching. Think of this: if you believe that the overall probability of winning the game is 0.5, then of those 50% losses, some (turns out, half) must be as a result of the host picking the car, and the other reason is because you got to the point in the game where you switch doors, and you lost as a result of that switch. Overall, you win the game by switching 50% of the time, lose by host 25% of the time, and lose by switching 25% of the time. Once we are past the host selecting, by random, those conditional numbers support what I was saying. I'm guessing though, that we were arguing two different points. I do agree that the overall chance of winning the whole game is 0.5 — Preceding unsigned comment added by 151.190.254.108 (talk • contribs)

I'm somewhat-confused, but let me try this:

1. With n doors, no matter how many goat doors we reveal, or by what method, it has no effect on the probability that you originally picked a car.

   This is not true. The method does matter. If the method is an omniscient host is opening doors known (beforehand) to have goats behind them then the probability you originally picked a car does not change. If the host is randomly picking, then the chance that your door has a car (assuming the host doesn't reveal the car) is 1/m where m is the number of doors left.

2. (as an aside, to relate this to the original problem) If the host chooses doors containing only goats, #1 still applies, but the probability that the remaining door is a car increases for each goat he reveals.

   This is the only case in which #1 applies.

3. If the host chooses doors randomly, without foreknowledge, the apparently increased chance to have chosen a car door, in violation of #1, is an illusion due to the increased chance of the host having opened the car door, and thus you having lost.

   No, it's no illusion. You're left with a random choice of a randomly selected set of doors. At this point, all doors are equal (including the one you originally picked).

Try it with n=3, where you can exhaustively list the possibilities. Or, more simply, since you've correctly stipulated that the doors eliminated randomly do not affect the experiment in any way, try it with n=2. q.e.d. Jouster  (whisper) 19:07, 17 July 2007 (UTC)

So, with n=3, your original chance is 1/3. If the host randomly opens a door 1/3 of the time you now lose (because the car is revealed). In the other 2/3 of the time you now have a 1/2 chance that the door you originally picked hides the car. Switch or not, you'll win the car 1/2 the time. The total chance of winning is 0 for the 1/3 of the time you immediately lose, plus 1/2 * 2/3 from the other case, i.e. a net total of 1/3 - whether you switch or not. -- Rick Block (talk) 01:17, 18 July 2007 (UTC)

We're not disagreeing here. That said, you are incorrect in your response to point #1—the probability that you originally chose a winning door can never change. Jouster  (whisper) 20:19, 18 July 2007 (UTC)

Hmm. I think we might be disagreeing. The probability that you originally chose a winning door does not change (if by "originally chose" means at the point you chose the door), but the probability that the car is behind this door (the one you chose) does change if the host is opening random doors. See below. -- Rick Block (talk) 16:12, 19 July 2007 (UTC)

I'm thinking now I am mistaken because I read another type of explanation that I could not seem to refute, but your point #3 is still somewhat confusing because it seems to contradict #1, not show how #1 doesn't work.

Alright, I have been convinced. Perhaps my confusion can lead to something positive, like some clearer wording about why this is true. What I gather now is that when the host picks randomly, 1/2 of the time that a car is not revealed randomly, the reason is not that you got lucky picks by the host but that the car is behind the door you picked. I followed this by a tree diagram. The other explanation I found online was just a different wording of the one you used to refute my 100 door argument. Thanks!

Is the earlier explanation, in the section titled "Why the probability is 2/3", more clear to you? -- Rick Block (talk) 01:17, 18 July 2007 (UTC)

Okay, I've spent entirely too much brainjuice on this. It probably has one or more glaring errors. But it's the algebra-ized version of the way I view the problem in my head (which has more to do with colors than with symbols; go figure). That said, I'm not so good at this whole color→algebra thing, so please fix it up:

Probabilities in an n-door problem with d non-contestant-selected doors still closed

	You select a winning door	Your door is now a winner	Remaining Door(s) contain a Winner	You've already lost
Host picking randomly	${\frac {1}{n}}$	${\frac {1}{n}}$	${\frac {d}{n}}$	$1-{\cfrac {n-d+2}{n}}$
Host picking only losing doors	${\frac {1}{n}}$	${\frac {1}{n}}$	${\cfrac {n-d}{n}}$	$\varnothing$

Jouster  (whisper) 23:48, 18 July 2007 (UTC)

No matter how many doors have been opened, the sum of the probabilities of your chosen door and the other unopened doors must be 1 if you haven't lost yet (it's 0 if you have lost). In your formulas this is not true (for either row). The probabilities are actually the following (perhaps I should find a reference for this :), but I'm reasonably certain). -- Rick Block (talk) 16:12, 19 July 2007 (UTC)

Probabilities in an n-door problem with d non-contestant-selected doors still closed

	You originally select a winning door	Your door is now a winner	All d remaining Door(s) contain a Winner	Each d remaining door contains a winner	You've already lost
Host picking randomly (absolute)	${\frac {1}{n}}$	${\frac {1}{n}}$	${\frac {d}{n}}$	${\frac {1}{n}}$	${\cfrac {n-1-d}{n}}$
Host picking randomly (conditional)	${\frac {1}{n}}$	${\frac {1}{d+1}}$	${\frac {d}{d+1}}$	${\frac {1}{d+1}}$	${\cfrac {n-1-d}{n}}$
Host picking only losing doors	${\frac {1}{n}}$	${\frac {1}{n}}$	${\cfrac {n-1}{n}}$	${\cfrac {n-1}{n*d}}$	$\varnothing$

I think I see at least part of the problem. You're apparently thinking about absolute probabilities while I'm thinking about conditional probabilities (conditioned on not having lost yet). This doesn't affect the "host picks only losing doors" row, but does affect the other one (I agree 1/n and d/n are the absolute probabilities). -- Rick Block (talk) 16:40, 19 July 2007 (UTC)

Correct, I'm trying to keep it to absolutes. I think I ran out of brainpower to update my table, though. I really need to do math more often; my "math muscles" have atrophied. Jouster  (whisper) 19:21, 19 July 2007 (UTC)

So, each row (in your table) should sum to 1. I've added an absolute row (above). Note that the conditional probability is the probability you haven't lost (1 - probability you have lost, which is (d+1)/n) times the absolute probability. -- Rick Block (talk) 23:23, 19 July 2007 (UTC)

I'm not sure what the final word here is, but I think it's wrong. Anyway, the article is still in error, stating

Possible host behaviors in unspecified problem
Host behavior	Result
The host does not know what lies behind the doors, and the player loses if the host reveals the car.	The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time.

I claim, as did whoever initiated this discussion, that

1. If the host does not know where the prize is, and

2. The host randomly opens one of the two doors not chosen,

then given the above, the stategy of not switching doors yeilds wins with probablity 1/3, exactly as it did when the host knew where the prize was!

This can be shown in a number of ways. The most convincing, using Bayes' theorem, appears last.

1. Ad absurdum: If you believe that there's a difference between the cases where the host knows where the prize is or guesses where the prize is (tosses a coin), then give some thought to the interim possibilities: The host knows that the prize had an uneven probability of being placed, say 1/4, 1/4, 1/2. Work out the probability of winning if you don't change your choice (host opens the door least likely to show the prize). Now consider what happens if the host thought he knew the probability ditribution of the prize placement, but was wrong - they used a different distribution! How can the contestant's strategy be affected by all this?!

2. If the host knows nothing, we don't need him. State the game as follows: Contestant picks a door (may as well be at random), and now opens one of the other two doors at random. Given that the opened door has no prize, what's the chance that the initial door does? I claim 1/3. If you still think that the answer is 1/2, consider the following variant:

Contestant opens one door at random. It has no prize. Now the contestant picks one of the remaining doors. Clearly the chance of winning is 1/2. Do you really believe that the two situations yeild the same probability of winning?

3. Bayes' theorem: I use the notation from the article: In Bayesian terms, probabilities are associated to propositions, and express a degree of belief in their truth, subject to whatever background information happens to be known. For this problem the background is the set of game rules, and the propositions of interest are:

$C_{i}\,$ : The car is behind Door i, for i equal to 1, 2 or 3.

$H_{ij}\,$ : The host opens Door j after the player has picked Door i, for i and j equal to 1, 2 or 3.

For example, $C_{1}\,$ denotes the proposition the car is behind Door 1, and $H_{12}\,$ denotes the proposition the host opens Door 2 after the player has picked Door 1. Indicating the background information with $I\,$, the assumptions are formally stated as follows.

First, the car can be behind any door, and all doors are a priori equally likely to hide the car. In this context a priori means before the game is played, or before seeing the goat. Hence, the prior probability of a proposition $C_{i}\,$ is:

$P(C_{i}|I)\,={\frac {1}{3}}\$ .

Second, the host will pick a door from the remaining two at random. This rule determines the conditional probability of a proposition $H_{ij}\,$ subject to where the car is, i.e. conditioned on a proposition $C_{k}\,$. Specifically, it is:

$P(H_{ij}|C_{k},\,I)\,\,=\,{\begin{cases}\,\\\,\\\,\\\,\end{cases}}$	$\,0\,$		if i = j, (the host cannot open the door picked by the player)
	$\,1/2\,$		if i ≠ j and j = k, (the host can open a door with a car behind it)
	$\,1/2\,$		if i ≠j and j ≠ k, (the host can open a door without a car behind it)

The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors. Without loss of generality assume, by re-numbering the doors if necessary, that the player picks Door 1, and the host then opens Door 3, showing him or her a goat. In other words, the host makes proposition $H_{13}\,$ true.

The posterior probability of winning by not switching doors, subject to the game rules and $H_{13}\,$, is then $P(C_{1}|H_{13},\,I)$. Using Bayes' theorem this is expressed as:

$P(C_{1}|H_{13},\,I)={\frac {P(H_{13}|C_{1},\,I)\,P(C_{1}|I)}{P(H_{13}|I)}}$ .

By the assumptions stated above, the numerator of the right-hand side is:

$P(H_{13}|C_{1},\,I)\,P(C_{1}|I)={\frac {1}{2}}\times {\frac {1}{3}}={\frac {1}{6}}$ .

The normalizing constant at the denominator is simply:

${\begin{array}{lcl}P(H_{13}|I)&{}={\displaystyle {\frac {1}{2}}\ .}\end{array}}$

as can be seen in the table above. Dividing the numerator by the normalizing constant yields:

$P(C_{1}|H_{13},\,I)={\frac {1}{6}}\,/\,{\frac {1}{2}}={\frac {1}{3}}$ .

This can be stated differently:

${\frac {P(H_{13}|C_{1},\,I)}{P(H_{13}|I)}}=1$ .

since the host doesn't know where the prize is! Therefore

$P(C_{1}|H_{13},\,I)=P(C_{1}|I)$ .

This shows that the probability that your initial choice was right remains 1/3 even after the host opens a door at random, given that the random door did not show the prize. You should switch doors, exactly as in the case where the host knew where the prize was. —Preceding unsigned comment added by Jasoncoop (talk • contribs) 13:24, 25 October 2007 (UTC)

All this is saying is that the total probability of winning by staying with your initial choice is 1/3 (no argument). If you also compute $P(C_{2}|H_{13},\,I)$ and $P(C_{3}|H_{13},\,I)$ you'll find they are each equal to 1/3 as well. What this means is you have a 1/3 chance of losing because the host opens the winning door, a 1/3 chance of winning if you stay with your initial choice, and a 1/3 chance of winning if you switch. Given that you don't immediately lose, switching doesn't matter. -- Rick Block (talk) 14:10, 25 October 2007 (UTC)

Wrong! of course $P(C_{2}|H_{13},\,I)$ and $P(C_{3}|H_{13},\,I)$ are each 1/3. These are the cases where your initial choice was wrong. The problem you're describing is not the right one! The question is what's the probablitity that you were initially right given that the third door is wrong. This posteriori information affects the probability in the way I described using Bayes' theorem. Can you find fault with the way I used it? Jasoncoop 14:47, 25 October 2007 (UTC)

You haven't included in your formalism that the third door is wrong, only that the host has opened it. You seem to agree that $P(C_{2}|H_{13},\,I)$ is 1/3. This is the same as $P(C_{1}|H_{13},\,I)$ and since door 2 is the only door you can switch to (the host opened door 3, right?) what difference does it make if you switch? Alternatively, you can extend the Bayesian analysis to include not only that the host opens door 3 but that this is not the winning door (exercise left to the reader). You'll find this introduces a factor of 2/3 in the denominator (corresponding to the probability that the car is behind either door 1 or door 2), making the result 1/2. -- Rick Block (talk) 18:42, 25 October 2007 (UTC)

Rick, thanks for your detailed and patient response. I think you must be a pretty good teacher. I substituted host opens door 3 and prize is not there for $H_{13}$ and came up with 1/2 as you promised.

I'm still a bit concerned about the question of what happens when the host has some knowlege about the way the prize door was chosen, but not certain information (some non uniform distribution). Maybe I'll work it out some time.

So in fact the two following situations are similar:

1. First pick a door at random and open it. Given that it's empty, each of the closed doors has probability 1/2 of containing the prize.

2. Pick a door at random, open one of the other two doors at random. Given that it's empty, each of the closed doors has probability 1/2 of containing the prize.

Somehow I find this a bit disturbing, but hopefully I'll get used to it...

Or maybe this is the key to understanding the problem - if neither the contestant nor the host knows where the prize is, it doesn't matter who goes first. Jasoncoop 23:03, 25 October 2007 (UTC)

No problem (and I'm not a teacher, just someone who likes to help). -- Rick Block (talk) 00:36, 26 October 2007 (UTC)

BTW - I assume you think the first of your similar situations above is intuitive but the second is less so. The critical point is whether there's a random choice involved. Consider 100 doors and 98 players each of whom selects a door (one after the other, no player being allowed to select a previously chosen door). First, do all the players have an equal chance? (yes, and Bayesian analysis is your friend) Second, if you now open one of the remaining two doors randomly and doing so doesn't reveal the prize, what happens? Alternatively, 100 doors and one player and we've managed to open 98 doors randomly without revealing the prize. Now what? -- Rick Block (talk) 04:36, 26 October 2007 (UTC)

Choice? or Strategy?

I believe that some who are confused by this problem are focused on the difference between a Strategy and a Choice. The Host presents the player with a Choice, and if it were truly acted on as a Choice then the probability would not be 2/3 winning. Only when the player examines the rules and develops a Strategy which is adhered to without fail, does the probability turn to hir favor. Those who have watched The Price is Right remember the indecision on the faces of players as they ponder what to do. In my (admittedly basic) understanding of probablilty, everything hinges upon a priori decisions. Thus, I posit that if a Player goes into the game with a Strategy of sticking to their original choice, they'll have 1/3 chance of winning. The Strategy of always switching when offered the inevitable Choice yields 2/3 Chance of winning. This has been established. But I think that it is important to note that if the Player has no Strategy, and truly decides to randomly Choose between switching and staying, then there is a 1/2 chance of success. I believe I am correct in this assessment, and I think it would make a good additional to the article to explain this, since I believe many of those who are confused are thinking in these terms. i.e. Explain the difference in probability between a true Strategy (Choice decided in advance) vs. random Choice. Of course, if I'm completely wrong on this (which is why I have placed this in discussion), I still think it might clarify the article to point out that this is a Strategy, and that the player is not really making a Choice - their outcome is predetermined when Monty opens a door. P.S. I note that one of the external links explains the exact case I present of 1/2 odds with a random Choice. Shouldn't that be mentioned somewhere in the article for completeness? BrianWilloughby 18:29, 11 July 2007 (UTC)

I think this is a very complicated way to look at the problem and is almost certainly not how most folks are approaching it. If a player chooses randomly to stay or switch, then (over numerous iterations) the aggregate probability will indeed be 1/2. The complication is that on any given iteration, the odds are 2/3 on switching vs. 1/3 on staying - regardless of when the player decides whether to switch. A player who randomly decides who only plays the game once does not have a 50/50 chance, but a 1/3 chance (if the random decision was stay) or a 2/3 chance (if the random decision was switch). I suspect this distinction between the outcome of a strategy and the outcome of an individual choice is quite frankly beyond most people's grasp of probability. An example might help clarify this. If there are 100 players who choose randomly, likely 50 will switch and 50 will stay. Of the 50 who switch, we'd expect about 2/3, i.e. roughly 33, to win the car. Of the 50 who stay only 1/3 will win, i.e. only 17. 17+33 equals 50, so on average these 100 players have a 50% chance of winning although no one of them individually has a 50/50 chance. Contrast this with a pool of 100 players who all switch. Each one of them has a 2/3 chance of winning and we'd expect out of 100 about 67 will win the car. The aggregate probability of this group is the same as the individual probability of each player in the group (unlike the previous case). -- Rick Block (talk) 18:59, 11 July 2007 (UTC)

You're not using the terms in a way that has any mathematical meaning. And it makes no difference what the player decides in advance. The fact is, there's a 1/3 chance that the car is behind the original door, and a 2/3 chance that it's behind Monty's door. But I think that it is important to note that if the Player has no Strategy, and truly decides to randomly Choose between switching and staying, then there is a 1/2 chance of success. This is meaningless. If you flip a coin you always have 1/2 chance of success. It has nothing to do with the problem.

If you call 100 coins tosses as a set, you have different odds of being right on the whole run. This is basic probability course material. Each coin toss is indeed 1/2 change on its own, but if you call a series of coin tosses in advance, your probably of getting it right is different than someone calling each coin toss individually. This is the reason I brought up the difference between choosing in advance, or making a last minute decision when Monty offers a choice. Those who decide their strategy and stick to it are calling multiple random events as a set. They are the only ones who gain the advantage in odds. Those who truly take Monty's offered choice as a new opportunity to make a decision have a 1/2 chance. To return to the coin analogy, nothing changes the physical odds of any single coin toss, but the probably changes when you take a group of coin tosses as a set versus taking each one individually. I'm merely saying that, taken individually, the second choice offered by Monty gives the user a 1/2 chance of winning.BrianWilloughby 22:06, 12 August 2007 (UTC)

Brian, the distinction you are trying to make between strategy and choice is actually a distinction between two different problems. Consider this: The player chooses a door, the host reveals a goat, the player then flips a coin to decide whether to stay or switch. This is a different problem and, as you have pointed out, the probability of winning the car is 1/2 in this case. However, your statement "... taken individually, the second choice offered by Monty gives the user a 1/2 chance of winning." is misleading at best. Those who truly take Monty's offered choice as a new opportunity to make a decision have new information available to them and so their odds of winning are 2/3. What you are describing is a game where Monty's offered choice is a new opportunity to make a random choice. That's a different problem. Alfred Centauri 02:04, 13 August 2007 (UTC)

I agree with "Debunked" and I think the confusion here is with not distinguishing "or" versus "exclusive or" in example 1 or in the Venn diagram. Also with the Venn diagram the probability of picking the box with the two boxes in it is 50%, exactly the same as the probability of picking the first box. In effect in drawing the two boxes circled in the Venn diagram you have changed the probability by making it 50% 1st box or 50% "2 box" box. This again confuses "exclusive or" with "or", eg the probability of the car being behind one door is 1/3, the probability of the car being behind one of the last two box doors is 1/2 therefore the probability of the car being in the venn diagram second half is 50%, not 2/3, since it can either be in the box containing the last two boxes or be in the first box.67.86.165.55 06:24, 1 August 2007 (UTC)

Nope. The contestant has to pick one door, not "I'll take the Venn diagram, Monty." The box surrounding Doors 2 and 3 is an abstraction, encapsulating two of the three doors that the contestant might pick, and 2/3ds of the possibilities.--Father Goose

Nope. The contestant actually picks 1 scenario (as per DEBUNKED), not 'one door'. Thus the chance of picking the car by switching is still 50%.— Preceding unsigned comment added by 203.49.196.163 (talk)

Take a look at the large image set in the Problem and solution section and explain to me why it's wrong.--Father Goose 07:49, 1 August 2007 (UTC)

It's wrong because Scenario 1 in the image set in reality consists of 2 scenarios. Now what if we consider it from different perspective: We know the host will always pick the goat. The host picks Goat A, so the player choice is really between Goat B & the Car. If the host picks Goat B, the player choice again is only Goat A & the Car. Switching does nothing to increase the chance of winning. Since the beginning of the game the chance of the player winning the car is 50%, not 33%. 203.49.196.163 00:33, 2 August 2007 (UTC)

So is the player twice as likely to pick the door on the left than the door in the middle or the door on the right?--Father Goose 01:12, 2 August 2007 (UTC)

Thanks, Father Goose. I can see the absurdity of my statement now. So: Player picks 1 door out of 3 --> 1/3 chance of winning. If the host picks Goat A/B and player stays, his chance of winning obviously is still the same (1/3), but the chance that the car is behind the door becomes 1/2? If player decides to switch, there are 4 scenarios:

• Host reveals Goat A, other door contains Car (25%).
• Host reveals Goat A, other door contains Goat B (25%).
• Host reveals Goat B, other door contains Car (25%).
• Host reveals Goat B, other door contains Goat A (25%).

Which means by switching, his chance of winning indeed increases and becomes 1/2? Not 2/3?! 203.49.196.163 05:01, 2 August 2007 (UTC)

Not quite. It's not the number of seemingly-unique scenarios that are important, it's the likelihood of each. Your list above should actually be:

• Host reveals Goat A, other door contains Car (33%).
• Host reveals Goat A, other door contains Goat B (17%).
• Host reveals Goat B, other door contains Car (33%).
• Host reveals Goat B, other door contains Goat A (17%).

This is because the two "host reveals goat/remaining door contains goat" scenarios can only occur when the player has initially picked the car (option 1 in the big diagram). There's a 1/3 chance of that, which splits into a 1/6 chance of Goat A revealed, Goat B remains; and a 1/6 chance of Goat B revealed, Goat A remains.

I recommend reading the "Why the probability is 2/3" section for a deeper understanding of why this is so. Beautifully deceptive problem, it is.--Father Goose 05:55, 2 August 2007 (UTC)

₠Thank you Father Goose, you are right and I withdraw my "I agree with Debunked" comment above. The example that makes it easier for me to understand is if there are 100 doors, you pick a door and then the host is told to remove 98 doors. The door he leaves has a 99% chance of being the Car, since the only scenario where he is not picking the cars is where you picked the car in the first place and the probability of that is only 1%, so it is much more likely when he opens the 98 doors that he has left closed the car. That to me is clearer for some reason than when only three doors are utilized in the problem. —The preceding unsigned comment was added by 70.181.21.230 (talk)

Though the point has been made, another explanation that reinforces the point is the "what is the probability that you are wrong?" approach.

You have 3 doors, and must choose one. The probability that you are wrong (or, the probability that it is one of the other doors) in your first choice is 2/3. When a door is revealed to show the donkey, it doesn't chang the fact that the probability that you were wrong in your first choice is still 2/3. Since you now know that one of the doors that you didn't initially pick is incorrect, the probability that you will be correct if you switch doors is 2/3.

New section - Not Switching

For a different perspective on the problem, consider not switching doors. In each of the three possible cases, there is at least one unchosen goat for the host to show, but only in the first of the three cases (1/3 of the time) does not switching win the car.

This means that 1/3 of the time, the car is the contestant's first choice, therefore 2/3 of the time the car is not the contestant's first choice, and switching to the other available choice will win the car.

---

I've deleted the new section with the giant image (above) pending discussion here. In my opinion, the giant image basically replicates the other giant image (with different visual representations of the doors, car, and goat), and the text essentially repeats the text already in the second paragraph in the Solution section, making this new section simply redundant with material elsewhere in the (already too long) article. -- Rick Block (talk) 13:56, 24 August 2007 (UTC)

My criticism would be that the diagram would be less confusing without the black hands (but possibly still not intuitive), and that if clip art was used, it might be non-free.--Father Goose 16:01, 24 August 2007 (UTC)

I made this image because I was confused by the original image like the "debunked" person above. I do not think the original image does a good job of explaining the solution, and additionally confuses things with the "goat A or goat B" area in the top right corner. It was only by creating this image that I was able to wrap my head around the solution, and thus I thought it could help other confused people understand the problem as well. There is no clip-art in this image; I drew everything (not very well, but I did). While including decision trees, variants, and bayesian proofs in the article appeals to statisticians, it does not help the average Wikipedia reader to understand the problem, while I believe that my illustration does. If it would help, I can add a final column to the image in which the contestant makes the switch, thus pointing to the car 2/3 of the time. Luvcraft 17:59, 24 August 2007 (UTC)

I'm glad you understand it now, but rather than add another image can we perhaps come up with some way to clarify the image in the Solution section? A while ago, I proposed replacing this image with something more like a table, e.g.

Player's initial choice	Probability	Host reveals	Outcome if switching
	1/3	Either goat
	1/3	Goat B
	1/3	Goat A

The images in this table aren't quite to my liking, but it's enough to get the general idea. Would a change of this nature clarify anything for you? -- Rick Block (talk) 19:02, 24 August 2007 (UTC)

Yes, I like this version a lot more than the current one. I look forward to seeing what Father Goose comes up with! Luvcraft 19:52, 24 August 2007 (UTC)

Correctly figuring out the problem via your own means is the best way to understand the problem, so creating your own diagram works better than looking at someone else's diagram. I'm not opposed to the diagram you created; the artwork's okay and it is simpler than the one I created, although I'm not sure it would be more readily understood by someone coming upon it for the first time.

I'd be willing to change my diagram set to use the "subtract one door" concept you came up with here, but it'll take me a few days before I have the time. Adapting my set to your idea has the advantage of keeping it in a preferred image format (SVG) and keeps the style consistent with the other diagrams in the article. The six-images-in-a-table format also allows for editable explanatory captions.

It might be possible to combine it with some of Rick's ideas for the table above as well.--Father Goose 19:14, 24 August 2007 (UTC)

That would be wonderful! Thank you! On further thought, I do agree that the black "anti-hands" in mine are confusing, and a diagram with them omitted would work much better. Luvcraft 19:49, 24 August 2007 (UTC)

Two Very Helpful Diagrams

I like the idea of having a diagram in the solution section of the article. However, with the current diagram it looks like if you replace the words "Host must reveal" with "Host luckily reveals" in parts 2 and 3, you would have an 'explanation' that switching wins 2/3 of the time, even if the host forgets what is behind the three doors but happens to luckily open a door with a goat.

Last year I brought up the version of the problem where Monty luckily reveals the goat when I noticed that the probability of winning upon switching does not increase in this version. This observation confused the heck out of me until I saw the diagrams in the "To Switch or Not To Switch" section on the following webpage.

http://math.ucsd.edu/~crypto/Monty/montybg.html

The diagrams there are a little confusing at first but they definitely capture the difference in these two versions of the problem. Can we use the two diagrams in the article? I think they are useful for visualizing why the host's knowledge of what is behind each door matters. This alternate version is probably why many people don't feel sure about the solution to the standard version after they are shown it. Synesthetic

In the "host luckily reveals" variation of the problem, the odds of winning if switching are 50/50, not 2/3. Read "Monty Hall problem#Why the probability is 2/3" for why this is so.--Father Goose 20:34, 15 September 2007 (UTC)

Ok, I didn't see that explanation by the tables. The explanation by the tables is the same as the one given by the diagrams above but the versions are presented in reverse order. I know that the probability is 1/2 in the "host luckily reveals case." It was just the word 'must' which was bothering me. I guess I just wanted to point out that 'must reveal' is completely different than 'always luckily reveals' in the solution section... but now I'm thinking that the 'always luckily reveals' version is a different problem and so it doesn't have to be in the solution section. Hmm, I don't know... should 'must' be defined more in the solution section? Should it be pointed out that 'must reveal a goat' does not mean 'do over until Monty reveals a goat?' Eh, I guess not. Oh well, I understand the solution from those diagrams or the tables in the article. I hope the readers of the article get it too. Thanks for the reply Father Goose.  :-) Synesthetic 21:53, 15 September 2007 (UTC)

Well, as mentioned in the section above this one, I've been intending to change the diagram anyway, and the 'must' wording might disappear as a result. I got started on that work but need to finish it.--Father Goose 23:21, 15 September 2007 (UTC)

I'd vote for keeping the word 'must' in versus simply saying 'the host reveals a goat.' The way in which the host reveals the goat is essential. I think that once people see the difference between the two versions, they are no longer confused. Perhaps there is a way to explain 'must' in a way that makes it entirely clear to the reader that 'luckily reveals' is completely different. I don't know. You could argue that 'must' means 'must' and I would agree with that. Good luck with the diagram. Synesthetic 04:37, 16 September 2007 (UTC)

Reference to Bayesian solution of Three Prisoner's Problem

A sentence about a Bayesian solution to the Three Prisoner's Problem has been added, deleted, added, deleted, and now added again. I agree with the deletion (this is mentioned, appropriately, in the article on the Three Prisoner's Problem, but has no particular relevance to Monty Hall), but would like others to comment as well before deleting it yet again. -- Rick Block (talk) 18:37, 10 September 2007 (UTC)

I will say, I don't really see what significance it has. If that was explained, it might be worth keeping, but right now, get rid of it. mattbuck 19:20, 10 September 2007 (UTC)

I agree, just having a reference in the "related problems" list will do.The Glopk 02:13, 12 September 2007 (UTC)

Mistake in the article?

"The player's chances of winning the car actually double by switching to the door the host offers."

This seems to be wrong. Should we just delete that sentence?

What's wrong about it? The probability of winning by staying is 1/3. The probability of winning by switching is 2/3, i.e. double 1/3. -- Rick Block (talk) 03:01, 10 October 2007 (UTC)

The probability of winning by switching is *to the other door* is 2/3 indeed. But here we say that you should choose *the door the host offers*, which, as far as I understand, will always be the wrong door, in accordance with the rule that the host always offers an empty door. -Michaël 07:49, 14 October 2007 (UTC)

No the host does not offer an empty door; (s)he opens an empty one, and offers the third one.--Niels Ø (noe) 09:02, 14 October 2007 (UTC)

Missing Assumption

The text of this article doesn't appear to state anywhere that the problem only works if we assume that the player would rather win a car than a goat. -- Mikeplokta 09:21, 14 October 2007 (UTC)

Had the contestant wanted the goat instead of the car, he or she could simply pick the door that Monty had already opened.--RLent (talk) 16:40, 20 December 2007 (UTC)

The unambiguous statement of the problem in the "Problem" section explicitly makes the question about increasing the chances of getting the car. -- Rick Block (talk) 15:20, 14 October 2007 (UTC)

Bayes' explanation

The discussion thus far creates the impression that the use of Bayes theorem is a safeguard against falling into the trap of the false answer or, equivalently, that people get the wrong answer because they don't use Bayes theorem in their heads. This is not the case. The added paragraph demonstrates that the use of Bayes theorem still leaves ample room for misformulating a problem. Moreover, it explains why the Monty Hall problem, by accentuating the contrast between "information revealed" and "total evidence", has become central to philosophical discussions about the adequacy of Bayesian reasoning in managing uncertainty.Kvihill 17:20, 16 October 2007 (UTC)

This section strikes me as being potentially useful, but is quite awkwardly worded at this point. For example, where does "P(a goat is behind door 3)" come from (given the setup, isn't this 1)? The last sentence "This difference ...has far reaching implications in reasoning under uncertainty." sounds like original research (unless it's a quote). I don't have a copy of Pearl's book, but the index (available at Amazon) doesn't mention the Monty Hall problem. Can someone who has this book verify the section that has been added reflects something that is actually in this book? -- Rick Block (talk) 18:53, 17 October 2007 (UTC)

I think this section should be removed. First, it is not an "aid to understanding" the correct solution, but rather an (attempted) explanation for why some people may answer incorrectly. Second, its formulation is quite poor: it shows "things" that look like mathematical symbols, but have no well defined meaning. Third, if it is, indeed, a claim that the preceding "Bayes Theorem" analysis is incomplete, it ought to show exactly why. A claim that Bayesian analysis may not sufice to describe rational reasoning from incomplete information is highly suspicious: there have been stronger (e.g. Popper's)

attacks in the past, and have been thoroughly debunked.The Glopk 20:28, 19 October 2007 (UTC)

I have removed this section and copied it here for the record.The Glopk 16:09, 25 October 2007 (UTC)

===Bayes' explanation===

Judea Pearl's book (1988) gives a Bayesian explanation for people's tendency to provide the (wrong) answer 1/2. After the hosts reveals that a goat is behind door 3, people tend to condition their beliefs on the revealed information "a goat is behind door 3" and obtain the answer:

$P(car~behind~door~2|a~goat~is~behind~door~3)=1/2$ .

The correct answer is obtained by conditioning on the total evidence available: "host revealed a goat behind door 3," giving:

$P(car~behind~door~2|host~revealed~goat~behind~door~3)=2/3$ .

The distinction between "information revealed" and "total evidence" has far reaching implications in reasoning under uncertainty [Pearl, 1990, 1992]

Relation to Bell's Inequality and Hidden Variable Theories

I get the feeling this is very reminiscent of Bell's inequality in relation to hidden variable theories in Quantum Mechanics - is the MH problem an analogy for that inequality perhaps? Fizzackerly 13:31, 19 October 2007 (UTC)

It is Quantum_entanglement! Simply amazing. Sizur 17:05, 30 October 2007 (UTC)

Generalizing to n Doors

I was surprised to see that the "switch at the very end" strategy for n doors isn't better supported. Is this because there's no suitable reference, or is it really an open problem?

FWIW I think the following sketches a proof: when there are two doors left the sum of the probabilities of success must be equal to 1; maximising one is equivalent to minimising the other; not switching until the last chance minimizes the probability of the initially chosen foor being correct. —Preceding unsigned comment added by 190.160.252.137 (talk) 19:49, 11 November 2007 (UTC)

Is this the same problem?

I don't understand why the problem is different from the following situation: "Suppose you walk up to two doors, one of which has a car behind it. Someone tells you that there was a third door that didn't have a car but that it was removed. You must now choose one of the two doors." Why is this problem different? Why is "switching" different from "choose one of two doors?" RobertM525 03:35, 13 November 2007 (UTC)

In thinking more about this, perhaps looking at it from the perspective of the host helps to make sense of it: "Situation 1: You have three doors. You cannot choose one because the player has chosen it. You cannot choose another because it has the car. You must choose the remaining goat door. Situation 2: You have three doors. You cannot choose one because the player has chosen it (and it has the car). You must now choose one of the other doors." I'm not sure it does help in any way, but it was a thought... :) RobertM525 03:48, 13 November 2007 (UTC)

The first situation, walk up to two doors ..., doesn't tell you anything about the two remaining doors. In the MH problem, the host must open a door and can't open yours (and can't reveal the car). The host is either in Situation 1 and, with a 2/3 chance, has to let you know which of the other two doors has the car, or in Situation 2 with a 1/3 chance and can show you either door. Situation 1 and the fact that the host is in this situation 2/3 of the time is the key. You pick one of three doors. 2/3 of the time when the host opens a door he's showing you the door with the car. -- Rick Block (talk) 04:46, 13 November 2007 (UTC)

Good job! Unfortunately, if this talkpage is ever refactored, this section will probably be deleted. What are the odds of all that, I wonder? ;) 216.169.163.106 (talk) 22:09, 10 December 2007 (UTC)

What if talk host doesn't know but is directed by some one who knows

I have a question. What if talk host doesn't know what is behind the doors, but is directed by some one, who knows what's behind, to open a specific door once the player selects a door? --Venkataramana vurity (talk) 16:45, 22 November 2007 (UTC)

If the person who's directing him knows where the car is and will never open that door, it's the same as the original "host knows" problem: 2/3ds odds of winning if the player switches. The act of deliberately avoiding the car is what introduces a non-random factor into it.--Father Goose (talk) 19:19, 22 November 2007 (UTC)

I don't understand why it's important for the host to know what's behind the doors. Don't the following premises have the same odds?

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

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

I understand that it is improbable that the host could repeat premise #2 without eventually revealing a car, however that is a completely different issue than premise 2 as stated. —Preceding unsigned comment added by Banderson1962 (talk • contribs) 22:38, 25 January 2008 (UTC)

This doesn't really answer the question "why is it important that the host knows what's behind the doors?" In fact, it makes no difference at all! Let's say Monty has no idea what door has the car and he's guessing as blindly as you are. Then the possibilities in the first round become 1. you picked the car and he opens a goat because that's all he has 2. you picked a goat, Monty has the car and a goat, but opens the goat randomly, and 3. you picked a goat, Monty has the car and a goat, and opens the door with the car. In scenario 3, the game is over and the second round, being moot, never occurs. However, in scenarios 1 and 2 the game goes on, and you are still confronted with the fact that when you went into the second round, Monty was twice as likely as you to have the car, and switching still doubles your odds of going home with that car. Drewtew (talk) 06:33, 29 January 2008 (UTC)drewtew

I asked this at the bottom of this very page today, then figured it out (I think.) In scenarios one and two, you're already lucky that Monty didn't open the 'car' door. The 2/3 chance that you picked a goat first time is multiplied by the 1/2 chance that Monty picked a goat too. (If you don't switch, you're left with the 1/3 chance that you picked the car.) Robin Johnson (talk) 20:47, 1 February 2008 (UTC)

Infinite doors

An anonymous user inserted an aid to understanding by considering infinite doors. In this case, the first door picked has zero probability of hiding the car, and switching yields the car with probability 1. Though perhaps still a bit unpolished, I found it helpful. However, Mattbuck (talk · contribs) undid the edit with the comment "doesn't help since it's not quite 0". How is the probability "not quite 0"? Phaunt (talk) 11:50, 29 November 2007 (UTC)

If you feel it's helpful, OK. But I personally don't see how this is any more helpful than the case with 100 doors. As a mathematician, I can say that as x -> inf, x^-1 -> 0, but is asymptotic to the x-axis. mattbuck (talk) 13:18, 29 November 2007 (UTC)

Hmm yes, it may not be more helpful than the 100-doors case. If somebody does not see that, will they accept the infinite case? Let's see what others think.

I believe that in probability theory, you can say that the probability is 0; I'm not sure though. To me, it seems very similar to a continuous random variable which has zero probability of assuming any single value. Phaunt (talk) 14:09, 29 November 2007 (UTC)

...or perhaps having an countably infinite sample space is just too unpleasant. Phaunt (talk) 14:16, 29 November 2007 (UTC)

Unless someone can find a published source for this particular argument, I'd suggest we not include it on grounds of WP:OR. It presupposes the host is able to open all but two of an infinite number of doors which I think is problematic (both intuitively and mathematically). I'm not suggesting we add this, but a similar problem could be formulated not involving doors but with the host telling the contestant "I'm thinking of a number, 1 or greater. After you tell me your guess, I'll tell you a different number that will either be my number or a random number if you've guessed my number (so one of the two numbers will be the number I'm thinking of). Do you want to switch?". -- Rick Block (talk) 14:54, 29 November 2007 (UTC)

Nice :-) Also, amusing mental image of Monty opening an infinite number of doors. As to whether this is WP:OR, yes, it probably would be. I'm too lazy to check the references and external links, but are the other aids to understanding we include properly sourced? Regards, Phaunt (talk) 16:11, 29 November 2007 (UTC)

I'm not doubting the conclusion's validity, but I'm a bit confused

Is the 2/3 probability supposed to be based on the initial door chosen, or the door chosen after there are 2 left and you're given the chance to switch? Because the diagram seems to indicate the former, while the jist I got from the article indicated the latter. Thanks. 71.127.243.28 (talk) 01:24, 1 January 2008 (UTC)

Well, both actually. When initially picked, the initial door has a 2/3 chance of not being the one hiding the car. Because of how the problem is constructed the host opening a losing door does not change this probability, so the initially chosen door still has a 2/3 chance of being a losing door when the contestant is given the opportunity to switch. After the losing door is opened, there's only one other door so it must have a 2/3 chance of being the winning door. -- Rick Block (talk) 17:19, 2 January 2008 (UTC)

Featured status? Really?

There is a citations tag at the top of this article. If this article requires more citations, how is this a featured article? Moreover, it's references section does not use the proper reference tags.

So how can this article possibly be a Featured article? --Son (talk) 04:52, 2 January 2008 (UTC)

Maybe because an anonymous vandal played the "Let's add some bogus tag" game? The Glopk (talk) 16:17, 2 January 2008 (UTC)

This article is written using Harvard referencing. Contrary to fairly widespread belief, the featured article criteria does not prescribe use of citation templates, or footnotes. -- Rick Block (talk) 16:48, 2 January 2008 (UTC)

If that is so, then how come I don't see any sentences with (author page #) at the end of it? hbdragon88 (talk) 07:32, 30 January 2008 (UTC)

There's an example of Harvard referencing in the very first paragraph: "A widely known statement of the problem appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (vos Savant 1990)". The reference vos Savant 1990 is expanded in the References section as "vos Savant, Marilyn (1990). "Ask Marilyn" column, Parade Magazine p. 12 (February 17, 1990)". I count at least 10 other examples of Harvard referencing in the rest of the article. Gandalf61 (talk) 15:10, 31 January 2008 (UTC)

OK, I got it now. I'm used to seeing paragraphs have references from multiple sources; these have fewer references per paragraph and I wasn't used to looking for the parentheses. Now I can pick out the Harvard ones. hbdragon88 (talk) 21:48, 1 February 2008 (UTC)

Featured Confusion. Really.

This article was more on-point when it was featured in July 2005, but even then it suffered the same flaw it exhibits today: matters of style aside, a "well written" article about a veridical paradox should successfully explain how common sense is led astray and convincingly explain the truth of the matter.

As discussion on the talk page indicates, subsequent elaborations such as detailed Bayesian analysis and digressions on variants, interesting though they may be, do not lend clarity. Perfectly intelligent people who fully understand the mathematics of probabilities can read the entire article and still be left suspecting some kind of elaborate academic hoax, like one of those fallacious proofs that 1+1=1.

The main problem is that Sources of confusion fails to address the fundamental misunderstanding that so easily ensnares common sense. The truth is in the article, but it is not clearly stated as the chief source of confusion: selective evidence.

I would be guilty of original research if I wrote an explanation of this, as I have not found a source with a clear and direct statement of the issue as it relates to this paradox in particular. A good article on evaluating selective evidence might merit more than low-importance and low-priority. As a cautionary example, this article's significance could be elevated as well, if only the source of confusion were not obscured by the elaborate explanations. 67.130.129.135 (talk) 03:27, 4 January 2008 (UTC)

To play "Devil's advocate", I will semi-seriously challenge one assertion of the article. In essence, I will agree that the problem, when correctly stated, yields 2/3 as the probability of winning by switching. However, I will contend that for the version stated in the Ask Marilyn column, the answer is not "technically ambiguous" but rather 1/2 - slightly muddy, perhaps, but not as ambiguous as all that. To understand why, it's important realize that "probability" is inevitably a manifestation of incomplete knowledge. Outside of the quantum realm, complete knowledge of a situation invariably yields prediction values of either 1.0 or 0. Probability, on the other hand, is based on the knowledge at hand, and so to claim that a probability estimate is ambiguous because important information is lacking is to misconstrue the meaning of probability. In the Ask Marilyn version, we really know only that the car is not behind door 3 and must therefore be behind one of the other two doors. This yields the value of 1/2. Of course, we also know that the host opened door 3 to reveal a goat, but for practical purposes, this is not usable knowledge. Based only on what is usable, we arrive at the 1/2 answer. I suppose one could take a Bayesian approach to interpreting the host's actions, in which case, one could estimate various probabilities based on guesses about the host's motivation. If so, that should be stated. It would be equally reasonable, however, to simply say switching or staying are equally likely to succeed as long as we don't know what the host is up to. Fmoolten (talk) 01:37, 20 January 2008 (UTC)

Mathematically speaking, it's not "equally likely", but undefined.--Father Goose (talk) 22:01, 25 January 2008 (UTC)

I would argue that the entire notion of probability is based on estimates of outcome under circumstances that are only partially defined. Completely defined circumstances yield probability values of 0 or 1 (excluding uncertainties in the quantum realm). "Defined" is therefore a matter of degree rather than an all-or-none concept. To say that the probabilies are equal when we don't know what the host is up to is consistent with this principle. Clearly, these circumstances are less well defined than in the correct version of the puzzle, but they still permit probability estimates based on the information available. An analogy exists with a coin toss. If all we know is that an ordinary coin is tossed, the probability of heads is estimated at 0.5. However, if the toss is better defined in terms of angle and height of toss, speed of rotation, center of gravity of the coin, and the nature of the surface below, the probability changes based on the greater degree of definition. In the Ask Marilyn puzzle, we at least know that the choice is between two doors. The puzzle would be more or less completely undefined if we didn't know how many doors were involved, and in that case, a probability estimate would be virtually meaningless.Fmoolten (talk) 17:06, 29 January 2008 (UTC)

What should the host do?

The host doesn't want to lose cars, but everyone watches the show and knows his pattern of behavior. If we assume all the players are smart, then there's no use for the host to only open a door if the player has picked the car, because then people won't switch. But if he always opens a door then all the players switch and win more often. He has nothing to win or lose if he opens a door 50% of the time when the player hasn't picked the car and 100% of the time when the player has picked the car, because then the player wins the car by switching as often as he loses it by switching. Nonetheless in the real world some people will always switch by mistake when given the option. So there's a trade-off between a notorious host who only opens the door when the player picks the car, but has very few mistaken switchers, and a subtle host who opens the door 40% of the time when the player hasn't picked the car, and gets many people who think it doesn't matter or even think it helps based on a few episodes. How to resolve that trade-off I can't tell say. But in a proper game against a cunning host it should be at least somewhat harmful to switch doors when the choice is offered. 70.15.116.59 (talk) 20:36, 24 January 2008 (UTC)

Assuming the host always offers you a switch when you've initially picked the car, this variant has three outcomes not two. You:

1. Win by staying (1/3 chance)
2. Win by switching (2/3 chance times probability host opens a door when you haven't initially picked the car)
3. Lose without a choice being offered (2/3 minus the previous number)

As you say, if the host gives you a chance to switch 50% of the time when you've initially picked a losing door switching is a toss up because you win by staying 1/3 of the time, win by switching 1/3 of the time (2/3 times 50%), and lose without being offered a chance to switch 1/3 of the time. So if the host offers the switch when you've picked a losing door more than 50% of the time, switching is better. If the host offers the switch less than 50% of the time, staying is better. If the host only offers the switch when the player initially picks the car, anyone who watches the show will never switch (2/3 of the time you'll lose with no switch offered, but if you're offered a switch you're 100% certain to win by staying). This host should give cars away 1/3 of the time. Assuming optimal play from the contestants, players don't switch unless the host offers the switch more than 50% of the time when they've picked a losing door - and until then the host gives cars away 1/3 of the time (and nobody ever switches, which makes for a dull show). Offering the switch more often increases the probability of winning by switching until it reaches 2/3, so with optimal play the host never does any better than giving cars away 1/3 of the time.

Of course, the way the real show worked was there was never an opportunity to switch doors but an opportunity to switch from your initial door to some other prize (something unknown in a box and/or a known amount of cash). Probability of winning by staying was 1/3. Probability of winning the car by switching was always 0 (because you weren't switching to the other door, but to an alternate prize) but the probability of winning something (like $500 cash) by switching was 100%. Statistically, this means you're better off switching only if the value of the alternate prize is more than 1/3 the value of the car, but if you don't switch then 2/3 of the time you end up with nothing. This is a much harder choice for most people (something of high value with low certainty vs. something of lower value with higher or even absolute certainty) -- Rick Block (talk) 15:07, 25 January 2008 (UTC)

Does the host have to know where the car is?

The article gives the impression that it's important that the host knew along where the car and goats are. Aren't the results the same whether he knew or not, since you've already eliminated the cases where he opens the 'car' door? Robin Johnson (talk) 14:07, 1 February 2008 (UTC)

Hm. For one thing, the article doesn't just "give the impression" that the host's knowledge matters, it states it outright, and for another... I'm just wrong. I've simulated this (and the ordinary problem). In the clueless-host version, if I've coded it right, you seem to win in 2/3 of the cases where the host didn't open the car door? Can that be right? The article seems to say it should be 1/2 and seems to produce those results with big enough samples. Hurrah.

I had to work this out for myself though... it miiiight be possible to express it more clearly in the article. Robin Johnson (talk) 18:34, 1 February 2008 (UTC)

Rigorous solution

This edit added a section titled "The Rigorous Solution" which includes a Bayesian analysis where the host's probability of picking which of the remaining two doors to open (in the event that the player originally picked the car) is allowed to vary, and shows switching is optimal regardless of the probability assigned. This analysis is claimed to include the effects of the host's strategy, thereby showing switching is optimal regardless of the host's strategy.

1. There's already a section on the host's strategy, so this text is at best misplaced.
2. Lacking a reference, I'd consider this analysis original research, which is prohibited.

Rather than debate the merits of this analysis, I'd suggest we not even consider including it unless a reliable source for it can be found. -- Rick Block (talk) 16:54, 10 February 2008 (UTC)

This elementary Bayesian calculation is not research. It would be better for the readers to see this. The probability if winning the car given switching is more general than 2/3. It is 1/(p+1), which equals 2/3 when p=1/2. It can be between 1/2 and 1, but always not less than the probabilty of winning car by not switching, which is equal to p/(p+1), and therefore between 0 and 1/2. —Preceding unsigned comment added by 70.137.168.95 (talk) 02:00, 11 February 2008 (UTC)

Given the number of papers and other material about this problem, I'd expect a reference for any analysis to be readily available (and, indeed, the existing section with a Bayesian analysis is referenced). The "Other host behaviors" section includes a number of specific host strategies where the success by switching varies from 0 to 1. In my opinion, including this specific variant where the host has a preference of probability p for one unpicked door over the other, assuming either can be opened, is simply not that helpful. Other editors' comments are welcome. -- Rick Block (talk) 02:36, 11 February 2008 (UTC)

The problem with this article is that the main point is cluttered by so many uninteresting generalizations, whereas the original game (where the host always opens a door other than the one picked by the player, which also does not have the car) is not analyzed rigorously. The most general stratgey of the host subject to the original rules of the game is characterized by a probability p, which is that probability with which the host opens door 2, given that the player chose 1 and the car is at 1. (There could be other probabilities for the cases of the player picking 2 and 3, but it suffices to conside one case.) To ignore the full analysis of the original problem is not wise, especially when the analysis reveals the surprising result that it is always optimal to switch. To emphasize the importance of this point, the player deos not need to know p in order to conclude that switching is optimal. This is true for every p. To assume p=1/2 is too restrictive! I do not understand why you want to censor this analysis using your admin power. —Preceding unsigned comment added by 70.137.145.250 (talk) 04:04, 11 February 2008 (UTC)

One other editor, user:Mattbuck, has also reverted this addition. This has nothing to do censorship or who's an admin and who's not (and, yes I am an admin, but this is a content dispute in which I am simply an editor like any other). Let's just wait for other editors' comments. -- Rick Block (talk) 04:50, 11 February 2008 (UTC)

Here is the rigorous solution: A rigorous solution requires addressing the question of the strategy of the host, namely, how the host picks a door to open, when more than one is possible. Following is a Bayesian analysis, which proves that switching is optimal regardless of the host's strategy:

The prior belief of the player is that the car is behind each door with probability 1/3. Suppose the player chooses door 1.

Denote by p the conditional probability with which the host opens door 3, given that the car is behind door 1. Thus, 1-p is the conditional probability with which the host opens door 2, given that the car is behind door 1. In the event the car is not behind door 1, there is only one door that the host can open.

Following are the probabilities and conditional probabilities that lead to the conclusion:

 Prob (Host opens door 3, given that the car is behind door 1) = p 
 Prob (Host opens door 3, given that the car is behind door 2) = 1
 Prob (Host opens door 3, given that the car is behind door 3) = 0
 Prob (Host opens door 3 ) = (1/3) x p + (1/3) x 1 + (1/3) x 0 = (p+1)/3.

The following equation holds:

  Prob (Car is behind door 1, given that the host opened door 3)   x  Prob(The host opens door 3) 
 =   
  Prob (Host opens door 3,    given that the car is behind door 1) x  Prob(The car is behind door 1)

Therefore,

  Prob(Car behind door 1, given that the host opened door 3)  x  (p+1)/3 = p/3

and thus

  Prob(Car behind door 1, given that the host opened door 3) = p/(p+1).

It follows that,

  Prob(Car behind door 2, given that the host opened door 3) = 1/(p+1).

Since p is not greater than 1,

 Prob(Car behind door 2, given that the host opened door 3) >=   Prob(Car behind door 1, given that the host opened door 3).  

Therefore, switching to door 2 is an an optimal strategy for every p. —Preceding unsigned comment added by 70.137.145.250 (talk) 05:44, 11 February 2008 (UTC)

I agree with editor Rick Block that this analysis based on the host's preference is not helpful. The language "a door, say No. 1" and "another door, say No. 3" is conventional mathematical usage for indicating that there is no a priori labeling of the doors. We are simply calling the door which the host opens No. 3. That is its defining characteristic. Had the host opened the other door, we would call it No. 3. The proposed analysis is creating a distinction that, in the statement of the problem, does not make a difference. 67.130.129.135 (talk) 23:16, 11 February 2008 (UTC)

User 67.130.129.135 is obviously mistaken. The two doors that were not picked are of course distinct. The host may follow a strategy that if the car is behind the door picked by the player, then the host opens the leftmost door that has not been picked. If this is the host's strategy, then the player may be able to infer with certainty where the car is. Specficially, suppose the doors are "left", "middle", and "right", and the player picked "left". If the car is behind "left" the host opens "middle" according to his strategy, and if the car is behind "middle" the host must open "right". Therefore, if the host opens "right" the player knows for sure that the car is behind "middle." The host's strategy 'does' affect the posterior probabilities for the player, however, as proven by editor 70.137.145.250 above, switching is always optimal (i.e., it is a "dominant strategy.") —Preceding unsigned comment added by 70.137.163.193 (talk) 03:52, 12 February 2008 (UTC)

To some extent, this sort of argument is what finding a sourced version of this analysis would avoid. The opinions of the editors of this article about the merits of this or that analysis is basically not that important. If the analysis is significant enough to warrant publication, then we can decide (among us) how or whether it should be reflected in the article. However, if it's not published we're on pretty dangerous ground. In the references I've seen, the notion of the host's strategy is distinctly not related to a preference for door 3 over door 2 - for example the "unambiguous" version of the problem statement by Mueser and Granberg ignores the possibility that the host does anything other than randomly pick (p=.5) between the remaining doors if the player happens to select the door hiding the car (and the host's strategy includes things like offering the opportunity to switch less often [or never] if the user happens to pick the door hiding the car). As a featured article, this article must meet the featured article criteria. Without a reference, I think this addition fails criteria 1C since it includes a claim that switching is optimal regardless of the host's strategy. I agree that if we define the host's strategy as a preference for one remaining door over another that switching is always better, however this is not what I think most references mean by the "host's strategy". My request to the user who wants this section added is to find a reference. -- Rick Block (talk) 04:24, 12 February 2008 (UTC)

Lots of people write a lot of false statements about this problem. A self-contained and valid mathematical proof stands on its own and does not need a reference. In addition, the lead section is incorrect. —Preceding unsigned comment added by 70.137.163.193 (talk) 06:15, 12 February 2008 (UTC)

Yes, lots of people write false statements about this problem which is one reason why attributing content to its source is so important. As I've tried to explain above, in this case I think a reference is warranted. I assume your issue with the lead is the "However, as long as the host ..." and subsequent sentences (which are true only if the host also doesn't have a preference for one door over another). This issue is mentioned in the "Problem" section and such a preference is explicitly excluded in the Bayes' theorem section. Do you really think it warrants being in the lead as well (again, Mueser and Granberg ignore it)? -- Rick Block (talk) 14:15, 12 February 2008 (UTC)

See Morgan et al. (1991). You have to consider the decision making problem of the PLAYER. The player does not know whether or not the host chooses either door with the same probability. If the host is a computer program, the host may pick the first permissible door. Therefore, it is very interesting to prove that the player does not need to assume anything because switching is a dominant strategy for the player. I meant to say that because lots of people wrote false statements about this problem, a reference to somebody else does not necessarily add authority. But see Morgan et al. (1991) if you wish. The proof I offered is self-contained and can be verified by people who took elementary probability. You may wish to consult with some mathematician at Columbia. When a qualified person reads the lead section, s/he sees the false statement and gets a very bad impression of Wikipedia. 70.137.163.193 (talk) 16:02, 12 February 2008 (UTC)

Morgan et al. discusses variant interpretations of the statement of the problem, and as such may be worthy of mention in the section on other host behaviors. They concur that under the standard interpretation the chance of winning for someone who switches is 2/3. I think the lead section is just fine in using that interpretation, given that the issue of ambiguity is raised within the lead and the standard disambiguation is prominently placed in the immediately following section. That the analysis is true of a more general interpretation of the problem is interesting, even informative, as are many of the digressions in the article. But I do not think it aids in understanding the veridical paradox that makes this such a notable riddle. 67.130.129.135 (talk) 18:52, 12 February 2008 (UTC)

The lead section should be corrected to include one more condition as follows (see capitalized). "However, as long as the host knows what is behind each door, always opens a door revealing a goat, OPENS EVERY NON-PICKED GOAT-REVEALING DOOR WITH EQUAL PROBABILITY, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door."198.4.83.52 (talk) 19:54, 12 February 2008 (UTC)

I don't even understand that.--Father Goose (talk) 20:50, 12 February 2008 (UTC)

Sorry, the word "DOOR" was missing. Without this additional assumption, the statement "opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door" is not true.198.4.83.52 (talk) 22:23, 12 February 2008 (UTC)

Could you run me through why that is?--Father Goose (talk) 01:25, 13 February 2008 (UTC)

(Re: changes by Rick Block made in response to this thread) Does it really matter whether the host makes a random or non-random choice when the two unpicked doors both contain goats?--Father Goose (talk) 02:06, 13 February 2008 (UTC)

Yes. Consider a host who always opens door 3 if possible and a succession of players who initially pick door 1. Then, if the host opens door 2 if the player switches a win is guaranteed - but if the host opens door 3 the odds on switching are 50/50. Against optimal play (all players switch), this host still loses 2/3 of the time (i.e. in the aggregate, players who switch win 2/3 of the time) but any individual player's chances of winning by switching are either 50% (no better than staying) or 100%. This host behavior is avoided in the Bayes' theorem section by setting Hij|Ck to 1/2 if i=k. Per the analysis above, we can set this to p for one of the j (say, door 3) and 1-p for the other j (e.g. door 2). With this setup, the probability of winning by switching if door j has been opened is p/(p+1) (anywhere between 1/2 and 1 depending on p - in the "always open door 3 if possible" case, p=1). This is precisely the point user:70.137.168.95 is making. -- Rick Block (talk) 04:17, 13 February 2008 (UTC)

Ohh, I see. I've added an entry to the "variants" section reflecting this scenario, as you suggested below.--Father Goose (talk) 08:27, 13 February 2008 (UTC)

RE: Rick Block (Father Goose, Please read this entire section "Rigorous Solution" in this Discussion, or the Morgan et al. paper.) What Mr. Rick Block is doing is harming the reputation of Wikipedia. He deleted the rigorous solution and put the following text as the solution: "If the player chooses to switch, the player wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three." As explained in (Morgan et al. 1991), this statement is "the most appealing of the false solutions." Mr. Block's solution, which is the current solution of the article, is the same as what is referred to as "false solution F1" in Morgan et al., who say about it the following (p.285):

"F1 is immediately appealing, and we found its advocates quite reluctant to capitulate. F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand. 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."

Mr. Block deleted a previous comment about the difference between unconditional and conditional probabilities. Mr. Block monopolizes this article and lowers the level of the article. It is not only the differece between "random" and "non-random." The player has no basis to believe that the host picks one of the two doors with probabilities 50:50. The host may do it with randomly with different probabilities, say 75:25 or even 100:0 (i.e., deterministically). The conditional probabilities are different in these cases, even though the unconditional probability is 2/3. Interestlingly, switching is optimal in every case. However, Mr. Block twice deleted this fact, which is cleverly explained in Morgan et al. What is the recourse against such behavior? 70.137.163.193 (talk) 04:42, 13 February 2008 (UTC)

Wikipedia policy encourages discussion on the Talk page to establish consensus. I attempted to go to the Morgan et al. paper, but I came to a purchase page at JSTOR. Flatscan (talk) 05:19, 13 February 2008 (UTC)

Rather than reword the entire article to make this distinction, as it currently stands the article specifies that the host makes an equal probability choice between two "goat" doors, which is (by far) the usual interpretation of the problem (so the player knows with certainty the host's behavior and can count on a 2/3 chance of winning by switching). In this case, the "conditional" and "unconditional" probabilities are the same (you agree with this, right?). IMO, this article is already nearly beyond the comprehension of most people and introducing this additional level of complexity is simply not necessary. I'd support adding a variant in the "Other host behaviors" section where the host's selection of the two goat doors is assigned a variable probability - but not complicating the main exposition (and not presenting this analysis as the one and only one "rigorous" analysis - it is simply a minor variant on the analysis already presented in the Bayes' theorem section). -- Rick Block (talk) 05:29, 13 February 2008 (UTC)

So, Mr. Block insists that this Wikipedia featured article presents, as the solution of this famous problem, the false solution that has been identified in the published refereed literature as the "most appealing false solution". Furthermore, Mr. Block deletes what other people write about it and threatens to suspend them if they try to revert his deletions. Of course, there cannot be a consensus with regard to a false solution. I find it outrageous and harmful to Wikipedia and a disservice to the intellectual community.70.137.163.193 (talk) 07:04, 13 February 2008 (UTC)

Okay, calm down now, he's started to acknowledge at least some of your points and introduce them into the article. He doesn't own the article, though he does take pains (along with other editors) to make sure it is understandable, well-presented, and well-sourced. I understand your frustration; it would be nice to just make your edits and have them stick, and not have to deal with other people and their viewpoints, but that is how Wikipedia works.

Since you're starting to win over various contributors to this article, now is not a good time to start bawling them out.--Father Goose (talk) 09:29, 13 February 2008 (UTC)

It is an interesting point, and some mention of it ought to be made here. I agree, though, that trying to work it into the main body of the article unnecessarily complicates what is, as it stands, a very clear and illuminating discussion of this problem for the general reader. Why the resistance to a (clearer) addition of the case(s) in question to the "Other host behaviours" section, as Rick Block suggests? That would seem to be the most suitable and agreeable solution here... --Wikiscient (talk) 12:45, 13 February 2008 (UTC)

The constraint that the host pick randomly, with equal probability, between two goat doors has been in the unambiguous statement of the problem in the article since July 2005. The solution to the problem as stated is that switching wins 2/3 of the time. This solution is indeed not the solution to other variants of this problem, of which there are many. Introducing the possibility of the host having a preference for one goat door over another with probability p is one of many potential complicating generalizations. To my knowledge, it is not the form generally analyzed. -- Rick Block (talk) 15:32, 13 February 2008 (UTC)

It is not "one of many complicating generalizations". It is the original problem and it does have a solution, as explained by Morgan et al. Adding the constraint that p=1/2 (rather than a general p) lowers the intellectual level of the analysis. The player cannot make this assumption and it is not in the original statement of the problem. The player must make a decision without assuming it. The host obviously does not toss a fair coin to implement it. Neither does a computer simulation. Therefore, this solution of the original problem should not be placed among the many complicating generalizations, most of which are not interesting and probably skipped by most readers. The main issue is that it is really unaccepatable for Wikipedia to present as the "solution of the problem", in the mostly read section, the explanation that Morgan et al. call the "most appealing false solution".70.137.163.193 (talk) 16:09, 13 February 2008 (UTC)

I think this generalization is irrelevant to the player's behavior, and as such only marginally relevant to the article as well. It is not correct to say that p=1/2 is a constraint, i.e. an unwarranted assumption, on the player's part. Rather, lacking further specifications, about the host's behavior, it is the _least informative_ prior the player can assume for the case in which car-hiding door is initially selected. And we could say "must assume" if the player subscribes to the maxent principle. This and related priors are represented by the symbol I in the Bayes Theorem section of the article, and are meant to summarize the "rules of the game" in the most commonly accepted specification of the game itself. Now, I agree that it is titillating, for the mathematically oriented, to show that the decision implied by the least informative prior actually generalizes to a wider class of priors (namely, the class in which the host can express a preference for which door to open when he has a choice), but this is something at best mentioned in the "other host's behaviors" section. It is definitely not, IMO, worth the adding further complexity to the main analysis.The Glopk (talk) 17:37, 13 February 2008 (UTC)

The problem as stated in the article is "Do the player's chances of getting the car increase by switching to Door 2?" This is a question about conditional probabilities. However, the solution section proves the unconditional probability is 2/3. This is a common mistake mistake. The highest grade I would give this solution is C+. So, even with the restriction of p=1/2, the current solution is a false answer to the problem posed.171.64.78.10 (talk) 19:40, 13 February 2008 (UTC)

The highest grade I'd give right now to your mathematical clarity is D-: talking about conditional probabilities makes no sense unless you specify the conditions. The analysis in the current version of the article (see, e.g. the Bayes Theorem section) is *entirely* about conditional probabilities. Specifically, it proves that that the values of the probabilities P(C_1 | H_12, I) and P(C_3 | H_12, I) for the propositions C_1 = "Car behind Door 1" and C_ 3 = "Car behind Door 3", both conditioned on the proposition H_12 = "Host opens Door 2 after the player has selected Door 1", and the game rules I, are respectively 1/3 and 2/3. The host behavior is completely specified by the assignment of the conditional probabilities P(H_12 | C_k, I) of the proposition "Host opens Door 2 after the player selects Door 1, conditioned on C_k = "Car behind door k", and the game rules I. For the case that we are discussing here, i.e. the one in which two doors are available for opening after the player's initial selection, the game rules in the most generally accepted version of the game specify that P(H_12 | C_1) = 1/2. As I pointed out above, even if the game rules did not specify that the two available doors are equally likely to be opened, so that the host could display an unknown-to-the-player preference "p" for one of the two doors, p = 1/2 would still be the least informative prior for the player to use in his analysis, hence the correct choice (according to the maxent principle). What the "novel" analysis shows is that the rational player would make the same decision by treating p as a nuisance (unobservable) variable. Fine, mathematically interesting. If you have a source, make a note and add it to the "other host behaviors" section, or at the end of the Bayes Theorem section. Just don't make it the focus of the whole article.The Glopk (talk) 20:45, 13 February 2008 (UTC)

Obviously, the "conditional probabilities" (of winning by sticking or by switching) refer to the probabilities of winning, given that the host has opened a certain losing door, whereas the unconditional ones refer to the probabilities of winning (using the strategies of always sticking or always switching) before the host has opened any door. I recommend to everybody to read Morgan et al.171.64.78.10 (talk) 21:54, 13 February 2008 (UTC)

It is truly unfortunate that the original statement of the problem in Ask Marilyn was imprecise. It is widely and notably considered to be ambiguous, so textual literalism is fruitless. The common and notable interpretation and disambiguation is clearly consistent with Marilyn vos Savant's original intent. Pointing out that a broader interpretation has a more general solution is interesting and constructive. Characterizing her answer as 'false' is unconstructive. I suggest reframing the presentation of this point as an expansion upon the topic, rather than as an attack on the solution of a simpler interpretation.

I have reverted an edit by user User:171.64.78.10 (see also User 171.64.78.10 talk) under section Problem and solution because the contentious criticism of the solution does not apply to the problem as restated and clarified in that section. 67.130.129.135 (talk) 01:02, 14 February 2008 (UTC)

Evidently, User 67.130.129.135 also does not understand that the unconditional probability argument is a false solution to the problem as stated in this very article. Please do not revert any more. —Preceding unsigned comment added by 198.4.83.52 (talk) 01:19, 14 February 2008 (UTC)

From the restatement of the problem in that section: "The problem as generally intended also assumes that the particular door the host opens conveys no special information about whether the player's initial choice is correct. The simplest way to make this explicit is to add a constraint that the host will open one of the remaining two doors randomly if the player initially picked the car." 67.130.129.135 (talk) 01:53, 14 February 2008 (UTC)

The fact that the current solution offered by the article is false because it offers the uncondtional probability rather the conditional one is not related to the constraint you mentioned.198.4.83.52 (talk) 02:12, 14 February 2008 (UTC)

In plain terms

So, am I to understand that the current brouhaha is over whether the host's choice of the two "unpicked" doors is non-random? If so, either:

• The player knows what pattern the host follows, making switching 100% likely to win when the host has deviated from his usual pattern; or
• The player doesn't know what pattern the host follows, making his aggregate odds of winning by switching to the remaining door 2/3ds, same as in the basic problem.

If it's the first case, that's such a huge omission from the basic statement of the problem, surely it cannot be treated as an inadvertent oversight or potential ambiguity.

To use a different example, if the host always told the player truthfully where the car was, that would change his odds of winning to 100% in all games. However, while the original phrasing doesn't preclude that as a possibility, the fact that such an unusual condition is not included in the phrasing makes it safe to assume that it is not in effect. I would say that "player knows host's predictable behavior" is a similarly unusual condition, and not something that could be treated as an ambiguity in the original problem.--Father Goose (talk) 03:11, 14 February 2008 (UTC)

First, a more important issue is that the solution offered in the article does not correctly answer the problem stated in the article because it confuses unconditional with conditional. See the section above. Second, instead of "non-random" think of random, but with unequal probabilities.198.4.83.52 (talk) 03:29, 14 February 2008 (UTC)

Do the odds for switching change for the player if he doesn't know which way the host's choice is biased? (Even if he knew that the host's choice was biased, but didn't know which way?)--Father Goose (talk) 04:44, 14 February 2008 (UTC)

Yes. "Host picks among two goat doors with equal probability" is a critical assumption to the solution presented in the article. It's the same as the difference between host knows where the car vs. not. The difference isn't what the player knows, but the host's behavior. -- Rick Block (talk) 04:54, 14 February 2008 (UTC)

All right, let's take the scenario that the host always opens the right-most unchosen door, except when it contains the car, in which case he opens the other door. The player doesn't know this. What are his odds of winning by switching to the unopened door?--Father Goose (talk) 05:04, 14 February 2008 (UTC)

If he picks door 1 and the host opens door 3, 50%. If he picks door 1 and the host opens door 2, 100%. If he picks door 2 and the host opens door 3, 50%. If he picks door 2 and the host opens door 1, 100%. (etc.). Overall, he has a 2/3 chance of winning by switching (50% in 3 cases, 100% in 3 cases), but in any given situation the chances are either 50% or 100% (whether he knows it or not). The "conditional" probability (with knowledge of the door the player picked and door the host opened) is either 50% or 100%. The "unconditional" probability (without yet knowing what door the player picks or what door the host opens) is 2/3. -- Rick Block (talk) 05:13, 14 February 2008 (UTC)

The problem is that the analysis you're making here is based on information you have, but that the player doesn't have. You know that if the host opens door 2, that signifies the car is behind door 3. The player, in this scenario, doesn't know what the host's door-choosing pattern is, and is thus still facing the original problem. It's only when the player gains some additional information from the host's choice -- i.e., the host has a pattern and the player knows what it is -- that the player's analysis changes to the one you describe above. That's a pretty drastic addition to the scenario, making it a variant of the game, not an ambiguity in the basic statement of the problem.

I continue to assert that adding the words "opens either of two doors concealing goats with equal probability" has no impact on what analysis the player must make of the basic variation of the game, and should thus be omitted.--Father Goose (talk) 09:09, 15 February 2008 (UTC)

To Father Goose [Re: removing the 1/6ths from the diagram; whether host picks randomly from the two "goat" doors is unspecified and does not change the analysis either way)] Now that you have removed it from the diagram, why don't you go ahead and remove the assumption of random pick throughout the article? It does not matter for the uncondtional probability, and some here say there is no need to calculate conditional probbailities.  ;)

-- 198.4.83.52 (talk) 00:50, 16 February 2008 (UTC)

There are lots of parties discussing that specific issue right now, so I'm not sure what the consensus is for it. I removed the "1/6" text from the diagram after having just added it yesterday when I realized that it's not necessarily right anyway, and just makes the labels more cluttered.

I'm still confused as to why people are asserting it's necessary to state that the host's choice of doors is perfectly random; leaving his which-of-two-goats choosing methodology unspecified introduces no ambiguities into the problem.--Father Goose (talk) 05:58, 16 February 2008 (UTC)

(outindent) Father Goose: Dropping the constraint that the host select from among two goat doors with equal probability introduces an ambiguity in the sense that the numerical probability of switching by winning is not a constant for all players regardless of what the host does. It creates a potential difference between the overall chance of winning by switching seen by players in the aggregate (the unconditional probability our anonymous friend of many IP addresses is talking about) and the chance of winning by switching as seen by an individual player at the point of being asked to switch knowing which specific door the host has opened (the conditional probability). In the problem statement (both vos Savant's and the unambiguous version in the article) the question is whether some individual player is better off switching, not "what is the aggregate chance of winning by switching". If we drop the constraint, and call the host's potential preference for one door over another p (the constraint forces p=1/2), the probability an individual player sees is 1/(p+1) which ranges from 1/2 to 1 (depending on p) although the unconditional probability seen by players in the aggregate is 2/3 regardless of p (indeed, a whimsical host might decide to pick the "rightmost goat door" on Mondays, Wednesdays, and Fridays and the leftmost on Tuesdays and Thursdays making p alternate between 0 and 1 depending on the day of the week with an aggregate value of .6). Again, this is exactly the point user:anon-many-ips is making. Note that without the constraint, the difference is there and affects the player's chances whether the player knows p or not, in exactly the same way that the player's probability of winning is 1/2 (both conditionally and unconditionally) if the host opens a remaining door randomly (the forgetful Monty version) whether the player knows how Monty has selected or not. -- Rick Block (talk) 16:49, 16 February 2008 (UTC)

Very Good, Rick. One more interpretation: the probability p is simply what the player believes rather than what Monty does. The decision-making problem is the player's not the outside observer's. If the player does not know all what Monty does, only that Monty follows the rules (open only an unpicked losing door), then the player may use p=1/2, reflecting equality of lack of knowledge. However, the analysis shows that whatever the player believes (i.e., for every p), the best decision is the same -- switch. 70.137.163.193 (talk) 17:19, 16 February 2008 (UTC)

What Morgan et al actually say

Here is verbatim what they say:

"We begin by enumerating and discussing the most appealing of the false solutions."...

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

Then they say:

"F1 is immediately appealing, and we found its advocates quite reluctant to capitulate. Fl's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand. 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."

The readers can verify that F1 is indeed the same as the solution currently in the Wikipedia article.70.137.163.193 (talk) 06:17, 14 February 2008 (UTC)

From the lead

However, as long as the host knows what is behind each door, always opens a door revealing a goat, opens either of two doors concealing goats with equal probability, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door. As there is only one other unopened door, the probability that this door conceals the car must be 2/3. It is therefore to the contestant's advantage, if they would rather have the car than the goat, to switch to door 2.

Does not the "however" provide the conditions you are so insistently seeking? If this solution summary is not to your liking, please propose one that is using terminology familiar to a general readership. If this is not the solution presentation you're objecting to, please identify precisely which one is. -- Rick Block (talk) 14:26, 14 February 2008 (UTC)

Ok. I coughed up the $14 and purchased the online Morgan et al article (from [1]). The version of the problem they initially analyze is essentially identical to the original statement of the problem from Parade (with no constraints on the host's behavior). They label as solution F1 the informal analysis that (paraphrased) "a player who doesn't switch wins 1/3 of the time, so switching wins 2/3 of the time" and call this a false solution. Indeed, it doesn't address the original problem as stated in Parade as it ignores the effects of the host's behavior. Rather than focus on ambiguities in the problem description they pursue an approach where the host's behavior is represented as a set of probabilities, leading to the possibility of different probabilities of winning by switching depending on the specific door the host opens. Although many host behavior variants are covered by their approach, not all are (they specifically exclude "host opens the player's door", "host opens the door revealing the car", and "host makes the offer to switch more or less often depending on the player's initial selection").

Given a problem statement where the host's actions are effectively described as variables requires distinguishing the player's initial chances (the "unconditional" problem) from the chances given the door the host reveals (the "conditional" problem). However given the unambiguous statement of the problem in the Problem section of this article, the specific door the host opens does not (cannot) affect the player's chances of winning. There's a fairly long standing consensus that the version of the problem discussed here should be an unambiguous version, permitting only a single solution which does not need to (and does not) distinguish "unconditional" vs. "conditional" (this difference is prohibited by the problem statement). IMO "Other host behaviors" would be a fine place to include a discussion of this issue. -- Rick Block (talk) 03:47, 14 February 2008 (UTC)

I am curiuos what kind of consensus is it that the solution does not distinguish "unconditional" vs. "conditional". Conditional probabilities constitute the central issue problems such as the Monty Hall paradox. 70.137.163.193 (talk) 06:35, 14 February 2008 (UTC)

Note that in the Solution section the bit that says:

More formally, 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:

• The player originally picked the door hiding the car. The game host has shown one of the two goats.
• The player originally picked the door hiding Goat A. The game host has shown the other goat.
• The player originally picked the door hiding Goat B. The game host has shown the other goat.

is true in the "conditional" case, i.e. regardless of which door the host has opened. The conditions in the problem statement enforce this. -- Rick Block (talk) 04:12, 14 February 2008 (UTC)

So, I assume you'd rather have this bit say:

More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are six possible situations, each with probability 1/6:

• The player originally picked the door hiding the car. The game host has opened door 2 showing a goat.
• The player originally picked the door hiding the car. The game host has opened door 3 showing a goat.
• The player originally picked the door hiding Goat A. The game host has opened door 2 showing the other goat.
• The player originally picked the door hiding Goat A. The game host has opened door 3 showing the other goat.
• The player originally picked the door hiding Goat B. The game host has opened door 2 showing the other goat.
• The player originally picked the door hiding Goat B. The game host has opened door 3 showing the other goat.

The host opens only one door, so only 3 of these cases apply and switching wins in 2 while staying wins in one. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.

Is this the kind of change you're asking for? -- Rick Block (talk) 14:26, 14 February 2008 (UTC)

Your six equally likely cases depend on the goats being distiguishable. How would your approach solve the same problem if there were no goats at all? 171.64.78.10 (talk) 23:15, 15 February 2008 (UTC)

Pretty clearly to maintain any connection to an intuitive 1/3-ish distribution, the primary cases have to be based on the location of the car which forces it closer to your (?) conditional probability analysis two sections down (obviously your point). We could perhaps still avoid describing this in terms of conditional probability, sort of like so:

---

More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:

• The player originally picked the door hiding the car. In this case the game host opens door 2 half the time and door 3 half the time, so there are two subcases each with probability 1/6
  • The player originally picked the car and the host opens door 2
  • The player originally picked the car and the host opens door 3
• The car is behind door 2 and the host opens door 3.
• The car is behind door 3 and the host opens door 2.

The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, playersa player who switches will win the car on average two times out of three.

The argument of the kind "players who switch will win the car on average two times out of three" is unconditional. On the other hand, once a door has been open, the player has posterior probabilities -- the 1/6 becomes 1/3 and the 1/3 becomes 2/3. This is important because one can easily come up with games where it would be optimal to switch if door 3 was opened, and to stick if door 2 was opened. The strategy of always switching in the Monty Hall problem gives 2/3 unconditional probability of winning, but unless the host randomizes (1/2,1/2), the conditional probabilities are not equal to 2/3, even though it is optimal to switch because the conditional probability of winning is at least 1/2 in any conditional case. This was already pointed out here but you rejected it. It would be more respectable for Wikipedia not to make restrictive assumptions if a simple analysis can handle the problem without them. 171.64.78.10 (talk) 03:22, 16 February 2008 (UTC)

---

Other than recasting this as an enumeration of cases, I don't think this is semantically different from your version (two sections below). Do you agree with this? -- Rick Block (talk) 02:32, 16 February 2008 (UTC)

I think the difference is unconditional vs. conditional. See above. 171.64.78.10 (talk) 03:22, 16 February 2008 (UTC)

Does the change above from plural to singular help? The goal here is a description accessible to a layperson, which I think needs to avoid use of technical terminology. We're not writing an article for a statistics textbook, but an article in a general encyclopedia. Responsive to your other comment (just above) we (not just me) have rejected unilateral changes made to a featured article without consensus. The extended discussion here has resulted in some changes already and will likely result in more. I think there's already agreement that the unrestricted version you want to see should be included, with most editors favoring putting it in the "Other host behaviors" section (a little hard to tell, but I think its unanimous other than you). -- Rick Block (talk) 17:15, 16 February 2008 (UTC)

Suggested Rigorous Proof

Suppose the player picked door 1. Before the host has opened any door, there are three equally likely possible situations, namely, the car is behind each door with probability 1/3. It is assumed that if the car is behind door 1, then the host opens door 2 or door 3 with equal probabilities. If the car is behind door 2, then the host opens door 3, and if the car is behind door 3, he opens door 2. Therefore, the probability that the car is behind door 1 and the host opens door 3 is (1/3)x(1/2)=1/6. The probability that the car is behind door 2 and the host opens door 3 is (1/3) x 1 = 1/3. These are the only scenarios under which the host opens door 3. Therefore, the probability that the host opens door 3 is 1/6 + 1/3 = 1/2. Given that host has opened door 3, the conditional probability that the car is behind door 1 is (1/6)/(1/2)=1/3, and the conditional probability that the car is behind door 2 is (1/3)/(1/2)=2/3. Therfore, given that the host opened door 3, switching wins the car with probability 2/3. Likewise, given that the host has opened door 2, switching wins the car with probability 2/3. 70.137.163.193 (talk) 16:44, 14 February 2008 (UTC)

Ok, I read this, and I do not see how it is any different from the standard Bayesian analysis that is already in the article. Do you think that is not "rigorous" enough? Please, read it, compare it to yours, and point exactly where they differ. Thanks.The Glopk (talk) 17:13, 14 February 2008 (UTC)

This is proposed (at the request of Rick Block)) as a replacement for the current "solution" in the very beginning of the article, because the current one does not solve the problem (it offers unconditional probabilities). Of course, this is a straightforward calculation of a conditional probabilities, direct from the definition. It is not even Bayesian.70.137.163.193 (talk) —Preceding comment was added at 17:50, 14 February 2008 (UTC)

To be clear, are you suggesting this for the lead section (before the table of contents, perhaps in place of the paragraph starting "Because there is no way..." ) or the "Solution" section (for example in place of the 2nd paragraph starting with "The chance of initially choosing ...")? In either case, are there specific other changes you're interested in? -- Rick Block (talk) 18:19, 14 February 2008 (UTC)

This is proposed to replace paragraphs 2 and 3 in the "Solution" section. I would also include there the simple extenstion where p replaces 1/2 and the same calculation gives 1/(1+p). 70.137.163.193 (talk) 18:37, 14 February 2008 (UTC)

Yes, that is clear. However, I find the current paragraph beginning 'More formally....' (just below the illustration) appealing because, in casting the host's action in terms of which goat rather than which door it is consistent with the interpretation that the labeling of the door ('say No. 3') is not intended to be significant. I strongly suspect that was Marilyn's intent, and is equivalent to stipulating equal probabilities among permissible doors. Under that interpretation, I think the paragraph is valid. Is this incorrect? IMHO this is a simpler analysis that is more accessible to the layperson, but I could be wrong because a good many laypersons are still stuck on the wrong answer regardless. 67.130.129.135 (talk) 20:14, 14 February 2008 (UTC)

Quoting Morgan et al.: "Its beauty as a false solution is that it is a true statement! It just does not solve the problem at hand."198.4.83.52 (talk) 21:32, 14 February 2008 (UTC)

I'm certain that Morgan et al. must have found it pedagogically entertaining to make that quip, but that just is not a true statement!

The "unconditional probablity" solution to the problem at hand is, in fact, as correct as the "conditional probability" solution. It is not the most mathematically rigorous and formal solution to the problem, it's true. But that seems entirely appropriate, given that its presentation in the "Problem and Solution" section of this article is meant primarily to help the general reader to understand why his/her "intuition" that "switching doors does't matter" is incorrect. The discussion in the "Bayes' Theorem" section then more rigorously addresses the point you (ie., the student of Morgan et al.) now seem to be making, ie. "The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors."

I see no reason to alter the article as you propose, and I wonder if your own sense of pedagogical humor is not becoming somewhat counterproductive at this point.

-- Wikiscient (talk) 23:18, 14 February 2008 (UTC)

Quoting further from Morgan et al.: "[This solution] is immediately appealing, and we found its advocates quite reluctant to capitulate." 198.4.83.52 (talk) 00:32, 15 February 2008 (UTC)

And what I'm saying is: for good reason, in both cases! --Wikiscient (talk) 00:42, 15 February 2008 (UTC)

(outindent) You keep saying this, but what exactly about it is false? You didn't respond to the suggestion (previous section) about expanding it from 3 to 6 cases (treating each possible door that can be opened separately). Even in the 3 case version it explicitly enumerates the cases in effect after the host has opened a door (either door). It doesn't use the words "conditional probability" and doesn't show the derivation of the equal probability cases, but is this necessary? How about:

More formally, after the host opens a door and the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with probability 1/3:

• The player originally picked the door hiding the car. The door the game host opened shows one of the two goats.
• The player originally picked the door hiding Goat A. The door the game host opened shows the other goat.
• The player originally picked the door hiding Goat B. The door the game host opened shows the other goat.

I'm not overly attached to this description, but just want to understand what the objection is. -- Rick Block (talk) 23:08, 14 February 2008 (UTC)

From the point of view of the player, the doors are distinguishable, not the goats. Therefore, to make her decision to switch or not to switch, she relies on her conditional probability of winning, given that the host opened door 3 (or the right-hand door, if you wish). In other words, she conditions on a certain door . The analysis should state such a conditional probability. Because the goats may be assumed indistinguishable, it suffices to look at only three situations (depending on where the car is) rather than six (including the placement of the goats). You might as well replace "goat" by "no car."

-- 171.64.78.10 (talk) 00:54, 15 February 2008 (UTC)

One need not use "conditional probability" to correctly solve this problem.

Consider the "Increasing the number of doors" section of this article:

"The same algorithm can be followed for any number of doors, N. The algorithm is "Choose a door, eliminate some number (0 < x < N-1) of the remaining, losing doors, decide to switch or not." This algorithm can be followed for N = 3 or N = 100. The higher N values demonstrate the same mathematical principle in a more obvious way."

That is a quite sufficiently general, rigorous, and valid solution to this problem.

--Wikiscient (talk) 02:08, 15 February 2008 (UTC)

The point many of us are trying to make is that given the constraints on the host's behavior making the problem unambiguous, which of the two doors the host opens makes no difference. Again, in the spirit of understanding (and I'm not being belligerent, honestly - just trying to understand your point of view), how about:

More formally, there are 6 possible arrangements of 2 goats and a car behind 3 doors - [C,G1,G2], [C,G2,G1], [G1,C,G2], [G1,G2,C], [G2,C,G1], [G2,G1,C]. Suppose the player picks door 1 and the host opens door 3. There are three possible situations corresponding to the player's initial choice, each with probability 1/3:

• The player originally picked the door hiding the car (arrangements [C,G1,G2] or [C,G2,G1]) and the host has opened door 3 showing one of the goats.
• The player originally picked the door hiding Goat 1 (arrangements [G1,C,G2] or [G1,G2,C]). The game host has shown Goat 2 behind door 3 (eliminating arrangement [G1,G2,C]).
• The player originally picked the door hiding Goat 2 (arrangements [G2,C,G1] or [G2,G1,C]). The game host has shown Goat 1 behind door 3 (eliminating arrangement [G2,G1,C]).

If the player chooses to switch, the player wins the car in the last two cases. blah blah blah

If we flip this from what's behind the door the player picks (which is clearly 1/3 car/goatA/goatB) to where the car is (which is also clearly 1/3 door 1, door 2, door 3) we can't keep the assumption constant that the player picks door 1 and the host opens door 3. Keeping an analysis that matches an intuitively obvious 1/3,1/3,1/3 distribution is extremely powerful and I would be reluctant to give this up unless there's simply no alternative. Comments? -- Rick Block (talk) 03:27, 15 February 2008 (UTC)

While I am in favor of keeping it as simple as possible, I am coming to see some merit in the critique that our friend of many IP addresses makes. I am not agreeing with the argument, but it has helped me see that the solution is incomplete.

First, allow me to apologize because some of my remarks were non-sequitor. By slip of the mouse, I was viewing Mueser et al. rather than Morgan et al., so I did not quite see the point being made. Both papers stress that the generally accepted solution 'assumes information that is not given in the problem.' This left me unclear on whether 'it just does not solve the problem at hand' referred to the interpretation of the problem or the adequacy of the solution. Sorry about that.

Still, to my way of thinking conditional probability is a means of analysis rather than a problem definition, so it has taken me a while to get the point. An equivalent case-by-case analysis, such as the six cases Rick suggested above, might still need to justify using equal weight by explicitly noting that all dependant atomic events have been multiplied out into independent composite events. I concur that the existing three case analysis is deficient, because it does not account for the dependant cases / conditional probabilities. Each with probability 1/3 is true, but not justified in the text.

I suspect that a case-by-case analysis is more accessible to the layperson than a conditional probability analysis, but I have no real evidence to justify that hunch. In my student days the former was taught before, and used to explain, the latter. No doubt that has colored my thinking. 67.130.129.135 (talk) 03:41, 15 February 2008 (UTC)

I strongly suspect the absolute probability of a layperson understanding a conditional probability analysis is near 0. -- Rick Block (talk) 04:05, 15 February 2008 (UTC)

Note that there is also a Wikipedia article on conditional probability. The question as stated in this article is: "Do the player's chances of getting the car increase by switching to Door 2?" By definition, this is a conditional-probability question because the setting is that a (possibly random) step has taken place (the host opened door 3), and the problem calls for comparing the posterior probabilities of "car at 1" and "car at 2" given that the host opened door 3. Thus, conditional probabilities are inevitable. If you do not calculate them, your solution does not solve the problem. A person who does not understand the concept of a conditional probability cannot appreciate this paradox, which is about mistakes in guessing conditional probabilities. Conditional probability is taught in high school. This problem is an excellent lesson on conditional probabilities. Perhaps the reader should be advised to read the article on conditional probability if not familiar with it.

- 70.137.163.193 (talk) 05:43, 15 February 2008 (UTC)

Please read the "Combining Doors" section of the present article.

In other words: this need NOT be by definition a "conditional probability" problem. It is, in fact (the more I think about it), an unnecessarily cumbersome and inelegant approach to calculating the probability distribution of final outcomes for this problem.

Why? Because the question here is whether or not to choose one probability distribution or another. Both of those probability distributions are trivially determined independently of which door the host happens to open. That the host opens some door has p = 1 -- and so it just is not relevant to this problem.

Again: the problem is that you are being asked, "do you prefer probability-distribution-A, or would you rather have probability-distribution-B?" You can (and the present article does) correctly calculate both probability distributions before any events have occurred. Both of the events which do occur before you make your choice -- a: you select a door, and b: the host selects a different door, with a donkey behind it -- always have outcomes that are irrelevant to your decision as to which probability distribution you would like to choose.

Once again: you do not need to use Bayes' theorem or any other treatment of conditional probability to arrive at a valid and accurate calulation of the two probability distributions between which you are to choose solely on the basis of whether you are interested in maximizing your chances of winning the car (switch) or whether you are interested in maximizing your chances of winning a goat (don't switch)!

If you disagree with this analysis (in support and summary of the present content of this article), please formally refute it with mathematical rigour -- and I for one will gladly "capitulate."

--Wikiscient (talk) 08:22, 15 February 2008 (UTC)

BTW: You know the famous (but apparently apochryphal?) story about Gauss and the sum of integers from 1 to 100, right?

Your argument here, so far, amounts to the same thing as if you were to have said to (the apochryphal) young Gauss,

"No! If you do not add them like this: '1 + 2 + 3 + 4 + ... + 100,' then you are WRONG!"

;) --Wikiscient (talk) 09:10, 15 February 2008 (UTC)

How does the 6 case analysis at the end of the previous section fail to address your concern (because the equal probabilities are asserted rather than computed)? -- Rick Block (talk) 15:31, 15 February 2008 (UTC)

Consider something along the lines of "Three possible choices for the player multiplied by two possible choices for the host gives a total of six independent possibilities with equal probability." 67.130.129.135 (talk) 18:18, 15 February 2008 (UTC)

Conditional-probability analysis for the layperson

It is all from the point of view of the player. The figure in the current solution section is very misleading for the layperson, because it does not show what the player knows (the player knows that he has chosen door 1):

• The three initial situations, with their initial probabilities:

CGG: 1/3

GCG: 1/3

GGC: 1/3

• The player opens door 1.

• The host will open a losing door, picking at random if there are two. The uncondtional probailities (before the host has acted):

CGO (Host opens Door 3)  : 1/6

COG (Host opens Door 2)  : 1/6

GCO (Host opens Door 3)  : 1/3

GOC (Host opens Door 2)  : 1/3

• The host opens door 3 using the above-mentioned potocol. Given that the host has opened door 3, only the following remain possible:

CGO (before: 1/6; after: (1/6)/[(1/6)+(1/3)] = 1/3, and

GCO (before: 1/3; after: (1/3)/[(1/6)+(1/3)] = 2/3.

• Conclusion: The car is behind door 2 with probability 2/3.

--198.4.83.52 (talk) 20:06, 15 February 2008 (UTC)

Now give the analysis for the scenario where the host's choice of losing doors (if there are two available) is unknown to the player.--Father Goose (talk) 05:28, 16 February 2008 (UTC)

If the host is not constrained to pick equally among goat doors by the problem statement, there is no basis for using the constant 1/6 in the analysis. Treating this aspect of the host's behavior as an unknown makes the analysis the same as one from somewhere above, where the host's preference is expressed as a probability p with the conclusion that the car is behind door 2 with probability 1/(p+1). The numbers change as follows:

(before host opens a door)

CGO (Host opens Door 3)  : 1/3*p

COG (Host opens Door 2)  : 1/3*(1-p)

GCO (Host opens Door 3)  : 1/3

GOC (Host opens Door 2)  : 1/3

(after)

CGO (before: 1/3*p; after: (1/3*p)/[(1/3*p)+(1/3)] = p/(p+1), and

GCO (before: 1/3; after: (1/3)/[(1/3*p)+(1/3)] = 1/(p+1).

This is exactly the analysis user:anon-many-ips wants to see in the article. Switching is better (actually no worse) than staying regardless of p, so the overall conclusion remains the player should switch. However, the exact probability is not necessarily 2/3. -- Rick Block (talk) 17:56, 16 February 2008 (UTC)

Indeed, this is where the "Rigorous Solution" section started. Are we getting close to consensus?

- 70.137.163.193 (talk) 18:48, 16 February 2008 (UTC)

The exact probability is 100% or 0%, depending on whether the car is behind the player's original door or not. A conditional probability is p/(p+1) or 1/(p+1), depending on whether the host opened his "favored" door or the other one. However, the player does not have such information in either case. Is he incapable of calculating an exact probability of his chances of winning by switching if he does not have that information?--Father Goose (talk) 20:22, 16 February 2008 (UTC)

Right (and the 0% or 100% is the conditional probability following the event of opening the player's final choice of door).

To user:anon-many-ips - I think there hasn't been significant opposition to including this analysis somewhere in the article ("Other host bahaviors" or after the existing analysis in the Bayes' theorem section being the most often suggested spots), but I don't think anyone other than you is in favor of making it (with the preference p as a variable) the main explanation in the Solution section. Many folks have been kind of quiet lately so additional comments at this point would be good. -- Rick Block (talk) 20:39, 16 February 2008 (UTC)

Okay, I buy it... "if the host is not constrained to pick equally among goat doors by the problem statement." The consensus seems to be that, in the most widely "acceptable" definition of the problem, the host is so constrained. Hence my original support for the conditional case in question under the "Other host behaviors" section. —Preceding unsigned comment added by Wikiscient (talk • contribs) 21:13, 16 February 2008 (UTC)

Current statement of the solution

More formally, 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:

• The player originally picked the door hiding the car. The game host has shown one of the two goats.
• The player originally picked the door hiding Goat A. The game host has shown the other goat.
• The player originally picked the door hiding Goat B. The game host has shown the other goat.

If the player chooses to switch, the player wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.

So far so good.

The solution would be different, for example, if the host did not know what was behind each door, or if the host sometimes had the option of not offering the player the chance to switch, or if the host chooses to open one of two doors concealing a goat non-randomly.

How would the solution described in the text above "so far so good" change if the host opened one of two goat doors non-randomly?--Father Goose (talk) 22:17, 16 February 2008 (UTC)

It would be exactly the same description. This proof is only about the unconditional probabilities. The 2/3 does not change. This proof does not use the fact that the player knows which door (i.e, left, middle, or right) the host has opened. To argue about the probabilities of winning of the remaining two doors, the doors should be referred to as L, M, R or 1,2,3 rather than "the door that concealls the car," etc. In my mind, the indirect referencing is more confusing for the layperson.

-70.137.163.193 (talk) 23:01, 16 February 2008 (UTC)

How does knowing which specific door the host has opened change the player's analysis, given that he is unable to distinguish any door from any other (aside from "the door I picked", "the door the host opened", and "the door that remains")?--Father Goose (talk) 01:18, 17 February 2008 (UTC)

I think there are two separable suggestions at play here. One is to reword the solution to address the conditional situation (the position of a specific player after the host has opened a specific door). The other is to drop the constraint that the host open one of two goat doors with equal probability. I think a solution worded similarly to the existing one that addresses the first concern would be like the one above, at the end of #What Morgan et al actually say (although anon-many-ips apparently still thinks this is an unconditional solution). Addressing both requires a more elaborate explanation, and (I think) has to introduce a variable representing the probability of the host's preference (the p we've been talking about). I'm not seriously suggesting this one, but I think it would perhaps be like the following. -- Rick Block (talk) 02:02, 17 February 2008 (UTC)

---

More formally, when the player is asked whether to switch there are three possible situations corresponding to the location of the car, each with probability 1/3. Assuming the player initially picked door 1:

• The player originally picked the door hiding the car. In this case if the game host opens door 3 with probability p and door 2 with probability (1-p) there are two subcases, with probabilities 1/3*p and 1/3*(1-p)
  • The player originally picked the car and the host opens door 2 - probability 1/3*(1-p)
  • The player originally picked the car and the host opens door 3 - probability 1/3*p
• The car is behind door 2 and the host opens door 3.
• The car is behind door 3 and the host opens door 2.

The host opens only one door, so only one of the first subcases and one other case with probability 1/3 apply. In the case where the player has initially picked the car, staying wins with probability 1/3*p or 1/3*(1-p) depending on whether the host opens door 3 or door 2. Switching wins in the 1/3 case where the player has not. Since p is a probability it ranges between 0 and 1, so switching always wins with at least the same probability as staying regardless of which door the host opens — and depending on p and which door the host opens might guarantee winning the car. In other words, a player will win the car more often by switching than by staying with the initially picked door.

If all players switch regardless of p and what door the host opens, their average probability of winning is 2/3 since 1/3 of these players will initially pick the car and lose while 2/3 will initially pick a goat and win.

---

"The other is to drop the constraint that the host open one of two goat doors with equal probability."

*Goose jumps up and down eagerly at this suggestion*

I don't mind including a conditional analysis of the problem in the article, but I think it would best be included in the Bayesian section. It's not necessary to analyze the problem as a conditional problem, and the conditional analysis introduces complexity that is not of use to the lay reader.--Father Goose (talk) 04:53, 17 February 2008 (UTC)

Now I'm confused. Including this constraint allows us to say, with certainty, that the probability of winning by switching is 2/3. This constraint (and the others) makes this the probability a player faces at the beginning before choosing a door, after choosing an initial door before the host opens a door, and after the host opens a door (either door). With these constraints, the unconditional and conditional probabilities are the same so the distinction between these analyses (although still there in a purely academic sense) are of mostly theoretical interest. I'm not sure where anyone else is on this, but I'd strongly prefer we

1. keep the constraint so the problem definition forces a single numerical answer
2. reword the solution description so it doesn't offend any absolute purists (this would be you, anon-many-ips)
3. include the more general problem (with p, with a more mathematical exposition explicitly distinguishing the unconditional and conditional probabilities) in a later section (possibly the Bayes' theorem section)

-- Rick Block (talk) 05:37, 17 February 2008 (UTC)

Here's a variant I'd like to discuss:

Whenever the player picks the "car" door, the host will always open door #3, unless that is the player's chosen door, in which case the host will open door #2.

The player knows that the host will always reveal one goat, and even knows that the host favors one of the doors (when given a choice between two goats) although the player doesn't know which door the host favors.

The player has made his choice, and the host has revealed a goat. The player reasons that his likelihood of initially picking the "car" door is 1/3, and that his likelihood of winning by switching is 2/3 -- i.e., he uses the unconditional analysis. Has he analyzed the problem wrongly?

If not, then we don't need to mention the "host chooses randomly" constraint in the basic statement of the problem. Meanwhile, I'd be fine with your third suggestion, putting the more general problem (including p) in the Bayes' theorem section.--Father Goose (talk) 10:40, 17 February 2008 (UTC)

Are you suggesting discussing this variant here (on the talk page) or in the article? If in the article, I'd suggest we find a reference that discusses this variant (not that we would be unable to solve it ourselves, but articles are meant to present nothing that can't be independently verified against a published source). I failed to mention this with regard to including the constraint, but (as far as I know) the "standard" version of the problem in the published literature (technical and not) includes this constraint - so the article should as well.

WRT the variant, I'd guess that if anon-many-ips is a professor and you were in his class, you'd get a C- since 2/3 is the correct numerical answer but you haven't shown his preferred analysis (which would be that you win by switching with a 50% chance if the host has opened his preferred door and a 100% chance if the host has opened his non-preferred door, and there's a 2/3 chance the host has opened his preferred door, so your overall chances are 2/3*50% plus 1/3*100% which is 2/3 - not to be confused with your 2/3 which is the proportion of players who switch that will win). -- Rick Block (talk) 17:16, 17 February 2008 (UTC)

Not to further belabor this point, but it might help to consider a variant that forces the conditional analysis. Rather than the host open a door and offer the chance to switch, the host gives the player the following choice. Keep your door or let me pick a door for you. How I'll pick is by first selecting a random number between 1 and 3. If the number is 1 the door I'll pick is the one where the car is (even if it's your original door). If the number is not one I'll pick randomly (equally) between your door and one more where the other one will be the car door unless you've picked it. The door I've picked for you is door X (could be 1, 2, or 3). With these rules, if you take the offer you don't necessarily win if you pick a goat first and you don't necessarily lose if you picked the car first, but you still end up winning 2/3 of the time (1/3*100% + 2/3*50%). This 2/3 is difficult to get to without doing the conditional analysis, but it would be exhibited by a simulator running hundreds or thousands of trials. This is a variant where the existing unconditional analysis would be manifestly wrong, even though it ends up with the same 2/3 answer. I'm guessing a bit (and not saying I agree), but I suspect anon-many-ips considers the unconditional analysis of the existing problem just as wrong. -- Rick Block (talk) 20:57, 17 February 2008 (UTC)

Why so quiet

Moved from User talk:Father Goose

Any idea why Talk:Monty Hall problem got so quiet all of a sudden? I thought we were almost there. -- Rick Block (talk) 01:01, 20 February 2008 (UTC)

Oh, frustration on my part. You're right that both Morgan and Mueser both point out that the problem, phrased a particular way, requires the conditional probability to be computed. The solution would be to use an unconditional statement of the problem, only the one that Morgan gives (according to the excerpt on the talk page) -- "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?" -- is not a complete statement of the problem, and is thus useless.

Left in a pedantic cul-de-sac, I chose to back out and drive away.--Father Goose (talk) 01:30, 20 February 2008 (UTC)

Where I thought it was going is that we'd end up with the constraint, and with a case-based analysis equivalent to the conditional analysis (like the one I wrote up in text, at the end of Talk:Monty Hall problem#What Morgan et al actually say). This doesn't match the illustration exactly, but it's not tremendously different. A possible alternative we haven't really talked about is asserting that the constraints make the conditional and unconditional analyses the same, deferring the proof to the Bayes' theorem section, and presenting the unconditional analysis. I'd guess this is more of a cop-out than anon-many-ips would be comfortable with, but if he's not returning to the talk page his preference probably won't count for much. In any event, I look forward to your further comments on the talk page. -- Rick Block (talk) 03:09, 20 February 2008 (UTC)

I am unable to determine how the conditional and unconditional statements of the problem differ, so these claims of the unconditional analysis being a "false solution" drive me to distraction. Treated as a logic problem, the unconditional analysis is plainly correct, and not simply by coincidence. And what is most important for the sake of a general-readership encyclopedia is that it easy to grasp. If for some reason Morgan et al. and Mr. IPs insist otherwise, well, I can't argue the math of it, but I know there's some blind men and an elephant problem going on.--Father Goose (talk) 03:29, 20 February 2008 (UTC)

The "false solution" bit is specifically about the unconditional solution applied to a version of the problem for which 2/3 may not be the correct answer (where the equal choice among goat doors constraint is not present). If we constrain the host so that nothing he does can affect the 2/3 outcome, (IMO) the difference between the unconditional solution and the conditional solution becomes an exercise in pedantry. The difference in the problem statements is "what is the general probability switching wins, over all players" vs. "given the host has opened door 3 (or 2), what is this specific player's probability of winning by switching". Constraint or not, the general probability is 2/3. However without the constraint, if the host has an absolute preference for door 3, the player specific probability is not 2/3 for all players - it's 1/2 for players when the host opens door 3 and 100% for players when the host opens door 2. With the constraint, it doesn't matter which door the host opens - each and every player's probability of winning by switching is 2/3. -- Rick Block (talk) 04:19, 20 February 2008 (UTC)

The problem is that if the analysis we present in the "Solution" section is correct, it is correct no matter what the host's behavior is when choosing between two goats -- switching away from the car never wins the car. We will either have to ditch our non-Bayesian analysis or ditch the constraint on it that the Bayesian model must use, since other means of analyzing the problem do not necessarily require such a constraint. If need be, in the Bayes section of the article, we can say something like "the analysis given in the Solution section above corresponds to the unconditional probability, which in the Bayesian model requires that p be assumed to be 1/2... If p is not assumed to be 1/2, a conditional probability must be calculated, using the following formula: (modified formula).

If for some reason we cannot take an approach like this in the article, I will mathematically model the placement of a printed copy of Morgan et al into the posterior region of the esteemed statisticians' posterior regions.--Father Goose (talk) 05:45, 20 February 2008 (UTC)

It is a correct analysis for the question "what is the general probability switching wins, over all players" (no matter what the host's behavior is when choosing between two goats) and arrives at the (correct) answer 2/3. With the constraint the host pick equally between two goats, 2/3 is also the answer to the question "given the host has opened door 3 (or 2), what is this specific player's probability of winning by switching". This means every single player, at the point he or she is asked if they want to switch, has a 2/3 chance of winning by switching (which is of course consistent with the analysis that says the general probability, for all players, is 2/3). The trick is that without the constraint, the general probability remains 2/3, but the host can act in ways where the individual (per player) probability at the point they're offered the chance to switch is not. So an analysis that says 2/3 is the specific player's probability regardless of this constraint must be wrong.

If it's wrong, how is it wrong?

The basic issue is that at the point the player picks, without the constraint not all combinations of car placements and open doors are equally likely. Specifically, assume the host has a 100% bias toward door 3 and you've originally picked door one. The host opens door 3 whenever the car is not behind door 3 (2/3 of the time), but then this says nothing at all about the chances of the car being behind door 1 or 2. It says only "the car is not behind door 3". Switching in this case is a 50/50 chance. Alternatively the host has to open door 2 the other 1/3 of the time, and this says "the car is without a doubt behind door 3". Switching in this case is a 100% win.

It's a little hard to see this problem given the current diagram, but it's fairly easy if in row 2 the player still picks door 1 and the car is behind door 2 (forcing the host to open door 3, as is shown) and in row 3 the player still picks door 1 and the car is behind door 3 (forcing the host to open door 2, as is shown). With this change the 1/3 per row is the equal probability location of the car, not the player's pick of car/goat a/goat b. Based on the door the host opens, one of row 2 or 3 applies (1/3 probability) and one of the subcases from row 1. Without the constraint we don't know the probability of these subcases - could be it was a equally random pick making them each 1/6 or (in the extreme) it could be 100% biased toward one door making one 0 and the other 1/3. If we keep the constraint, whichever door the host opens only one of the 1/6 subscases (where we picked the car) and one of the 1/3 cases (where we didn't pick the car) still apply, so we win by switching twice as often (i.e. with probability 2/3). If we don't keep the constraint we don't know the probability of the subcases - they could be anywhere between 0 and 1/3. 0 means the host has a 100% bias and was forced to open his non-favorite door, and switching is guaranteed to win. 1/3 means the host has a 100% bias and opened his favorite door, and switching has the same chance of winning as staying.

I think I've already said this, but I suggest we keep the constraint and change the solution description (and the figure) to be consistent with the explanation above. -- Rick Block (talk) 14:44, 20 February 2008 (UTC)

The player's odds of winning are never 2/3, 1/3, or otherwise. They are always 100% or 0%, depending on whether the door they are considering switching to contains the car or not. All other solutions are false.--Father Goose (talk) 20:51, 20 February 2008 (UTC)

Well, actually, that's just taking the conditional analysis somewhat further. If I flip a coin and don't show you the result, yes, it's either heads or tails with 100% certainty, but until you know the outcome your chances of picking the result are 50%. However, if I flip a coin and half the time after I flip it blink rapidly twice if it's heads (I can see it and you can't), then your 50/50 chance turns into something else entirely depending on whether I've blinked twice rapidly. You still don't know what it is, but you have a better clue. Same thing with the doors. Without the equal goat door constraint, the initial 2/3 can turn into something else depending on what the host does.

Are you seriously saying you don't understand the above explanation? Anybody else want to give it a try? -- Rick Block (talk) 00:09, 21 February 2008 (UTC)

In the above scenario, you are saying that half of the time when it's heads, you will blink, and I can perceive that, which allows me to know with complete certainty that it's heads?--Father Goose (talk) 01:36, 21 February 2008 (UTC)

Yes, but I'm not asking if you understand the coin example. Do you understand the door explanation above, from 14:44, 20 February? -- Rick Block (talk) 02:10, 21 February 2008 (UTC)

I do, and I've understood it every time it's been presented on this page (about a half a dozen times now). Now, as for the coin example, the action you must take to maximize your chances of winning has changed because the host's blinking has conveyed information that you did not have at the start of the game. The same cannot be said of the Monty Hall problem; after the host opens a goat door, you know nothing more of the probabilities you face than you did before the game started. You knew he was going to open a goat door, and you knew his choice of doors would not convey to you any information about whether you picked the "car" door in the first place.--Father Goose (talk) 04:02, 21 February 2008 (UTC)

With the equal goat constraint I agree. Without the constraint, the host may or may not be conveying more information. So an analysis that does not distinguish these cases must be at least incomplete (many-anon-ips says "false"). The problem with the current analysis is the assumption that the host's actions don't affect the probability of winning (they can, in the case where the host has a preference for one goat door over another). With the current diagram, work out the scenarios where the host has a 100% preference for door 3. In this case the player's chances of winning by switching are either 50% or 100% depending on the door the host opens (rows 2 and 3 each apply half the time). The 100% case happens 1/3 of the time and the 50% case 2/3 of the time, so the overall chances are still 2/3, but if the question is "given the host has opened door 3 (or 2), what is this specific player's probability of winning by switching" 2/3 is not the right answer. The equal goat constraint ensures 2/3 is the right answer, but since the analysis we provide doesn't differ depending on whether we include this constraint our analysis must be wrong.

What about this aren't you seeing? -- Rick Block (talk) 05:39, 21 February 2008 (UTC)

By what means would the player be able to gain extra information from a host's non-random pick amongst two goats?

I'm the player, I don't know what's behind any of the doors, I've picked door 1, the host has revealed a goat behind door 3, and I'll even add the constraint that I know the host always opens one door preferentially -- I just don't know which door. When he reveals a goat, do I have any greater insight into where the car might be than I would if he always picked randomly? What has tipped me off? What perceptible, real-world event have I just witnessed that makes me think, aha, my odds of winning by switching are no longer 2/3ds?--Father Goose (talk) 07:05, 21 February 2008 (UTC)

This is yet another question - what is the player's perceived probability of winning by switching? Question for you - in the regular set up, if Monty forgets where the car is and opens a door at random (without letting on to anyone that he forgot) but fortuitously reveals a goat, what would you say is the player's chance of winning by switching? Does it matter whether the player knows Monty forgot? -- Rick Block (talk) 14:25, 21 February 2008 (UTC)

(reset indent)All right, I'll take a different tack. Is it not possible to formulate a unconditional conditional analysis that represents the fact that any time the player picks the car first, he will lose when switching, no matter which behavior the host adopts when choosing between two goats?--Father Goose (talk) 20:50, 21 February 2008 (UTC)

Excuse me, I meant to say a conditional analysis.--Father Goose (talk) 00:36, 22 February 2008 (UTC)

Isn't that the one involving p? Can you tell me what you dislike about the suggested text at the end of #What Morgan et al actually say? It seems to me that this is not that different than what we have and addresses the original concern raised by anon-many-ips. -- Rick Block (talk) 01:42, 22 February 2008 (UTC)

That appears okay, but my concern is with the specific sentence "opens either of two doors concealing goats with equal probability". If it is possible, using a conditional analysis, to show that any time the player picks the car first, he will lose when switching, no matter which behavior the host adopts when choosing between two goats, then we do not need to specify what behavior the host adopts when choosing between two goats.--Father Goose (talk) 02:18, 22 February 2008 (UTC)

I've made this point before as well - we specify the behavior and have since July 2005 because it is how the problem is generally presented in the literature (although there are certainly some exceptions). Without specifying the behavior, the answer to "what is the player's chance of winning by switching" (specific player, not all players in general) is "it depends - anywhere from 1/2 to 1/1". The player is no worse off switching (and always loses if he picks the car first), but the specific answer of "2/3 chance of winning by switching" is the answer for each and every player only if the constraint is there. IMO, omitting the constraint and presenting a primary solution in this article that says "it depends" is not a reasonable option.

Has everyone else completely lost interest in this? I don't mind our back and forth, but even if we end up in agreement we'll only have a consensus of 2. -- Rick Block (talk) 03:37, 22 February 2008 (UTC)

Fuck it, I'm spending too much time on this with nothing to show for it. See you later.--Father Goose (talk) 09:48, 22 February 2008 (UTC)

Ahh, I feel better already.--Father Goose (talk) 09:52, 22 February 2008 (UTC)

Hello again Rick. No, this anon-single-ip has not lost interest. Just a little weary of going in circles and, well, I do have a day job. Part of my interest comes from having used the problem many times as a parlor game, suitable for teenagers through centenarians. It never fails to spark lively discussion and provide pedagogic opportunities. Regarding some of the main issues of recent extended discussion:

. I already stated I agree with the view that the focus of the article should be the notably conventional constrained version. As a problem of 'simple' probabilities it is sufficiently interesting and important, given that the average person 'guesses' 1/2 and is not easily persuaded of 2/3. I assume that is settled.

. I also agree that the game-theoretic version where host strategy must be taken into account is also interesting and notable. I think a brief description under Variants with a link or two to articles on concepts from game theory would be appropriate.

. To clarify the distinction, in the constrained version the host is really just a referee, not a player, and the guest is playing solitaire. The distinction between a priori strategy and a posteriori choice is moot. To use the coin flipping analogy, whether one chooses before the coin is flipped or after or has been flipped but before it is revealed, not only is the answer the same, it is the same for the same reason. The coin’s actions are known to be random with known distribution. It is not choosing a strategy based on expectation of your actions, therefore it is not a player. The answer is a single-valued probability, not a multi-valued game matrix with 0% for one strategy and 100% for the other.

. Because the host’s action is constrained by the guest’s choice, the notion of conditional probability is applicable. This does not mean the problem is no longer a matter of single-valued probability. The notions of conditional probability and strategic choice should not be conflated. 67.130.129.135 (talk) 22:09, 22 February 2008 (UTC)

RfC: Consensus for rephrased solution

An anonymous user (editing from a variety of IP addresses) has expressed a concern about the solution presented in the article. All of the discussion above, starting with the section #Rigorous solution relates to this issue. Should the wording of the primary solution be changed and is the following suggested wording satisfactory?

---

More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:

• The player originally picked the door hiding the car. In this case the game host opens door 2 half the time and door 3 half the time, so there are two subcases each with probability 1/6
  • The player originally picked the car and the host opens door 2
  • The player originally picked the car and the host opens door 3
• The car is behind door 2 and the host opens door 3.
• The car is behind door 3 and the host opens door 2.

The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, a player who switches will win the car on average two times out of three.

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