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.

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Very important topic

{{technical (expert)}}

{{technical}}

The below suggested simple "frequency" solution should replace the current confusing conditional probability solution
See discussion further down for reasons why it should replace the current solution.

Since these have had plenty of expert attention, I am going to disable the templates. — Carl (CBM · talk) 12:16, 16 September 2008 (UTC)

Summary and Solution

There are three doors. Two doors have a goat behind them and one door has a car behind.
(Remember goat is bad and car is good)

1) Choose one door
2) Game host opens one door containing a goat
3) Do you change door?

For example:

Door 1	Door 2	Door 3
Goat	Goat	Car

Now you have two opions: Not switch or Switch

Not switch

pick	show	outcome
1	2	lose
2	1	lose
3	2 or 1	win

If you do not switch the probability of winning is 1/3

Switch

pick	show	outcome
1	2	win
2	1	win
3	2 or 1	lose

If you switch the probability of winning is 2/3

You should therefore choose to switch doors

---

Note that there also exist an simpler and more elegant solution to the above problem.

From above we know that the probability of winning if we do not switch doors is 1/3.

We also know that we only have two options: Not switch or Switch

This means that:

P(switch) + P(not switch) = 1

which means that

P(switch) = 1 - P(not switch)= 1 - 1/3 = 2/3

Which is the same solution we had previously

---

tildes ( 82.39.51.194 (talk) 10:40, 17 August 2008 (UTC) )

I propose a very compact, straightforward solution with no mention of probability per se. The problem is, in fact, simple so we should treat it that way. After reading most of this discussion and some of the references about how to explain it in an understandable way, I've yet to see the following type of explanation.

The problem is easy to solve - every time the game is played, one of two sequences occur:

A)Contestant picks a door that has the car behind it; host opens one of the other doors revealing a goat; behind the remaining door is the remaining goat.

B)Contestant picks a door that has a goat behind it; host opens the other door with a goat behind it; behind the remaining door is the car.

Sequence A happens 1 time in 3, so, doing the math, sequence B happens 2 times in 3.

Stickers win in sequence A and switchers win in sequence B.

I told you it was easy.

Millbast5 (talk) 22:03, 17 September 2008 (UTC)

Hang on a second.

Not switch

pick	show	outcome
1	2	lose
2	1	lose
3	2 or 1	win

3 win? 3 is the goat, which is a known lose... Did I miss something? And this seems to presume knowing the outcome. If I pick 2, & switch, & 2 is the winning door, how, exactly, is this a win? TREKphiler ^{hit me ♠} 05:36, 22 October 2008 (UTC)

Yes you are missing something and that is your brain! :-) Door 1 & 2 are goats —Preceding unsigned comment added by 92.41.207.32 (talk) 10:42, 6 February 2009 (UTC)

Repetitive new solution section

I've undone this change twice. It renames the existing "Solution" section as "Discussion" and adds an additional "Solution" section that basically repeats what's in the existing Solution section - without references, and with a variety of style issues per WP:MOS (such as directly addressing the reader). If there's some shortcoming in the current Solution section please say what it is and we can work on addressing the concern. -- Rick Block (talk) 18:57, 14 August 2008 (UTC)

I've undone it myself too, and it's back. Can we perhaps lock the article until this user responds in the talk page?The Glopk (talk) 14:13, 15 August 2008 (UTC)

--

The problem I have with your so called solution section is that it is general confussing! You are confussed, your glossy matrix and tree are confussed. It is not about where the car is !! The solution will look the same irregardles. It is a question whether you choose to switch or not and which door the contestant choose.
The whole point of the Montey hall excercice is to answer the question wheter you switch or not and still you refuse to take that into account !

Therefor you should divide your matrix it into

Switch
Not switch

and then review

1)all the possible doors the contestant can choose

2)the response of the host

3)the outcome of such a choice

Your original matrix might be glossy and fancy but regret to inform you that it is inaccurate

It is better if you put your glossy matrix under the section "Sources of confusion"

I can also inform you that I am going to change back my solution until you either, stop removing my solution or redesign your matrix.

Ps I found it quite strange that I have to fight for such an obvious point ! It just reaffirm the notion that wikipedia is good in theory but does not work in practice. Now you have to fight a war with every moron with a computer ! Where are the credentials, expertise and self critic ? Just because you have a computer doesn mean that you are an expert !
I have attached the original post
tildes ( 82.39.51.194 (talk) 10:40, 17 August 2008 (UTC) )

---

Discussion

—Preceding unsigned comment added by 82.39.51.194 (talk) 08:35, 16 August 2008 (UTC)

The existing "solution" section is clearer than the above explanation. Cretog8 (talk) 15:04, 16 August 2008 (UTC)

The solution proposed above keeps the arrangement of goats and car the same and varies the player's initial pick rather than keeping the player's initial pick the same and varying the location of the car. Both of these approaches simplify the actual 9 cases (or 18 assuming the goats can be distinguished) down to three to better show the 2/3 probability of winning. In the rest of the article (in particular, in the Bayesian analysis section) the player's pick is kept as door 1 rather than the alternative approach of examining all picks given a single arrangement of goats and cars - and this approach is the most commonly presented in the cited references (and matches the presentation of the problem published in the vos Savant Parade column).

Another reason this approach is used is that the problem as generally presented can be considered a conditional probability problem given both a specific initial pick and a specific door the host has opened. This interpretation of the problem (see the Morgan et al reference) is discussed in the last two paragraphs of the solution section and involves examining the situation after the player's initial pick and given which door the host opens. Initially using the approach where the goat/car configuration is constant but the player's pick varies makes the connection to this conditional analysis extremely difficult to see.

The solution section was actually changed fairly recently to use the "fixed pick, varying car location" approach (rather than the "fixed configuration, varying pick" approach) to be consistent with the majority of the references and with the rest of the article. We can certainly discuss whether including the alternative solution is worthwhile, however given that they are effectively equivalent (both relying on a "without loss of generality" assumption), and that most sources use the "fixed pick, varying car location" approach, and that the approach currently used easily extends to the conditional analysis, and that the article is quite long already, I don't think you'll find much support for including this alternate solution. -- Rick Block (talk) 19:14, 16 August 2008 (UTC)

---

I dont agree with what you are saying ! It is about presenting an easy to understand and approachable article. as of now none of this is taking place. Who gives a crap about "keeping the player's initial pick the same and varying the location of the car". That is not what the Monty Hall problem is about. The sooner you will realize this the better of this article will be! You need to keep the most critical stuff and just remove everything ells ! Howevere I am starting to lose interest in all this bull crap ! This is all about you ! You have written something that you protect like hamster. This is exacly the reason why wikipedia will never gain any credential in academic circles because it is not about the optimal soluton but rather individual ego! —Preceding unsigned comment added by 82.39.51.194 (talk) 19:53, 16 August 2008 (UTC)

Please 82.39.51.194, no personal attacks. Feel free to discuss your displeasure and/or disagreement with the article but refrain from attacking the editors. -hydnjo talk 20:41, 16 August 2008 (UTC)

yes, I agree no personal attacks! It is just that I get so frustrated when the solution is staring them in the face and they still refuse to accept it because they are biased towards their own writing (which apparently is a lot). Just because someone wrote something first doesn't mean that such a person has the right to "validate" (oohh daddy please can I) or delete everything that comes after. You seam to have a lot of rules! Dont you have a rule for that? —Preceding unsigned comment added by 82.39.51.194 (talk) 21:26, 16 August 2008 (UTC)

There is indeed a rule for that: WP:OWN says that nobody should behave as if they "own" a particular article. I've put a welcome message at your talk page which you can look over to find out more stuff. All the same, you have to have some humility, too, and accept that maybe the reason your suggestions aren't being implemented is because several people don't think they're an improvement. Cretog8 (talk) 22:47, 16 August 2008 (UTC)

As a Featured (July 23, 2005) and high-traffic (392 thousand views, Jan-Jun '08) and heavily edited (500 edits to date in '08) article, there are lots of inputs for further improvement. The article has been scrutinized by the community at large in a process called Featured article review and the criticisms from that review have been successfully addressed so that the article meets today's FA standards. Given that bit of perspective, the article is absolutely not frozen and the editors that who are watching continue to to demonstrate (IMO) an open mind when it comes to addressing suggestions for improvement. It may be helpful to scan the archives of this talk page to gain further perspective about the responses to your suggestions. -hydnjo talk 23:26, 16 August 2008 (UTC)

Moreover, the current version attempts to stay as close as possible to the academic references. Have you read the references that are in the article and do you have other references that present the solution using your preferred approach? Even if you're not going to respond to the explanation above about why the current approach is used, a suggestion that you find "so and so's explanation from this paper or book" easier to understand would be received quite differently from "I like my explanation better". -- Rick Block (talk) 23:33, 16 August 2008 (UTC)

With all due respect I am not convinced that you (any of you) fully understand the rules of the Monty Hall Game.

1) First the allocation of goats and the car is taking place (remains fixed for the rest of the game).

2) Then the contestant choose one door.

Your "fixed pick, varying car location" approach totaly contradict this line of reasoning. The location of the car IS NOT changing. The allocation is stationary, it does NOT change over the period of the game. Again the first thing that happens is the allocation of the car. Again that location remaine FIXED for the ENTIRE game. But still you in insist like some stubborn child that the location is varying. It is NOT ! and it never will be ! The Monty Hall game is a SEQUENTIAL GAME which means that you CAN NOT just switch around the location of the car based upon your preferences in a later stage of the game. This means that the ONLY valid way of approaching this is the way I have put forward which means that you keep the allocation fixed and you evaluate each pick individually (the pick, the game show respons, and the outcome).

Also the argument that "majority of the references" do it one way isn't valid either. This is not an excercise in burping up some solution from some jerk off text book. Fine! you should have references to other articles but in the end it is about evaluating which approach is the most correct and not the least which approach is the easiest to understand. I hate to burst you bubble but the current approach dosent fulfill neither of these criterias irregardles if you have won a nobel price for the article.

Further, it is not about Bayesian statistics ! To allow Bayesian statistics to dictate how the original problem and its solution is presented is WRONG. Especially when such a solution contradict(is incorrect)the original game and the sequential steps such a game is based upon. If you want to have a section of Bayesian statistics that is fine! But you need to very carefully point out the different assumption such an analysis is based on! This is not taking place at the moment! The way I see it is that you have all been so blinded by the complexity of the Bayesian analysis that you have completely surrendered to its preachings! --82.39.51.194 (talk) 10:37, 17 August 2008 (UTC)

Perhaps it is not clear to you what is the meaning of a clause like "Assume, without loss of generality, X.". To put it simply, it means: "I could repeat the same argument for cases W,Y,Z,etc., But since these are obviously equivalent to X, as they differ only by a change of names, I'll just write down the case for X and be done with it". For example, consider the proof of the Pythagorean theorem: the logic of the proof would be unchanged if you rotated the names of the triangle's vertexes so that A becomes B, B becomes C, and C becomes A. That is, the particular labeling of the vertexes is irrelevant to the proof. Similarly, in the analysis of the Monty Hall problem, the naming of the doors is unimportant, so long as it is kept consistent within the reasoning. Because of this, for example, the Bayesian analysis of the problem is written concisely using the "Assume, without loss of generality" clause.The Glopk (talk) 16:48, 17 August 2008 (UTC)

(to 82.39.51.194) First, please read Wikipedia:No original research. Taken to its extreme what this policy says is that even if the preponderance of references are provably incorrect (which, just to be clear, is not at all the case here), the Wikipedia article must summarize what they say rather than presenting something we make up ourselves. Second, although you're perfectly correct about the sequencing of the problem we're talking about the probability of winning by switching, either over all iterations of the game or (in the Morgan et al interpretation) given a specific player's initial pick and the host's specific response. Also note that the Parade description of the problem specifically includes the words:

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

If the goal is to determine the overall chance of winning by switching, over all possible scenarios, in a strict sense we should enumerate all possible car locations, player picks, and host responses (per the fully expanded decision tree as presented in Grinstead and Snell). On the other hand, from the player's point of view, the player (who doesn't know where the car is) picks a door and then the host opens a door, so following through what happens when the player picks a specific door (say #1) over all possible car locations (the approach currently in the article and in most references) is entirely sufficient. This is not varying the location of the car after the player has picked, but examining all possibilities of where the car might be to determine the probability of winning. Indeed, given the Parade version of the problem your preferred approach makes almost no sense at all (why are we considering what happens if the player picks door 2 when it's stated the player has picked door 1?). Morgan et al take this one step further, and consider only the scenario where the player has picked door 1 and the host has opened door 3 (eliminating the possibility that the car is behind door 3! - and, even in only this one case, the probability of winning by switching is still 2/3 [if the host chooses which of two goat doors to open with equal probability]).

I suspect you're thinking the problem is asking about the probability of winning over all possible car locations, player picks, and host responses. What would your analysis be if the problem is about the chances of winning for a player who's literally initially picked door #1 followed by the host opening door #3? -- Rick Block (talk) 17:38, 17 August 2008 (UTC)

You dont seem to understand! The most critical question we should answer is NOT what happens if the contestant chooses one specific door and the the host opens another door! The problem is more general than that ! The whole point with the Monty Hall exercise is to answer the question whether or not the contestant should switch doors in GENERAL. This is also indicated in your quote

"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?"

The word "say" that appear before the door numbers indicate to me that this is just an example. The individual case of door 1 and door 3 is not that important! It could have been any configuration! Again the important question is whether or not the contestant should switch doors in GENERAL. Also note the sequential nature of your quote. 1)first the allocation has taken place (which remain fixed for the rest of the game 2)Then the contestant choose one door. When you start to shuffle around the location of the car it CONTRADICT the sequential nature of the game. It also obscure the whole purpose of the Monty Hall exercise which is to prove that the contestant will benefit from switching doors in ANY situation not just "for a player who's literally initially picked door #1 followed by the host opening door #3? "--82.39.51.194 (talk) 08:39, 18 August 2008 (UTC)

My points are:

1. If the goal is to answer the general question then all goat/car configurations, all initial picks, and all host responses should be considered. Your approach is considering only one configuration using an assumption that the same logic pertains to all other configurations. The approach currently in the article considers only one initial pick and uses an assumption that the same logic applies to all other picks. Your assertion seems to be that the latter approach is wrong. Either one of these is valid (they are like looking at two sides of the same coin), although most references use the approach currently in the article. If you're not seeing that these approaches are fundamentally equivalent, I think it is you who is not understanding the problem. In addition the current approach is (IMO) more consistent with the form of the Parade problem statement which uses the "say No. 1" terminology (which encourages thinking through scenarios involving a given initial pick).
2. Some references (e.g. Morgan et al) explicitly consider the situation at the point the player has picked a door and after the host has opened a door. Your approach makes this analysis difficult, however the current approach makes this a relatively simple extension of the analysis.

The approach currently in the article is used because most references use it and this approach easily extends to cover the "conditional" (Morgan et al) interpretation of the problem. Even though you think the critical question is the general question, the conditional interpretation exists in the literature (the Morgan et al and Gillman references) so it is considered in the article. I'll ask again - how would you address this interpretation? -- Rick Block (talk) 19:15, 18 August 2008 (UTC)

---

First of all the current solution is NOT more consistent with the form of the Parade problem statement. Actually my proposed solution is MORE consistent with the Parade problem! The reason for that is that it is more logically consistent (sequential game nature which means that the location of the car remain fixed), more general (consider a larger amount of cases) and easier to understand (iteration nerver suns out of styl).

Secondly, How I would deal with the conditional interpretation? For me that is an interpretation that is not critical for the Monty Hall problem nor its solution. You can and should understand the Monty Hall problem and its solution only with the basic frequency statistics approach (without Bayesian statistics). Note that my proposed solution uses frequency which is the most consistent with the original formulation of the Mony Hall game since again

1) It DOES NOT contradict the logic and sequential nature of the original Monty Hall game. The location of the car (and the goats for that matter) remain fixed for the entire game.
The logic is intact
2) Is not based upon any prior assumtions about the door selection. what you see it what you get.
3) Does not start in the middle of the game (door selecton), again sequential nature
3) The formulation is much more general (consider a larger amount of cases)
4) Much easier to extend to alternative configurations for example D1=C,D2=G,D3=G or D1=G,D2=C,D3=C or D1=G,D2=G,D3=C

As I said previously if you want to include a section on conditional probabilities as an extra bonus you have to very carefully point out the different assumption such an analysis is based on. If you dont correctly point out such differences the Bayesian section will actualy contradict the original sequential reasong of the Monty Hall game.

1) Firstly you you need to explain that in order for a conditional probability approach to work we have to take the steep from frequency statistics to Bayesian statistics. This means that we have to depart form the sequential nature of the problem where we first have the allocation of the car and goats and then we we have the door selection. Now instead we are going to assume that the player already has done his door selection. In order to evaluate such a decision in the middle of the game we have to vary the location of the car. So basically we are inverting the origial order of the game. 1) door selection 2) allocation

2)Secondly you need to explain the reason for that). You need point out that Bayesian statistics in the form of conditional probabilities are based upon the assumtion of serial correlation (normal distribution with fat tails) which means that the observations are not independent. For example

WHOA THERE!!! The above paragraph is pure nonsense. The Bayesian formulation of probability theory, based on Cox's axioms, is completely general and has been shown to be equivalent to any other standard formulation (e.g. Kolmogorov's). See the references ited in that section, e.g., E.T. Jaynes's "Probability Theory as Logic". Please do not make absurd statements to (try to) make a point. Also, please stop talking about "Bayesian statistics". There is not statistics involved in the Monty Hall problem: rather, it is purely a problem of probability theory.The Glopk (talk) 01:25, 21 August 2008 (UTC)

P(A∩B) is the probability of A and B happening
P(B) is the probability of B happening
P(A|B) is the probability of A happening given that event B has happened

Bayesian expression: P(A∩B)=P(B)*P(A|B)

When we have serial independence ( A and B are independent) then the P(A|B) expression is reduced to P(A)

Which means that our Bayesian expression is reduced to the frequency expression

P(A∩B)=P(B)*P(A)

These two explanations are important because it helps the reader to understand the some what contradicting and confussing set up! Here I am open for suggestions! Any solid and easy to understand explanations that can simplify the transition is most welcomed !

I now feel that I have put forward a solid argument why the current solution is not optimal. If you still fail to take my points into consideration I suggest that we seek outside help on the matter since otherwise we will continue this discussion for all eternity. To tell you the truth I have more productive things to do that arguing with you about these things!
--82.39.51.194 (talk) 10:50, 19 August 2008 (UTC)

I will not go into the details of what you wrote here; others can do that much better than I can.

First, I want to applaud your change of attitude instead of changing the article time and time again, you are now discussing the matter.

However, you should note that you seem to be the only one arguing this point. That doesn't in itself mean you are wrong, of course; but you should appreciate the fact that many editors, some of them very well versed in this theory, have already been over this article in great detail. That in turn means you should consider the possibility that you are in fact mistaken. In your first posts here you seemed to take the stance that you were inquestionably correct, but as I said before, you have changed your attitude a bit, instead putting forward arguments to support your opinion. Kudos.

One point that you haven't responded to yet is the tenet of no original research. The references cited in the article support the problem statement we currently use, and so far you have proposed no references to support your version. Oliphaunt (talk) 11:57, 19 August 2008 (UTC)

You are right I have to considered the possibility that I might be wrong! I have done that and the answer was that I am not! It is not that much to wrong about! Further, the best references is the original Monty Hall game and its corresponding rules! I can probably dig up a reference on the frequency statistics approach (which I assume is my suggested approach) but I am not sure it is necessary. It is like asking for a reference for why 2+2=4. It is a simple exercise on calculating probabilities! If you find that difficult then you should probably not take on the conditional probability approach which dominates the current article ! --82.39.51.194 (talk) 13:31, 19 August 2008 (UTC)

What you're not right about is your assertion that the current approach contradicts the sequence of the game. It absolutely follows the sequence, from the player's perspective. First two goats and a car are placed behind 3 closed doors, but the player doesn't know what is behind each door. In your solution, the configuration is given. Why doesn't the player just pick door 3 and stick with their initial choice? The reason of course is that the player does not know the configuration. In the version currently in the article, the configuration is left unknown (matching the situation from the player's point of view). The configuration is certainly fixed before the player picks a door, but the player doesn't know the configuration. What the player does know is the door he or she initially picks. The current solution, most sources, and the Parade problem description implicitly say (and the Bayesian analysis section explicitly says) "let's call the door the player picks door 1 (renumbering the doors if necessary)". The analysis proceeds given the unknown, but fixed, configuration of goats and car with the (now fixed) initial player choice. We examine all possible configurations of the goats and car not because we're moving them around after the player has picked (which would indeed violate the sequence of the game) but to enumerate all possible scenarios using the same frequency based approach your solution uses. This approach

1) exactly matches the player's view of the game. The location of the car is fixed, but unknown (in your solution, the car location is known but the player's pick is treated as variable - this is distinctly not the player's view and, literally shows only that switching wins with 2/3 probability if the car is behind door 3)

2) is not based on a specific arrangement of goats and car, which (again) matches the player's view of the game.

3) does not start in the middle of the game (if the current section is at all confusing in this regard, we could certainly work on clarifying it)

4) easily extends to the conditional analysis, without needing to switch to Bayesian logic (indeed, the current solution section includes the conditional analysis, without mentioning anything about Bayesian analysis)

5) considers all initial configurations (and, through renumbering of the doors, all initial picks and host responses)

6) is consistent with the view of the problem throughout the article (all the images and all the text consistently call the player's initial choice "door 1" and the door the host opens "door 3")

You're quite welcome to seek outside help, although I would suggest the best help would be sources. The existing references are (as far as I know) the best, most authoritative sources on the topic of the Monty Hall problem. The existing approach follows the solution typically presented in these sources. I've said this about 3 times already, but your solution and the solution currently in the article are essentially equivalent (both rely on a "without loss of generality" assumption). Given two essentially equivalent approaches, using the one that more closely matches the references, more closely matches the player's view of the game, and more easily extends to cover the conditional analysis seems like the obvious choice. -- Rick Block (talk) 16:20, 19 August 2008 (UTC)

yeahh yeahh what ever !--82.39.51.194 (talk) 17:29, 19 August 2008 (UTC)

I've clarified the sequencing in the Solution section. I hope this helps to address your concerns. -- Rick Block (talk) 02:54, 21 August 2008 (UTC)

I actually agree with 82.39.51.194 I also find the current article difficult to understand. I think a frequency statistic approach is easier to understand and more consistent with the original game.--92.41.172.75 (talk) 08:50, 21 August 2008 (UTC)

Suggestion to replace existing solution section

It appears 92.41.17.172 is suggesting that the current solution section be replaced with the contents from #Summary and Solution (above). My understanding was that per #Discussion (also above) we basically came to an agreement, albeit not very enthusiastic on the part of 82.39.51.194, about this. I won't repeat the discussion from above, but does anyone have anything more to say about this? -- Rick Block (talk) 18:30, 28 August 2008 (UTC)

The proposed change (diff) replaces the existing solution with text which is less detailed, has no supporting references, and which includes less appealing figures. The existing section should remain. TenOfAllTrades(talk) 18:56, 28 August 2008 (UTC)

I tend to agree with Rick, too, that the existing solution should remain. In my 35 years of teaching mathematics at a major university, including many, many probability courses, I've dealt with the Monty Hall problem many times in classes. What intrigues me most, though, is not so much the fact that the direct solution given seems to bother so many (perhaps because of the supposed veridical nature of the statement of the problem), but the fact that if one merely steps back and looks at the entire problem, there is a remarkably simple solution. The key to this solution is not to focus on switching but instead to focus on not switching. Choosing the strategy of not switching is tantamount to ignoring any and all extra information with the problem, and it is thus manifestly clear that:

P(not switching) = 1/3

and thus we have

P(switching) = 1 - P(not switching) = 2/3.

It uses nothing more than P(A) + P(not A) = 1. This is no more (and perhaps a lot less) counterintuitive than all the rest of the considerations which focus directly on choosing the switching option. I have always given perfect marks for this elegant solution and wonder why it is not mentioned in the article. Focussing on the switching strategy directly involves something like a probability tree or other construct (entailing conditional probability), fraught with potholes that can trap or fool the not-so-careful reader.

Some of the most elegant solutions to problems in mathematics arise by looking at such problems from a different perspective. This is just one such example. -- Chuck (talk) 19:29, 28 August 2008 (UTC)

Wait a minute. The expression P(A) + P(not A) = 1 refers to set complementation in a universal set - which has probability 1. And the not-switch strategy is not the complement of the set consisting of the switch strategy in the universe consisting of all possible strategies. For one thing, if there a two strategies there an infinite number of mixed strategies in that universe. While it is true that the probability of winning by switching plus the probability of winning by not switching is 1, it is something that you have to demonstrate by analyzing the problem. Its not hard but it sure doesn't follow from the tautology given. Just think, changing the name of the not-switch strategy to 'sticking strategy' sends the argument down the tubes.

In this particular game, the contestant always has the opportunity to switch so the switch strategy and the not-switch strategy always end up selecting two different doors and since the the other door is open and showing a goat, the car must be behind one of those two doors. Thus probability of winning by switching plus the probability of winning by not switching is 1. However in a variant of the game where the host did not always present an opportunity to switch then some of the time the switch strategy and the not switch strategy pick the same door, hence the sum of their probabilities of winning would be less than 1. Is that an example that shows P(A) + P(not A) = 1 is not true?Millbast5 (talk) 07:43, 18 September 2008 (UTC)

Chuck, my hat is off for you! I like your way of thinking.
"P(not switching) = 1/3 and P(switching) = 1 - P(not switching) = 2/3. It uses nothing more than P(A) + P(not A) = 1"
So simple but still so powerful! Only that well crafted sentence says more than the complete article in my opinion.
I think that reasoning in combination with the new suggested "frequency" solution would greatly improve the article--92.41.17.172 (talk) 20:53, 28 August 2008 (UTC)

I don't mean to be rude, but have you read the article? The solution section says Players who choose to switch win if the car is behind either of the two unchosen doors rather than the one that was originally picked. In two cases with 1/3 probability switching wins, so the overall probability of winning by switching is 2/3 as shown in the diagram below. The diagram then shows (visually) what happens if you switch (Chuck's point is simply the inverse of this). The killer argument against this simplistic analysis (and Chuck's) is that it completely ignores the fact that the host opens a door. Why should this same argument apply both before and after the host opens a door? The initial pick is 1/3 (surely), but then the host opens a door. Why is this any different from Howie opening a case on Deal or No Deal (or would you say never take the deal because the chances of your case being the grand prize go up every time a case is opened)? I don't teach math at a major university, but this solution simply fails to address the problem (per Morgan et al.) and doesn't seem worth more than a B- (correct answer, completely inadequate reasoning). Consider three different hosts - host 1 who opens a randomly selected "goat door" if the player initially picks the car, host 2 who always opens the rightmost "goat door" if the player initially picks the car and host 3 who opens one of the unpicked doors randomly. With host 1 the correct answer is 1/3 chance of winning if you don't switch. With host 2, assuming the player stays with the initial pick (and it's door 1), the correct answer is 0 chance of winning if the host opens door 2 but 50% chance of winning if the host opens door 3. With host 3 the correct answer is 1/2 chance of winning (assuming the host didn't reveal the car). What this means is that the argument that the initial pick results in a 1/3 chance of winning is, um, not quite correct. If you don't also consider how the host selects what door to open the answer may (coincidentally) end up with the right answer (if it's host 1) but this is fundamentally a coincidence. Given any possible host behavior the average chance of winning if you don't switch is 1/3, but if you pay attention to which door the host opens (and how the host chooses to open doors) the chances of winning by staying are anywhere from 0 to 50% depending on which door the host opens. The point is the real solution has to include a consideration of how the host chooses which door to open. If it doesn't, then the answer applies regardless of which host we're talking about (host 1, host 2, or host 3) - i.e. you arrived at the right numerical answer but your reasoning is faulty. -- Rick Block (talk) 04:45, 29 August 2008 (UTC)

Sorry, Rick, but you are being rude (which is out of place here) and it seems you have been very much caught up in the veridical nature of the posed problem. Yes, I have indeed read the article, and it beats around the bush with various possible host actions (which is another matter that I'm not addressing) ... but of course I am observing the generally agreed on rule that says the host always opens a goat door different from the door chosen by the contestant. What you don't see is that there are two and only two options which under the host assumption exhaust all possibilities: (A) be obstinate (not switching no matter what is shown) or (not A) always switch when a door (different from the one selected by the contestant) showing a goat is opened. While (not A) has all the folderal about choosing to switch), option (A) does not (if you just think about it for a moment). Moreover, (A) union (not A) is the entire event space, so P(A) + P(not A) = 1. And since P(A) is manifestly eequal to 1/3, P(not A) = 2/3.

You complain that my solution ignores the fact that the host opens a door showing a goat. Note that it makes no difference in the outcome if the host opens either of the doors not initially selected. In the obstinate case (never switch no matter what is opened) the probability of winning the car is 1/3. If the contestant has initially chosen a goat door and her strategy is always to switch, then no matter which of the other two doors is opened, she will always with the car by switching to the opened door if it reveals a car or to the other closed door if it reveals a goat.

In either of these two scenarios, the problem you and many have is in realizaing that (always switching) and (never switching) exhaust all possibilities. That is precisely where the problem seems like a (veridical) paradox ... and that is why the solution I give is so elegant and, at first glance, seems that it must be wrong - even though it is completely correct.

I am reminded how, in teaching how to find areas of regions to students, we tell them that one way is to decompose the region into simpler regions and add up the areas of the pieces. Then we watch the solutions come in for a problem such as: find the area of the shaded region shown - which is nothing more than a large rectangle with a smaller rectangle removed - and students slavishly decompose the region into four rectangles, compute the area of each, and add them up, correctly of course - but completely missing the obvious solution to subtract the area of the smaller rectangle from that of the enclosing rectangle! Too bad people can't think outside the box or (in this case) inside the box.

People's dogged insistence that the result for never switching has something to do with opening a different door from the one selected is a bit like this, the only point to the problem is to see that since the host opens a different door from that chosen by the contestant, then the obstinate and always switch options are discjoint and do indeed exhaust all possibliities.

Oh well, some people look at a sphere and see perfection - others look at one and see a snowball and duck for cover. -- Chuck (talk) 14:11, 29 August 2008 (UTC)

Sorry for the rudeness. However, see below as well. -- Rick Block (talk) 15:26, 29 August 2008 (UTC)

ha, ha. I and many with me prefer Chuck's simple solution plus 82.39.51.194 "frequency" solution anyday over that lengthy and confused outburst by Rick Block!
There are no such thing as host-1,host-2 and host-3. The only host that exist is host-1 = Random selection of "goat door". Why would I consider an example that does not apply?
With all due respect there might be a reason for why you are not teaching math at a major university. --92.41.46.220 (talk) 09:06, 29 August 2008 (UTC)

Respect acknowledged, however your reasoning is still inadequate. Yes host 1 is our host, but there's nothing in the simple solution that makes use of this fact. Assume for the moment we're talking about host 3 (makes it more like Deal or No Deal), not host 1. The same simple argument can be used, but in this case this argument results in an incorrect solution. The flaw is the implied claim that because P(not switching) = 1/3 when the player initially selects a door it remains so after the host opens a door, i.e. that how the host chooses to open a door is "extra information" that can be ignored. Not so.

You have to at least explicitly say that P(not switching) doesn't change when the host (our host, host 1) opens a door because of the constraints on the host's behavior, but then the question becomes how do you know this? Asserting it to be true with no reasoning doesn't seem sufficient.

If you have access to it, please read the Morgan et al. paper. This paper distinguishes host 1 from host 2 which is a much more subtle distinction than the difference between host 1 and host 3. With host 2, the average chance of winning by not switching (ignoring which door the host opens) is 1/3 - just like host 1. However, with host 2 there are two distinct probabilities depending on whether the host opens the rightmost or leftmost door. If the host opens the rightmost door (which he does unless the car is behind it) the chance is 1/2. If the host opens the leftmost door (which he does only when the car is behind the rightmost door) the chance is 0 (of winning by not switching). You can in some sense say the probability of winning by not switching in this case is 1/3, however the probability for any given player (with knowledge of which door the host opens) is either 0 or 1/2.

So, back to the article. I believe the solution as stated is as simple as possible without being too simple. -- Rick Block (talk) 15:26, 29 August 2008 (UTC)

I also agree with 82.39.51.194 and 92.41.46.220. I think we have consensus now to change the solution section!
--Pello-500 (talk) 13:51, 29 August 2008 (UTC)

1. It appears that Chuck actually doesn't agree with your change;
2. It's incredibly obvious that Pello-500 is just a new account created by 92.41.

If you're not interested in engaging in serious discussion here, please stop editing the article. TenOfAllTrades(talk) 14:00, 29 August 2008 (UTC)

In addition to edit warring over the article, Pello/92.41/82.39 also erased my comment calling him on it. I've restored it above. Note that he had originally (and erroneously) claimed that Chuck had agreed with his position. TenOfAllTrades(talk) 21:43, 29 August 2008 (UTC)

I also agree with 82.39.51.194 and 92.41.46.220. The frequency solution is most optimal.--84.92.246.30 (talk) 17:26, 30 August 2008 (UTC)

RfC: New or old solution section?

The new suggested "frequency" + Chuck's solution is easier to understand. Current editors refuse to accept this due to ownership. They won't let anyone edit the article !

Initial thought - if I understand correctly the question is not about what the actual solution is, that is agreed, but it is about the best way to explain it. From my experience different people have different way of looking at things. When I am trying to get to grips with a probability problem I usually find that one particular way of looking at it helps me to get it. The may not be the way that helps other people. My suggestion, therefore, is to have both solutions. Maybe the second could be added under the heading 'Another way to understand the solution'. Martin Hogbin (talk) 17:56, 29 August 2008 (UTC)

Disagree, please leave the old solution. The new proposed "frequency" solution is poorly written and referenced. Chuck's solution is just the "naive Bayes" one (uniform prior on the host behavior). As remarked by Rich Block, the "old" solution section, i.e. the one that passed the F.A. review, offers a more comprehensive treatment that is supported by the refrences and does cover more sophisticated host behaviors.The Glopk (talk) 00:29, 30 August 2008 (UTC)

My suggestion as to what should be done depends on what people believe that the purpose of this article is. For example, is the purpose purely to consider the problem academically and with some degree of rigor or is there a secondary function of explaining to as many readers as possible what the solution is and why in a way that makes sense to them. Martin Hogbin (talk) 08:53, 30 August 2008 (UTC)

I don't know if it's necessary to repeat it, but in case someone doesn't read the section above, I'll note that the proposed replacement section is formatted poorly, lacks references, and contains less explanatory text (both quality and quantity) than the proposed replacement. TenOfAllTrades(talk) 17:48, 30 August 2008 (UTC)

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Can I make sure that I now understand what the current debate is about? Some editors want to keep the page as it is but recognise that there is a rather subtle problem with it. Others wish to replace it with a more rigorous approach which some feel has been badly presented and is not necessary. Martin Hogbin (talk) 09:39, 31 August 2008 (UTC) On a second look at things I can see that I got that wrong, I was confusing the current dispute with and earlier discussion that Rick pointed me to.Martin Hogbin (talk) 15:01, 31 August 2008 (UTC)

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Now that I have some idea of what is going on here, I can give my views on the current dispute. I do not think that the current solution should be replaced with the proposed one. However, I do think that it is more natural to have a fixed car position and vary the pick, but I also understand the reasons that the solution is the way that it is. I do have some criticisms of the article as it is and hopefully I will have some suggestions for improvements, but scrapping what has been done and replacing it with the proposed replacements not the way to improve in my opinion. Martin Hogbin (talk) 18:08, 31 August 2008 (UTC)

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I'm in favour of including both solutions. Because one method, to quote, "is formatted poorly, lacks references, and contains less explanatory text" is not grounds to remove it altogether, but to bring it up to standards. If we removed everything that was poorly written rather than improved them, Wikipedia would have about 20 articles. Besides, other topics like Zeno's paradoxes have entire articles dedicated to the various proposed solutions, so the "we already have a solution" argument doesn't seem valid here. Android 93 (talk) 06:14, 6 September 2008 (UTC)

As you will see I have made a similar suggestion here but any proposals for change get a very frosty reception. Have a look at the history of this page and you will see that an IP editor recently added an extra explanatory section only to have it immediately deleted, with no explanation and with the deletion being marked as a minor edit!

Another person tried to start a separate article only to find it set upon with a voracity I have never seen before in WP and deleted with days.

I have been trying hard to work with the existing/historical editors to find way to improve the page without compromising the integrity of the current article. So far my advice has been that I can make minor edits but if they are reverted I must immediately discuss them.

I Strongly believe that the current article fails to be convincing in respect of the basic paradox and that this must be improved. My view is supported by the fact that since the RFC there have been two people on the talk page who did not believe the article. One was eventually convinced by discussions here. As you say, there are other articles which make a better job of explaining their central paradoxes.Martin Hogbin (talk) 10:09, 6 September 2008 (UTC)

In an effort to allow this RFC to proceed unimpeded by "current editors" I have refrained from commenting, however I feel compelled to respond to this. I can see how editors new to this page might interpret responses to suggested changes as frosty, but in defense of the current editors this article is a featured article that has been through two featured article reviews [1] [2]. This certainly doesn't mean it's perfect, but it does mean there's been a significant amount of effort by multiple people spent on making the article not only mathematically correct with all major claims referenced to reliable sources but also comply with all other featured article criteria which include adherence to the WP:manual of style and a "professional standard" of writing. The recent addition by user:Technicalmayhem (a new, but not an IP, editor) was first deleted (by me) with this edit summary: revert - many WP:MOS issues with this addition. Technicalmayhem re-added it and this repeat addition was reverted by another editor also apparently watching the page. The re-addition was done with no explanation and marked as a minor edit (!), as well as the re-deletion. The separate article, "Introduction to the Monty Hall Problem", was created by user:Pello-500 (arguably the same user as 82.39.51.194 and 92.41.46.220) for the apparent purpose of including this user's preferred "solution" without gaining consensus to include it here. This is not how content disputes should be handled per Wikipedia:Dispute resolution, and since this user previously initiated this RFC he or she is clearly aware of the proper steps for handling content disputes but for whatever reason chose to ignore them. It's hard to interpret creating this article as anything other than bad faith tendentious editing.

Rick, I was not referring to your reversion but this one: 4 September 2008 Jonobennett m (48,223 bytes) (Reverted 1 edit by 68.2.55.158.)Martin Hogbin (talk) 20:24, 6 September 2008 (UTC)

From my point of view, the discussion below of Martin Hogbin's suggestions for improvement has been proceeding pretty much exactly how such discussions should go. I'm not, and I don't think anyone else is, saying no changes can be made. The caution about making a change once and if it's reverted then discussing it was not intended to inhibit changes but to encourage them (per WP:BOLD). I partially agree with the point that the article fails to be convincing (not just one, but both "doubters" have apparently been convinced) however I'm highly skeptical that ANY explanation or even combination of explanations would convince all readers. To be clear, I'm not saying it can't be made more convincing but like it says in the lead - even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief. -- Rick Block (talk) 18:09, 6 September 2008 (UTC)

I agree it will not be easy but surely it is worth a go. As with similar problems, explanations from many different angles could be a good place to start. What is the best way to proceed? I do not want to start editing the current article by adding a new section, which will start off badly written, formatted, and referenced and may well get removed. Do you agree that a development version of the article might be a good place to start, with the intention of a merger when agreed, or perhaps leave it as a separate article with a link from this one? Martin Hogbin (talk) 20:24, 6 September 2008 (UTC)

Creating a new article to be merged later is definitely not the way to proceed. Proposing the text for a change, or even a new section, here (well, probably not here but in a section below) on the talk page would be perfectly reasonable. If you're talking about major structural changes, you could create a sandbox version, perhaps on a page in your user space like user:Martin Hogbin/Monty Hall problem (draft). -- Rick Block (talk) 20:54, 6 September 2008 (UTC)

I have set up a user page as described - see section below.Martin Hogbin (talk) 10:14, 7 September 2008 (UTC)

• keep original. If I understand correctly the two solutions that are being offered (no one has bothered to supply diffs), then I think the current (original) version is better. having taught this problem in the past, however, I think it needs some revisions. does anyone object if I do some editing? --Ludwigs2 03:07, 7 September 2008 (UTC)

Bayes was right, but I think the logic of the existing solution is flawed

The Bayesian analysis is correct to the extent that it demonstrates that the probability of the car being behind Door 1 remains 1/3 after the decisions are made by the player and the host. That is the "decisions" by the player and the host do not affect the probability of the car being behind Door 1.

However, I believe the interpretation of the Bayesian analysis is flawed.

In particular, although the Bayesian analysis correctly accounts for the fact that the host knows the location of the car it does not reflect knowledge gained by the player when it is revealed that the car is not behind Door 3.

Bayes's Theorem can be used to take account of the fact that the player learns the car is not behind door 3 by calculating the probability of the car being behind Door 1 given that it is not behind Door 3.

Let's do it in English.

The probability that the car is not behind Door 1 given that the car is not behind Door 3

= [(The probability that the car is not behind Door 3 given it is behind Door 1) x (The probability that the car is behind Door 1)] / (The probability that the car is not behind Door 3)

= [1 x 1/3] / 2/3

= 1/2

This result is consistent with many people's intuition that if there are two doors and a car is behind one of them then there is a 50:50 chance of the car being behind either door.

That is, I think conclusions shown on the website are wrong.

Given the debate that has already taken place I guess there will be some opposition to my view. --Paul Gerrard at APT (talk) 11:51, 30 August 2008 (UTC)

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Paul, I would be happy to try and persuade you that you are wrong but this is not the place. You can email me at martin003@hogbin.org. Martin Hogbin (talk) 14:02, 30 August 2008 (UTC)

The basic issue is that "the probability that the car is not behind Door 1 given that it is not behind Door 3" (which is indeed 1/2) is NOT the question posed by the Monty Hall problem. In English, what we're looking for (flipping "is not behind Door 1" to "is behind Door 1" - seems a little clearer this way) is "the probability that the car is behind Door 1 given that the host opens Door 3". With the standard rules, this probability ("host opens Door 3") is 1/2 given the car is behind Door 1 (from the problem statement: If both remaining doors have goats behind them, he chooses one randomly). This probability is also 1/2 given the car is not behind Door 1, since if the car is not behind Door 1 it's equally likely to be behind Door 2 or Door 3. Combined, what this means is the probability the host opens Door 3 is 1/2 whether or not we're also given the car is behind Door 1. This makes the Bayes expansion be (1/2) x (1/3) / (1/2), i.e. 1/3. -- Rick Block (talk) 16:28, 30 August 2008 (UTC)

Looks like it is OK to answer here. Here is a very simple way to see that the probability of getting the car if you switch is 2/3. If you switch, you always get the opposite of what you started with. You have a 2/3 probability of starting with a goat. Martin Hogbin (talk) 16:51, 30 August 2008 (UTC)

This is simple, but it relies on an assumption that when the host opens a door nothing changes (or changes the question from "what is the probability of winning after the host opens a door given the Monty Hall rules" to "what is the probability of picking a car from among 3 doors"). Under the standard rules it turns out that the host opening a door doesn't affect the player's initial chances of having selected the car (this is reflected in the Bayes expansion above with the two 1/2's that cancel each other out), but this simple explanation does not use those rules so it is fundamentally incomplete (there is much more discussion of this in the #Suggestion to replace existing solution section thread just above). Why does this explanation not apply to Deal or No Deal? -- Rick Block (talk) 17:25, 30 August 2008 (UTC)

Yes Rick, you are quite right. However if I add to my explanation the observation that, with the standard rules, no information about the location of the car is given when the host opens a door and thus the original probability of getting a goat holds after the door has been opened, then my explanation is a good one for the standard rules case. However it is then not such a simple explanation and cannot be directly applied to other cases. At least I understand what the RFC is all about now. Martin Hogbin (talk) 18:01, 30 August 2008 (UTC)

Here is rather extreme example to show what Rick means and why my original answer was incomplete. Suppose we work to the standard rules, so you would expect my explanation to apply, but, after he as opened a door, the host tells the contestant whether he has initially chosen a car or a goat (rather spoils the game) then the probability of getting a car if the contestant swaps is clearly not 2/3 but either 1 or 0 depending on what the host tells the contestant. Martin Hogbin (talk) 18:12, 30 August 2008 (UTC)

The question still remains how we know given the standard rules that "no information about the location of the car is given when the host opens a door". This is an assertion, not reasoning, and (as it turns out) showing this with reasoning is not overly simple. Again, I think the best reference for this is the Morgan et al. paper (in the references), or the current version of this article. This topic was discussed at length several months ago, shortly before the latest featured article review - starting with Talk:Monty Hall problem/Archive 6#Rigorous solution. In my opinion the ensuing discussion (that continues in the archives into Talk:Monty Hall problem/Archive 7), somewhat contentiously spearheaded by an anonymous user (using multiple IP addresses in the range of user:70.137.168.95) who originally pointed us to the Morgan et al. paper, in conjunction with scrupulous referencing improvements to satisfy concerns raised during the last FARC have led to significant improvements in this article (as well as to my personal understanding of the subtleties involved in this problem). One of the key points this user brought to the discussion (backed up by the Morgan et al. paper) was that most solutions presented to the problem are overly simplified and include but don't justify the assertion that the host opening a door doesn't change the initial 1/3 probability. Another very good reference about this is the Falk paper (also in the references). According to Falk the answer that most people present (after the host opens a door, it's a 50/50 chance) is rooted in a deeply intuitive "equal probability" assumption; however many "correct" answers commonly presented simply replace this intuition with an appeal to another one, i.e. the "belief that exposing information that is already known does not affect probabilities". Neither of these is based on particularly sound reasoning and problems can be constructed where either one leads to an incorrect answer. This problem is contentious at least in part because it's not simple. -- Rick Block (talk) 21:21, 30 August 2008 (UTC)

There certainly is more to this than meets the eye. Might a way forward be as follows? Formulate a version of the problem where a simple analysis, such as I gave, is correct. We are at this stage picking a problem to suit our desired answer. In other words construct a problem where what has been called the most attractive false solution is actually a good solution. Now clearly state our reformulated problem as an idealised version of the Monty Hall problem and give the, now correct, simple solution. My guess is that this will be what 90% of readers will want and expect. It is just a continuation of the process started by Krauss and Wang.

The discussion about the hosts behaviour and the contestants understanding of this then can become a secondary, much more complicated, and perhaps contentious, discussion for those interested in the fine details.Martin Hogbin (talk) 22:25, 30 August 2008 (UTC)

This has been suggested before - the problem is that if we do this we drift into WP:OR unless the version of the problem we present clearly meets WP:RS. I think this boils down to whether the solution currently presented is accessible to a lay person. I think the answer is yes (but I am sufficiently "into" this problem that I realize my opinion on this is basically worthless). -- Rick Block (talk) 04:14, 31 August 2008 (UTC)

Maybe we should consider the occasional exception (WP:IGNORE) here. The dividing line between OR and verifiable is rather vague anyway; not all third party references should carry equal weight and the process of assessing how authoritative any given reference is is a form of OR. With this particular problem we could probably find more references, from normally authoritative sources, that give the completely wrong answer than we could good ones. On the other hand experience with the problem has shown that our intuition is not to be trusted either. Regarding the RFC, I have continued above.Martin Hogbin (talk) 09:35, 31 August 2008 (UTC)

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Hi Martin, thanks for the information.
I agree with much of what you say, but I disagree with the assertion that:

"no information about the location of the car is given when the host opens a door".

It seems to me that when the host opens a door the player is given information about the location of the car. In particular, the host's action informs the player that the car is not behind Door 3 and that the car must therefore be behind either Door 1 or Door 2.

Yes, you are right, I put it badly.Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)

This information does not alter the fact that at the outset of the game there was a 1/3 chance that the car could have been behind any particular door.

That is right but also the chance a car is behind the door that I initially chose remains at 1/3 if the host must open a goat door and, if given a choice, must chose randomly, and I know all this. This is generally agreed to be true in the above circumstances but my mistake was to take this as being self evident. If the host acts differently, then the above statement is often not true. If you agree that, after a door is opened, the chances of the car being behind your initial door is still 1/3 then my argument still holds. Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)

I can't see the logic of the probability remaining 1/3. If it were 1/3 and it is known that there was a goat behind Door 3 this would mean that the probability of the car being behind Door 2 had risen to 2/3. I know that is what the website says, but in my response to Glopk below I explain why I believe that probabilities can't be transferred as suggested on the website. Can you point me to any of the previous discussion about transferring probabilities? --Paul Gerrard at APT (talk) 12:23, 1 September 2008 (UTC)

Let me try a slightly different version of the problem. The contestant picks a door. We agree that he has a 2/3 chance of picking a goat. We also know that if he switches he will get the opposite of his original choice. Suppose, therefore, that this person decides, before a door is opened, that he will switch. Do you agree that he then must have a 2/3 chance of getting a car? Martin Hogbin (talk) 18:08, 1 September 2008 (UTC)

Hi Martin - I agree that at the outset there is a 2/3 chance of picking a goat, but I am not sure what you mean when you say that "if he switches he will get the opposite of his original choice". Also, if the player decides to switch why would the probability of picking a car increase to 2/3. If you extend the logic to switching a second time would you say that the probability of picking the car was 1 (i.e. a certainty). What would be the reason for suggesting that deciding to switch would increase the probability of picking the car? —Preceding unsigned comment added by Paul Gerrard at APT (talk • contribs) 00:11, 2 September 2008 (UTC)

By "if he switches he will get the opposite of his original choice" I refer to the simple fact that , if the player initially chooses a goat and the switches he must get a car(because his original choice is one of the goats and the host has removed the other one from play leaving the car as the only other door) and if the player initially chooses a car and switches he must get a goat (because there is only one car and he has it). The above always applies, without exception, regardless of the strategy of the player or the host (we are here assuming that the game rules require the host to always open a goat door and offer a swap).

Regarding a second swap, no of course, the probability does not rise to 1. The whole point of this problem is that after a door has been opened, the player has more information as to the whereabouts of the car, making the unchosen door more likely to have it. I have to say, and indeed am saying, that the current explanation is not as convincing as it should be. I am currently trying to improve it. Martin Hogbin (talk) 08:46, 2 September 2008 (UTC)

However, while the original probabilities remain unchanged the information that the car was not placed behind Door 3 effectively changes the game from a "3 door game" to a "2 door game".
The effect of knowing that the car is not behind Door 3 on the probability that the car is behind Door 1 can be demonstrated as I have done above using Bayes' Theorem.
Bayes Theorem can also be used to calculate the probability that the car is behind Door 2 given that it is not behind Door 3. (The answer is 1/2.)
Our different perspectives are as intriguing as the puzzle itself.
What am I missing? (I have tried to make sense of the references with the exception of Morgan et al which costs $14)--Paul Gerrard at APT (talk) 11:29, 31 August 2008 (UTC)

No, because the opening of a door does give you extra information about the chances of the car being behind the other door. Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)

Paul, please re-read the "Bayesian Analysis" section. As others have already pointed out, the proposition of interest for the MHP is not $\qquad C_{i}|\neg C_{3}\,=\,$ "The car is behind door i, given that it is not behind door 3", for $i=1,2\,$. Rather, the player is interested in the truth of $\qquad C_{i}|H_{j3},I\,=\,$ "The car is behind door i, given that the host opens door 3 after the the player initially chooses door j", for $i,j=1,2\,$, and where $I\,$ represents everything the player knows prior to making the first choice, including the host's behavior. The two propositions are obviously different, and therefore it is to be expected that they evaluate to different numbers. But only the latter one matters. Further, the evaluation of $\qquad P(C_{i}|H_{j3},I)\,$ using the Bayes theorem shows the need to make explicit the assumptions concerning the host's behavior through the value of $\qquad P(H_{j3}|C_{k},I)\,$, the probability of the proposition "The host opens door 3 after the player chooses door j, given that the car is behind door k". All possible host's behaviors are specified by assigning values to this function of j and k. The Bayesian Analysis section in the article shows only one particular such function, the one consistent with the "standard" interpretation of the MHP, i.e. "The host shows a goat behind one of the two doors not chosen by the player. If two such doors are available, they are equally likely to be opened".The Glopk (talk) 16:46, 31 August 2008 (UTC)

Thanks The Glopk, it is good to be reminded that $\qquad P(C_{i}|H_{j3},I)\,$ is not the same as $\qquad P(C_{i}|\neg C_{3})\,$. This a very important distinction and I agree that these expressions are very different and that there is no reason to believe they should be equal.
It is therefore important to reconsider the problem to make sure we are trying to answer the same question. My understanding of the problem is that we are trying to determine whether it is to the advantage of the player to switch after the host opens a door.
We also need to review how the expressions relate to the issue of whether it is to the advantage of the player to switch after the host opens a door.
The differences between the expressions are subtle.
My understanding is that $\qquad P(C_{1}|H_{13},I)\,$ = 1/3 shows that the rules the host follows to select a door do not alter the probabilities - they are still 1/3.
However, $\qquad P(C_{1}|H_{13},I)\,$ does not take into the fact that the host has revealed that there is a goat behind Door 3 even though the rules of the game require the host to choose a goat door.
It is really hard to find words which explain why this is a problem, but please consider $\qquad P(C_{2}|H_{13},I)\,$. If you evaluate this expression in the same way that website evaluates $\qquad P(C_{1}|H_{13},I)\,$ you will find that:

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

Similarly, you will find $\qquad P(C_{3}|H_{13},I)\,$ = 1/3 unless you change the probabilities to reflect the fact that it is known that a goat is behind Door 3.

However, if you change the probability of a goat being behind Door 3 to zero then you must also change the probabilities of the car being behind the other 2 doors such that their sum is 1. This raises the question of what would the new probabilities become. I see no alternative other than 50% chance for both Door 1 and Door 2.

Also, I know the existing website suggests that the host's showing of the goat behind Door 3 has the effect of "transferring" the 1/3 of probability to Door 2. The following example helps demonstrate why I believe such transfers do not occur. Imagine a similar game with 3 doors, two goats, and a car. The probabilities for each door are 1/3. Also the probability of the car being behind any pair of doors is 2/3 (e.g. the probability that the car is behind either Door 1 or Door 2 is 2/3). Now imagine that any goat door is opened - what then happens to the probability that the car is behind either of the other doors? If we assume that probabilities are transfered as per the website then for each of the remaining two doors the probability that the car is behind that door becomes 2/3 and the sum of these 2 probabilities is 4/3 whereas the sum of the probabilities must be 1. Thus, there is a problem in transferring probabilities.

Now let's consider $\qquad P(C_{1}|\neg C_{3})\,$. It reflects everything that is known by the player. The process by which the host chose Door 3 has given no more information to the player than that a goat was behind Door 3. Thus, $\qquad P(C_{1}|\neg C_{3})\,$ = 1/2 seems to me to indicate that it is the true probability of the car being behind Door 1 and there is no advantage in switching.
Finally, lets consider what happens to $\qquad P(C_{1}|H_{13},I)\,$ if we adjust the probabilities to reflect the fact that we know the car is not behind door 3 as suggested above.
That is $\qquad P(C_{1})=P(C_{2})=1/2\,$ and $\qquad P(C_{3})=0\,$. If you do this you will find that:

$\qquad P(C_{1}|H_{13},I)\ =P(C_{1}|\neg C_{3})\ =1/2,$

That is, the Bayesian Theorem actually gives the same result as $\qquad P(C_{1}|\neg C_{3})\,$ providing the probabilities used in the formula are adjusted to take account of the fact that a goat is behind Door 3.
Thus, it seems to me that there is no advantage in switching. --Paul Gerrard at APT (talk) 11:47, 1 September 2008 (UTC)

$\qquad P(C_{i}|H_{j3},I)\,$ is the probability that the car is behind door i given the host opens door 3 after the player initially selected door j (under the rules of the game). This is exactly the question we're interested in. The rules of the game ensure the host opens a goat door. You say this "does not take into the fact that the host has revealed that there is a goat behind Door 3" and admit that it "is really hard to find words which explain why this is a problem". That's good, because it is not a problem.

$\qquad P(C_{2}|H_{13},I)\,$ evaluates to 2/3, not 1/3, since

• $\qquad P(H_{13}|C_{2},I)=1\,$ (if the car is behind Door 2 and the player picks Door 1 the host is forced to open Door 3)

Similarly, $\qquad P(C_{3}|H_{13},I)\,$ is 0, not 1/3, since

• $\qquad P(H_{13}|C_{3},I)=0\,$ (if the car is behind Door 3 and the player picks Door 1 the host never opens Door 3)

No "changing of probabilities" is required to get these to sum to 1. Before the host opens Door 3 the probabilities are 1/3, 1/3, 1/3. After the host opens Door 3 (under the rules of the game) the probabilities are 1/3, 2/3, 0. The effect (in this case, under these rules) is that Door 3's probability has been "transferred" to Door 2. This doesn't say anything like you can group any two arbitrary doors and probabilities transfer between them.

Our differences of opinion seem to hinge around what probabilities we feed into the Bayesian formula.

If it is correct to say that all probabilities are 1/3 then I agree with your calculations.

If it is correct to adjust probabilities to 1/2, 1/2, 0 for Doors 1, 2, 3 respectively then I believe my calculations are correct.

I believe you are right if we are talking about probabilities prior to the door being opened and I am right if we are talking about probabilies after the host opens a door. As we are trying to take account the information available to the player at the time the player must decide to switch or not not to switch, we must use the information available to the player and at this time the player knows only that there is a goat behind Door 3. The rules of the game provide no more information than this. Thus I believe the probabilities must be 1/2, 1/2, 0 for Doors 1, 2, 3 respectively rather than 1/3 for each door.

With respect to the grouping of doors I agree that one cannot group doors and probabilities transfer between them. My example was to prove that one cannot do it. However, it seems to me that your solution does exactly this. Isn't your solution efectively saying that:

The probability of the car being behind Door 1, 2, or, 3 is 1/3 for each door.

You then group the 2 doors that are not chosen by the player and say there is a 2/3 chance the care is behind one or other of those doors.

The host then selects a goat door from the two doors that were not chosen by the player.

You have then concluded that the entire 2/3 probablity that the car was behind either of these 2 doors must therefore be the probability of the car being behind the door that was not selected by either the player or the host.

I disagree with the logic for two reasons.

Firstly, in the same way as you have grouped the unselected doors you could have said that the probability of the car being behind either Door 1 or Door 3 is 2/3. Then when the host opened Door 3 I think your logic would suggest that the probability of the goat being behind Door 1 would become 2/3.

Secondly, if we go back to Bayes and consider the probability of there being a goat behind Door 2 if it is revealed that a goat is behind door 3 assuming that the probability of a goat being behind either door is 1/3. If we calculate:

$\qquad P(C_{2}|\neg C_{3})=P(\neg C_{3}|C_{2}).P(C_{2})/P(\neg C_{3})\,$

we get $\qquad P(C_{2}|\neg C_{3})=\,$ 1 x 1/3 / 2/3 = 1/2
I don't see how the rules of the game can push this to 2/3 as the constraints on the host do not reveal any information other than the car is not behind Door 3.
--Paul Gerrard at APT (talk) 02:36, 2 September 2008 (UTC)

---

$\qquad P(C_{1}|\neg C_{3})\,$ is indeed 1/2. The difference is this ignores the constraints on the host (in particular, that the host MUST open Door 3 if the player picks Door 1 and the car is behind Door 2). This would be the probability of winning if the host happens to open Door 3 choosing randomly between Door 2 and Door 3 (this is the "forgetful Monty" variant discussed in the article). -- Rick Block (talk) 16:30, 1 September 2008 (UTC)

Let's consider if the constraints of the game affect the probability of the host selecting Door 3. If we have no information about the location of the car the probability of the host opening Door 3 is 50% given that he is not going to open Door 1. This probability is true because:

There is a 1/3 chance the car is behind Door 1 and if it is there is a 50% chance the host picks Door 2 and a 50% chance the host will pick Door 3.

There is a 1/3 chance the car is behind Door 2 and if it is there is a 100% chance the host picks Door 3

There is a 1/3 chance the car is behind Door 3 and if it is there is a 100% chance the host picks Door 2

That is the probability of the host selecting Door 3 is the same as the probability of the host selecting Door 2. (This probability is 1/6 + 1/3 = 1/2)

I cannot see how the constaints on the host influence the probability that the car is behind Door 1.--Paul Gerrard at APT (talk) 03:04, 2 September 2008 (UTC)

Let's try this again. You're ignoring the effect of the game rules on the host's actions which is why saying we're given Door 3 has a goat is different from saying the host (under the rules of the game) has opened this door after the player initially picked Door 1. These are different givens, yielding (in this case) different probabilities. I'm not "adjusting" any probabilities or "grouping" any doors, only using what the probabilities are and plugging them into Bayes formula. I AGREE

$\qquad P(C_{2}|\neg C_{3})=1/2\,$

but disagree that this is the question we're answering.

Now it's time for you to say where you disagree with the following:

$\qquad P(H_{13}|C_{2})=1\,$ (probability of host opening Door 3 after the player initially picks Door 1 given the car is behind Door 2 is 1 - the game rules enforce this)

$\qquad P(C_{2})=1/3\,$ (probability of the car being behind Door 2 is 1/3)

$\qquad P(H_{13})=1/2\,$ (probability of the host opening Door 3 after the player initially picks Door 1 is 1/2 - the game rules enforce this as well and you correctly derived it, just above)

So, by Bayes

$\qquad P(C_{2}|H_{13},I)=(P(H_{13}|C_{2})\times P(C_{2}))/P(H_{13})\,$

$\qquad P(C_{2}|H_{13},I)=(1\times 1/3)/(1/2)\,$

$\qquad P(C_{2}|H_{13},I)=2/3\,$

Please note the big bold banner at the top of this discussion page which says the article is mathematically sound and the result has been experimentally verified numerous times. If you disagree, it's FAR more likely (I'd say certain) that you're incorrect rather than that the article is wrong. -- Rick Block (talk) 03:31, 2 September 2008 (UTC)

The player's initial choice is relevant to the extent that it restricts the choice of the host to the other remaining doors. However, is it relevant to whether there is any advantage in switching?

It is true that the player has a 2/3 chance of picking a goat and given the host must choose a goat it is very tempting to conclude that there is a 2/3 chance that the car will be behind the unselected door. That is, 2/3 of the time the rules of the game will cause the host to reveal with absolute certainty the location of the car. But, is it really true that the probability that the car is behind the unselected door is 2/3? Do the rules of the game really make this conclusion valid? It is tempting to say yes.

However, hmmmm ... BOTHER

I think you are right and I am wrong.

At the outset the player has a:

1/3 chance of selecting a door with the car behind it and if he switches he certainly loses.

2/3 chance of selecting a door with a goat behind it and if he switches he certainly wins.

That is, a switching strategy gives a 2/3 chance of winning.

This is a really good problem. Thanks for your patience in helping to me to work through this.

I WAS WRONG! THERE IS AN ADVANTAGE IN SWITCHING --Paul Gerrard at APT (talk) 13:14, 2 September 2008 (UTC)

And, to your original point, the Bayes expansion supports this as well. Please also note that you are in very good company - folks who initially got this wrong include Paul Erdős. -- Rick Block (talk) 13:54, 2 September 2008 (UTC)

Player's initial choice is irrelevant

The first choice of the player is irrelevant. The host will ALLWAYS open the door with a goat behind it leaving the player with two doors of which either one has the car behind it. The palyer then picks one of these two doors and so has a 50-50 chance of winning the car. Mathematically you should not count the hosts action because the propability that he opens the door with the car behind it is 0. So the number of actual choices are four of which two leads to the car and two to the goat. —Preceding unsigned comment added by 81.197.22.58 (talk) 21:12, 1 September 2008 (UTC)

The player's choice absolutely matters. The host can't open this door, and must open a door with goat. Yes, the host always opens a door with a goat behind it but because the host can't open the player's door and must open a door with a goat, if the player initially picks a goat (there's a 2/3 chance of this happening) the host is showing the player which of the other two doors (the unselected ones) has the car. Please read the Solution section and look at the diagram. If you're still confused we can talk about it. -- Rick Block (talk) 01:02, 2 September 2008 (UTC)

Allright, let's then ask what is the propability for the player to win the car AFTER (for the rest of the game) the host has opened the door? That must most certainly be 50%. Then we only have left the discussion about the relevance of the player's first choice. —Preceding unsigned comment added by 85.156.86.237 (talk) 09:30, 2 September 2008 (UTC)

Just because there are two doors left after the host opens one does not mean they're equally likely to have the car. Under the rules of the MH problem, they're not. Do you agree the playing card simulation is equivalent, i.e. shuffle two red twos (losers) and the ace of spades (the winner), deal one to the "player", and you (playing the part of the host) look at your two cards and turn up a red two? If so, try this, with playing cards, maybe 20 or 30 times and keep a record of how often you (the host) end up with the ace (in which case, switching would win). I suspect if you actually do this you'll discover not only that switching wins more often than not, but that you can take a shortcut which is that when you (the host) look at your hand you don't even have to turn up a two to decide if the player will win by switching. If the ace is one of the two cards (2/3 chance), switching wins. The same thing is true with the MH problem. After the player picks, if the car is behind either of the two unpicked doors then ... switching will win (and this is known before the host opens a door, based only on whether the player initially picked the car!). -- Rick Block (talk) 10:03, 2 September 2008 (UTC)

You seem to be right. Now when I get the idea, it leads me thinking that this problem would be easier to understand seen from the host's point of view. When the player doesn't switch his choice, the host can lose the car only if the player has picked the door with the car behind it in the first time. This is 1/3 chance so the host has 2/3 chance for keeping the car. By switcing his choice the player lets the host keep the car if it is behind the door he picked first (with 1/3 chance) and the player gets the 2/3 odds which the host had before switching. I think this pretty much complies with Chuck's (?) P(not switching) = 1/3 and thus we have P(switching) = 1 - P(not switching) = 2/3. I thank You very much for your time and comments. —Preceding unsigned comment added by 81.197.28.197 (talk) 22:09, 3 September 2008 (UTC)

JUST ONE MORE THING. To make it even clearer, you could present an analogyous case where there are 100 doors with 99 goats and one car behind them. First you let the player choose one door and then the host opens one door which he knows there is a goat behind it. Then the host would ask if the player wants to switch the door he picket first to all the 98 doors that are left. The most favourable choice must be obvious! The cunning thing in the original problem, is the number three, which reduces the choice to two doors in the second phase. It would though be interesting if someone could present a similar (simple and without too much mathematics) example in the three-prisoner case! —Preceding unsigned comment added by 80.186.57.76 (talk) 08:53, 4 September 2008 (UTC)

The way results are recorded in the playing card simulation makes it equivalent to a game in which the player chooses door #1, and then Monty says, “In a minute I’ll open one of the doors. Would you like to stick with #1 or switch to whichever of the two doors remains closed?”

A better simulation for the game, as stated, would be to keep track of four sets of numbers: wins by switching to #2, wins by switching to #3, wins by sticking with #1 when #2 is the remaining choice, and wins by sticking with #1 when #3 is the remaining choice. Out of all runs, the player will tend to win 1/3 of the time by switching to #2 and 1/6 of the time by sticking to #1 when #2 is the remaining choice.

Keeping track of these numbers might help the player understand not just that switching doubles ones chances, but why switching doubles one chances even when the puzzle uses a scenario that has only a 50% chance of occurring once door #1 is selected. Since the odds of winning by switching have been split into two, with one half being ignored, the odds of winning by sticking should be split as well: compare the 1/3 chance of winning by switching to #2 to the 1/6 chance of winning by sticking with #1 when #2 is the door that remains closed. Simple314 (talk) 02:10, 26 January 2009 (UTC)

Confusing text

In the Sources of Confusion section, the last paragraph is confusing.

Another source of confusion is that the usual wording of the problem statement asks about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open if the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3.

The example of the difference between conditional and unconditional is not illuminating. Specifically, why the probability is 1/2 if the host opens Door 3. What probability are we referring to here? --dhawk (talk) 17:06, 31 August 2008 (UTC)

The probability of winning by switching is reduced in the above circumstances because the fact that the host has opened door 3 indicates a greater likelihood (1/2 rather than 1/3) that you have chosen the car. This is because, if you had chosen a goat there is the possibility that he might have had to open door 2 because the car was in 3. If, on the other hand, he opens door 2, you know it is your lucky day because the car must be in 3 as otherwise he would have had to open it. Martin Hogbin (talk) 17:55, 31 August 2008 (UTC)

Ahh. That's interesting enough that I think it should be expanded, perhaps to include the fact that you would be certain that switching would get you the car if the host opened door 2. Maybe it's just me, but that paragraph by itself doesn't make those details clear. dhawk (talk) 05:43, 1 September 2008 (UTC)

Yes, perhaps further explanation would be in order but I believe that the issue of different host strategies is a sideline that greatly and unnecessarily complicates the the main problem, which is that nearly everybody does not believe that you get a 2/3 chance of the car if you swap (conditions apply). Martin Hogbin (talk) 18:47, 1 September 2008 (UTC)

Suggested improvements

Although I agree that the solution given should not be replaced with the suggested one I do think that the proposer has a point in that the page does not do a very good job of addressing the fundamental issue that the chances of getting a car if you swap is 2/3 and not 1/2 (Morgan et al. conditions plus the constraint that the host picks randomly when he has a choice, and the contestant knows this).

I do understand that there are issues regarding verifiability in deviating from the current solution but I believe that we should make every effort to make the current solution more convincing. I am sure that there are ways in which issues of verifiability can be addressed if we try.

I think the issue of different host actions is a red herring that complicates and already exceedingly difficult problem and that this should be omitted from the main part of the solution and from 'Sources of confusion' - the last thing that section needs is another source of confusion.

Two starting suggestions are:

1 Describe the problem at the start of the solution in sufficient detail that only one answer exists. This would include describing the host's exact policy.

2 Describe the problem in a way that eliminates the issue of host startegy, for example by including the quote from Morgan et al,'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?'. Martin Hogbin (talk) 18:59, 1 September 2008 (UTC)

The various comments on this page since I have been here indicate to me that the solution needs to be better explained. Martin Hogbin (talk) 22:09, 1 September 2008 (UTC)

Certainly no article is ever perfect. The existing "Problem" section provides both the well-known (ambiguous) Parade version and a fully qualified version that permits only one possible (numerical) answer. Given this, I'm not sure what your first suggestion is.

Your second suggestion (to use the Morgan et al. phrasing) would be odd, since they use this as a description of what they consider to be a different problem. Their issue is not so much "host strategy", but how to interpret the question. Nearly all phrasings of the Monty Hall problem put the player at the point the host has already opened a door and ask whether switching to the other door is better. This phrasing is consistent with Morgan et al.'s interpretation of the problem as a conditional probability question, which leads to a 2/3 chance of winning by switching only if the host is constrained to randomly (with equal probability) select from among two goat doors in the case the player initially happens to select the car. The unambiguous problem statement in the Problem section includes this constraint. Without this constraint what Morgan et al. call the "unconditional" question (corresponding to the "overall" chance of winning across all hypothetical players, i.e. if 1000 players play this game and all of them choose to switch, how many would we expect to win the car) still has the same 2/3 win by switching solution - but (as they show in their paper) permits other host behaviors where knowing which specific door the host opens yields a probability of winning by switching anywhere from 1/2 to 1. This is in some sense a technicality, but it's a significant enough technicality that it needs to be mentioned here. Nearly all people have a VERY hard time understanding that these are two different questions and that they require slightly different reasoning. We had an extended discussion about this some time ago (it started with the thread I referred you to - and continued for weeks if not months) and ultimately ended up with what's in the article now. The goal is to be pedantically (mathematically) correct but to avoid introducing the concepts of conditional vs. unconditional probability until after the (usual) unconditional solution is presented.

As an encyclopedia article, I think we're certainly in the ballpark of a reasonable approach. The goal is NOT to convince everyone that the 2/3 answer is correct but to provide the "normal" answer (in a way that most people can understand) and then to expand on this to cover the more technically accurate Morgan et al. approach (hopefully without losing too many people). That it's possible to have different "conditional" and "unconditional" answers for the same problem (with the same host constraints) is (I believe) extraordinarily unusual (I've tried, but have been unable to come up with other examples). This isn't exactly the crux of the Monty Hall problem, but I think putting the player in front of two closed doors and one open door is. The "normal" solutions certainly depend on an assumption that the host does nothing that changes the probability of the player's first pick, i.e. the door the host opens is a goat so the other door has a (1 - 1/3) chance of being the car or if you pick a goat first with a 2/3 chance and then switch you now have a 2/3 chance of winning the car. These are very common, but (per Morgan et al.) mathematically insufficient ways to "solve" the problem. I don't think the slightly more rigorous approach presented in the Solution section is tremendously more difficult to comprehend.

Again, no article is ever perfect. If you have specific suggestions for wording improvements please offer them up (or boldly make them - but only once without discussing them here if they're reverted). -- Rick Block (talk) 02:32, 2 September 2008 (UTC)

I think you are missing the point of what I am trying to do. I know much thought and hard work has gone into the page and those editors responsible would not be happy to see their work replaced but there has been a criticism of this page and a suggestion that the solution is replaced with another. I do not support this, but I do think that some of the criticisms are valid and should be addressed. As I originally said in response to the RFC, my basic suggestion is to add rather than replace.

Wikipedia should reliably inform, but it cannot do this if the reader does not believe what they are being told. The current attitude seems to be, 'We have done a good and rigorous job of explaining the paradox, that is all we can do. If readers do not believe the answer that is just too bad'. In this respect some of the articles on the related problems do a better job even though the articles are of poorer quality by Wikipedia standards.

What about a separate section called, 'Still not convinced? or even a separate article. We could then try different approaches to help people understand the paradox. We could state at the top that this section ignores some of the subtleties of the problem in favour of clarity. We could also consider my suggestion 2 above. It is not uncommon in mathematics to first answer a different question from the one originally posed and then to consider how the two questions differ. This approach might lead to a better discussion of the conditional/unconditional issue.

I would like to do some serious addition to this subject in cooperation with others, but in such a way that the integrity of the current article is not compromised. Perhaps a development article called 'Explanations of the Monty Hall problem', with consideration to a possible future merger - it might fall flat on its face but that way the current article is unharmed. Martin Hogbin (talk) 09:24, 3 September 2008 (UTC)

I feel WP:NOR should be pretty much sacrosanct; what we write in Wikipedia must be verifiable. What we could do, however, is to add a page in the Talk namespace with some more explanations, or even a Wikibooks article (are there any guidelines on referring to Wikibooks?)

Also, your suggestion of "Still not convinced?" immediately made me think of Talk:0.999.../Arguments. We could follow a similar approach here for those who dispute the theoretical result. Oliphaunt (talk) 17:33, 3 September 2008 (UTC)

We may be able to achieve some of what I am hoping for without OR. For example, my suggestion to start with a different question was based on the statement of a similar question by Morgan. Regarding WP:NOR there are reasons why we should also consider WP:IGNORE and WP:COMMON SENSE. Because of the concentration on NOR, the article does not do as well as it might in terms of being convincing. Also, in this particular case there is in fact a degree of OR in the selection of references that editors have chosen to cite. A perverse editor could rewrite the article giving the wrong answer but with it still being fully verifiable, citing letters and statements from many eminent organisations.

I do not think "Still not convinced?" should be just relegated to the talk pages although I agree that a dedicated talk page is a good idea. Can we not also try to improve the article's explanatory power by adding something? Martin Hogbin (talk) 20:16, 3 September 2008 (UTC)

Article deletion discussion

A newly-created article related to this topic – Introduction to the Monty Hall Problem – has been nominated for deletion. Interested individuals may participate in the discussion at Wikipedia:Articles for deletion/Introduction to the Monty Hall Problem. TenOfAllTrades(talk) 11:59, 4 September 2008 (UTC)

It's gone now. Wow, that was fast... —Preceding unsigned comment added by 69.36.227.135 (talk) 23:45, 4 September 2008 (UTC)

User space for development of a new section

I have set up a user page user:Martin Hogbin/Monty Hall problem (draft) for the purpose of developing clearer and more convincing solutions and explanations of the problem. All editors please feel free to add your pet solutions there for discussion and futher editing.

All editors (current, historical, and prospective) please note that I hope this page will eventually be edited to a standard where all, or parts of it, can be added here under a separate section.Martin Hogbin (talk) 10:21, 7 September 2008 (UTC)

Section Aids to understanding in the article exists for this purpose, and I think it covers the best of the notable common sense explanations well. If you can find a notable explanation that is easier to grasp then more power to you, but the suggestion that we ignore rules in favor of common sense is ironical, for as a veridical paradox the very point of the problem is that common sense is wrong. The fact that some people remain unconvinced by any explanation is a testament to its excellence as a veridical paradox: common sense rebels against it. ~ Ningauble (talk) 20:23, 9 September 2008 (UTC)

I do not understand what you mean by 'notable common sense explanations'. I do not believe that many people remain unconvinced by any explanation, it may be that there is one that will do it for them. Martin Hogbin (talk) 20:54, 10 September 2008 (UTC)

By "common sense explanations" I was referring to those which sacrifice rigor in favor of clarity and ability to convince. Perhaps "informal explanations" would be a better term, but I was straining to emphasize a point about common sense being at issue. By "notable" I mean to emphasize that, since there are so many erroneous treatments in circulation, even an ordinarily WP:RS needs some vetting akin to WP:GNG. Perhaps that is an overstatement, but the situation calls for careful scrutiny.

I didn't say many people remain unconvinced, but that some do. I know several who are normal, functioning adults. Don't underestimate the cognitive dissonance that must be overcome when common sense is challenged. ~ Ningauble (talk) 20:51, 11 September 2008 (UTC)

Not ambiguous

The article says "Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example whether the host must open a door and must make the offer to switch." I don't think this is true. I don't think that the original version is ambiguous. It doesn't say, if host behaves every time the same, but that is not important, because the question is about a single decision. That single decision has probabilities for every choice you take. The host could behave differently at a different point in time, and the resulting probabilities might change. But that is not the question.

You can use a computer to simulate this, in which case you would have to put the car behind a different door every time, let the host open a different door all the time, and then let you decide to stick with the door you chose first or not. Then you can use the results of the simulation as an estimation of then probability of the original setup. When you simluate that with a computer, you of course have know if the host always behaves the same. But that was not the question. In the original question, the host *does* open a door, behind it is *not* the car, and he *does* let you choose again. --195.227.10.66 (talk) 17:49, 10 September 2008 (UTC)

If one assumes Monty opened a door at random then the odds are different. His intention, or the rule he follows, does matter. Vos Savant acknowledged the ambiguity by admitting in a follow up article that the constraint was implicit in her solution but had not been explicit in her statement of the problem:[3] "... remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. ... Anything else is a different question." ~ Ningauble (talk) 18:23, 10 September 2008 (UTC)

Of course it is "ambiguous", or, more accurately, "incorrect". Lacking knowledge of the rules of the host's behavior introduces an independent variable into the equation. I don't see how "technically incorrect" differs from "incorrect" in mathematics. As far as I can tell, the words "ambiguous" and "technically incorrect" are media spin, trying to hide the fact that the problem and solution given by vos Santos were wrong.Apollo (talk) 16:23, 11 September 2008 (UTC)

Further ambiguity if host behaviour not clear from start: If you are in a quix programme where you suspect the producers want to avoid having to pay a car for you, you may suspect the extra choice is given because you already picked the car. Or if you suspect they WANT you to win the car - to boost the popularity of the programme, say - the extra choice may be an indication you DIDN'T pick the car to begin with. So, without info on host behaviour, this is not a pure math problem; it's one about psychology, TV company economy, viewer behaviour, and other issues way outside math. What is the information content of the fact that you are given the second choice? Only if you knew in advance that you WOULD be given this second choice is the content nil, and the math of the standard solution to the problem correct.--Noe (talk) 16:36, 11 September 2008 (UTC)

"Ambiguous" doesn't mean a little bit wrong (as in a little bit pregnant). Rather, it conveys more information about the nature of the error than "incorrect." Her answer to the problem stated was incorrect because she failed to state unambiguously what she meant the problem to be. I find it amusing how she tried to spin it as a misinterpretation by the reader rather than a misstatement by herself, for ambiguity may always be construed against the author. ...but we digress. ~ Ningauble (talk) 18:39, 11 September 2008 (UTC)

Essentially none of the objections to her answer were based on grounds that the problem statement was ambiguous (there's a quote in one of the references about this - maybe we should add this). The problem statement IS in a technical sense ambiguous, but given the context of "math puzzles" the missing assumptions are both what most people intuitively assume and highly reasonable. -- Rick Block (talk) 19:15, 11 September 2008 (UTC)

I agree. Almost any statement of any problem leaves some questions unanswered. These issues were used as a means of avoiding embarrassment by people who got the answer wrong, based on reasonable assumptions as to what the question was.Martin Hogbin (talk) 20:06, 11 September 2008 (UTC)

I agree, most of the confusion does not arise from this ambiguity. Although the objection of Morgan et al. could (arguably) be dismissed in similar manner as "a different question" than that intended, it is an analysis too important to leave out. ~ Ningauble (talk) 21:28, 11 September 2008 (UTC)

The problem as stated is not mathematically ambiguous nor incorrect - because it describes a situation which could occur and asks a question which has an answer. Yes, the hosts behavior is not specified to the degree you might wish but as a math problem, you just have to solve it without knowing more. You have to deal with the possibility that the host opened the door at random, and it happened to have a goat behind it; you have to deal with the possibility that the host knows the contestant picked the car and is merely trying to convince him or her to switch; you have to deal with every behavior that is consistent with the information given.

For all the furor, once the hosts behavior is fully specified it is a straightforward math problem but it becomes much more interesting if you assume you're supposed to solve the problem exactly as it stands. (Almost certainly, that was not the intent of the person who asked the question originally, but that is another matter.) Yes, it does have an answer and the answer is that it is better to switch - how much better doesn't have a simple numerical answer but the form it takes shows that switching is superior. And that is all thats needed to answer the question as posed.

Millbast5 (talk) 10:42, 13 September 2008 (UTC)

I agree with you that it is possible to give a solution based on reasonable assumptions. However, I think that in your statement, 'once the hosts behavior is fully specified it is a straightforward math problem', you completely miss the point. There are interesting and relevant issues concerning the host's behaviour but what makes this problem special is that even when fully specified, nearly everyone gets it wrong, including excellent mathematicians and statisticians. Some of them have tried to use the issue of the host's behaviour to cover up the fact that they have made a mistake.Martin Hogbin (talk) 12:07, 13 September 2008 (UTC)

The problem may be tricky as a word problem but as a probability problem it is one that most students in an elementary course in probability would get right. Stated as a problem in terms of the classic balls and urn it is: an urn with three balls in it, known by everyone concerned to be two white balls and one red. One person reaches into the urn and withdraws a ball at random which he keeps concealed in his hand. A second person announces that she will withdraw a white ball then looks into the urn and lifts out a white ball, as promised. What is the probability that the ball left in the urn is white? What is the probability the ball left in the urn is red?

Stated this way it is transparent is it not? The ball left in the urn is white is if and only if the first person withdrew the red ball - which happens with probability 1/3. The ball left in the urn is red with probability 1 - 1/3, or 2/3.

I can't say why the statement in terms of doors, etc., hides the fact that the initial action determines everything that follows. But here its quite clear - the guy draws a red ball, the second person withdraws one of the two white balls, white ball left - automatic. The guy draws a white ball, the second person withdraws the other white ball, red ball left - automatic.

I stand by my claim the that as a math problem it is straightforward. (Actually, a piece of cake relative to some other ball and urn problems students routinely solve.)

Now to the interesting form of the problem - where nothing about the hosts behavior is known beyond what was explicitly stated in the given problem. I did not say that the problem could be solved given reasonable assumptions - I said it could be solved as it stood. It does require methods that are accepted as good probabilistic techniques - the use of Bayes Theorem for example - but no assumptions.

How does one deal with the possibility that the host might have opened a door at random and, by chance, a goat was revealed? That part is relatively easy. Opening a door at random leaves the switch and non-switch strategies equal, whether a goat or a car is revealed, hence if the other possibilities for the hosts behavior put the switch strategy ahead it is better overall.

I should amend my statement a bit - I should not have said the problem had no numeric solution for the amount of the difference between the switch and non-switch strategies. I should have said that I can determine that the switch strategy is better but the method I used does not quantify the difference. The way I dealt with the possibility of random door opening is what stands in the way of quantification - happily the problem doesn't require it.

The rest of the solution is fairly complicated but I will glad to post the details if someone wants to see them. The overall method is to parameterize all the possibilities for the hosts behavior with two continuous random variables, p and q, where p is the probability the host will open a door he knows has a goat behind it when the contestants initial choice is the winning one and q is the probability he'll do that when the contestants initial choice is a loser. To arrive at a distribution for these random variables, Bayes Theorem is used with an equal-likelihood apriori distribution and the data that in the one known trial, a door was opened revealing a goat. Using that distribution one averages over all possibilities to arrive at a number for the probability of getting the car using the switch strategy.Millbast5 (talk) 08:01, 14 September 2008 (UTC)

You should try asking your urn version of the problem to people that you know. You will be amazed how many get it wrong. There is nothing magic about the door-and-goat statement that makes it especially difficult - the difficulty is in understanding how the probability works. Of course, once you know the answer it is easy!

It makes no difference whether the host knows where the goat is, all that matters is that he does, in fact, reveal a goat. Let me restate this in terms of your urn problem: an urn with three balls in it, known by everyone concerned to be two white balls and one red. One person reaches into the urn and withdraws a ball at random which he keeps concealed in his hand. A second person withdraws a bell at random which proves to be white. What is the probability that the ball left in the urn is white? What is the probability the ball left in the urn is red? What is your answer to the above question?Martin Hogbin (talk) 09:18, 14 September 2008 (UTC)

It is of course possible to "solve" the problem without making any assumptions about the host's behavior as User:Millbast5 suggests by assigning probabilities to all possible variants. Morgan et al. do this with the host's preference for choosing between two goat doors while keeping the other constraints intact (host must open a goat door and must make the offer to switch). This analysis says that as long as the host must open a door to show a goat and must make the offer to switch, the probability of winning by switching varies from 1/2 to 1 (depending on how the host picks between two goats), i.e. switching is always at least as good as staying but unless we further constrain the host (for example, adding that the host picks between two goats randomly with equal probability) we can't know a numerical answer.

There's a paper by Chun presenting a formal way to describe the fully generalized problem, "On the Information Economics Approach to the Generalized Game Show Problem" published in The American Statistician in 1999 Vol 53(1). Without digging out the paper I forget whether Chun uses the description to net out the bottom line for the probability of winning by switching, but in a similar fashion to Morgan et al. one could clearly assign probabilities to the variants and end up with a formula involving these probabilities. The "Other host behaviors" section hints at this, listing some of the published variants and the resultant probabilities of winning, but doesn't "sum it up". Note that given the variants listed in this section, the probability of winning by switching is between 0 and 1, and an analysis that concludes you're better off switching is going to rely on how probabilities are assigned to the variants. If there's a paper that shows an assumption-less analysis (Chun's or some other paper), we could try to present it here, but without a source to work from I think we're definitely in WP:OR territory. Another question is whether such an analysis would be appropriate to include in this article, which is often criticized for being too long already. I'd suggest that it might be appropriate to mention such a analysis (if one has been published) but not attempt to include it in the article. -- Rick Block (talk) 16:42, 14 September 2008 (UTC)

Are you saying that certain host actions (subject to host must always show a goat and offer the switch) can reduce the odds of winning if you switch to less than 2/3?Martin Hogbin (talk) 19:40, 14 September 2008 (UTC)

Depending on what you mean by "the odds", yes. This is the whole point of the Morgan et al. paper. Must show a goat and must make the offer makes the overall (unconditional) probability, across all players, 2/3. But, if the host opens the rightmost door whenever possible (or has any other identifiable preference for one door or another) there are two subsets of players. Those who switch when the host opens the rightmost (absolutely preferred) door win 1/2 the time. Those who switch when the host opens the leftmost (absolutely unpreferred) door always win. The usual presentation of the problem ("say you've picked Door 1 and the host opens Door 3") implies we're asking about the probability given the knowledge of which door the host has opened, i.e. the conditional probability as opposed to the unconditional probability. If we don't constrain the host to pick between two goat doors with equal probability, the best we can say about the chances of winning by switching if we already know which door the host has opened is that it's somewhere between 1/2 and 1, with an "average" probability (across all players) of 2/3. Morgan et al. assign the host's preference for the door he opens a probability p - switching wins with probability 1/(1+p). -- Rick Block (talk) 00:16, 15 September 2008 (UTC)

Thanks for that explanation.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

To answer the question Martin asked: the probability that the first person draws a white ball is 2/3 and nothing can change that, though information about whether he has in fact drawn the red or a white ball in specific instances can be obtained by drawing another ball at random from the urn. When the second random ball is red, which happens with probability 1/3, the conditional probability is 1 that the first ball drawn was white; when the second random ball is white, which happens with probability 2/3, we let x be the conditional probability the first ball drawn was white. Combining the two cases we get 1/3 * 1 + 2/3 * x as the overall probability of the first ball drawn being white. Since we know the overall probability is 2/3, we can solve for x: 1/3 * 1 + 2/3 * x = 2/3 <=> x = 1/2. Happily, that agrees with our intuition.

Yes, after some thought, I agree with you. There is a difference between does draw a white ball and must draw a white ball. Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

We can get the conditional probability for the case when a white ball is intentionally extracted from the urn (my statement of the problem) as follows: when the second drawn ball is red, which happens with probability 0, the first drawn ball was white with conditional probability 1; when the second drawn ball is white, which happens with probability 1, let the conditional probability be x as before. Combining the cases: 0 * 1 + 1 * x = 2/3 <=> x = 2/3.

Did you really say "It makes no difference... he does, in fact, reveal a goat."? I keep reading it and it keeps coming up the same. You must have seen the argument above in some form and rationalized that away. How about changing the rule for the second person to be that she withdraws the red ball if its in the urn and only when there are two white balls in it does she withdraw a white. Thus whenever she withdraws a white the first person must have drawn the red ball, so the conditional probability of the first person having drawn white is 0. Hopefully this is sufficiently clear that you can see, regardless of what your intuition tells you, that the method that leads to the event of a white being drawn by the second person does make a difference in the conditional probability of the first person having drawn a white ball.

Yes you are right.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

I think we might have stumbled onto something here explaining the population's difficulty with this problem - they somehow twist it around in their minds until it depends on grasping how two events that are exactly alike in appearance can result in different effects on a conditional probability. It seems so obvious to them that they cannot have different effects that simple, straightforward derivations of the solution are rejected, even though they can't point to a flaw.

Could be.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

Yes, I think you are right. My mistake was to assume, without thinking, that a white ball being removed by chance has exactly the same effect as its being removed intentionally. I therefore think it is important in our explanation to cover this point in some way. In the current solution the probability of losing by switching, if the player initially chooses a car, is split into two lots of 1/6, with no explanation of why this is so and in what circumstances it would not be.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)

I don't believe we should go into this level of detail in giving a solution for the general reader - you have to be into 'philosophy' to even worry about the information aspect of various actions. I reworded my description of the game (in terms of urns and balls), into doors, goats and cars. It came out just as clear to me - clearer in fact because I avoid probability altogether. I read in one of the references that people understand frequency better than odds, so I stated it in frequency terms. Think the guy was right. I inserted it into the solutions sections above. Seven lines, no pictures, no probability and only two symbols - 'A' and 'B'. Almost anyone can read it over several times in five minute and get the whole thing in his head at one time.Millbast5 (talk) 23:49, 17 September 2008 (UTC)

I suppose it might seem a bit childish, but one can also settle this by playing the game numerous times, applying different methods for the extraction of the second white ball and keeping track of the results for the different ones.Millbast5 (talk) 01:40, 15 September 2008 (UTC)

Oh rats, I didn't answer exactly the question Martin asked - he asked about the ball in the urn and I answered as though he asked about the ball in the first persons hand. It should make no difference because I did state what I was calculating. —Preceding unsigned comment added by Millbast5 (talk • contribs) 02:05, 15 September 2008 (UTC)

I assume the statement of the problem discussed in this thread is the one from Parade, as quoted in the lead:

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?

As I stated earlier in this thread, it is ambiguous because we do not know whether there is any information content in the fact that the host is opening another door - did he choose to do so BECAUSE the guest either hit or didn't hit the car with his first choice? If the cost of the prize won by the guest is deducted from the host's salary, say, he may simply be trying to direct the guest away from the car, so THE FACT that he was given a second choice is an indication that he DID in fact choose the car from the start, i.e. DON'T SWITCH! Or, if the advertising income for the quiz show is boosted whenever a guest wins a car, the producers may want to direct the guest to the car, so that the fact he was given a second choice is an indication he did NOT choose the car to begin with, i.e. DO SWITCH! We are not given enough information to decide whether one of these scenarios may be taking place, so in order to attack the problem as a math problem, we need to make certain assumptions to begin with, e.g. that the host ALWAYS gives the guest a second choice after opening a door with a goat behind it. This assumption may be a natural one to make, and the fact that many people defend "wrong" solutions to the problem is NOT simply caused by this ambiguity, but still, it IS ambiguous.--Noe (talk) 07:41, 15 September 2008 (UTC)

The point user:Millbast5 is making is that the problem is not ambiguous (meaning it has two different, conflicting interpretations) but rather it simply has more uncertainties than we'd like. We can solve the problem even not knowing any of these things by assigning them each a probability and developing an equation for the overall probability of winning based on these probabilities. For example, we could start with three probabilities p1, p2, and p3 representing the probabilities that the host made the offer to switch because the player initially picked the car, because the player didn't initially pick the car, and because it's simply the rules of the game (respectively). Clearly,

p1+p2+p3 = 1

and a player who switches always loses in the p1 case, always wins in the p2 case, and may or may not win in the p3 case. So, we have (so far)

p(winning by switching) = p2 + x * p3

where x is the conditional probability of winning given that we're in the p3 case (i.e. neither the p1 or p2 cases). This sort of analysis can continue and we'll end up with the probability of winning being a function of 5 to 10 probabilities expressing the likelihood of a set of various scenarios. The range of this function will be [0,1]. If we assume each of the mutually exclusive probabilities are equally likely (for example, p1, p2, p3 above are all 1/3) we can compute a number - but this number would depend on how we break down the possibilities. Instead of the 3 possibilities for "host intent" above, we could say there are two possibilities - host "plays fair" (p3) or not (p1 or p2). Breaking it down this way and assuming equal likelihood we'll end up with p1=p2=1/4, and p3=1/2 (so our final numerical answer will be different).

The point is a conditional probability analysis is sufficient to analyze a problem with as many uncertainties as you'd like, however I'm not sure anyone other than a mathematician would think such a solution would be satisfying. -- Rick Block (talk) 14:11, 15 September 2008 (UTC)

Thanks for the assist Rick, but I meant something simpler - I meant the question of whether the switch or not-switch strategy is better does have a definite answer. (The answer is that the switch strategy is better.) Assigning probabilities in the way you're describing comes under the heading of making assumptions. Statisticians, mathematicians and others have developed methods to deal with this type of shortage of information which have been borne out as accurate and those methods are what I used - I would expect another mathematician solving this problem to do the same.

As Noe mentions one might be in a situation where the non-switch strategy is better even though the switch strategy is better over the full gamut of possibilities. (In game theory, a strategy which is better in every case than another strategy is said to dominate the other one - which isn't true here.) Of course, it takes information that is not given in this problem to know you are in a situation where you should use the non-switch strategy - but if you do have some other information, you use it. Or if you have a hunch about that evil looking host that you just can't ignore. If you got a peek behind one of the doors and saw a goat, use that too. (If you're using the switch strategy, of course your original choice will be that very door so you're certain to win if you get a chance to switch.)

Can I get this straight. There are two types of issue that can affect the conditional probability. The first is really down to the rules of the game (rather than host behaviour but there is no clear distinction) and is that the host must open a door to reveal a goat, in other words by the rules of the game the host can never open a door to reveal the car.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)

The second type of issue is the way in which the host chooses which goat to reveal when he has a choice.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)

I agree with Martin that this latter type of issue is far different than the rules of the game.

I'm of the opinion that bringing up the possibility of the host having a known preference for which door he opens to reveal a goat is deleterious to the article. Of course, its possible but nothing in the problem statement hinted at the possibility that contestants watched the show long enough to spot his preference - the original question applies as well to the very first time the game show was held as any other. It is not a question that can even come up if the doors are not identified and the doors were identified as for expository reasons - making things more concrete. The problem can be stated without such identification and statements of equivalent problems don't have anything corresponding to doors.

Taking an accidental feature of the problem statement plus adding an assumption that the process has been going on long enough for a preference to become known is basicly a different problem - at best it should be in a footnote. You could bring up the possiblity that host used his left hand to open the door more often than his right in some circumstances or blinked his eyes at a differing rates - distracting trivia is the category this sort of thing falls into.Millbast5 (talk) 06:33, 18 September 2008 (UTC)

I think I will clean up my computation of the probability of winning with the switch strategy and put it on my talk page so that anyone that wants can see exactly what I'm talking about. Everyone thats agreed with me up until now has misinterpreted what I was saying. Should be there by tomorrow night at the latest.

Maybe I've made a little progress because Noe did not mention the possibility of the host randomly opening doors as something making the problem ambiguous.

Millbast5 (talk) 01:44, 16 September 2008 (UTC)

My talk page contains the computation for the switch strategy.Millbast5 (talk) 09:17, 16 September 2008 (UTC)

@

Lead change

The two sentences added to the lead (see this diff) assume the host opening a door has no effect on the probability of initially selecting the car (which is true only because of the constraints on the host), and essentially explain the fairly trivial point that 1 - 1/3 = 2/3. How about if we state the major reason that the initially picked door remains a 1/3 chance? For example:

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In the usual interpretation of the problem the door the host opens is not randomly selected and opening this door does not affect the player's initial 1/3 chance of having picked the car. The car is behind one of the two unopened doors; if there is a 1/3 chance it is behind one there must a 2/3 chance it is behind the other. The player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

The critical point is that the host opening the door does not affect the player's initial 1/3 chance of picking the car - mostly because the host does not open a random door. I think assuming this to be true, without no explanation whatsoever, does not accurately summarize the solution. -- Rick Block (talk) 02:27, 13 September 2008 (UTC)

I prefer the version previous to the diff. It gives the lead section more punch, and gives the reader pause to think. To elaborate: (1) Stating the naive perspective and the correct result, with emphasis on the magnitude of the error, is an excellent way to explain why the subject is interesting. (2) While teasers in the lead are generally discouraged, it is appropriate and arguably necessary to make an exception for articles about puzzles or paradoxes. Explanation of the solution, even in summary, ought to be deferred to the body of the article in order to accommodate readers who are inclined to try reasoning it out for themselves. ~ Ningauble (talk) 14:14, 13 September 2008 (UTC)

Another objection - moved from main page

I moved this from the main article to here. It does not really belong here but it has been suggested that we set up a space for discussion of the subject so that it is kept separate from the article itself and discussion on how to improve it. I am not sure how to do that but I think it is a good idea. Can any one help?Martin Hogbin (talk) 08:50, 13 September 2008 (UTC)

I found this answer to this question on another website and am in no way taking credit for this answer...

"The solution given is that the selection should be switched because the door you chose has a 1/3 chance of success, while the other unopened door now has a 2/3 chance of success.

This solution is surprising, counter-intuitive... and wrong.

Marilyn vos Savant was wrong. The solution given by Mark Evanier was wrong. And the solution given by these fictional math geniuses of "21" was spectacularly wrong.

What am I talking about?! How can all these real and fictional geniuses be wrong?

Switching doors can double your odds of winning, but only if Monty Hall not only knows which door has the car, but is REQUIRED under the rules of the game to open a door that (i) is not the door you chose, and (ii) is not the door that has the car behind it. This must be stated as part of the puzzle in order for the correct solution to be that there is an advantage to switching doors.

This is not merely a semantic distinction -- it is critical to the problem. If Monty Hall is operating under those rigid rules, by opening a goat door after you've chosen, he is giving you valuable information. If you guessed correctly (a 1/3 chance), he can open either of the other two doors. But if you guessed incorrectly (a 2/3 chance), he is FORCED to open up the one door that is neither your choice nor the door with the car.

Now, the puzzle, as presented by Kevin Spacey, does not state that Monty Hall is required to open up any door. He simply says that after you choose your door, Monty "decides" to open up another one that reveals a goat.

If the rules of the game are that he opens a randomly selected door after you make your choice, then the logic of switching doors break down -- there is no advantage to switching whatsoever.

If he simply opens up a door without regard to any pre-determined rules of the game, there is also no advantage to switching. You don't know why he opened a door, so you haven't gained any information.

And in fact, in the clip from the movie "21," the solution given is spectacularly wrong because Kevin Spacey's genius character explicitly posits that you don't know why Monty Hall is opening up a door -- he says that the game show host may be trying to "playing a trick on you, trying to use reverse psychology to get you to pick a goat." Despite what these fictional math geniuses say, there surely is no advantage to switching in this case, because, again, you've gained no useful information. It may be to your disadvantage to switch under these circumstances.

Let's say you play this game 100 times, and Monty reveals a goat door only when you've initially picked the car door (and does nothing when you initially pick the goat door -- just opens your door and tells you that you lost). 67 times you'll pick a goat door and automatically lose. If you adopt a switching strategy, the 33 times you pick a car door, you'll then be shown a goat door by Monty, get "tricked" into switching, and lose. The switching strategy will yield you exactly zero wins out of 100. (Refusing to switch will give you 33 wins.)

The same holds true if you play only once. You pick a door, and then Monty reveals a goat door. Unless his actions are explicitly part of the rules of the game, you have no idea why he showed you the goat door. He may have only showed it to you because your first pick was the car door, and he is in fact trying to trick you into switching. You're now in a mind game with Monty Hall that has nothing to do with the "statistics" and "variable change," that Kevin Spacey's super-student refers to.

This distinction is part of the reason the commonly-given solution is so counter-intuitive. The way it's presented, you're playing this game, and Monty Hall suddenly and without explanation reveals a goat door and gives you a chance to switch -- it doesn't seem as though it would be to your advantage to do so, and in fact it isn't. But once you say that Monty Hall is required under the rules of the game to reveal a goat door and give you a chance to switch, it makes a bit more intuitive sense that there may be an advantage in switching (and in fact there would be)."

source -->(http://gocomics.typepad.com/tomthedancingbugblog/)

The objection above results from the way that probability problems are often stated.

In the Monty Hall problem, if Monty picks a door at random, there is the possibility that he will pick the car. But, he is always seen to pick a goat (and this is said in the statement of the problem). Now it could be that Monty picked randomly but the TV company only showed the times when he picked a goat, however, it is far more likely that he knew where the goat was. In both cases the results are the same, of the screened shows the player who swapped won more often than the one who did not. The problem states that Monty reveals a goat thus we are expected to consider only the cases where this is so.Martin Hogbin (talk) 10:08, 13 September 2008 (UTC)

This objection is already addressed in the article (2nd to last paragraph in the lead, in the Problem section, in the Sources of Confusion section, and in the Other host behaviors section). The topic is already covered (with references), and this material is copyrighted (so can't be used directly, and isn't what I'd call a reliable source anyway). Is there a suggestion here for some change to the article? -- Rick Block (talk) 20:07, 13 September 2008 (UTC)

Not by me. I moved the text above from the article (where it hand been added by an IP editor) to here.Martin Hogbin (talk) 20:40, 13 September 2008 (UTC)

Language clean up

In the introdudtion, it currently states:

" Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty."

Two things need changing in this. The first is the reduction of words. Not advantageous to disadvantageous to begin with. The second thing is that the sentence is misleading imho. The point of the Monty Hall problem is that it is a counter-intuitive probability based problem. The point is not at what point is it disadvantageous but that probability shows that in the long run it is advantageous to switch. The addition of the statement I quoted above seems to distract from the point the article is trying to make and should not be in the introduction. --Candy (talk) 18:57, 22 September 2008 (UTC)

Agreed. Rick had suggested a clearer explanation two sections previous in this discussion (Lead change) but, for reasons given there, I favor simply removing these two sentences from the lead section. Does anyone object to this proposed revert? ~ Ningauble (talk) 19:03, 24 September 2008 (UTC)

Agreed. Anything that makes things clearer.Martin Hogbin (talk) 17:49, 25 September 2008 (UTC)

I agree with the suggestion of using the word 'disadvantageous' but not with removing the two sentences - to me they are the best explanation in the whole article of why the switch strategy wins with probability 2/3 - I believe that compactness is a big plus and that pictures are a distraction. I would favor emphasizing the idea at this stage that the host is intentionally opening a door with a goat behind it. So something like 'when the host opens the other door he knows conceals a goat'. Grammatically it is incorrect as it is because it isn't the 'losing door' that is revealed but the goat behind it.Millbast5 (talk) 00:45, 27 September 2008 (UTC)

Random host variant

Regarding the variants on host behaviors, I don't find the example of:

"The host does not know what lies behind the doors, and opens one at random without revealing the car (Granberg and Brown, 1995:712)."

to be very useful. This conditional ignores the outcome that occurs 1/2 the time when the car is behind door 2 or 3 and the host must choose a random door to open - he accidentally chooses the car. Since this example of host behavior ignores these potential outcomes of the game and the idea that host chooses randomly is covered well in the previous "aids to understanding" section, I would advocate deleting this variant from the table. However if it is to be included, shouldn't the odds of getting the car by switching still be 2/3? Of course the overall odds of the game if the host does not know the location of the prize has been shown to be 1/2, with the car being prematurely revealed in 2 of the 6 overall outcomes and the win/lose status of the game to be unknown. But in this specific case, it is restricted to the condition where the host "opens one at random without revealing the car." Isn't it correct that once you are to this point in the game, the probability is exactly the same as if the host had known the location of the car and deliberately exposed a goat? This variant eliminates the option of the host exposing the car, so he must act as if he has knowledge of the car's location even though it is stated that he does not. In other words, once you have gotten to this point, you have one of three scenarios (sticking with the idea the player has chosen door number1)

1. car behind door1 and host has opened door 2 or 3
2. car behind door2 and host has opened door 3
3. car behind door3 and host has opened door 2

Each of these scenarios has a 1/3 chance, in 1 you loose by switching and in 2 or 3 you win by switching thus the odds are the same as with a knowledgeable host switching gives a 2/3 chance of winning. Clarence3456 (talk) 20:03, 25 September 2008 (UTC)

This is a notable variant (many references discuss the difference between this variant and the standard one), so I think it should remain in the table. The odds are 1/2 because there are 6 equally likely events:

1. car behind door1 and host has opened door 2
2. car behind door1 and host has opened door 3
3. car behind door2 and host has opened door 1
4. car behind door2 and host has opened door 3
5. car behind door3 and host has opened door 1
6. car behind door3 and host has opened door 2

but because we're given that the host happened to open a door that didn't show the car we know we must be in one of (still equally likely) cases 1,2,4, or 6. We don't know what would happen in cases 3 or 5 so we simply ignore them, but that doesn't increase the probabilities of cases 4 and 6. If we're given the host didn't show the car, we're only looking at 4/6 of the total cases and among those there's a 50/50 chance of winning by switching. -- Rick Block (talk) 13:39, 26 September 2008 (UTC)

I'm not disagreeing with your explanation Rick, but I think the best method of showing the probability is 1/2 is to compute the number from an indisputable probability - that being: the probability is 0 that the winning door was chosen if the car is revealed. Randomly opening a door entails two results - revealing a goat and revealing a car. Neither of these can possibly alter the fact that the winning door was chosen with overall probability of 1/3 but each result does alter the probability in a particular case. Since the overall probability cannot change these different results must cancel each other out. One third of the time the randomly opened door reveals a car and when that happens the probability that the chosen door conceals the car is 0. Two thirds of the time the randomly opened door reveals a goat and when that happens the probability the chosen door conceals the car is x. 1/3 * 0 + 2/3 * x = 1/3 is the equation that describes the state of affairs. Solving for x one gets 1/2.

It is generally treacherous to consider one outcome of a random event in isolation because its easy to lose sight of the importance of other outcomes. Even if you avoid the pitfalls you can often use the others to verify your computations.Millbast5 (talk) 01:41, 27 September 2008 (UTC)

You can use this same argument to show that the host intentionally opening a door he knows conceals a goat cannot affect the probability in a particular case. As before let x be that probability. The probability is 1 that he will reveal a goat so the equation is: 1 * x = 1/3, i.e. x = 1/3. Or you could say that since there is no alternate outcome to cancel whatever change you might hypothesize, the change must be zero.Millbast5 (talk) 01:58, 27 September 2008 (UTC)

Milsbast5 I see you and Rick are correct. An alternate to Milsbast's equation that helps me understand is that the total probability must of course be 1, and as you state, if the host reveals a goat, the chance that door contains the car is 0. The chance the each other two contained the car was equal at the start of the game and no actions have changed that, so call the odds that either door contains the car x. So overall: 0 + 2*x = 1. So x must be 1/2. If you guys want, you can delete this section of the discussion to keep it clean or leave for others.Clarence3456 (talk) 19:02, 27 September 2008 (UTC)

Clarence your argument is simpler and simplicity is always good. The reason for my more circuitous argument is that it ties together two things which should be tied together - the different results that come from one random event. It also explains why opening a door at random and revealing a goat by chance is not the same thing as opening a door known to conceal a goat, even though to the casual observer they are identical. —Preceding unsigned comment added by Millbast5 (talk • contribs) 01:14, 28 September 2008 (UTC)

I keep trying to think of new ways to explain this problem so that even someone who firmly believes that the two strategies are equally good can finally get it. Here is another go at it.

Imagine there are two players involved, named N and S - initially they have to guess the same door but after the host has opened a door, S has to switch to the remaining door. Also imagine, to begin with, that the host opens a door at random. That is extremely easy to analyze - each of the three participants has a 1/3 chance of having or opening the door that conceals the car. You will note that in the cases where the host opens a door concealing a goat, one of the other two wins and they win equally often, of course, since they each get the winning door 1/3 of the time.

Those who believe that it matters not a whit whether the host opens a door he knows hides a goat or he opens a door at random and reveals a goat by chance, can point out that none of the trials where the host did the latter would have come out any different if he picked the same door on purpose. Couldn't possibly get a different result in those cases - N has same door, Host opens same door so S is forced to take the same door. Case closed - it doesn't matter.

Now we consider the cases where the host opens a door at random to reveal a car, which happens 1/3 of the time. Can't forget those cases if the host is going around opening doors at random. So N has a goat, host shows car, S stuck with the other goat. Now pretend nothing changes except that the host is not allowed to show the car. He's also not allowed to take N/S's initial choice so he has to take the door S was eventually forced to take in the random play - there is no other door to consider. Well, now S is forced to take the door that hides the car. Clearly not fair - S gets every car the host showed in the random play in addition to the ones he got when the host randomly revealed a goat. N gets exactly what he got before.

So if one focuses on the cases where the host opens a door randomly and reveals a goat then the result is the same if he opened the same door because it hid a goat. On the other hand, if you consider all the cases where the host could open a door he knew concealed a goat - which is every case - then there is huge difference. S wins twice as many cars as N does.

What should you do if you don't know why the host opened the door to show a goat (as you don't know in the original problem statement), you still jump at the opportunity to switch because you never gain by not switching and you might gain big time by switching. Heck, even if you still believe that switching and not-switching always end up of the same you ought to switch because of the small chance you are wrong and the, so-called, experts are right - stranger things have happened. Of course, if you have formed a strong attachment to the door you chose at the outset or if you have a lurking suspicion the host knows you picked the right door and is trying to trick you into switching, then defy the odds and stick with it. Imagine the mental anguish if you chose the winning door originally and switched - the experts haven't factored that into their analysis!Millbast5 (talk) 00:47, 28 September 2008 (UTC)

Correction! While its true that you never gain on your 1/3 chance of winning by not switching, its possible to fall below winning 1/3 of the time by switching - whenever the host is allowed the option of not opening a door at all. He may, for example, open a door only when your first guess was correct. The general rule governing it is: if the host opens a door known to hide a goat P percent of the time when he knows you've picked the winning door and does the same less than P/2 percent of the time when he knows you've picked a losing door, then the switch strategy wins less than 1/3 of the time.Millbast5 (talk) 05:39, 28 September 2008 (UTC)

I agree that a more convincing explanation is required. Why not add your suggestion to this page user:Martin Hogbin/Monty Hall problem (draft), where it can be discussed and improved with a view to adding it here when it is ready and acceptable.Martin Hogbin (talk) 09:13, 28 September 2008 (UTC)

I see that you have.Martin Hogbin (talk) 11:05, 28 September 2008 (UTC)

---

I will try to prove that the version in which the host opens doors randomly, but never picks a door with a car behind it, does not decrease the player’s chance to win the car when switching doors, compared to the original situation.

The original situation, in which a player picks door 1, has six theoretical possibilities:

player	host	host	player	host	total
door1	door2	door3	chance	chance	chance
a	g		1/3	1/2	1/6
a		g	1/3	1/2	1/6
g	a		2/3	0	0
g		g	2/3	1/2	1/3
g	g		2/3	1/2	1/3
g		a	2/3	0	0

a = auto; g = goat; empty spaces are unopened doors with goats behind them

Explanation: if door1 hides a goat (player’s chance = 2/3), the host has two options: reveal a goat behind door2 or reveal a goat behind door3. Since the goats are placed randomly, these have even chances.

The version in which a host randomly opens door 2 or 3 has the same possibilities with different chances:

player	host	host	player	host	total
door1	door2	door3	chance	chance	chance
a	g		1/3	1/2	1/6 *
a		g	1/3	1/2	1/6 *
g	a		2/3	1/4	1/6
g		g	2/3	1/4	1/6 *
g	g		2/3	1/4	1/6 *
g		a	2/3	1/4	1/6

* In 4/6 = 2/3 of the possibilities the host does not reveal the car. If we would have to statistically test this, we could do the game e.g. 999 times and create about 666 situations in which no car was revealed. Then we take a look into only those 666 cases, and pick one of those. Now there are four options that could have created this situation:

player	host	host	player	host	total
door1	door2	door3	chance	chance	chance
a	g		1/3	1/2	1/6
a		g	1/3	1/2	1/6
g		g	2/3	1/4	1/6
g	g		2/3	1/4	1/6

Indeed, they add up to a total chance of 4/6 = 2/3, which is exactly the part we are in now. But these chances are old chances that created this situation. What is the chance that any of these four possibilities is now true? Of course it is 1. The sum of chances changed from 2/3 to 1, because we stepped out of the 999 options into 666, carefully selected by one variable. In fact, the ‘knowledge’ of the host in the original problem has been replaced by the ‘knowledge’ of the creator of this new situation. By removing all ‘cars-revealed-by-host-possibilities’ afterwards, we create the same situation as when ‘not-reveal-cars-in-the-first-place’.

If the sum of four possibilities is 1, what are the individual chances? Now the player finds himself in a new situation in which one of those four possibilities is true. This is obvious because no car but a goat has been revealed. It doesn’t matter to him why the host opened a door, random or by knowledge; his behaviour is in both cases the same, leading to the same new knowledge of the player. Could the host have changed the chances of the two possibilities in which the car is behind door1? No, he could only prefer one ‘goat’ above another, but they share no relevant difference. So he should have changed the chances of the other possibilities. Is that possible? Of course, revealing a goat behind doorX disables the other option which has the car behind doorX, so these chances become one, in number. These are the new chances:

player	host	host	player	host	total
door1	door2	door3	chance	chance	chance
a	g		1/3	1/2	1/6
a		g	1/3	1/2	1/6
g		g	2/3	1/2	1/3
g	g		2/3	1/2	1/3

Which are the same as the chances in the original problem. Could I have explained this easier? Probably. Test it yourself by playing three cards of which two are identical and one is unique. Let a player pick one and turn around another one yourself. Every time you turn around the unique card, stop the game and start again. It is not different from knowing the cards and turn around the twin card in the first place. Heptalogos (talk) 23:23, 31 January 2009 (UTC)

"The conclusions of this article have been confirmed by experiment" Does anyone have information about experiments on the 'random host variant'? Heptalogos (talk) 16:48, 1 February 2009 (UTC)

The primary purpose of the bold disclaimer at the top of this page is to encourage folks who are unconvinced switching wins 2/3 of the time in the regular version to think about it before posting comments here doubting the solution. Many experiments and simulations have shown the 2/3 result. I am not aware of similar experiments regarding the random door variant.

However, I think you may be slightly misunderstanding this variant. If the host reveals the car accidentally we don't keep the same initial player pick and have the host pick another door (until he manages to not reveal the car), we ignore this case completely. In your playing card version if you reveal the unique card, stop the game and start again and don't count this time, i.e. re-randomize the cards and have the player pick one. I think you understand that if you do this 999 times, roughly 333 times you'll reveal the unique card and roughly 333 of the other 666 times the player will have picked the unique card (right?). In this variant we're only counting those 666 times (not the entire 999), so the chance of winning by switching is 1/2. Your tables above are confusing because of the "player chance" and "host chance" columns. If you remove these columns and leave the equally possible 6 cases (your second table), the total chance of each is 1/6 as you indicate. We can count the cases where the player picks the car and see the chance of this is 2/6. Similarly, counting the cases where the player picks the goat indicate a chance of this of 4/6 (2/6 plus 4/6 better be 1, since the player either picks the car or a goat!). What happens if we're given that the host didn't reveal the car is that there aren't 6 cases, but only 4. In 2 of these the player picked the car. The probability of the player having picked the car changes from 1/3 to 1/2 because we've eliminated 2 of the possible 6 cases (and in both of them the player initially picked a goat). -- Rick Block (talk) 18:54, 1 February 2009 (UTC)

I understand. The 333 cases which we threw away are of course all car picks which now indirectly influence the outcome of the remaining group, as a whole. Now I understand why I cannot accept the issue: it seems from the description of the situation that Monty is playing his usual game, remembering the door he should open, but then in one case forgets. In this situation there is no way to define the chances, just because it's a single event. Actually, I would say that in case he never in his career accidentally picks a car, nothing changes the odds because there is no behaviour change. In other words: the chances only change to 50/50 if the game changes statistically, which would mean that Monty always picks randomly, revealing cars 1/3 of the time. Only then it would be true.

Now in the original problem description it says somewhere that "This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car." Only this is true, and it is actually not true, as stated above, that the player should switch (in any individual case). So it can be a good advise to all players of the game generally, to follow consequently, in a consequent game. The statement about a clueless host is correct if this is either a consequent, repeating situation, or if it's defined as generally the best choice, but in the latter case a car should be revealed a significant number of times. I suggest to change the idea of Monty having a memory leak on a particular day, into a consequent situation from which chances can be correctly analyzed. Heptalogos (talk) 20:13, 1 February 2009 (UTC)

I still stand with my attempted proof of "the version in which the host opens doors randomly, but never picks a door with a car behind it", because this is something else than my 'easier explanation' below with the card game. The card game is statistically realistic, revealing the 'wrong' card and stopping the game repeatedly. But playing random and yet never open the 'wrong' door is a small chance on it's own! (Which becomes smaller when more reps are done.) Hence this is a different world, starting with the assumption that the right door always opens. The 333 expected options do not exist, are not excluded, and do not change the chances of the remaning part. Heptalogos (talk) 20:34, 1 February 2009 (UTC)

Even if the situation is a single game where Monty forgets we can talk about the chances, although in this case in a strict sense it would be a Bayesian probability rather than a frequency probability. And, although there is a small chance playing randomly that a wrong door is not opened, this doesn't mean much about the frequency probability which by definition is the limit over a large number of trials. If the host hasn't revealed the car roughly 1/3 of the time you simply haven't run enough trials. -- Rick Block (talk) 00:07, 2 February 2009 (UTC)

Actually we can never define chances in any single specific case. We can only say that in indentical situations chances are..., which means that the outcome is... x% of the time. In case Monty only sometimes forgets, I see two possibilities:

1. Monty forgets once every x shows. (We even need to know x.)

2. In all series of identical shows, the host will forget once. (number of shows in a series is irrelevant)

We need this population of possibilities to define chances within, one of which could be a random single event's chance, but not a specific single event's chance. How many times Monty forgets? What is the chance he forgets? We only don't need to know this when he always forgets, or in other words, when the host acts randomly, consequently.

I think we are reversing the essentials here. To define chances, we basically need a significant group of events, from which we know the overall outcomes. Then we can group similar outcomes and calculate the chance of one of those. Furthermore we may try to understand what causes such groups. A probable cause can be the behaviour of a host randomly opening one of two doors. Now if we presume this behaviour, it is only that we can define the overall chances because we predict the outcome of this behaviour, which is the given group. We assume that the behaviour will create a significant group of outcomes, but it should really reach this number of events before we can be sure that the overall reality looks like it. If this behaviour only creates one single event we cannot say anything about it!

Not only do we miss a big enough control group to define solid chances; even if we define such chances, it makes no sense assuming that it will predict a single specific event. The fun part is that one cannot reject this by experiment, because in those cases a control group will already be created. We can only prove that behaviour is random when we create a big enough reality with it. But it is more than that: without this reality there actually is no random behaviour. I apologize for the generality of this discussion, but I only started this because in this case even a description of such a reality is missing! Random behaviour is assumed as a single exception! This is totally unreal and cannot even be closely proved experimentally. As soon as you do, you create reality. Most realities which are big enough to be near-significant, will confirm the chances, just because of the 'size' of the reality. How many small realities deviate? We never know, but if when we can group them, we can create the bigger reality that we need. That's why we need a bigger control group like one of the two possibilities presented above.

This is not just a formal filosophical argumentation. There is no such thing as random behaviour in single specific events. Every single event has it's own unique variables which may or may not be known by anybody. That is why we never can predict, by experiment, a single specific outcome of random behaviour. It is only in a significant group of events that all of the uncontrolled variables wipe out each other's relevant influence on the given situation. Heptalogos (talk) 22:17, 2 February 2009 (UTC)

If I put 17 marbles in an opaque urn, 16 of which are white and one black but otherwise identical, you put your hand in and grab one, I think it's common to say you have a 1 in 17 chance of picking the black one (even if you only do this once). We don't have to do this a large number of times to analyze what the chances are using probability theory. Similarly, we can say one individual player in the Monty Hall problem has a 1/3 chance of initially picking the door hiding the car (which we assert is placed "randomly" behind one of the 3 doors). If the host forgets where the car is and opens a door "randomly" (and if the car was actually placed randomly but the host forgets where it is, opening either door is equivalent to opening a door randomly) and fortuitously doesn't reveal the car, we can (using probability theory) say the player has a 1/2 chance of winning by switching because there are 4 equally likely scenarios with 2 where switching wins. Could we design a test to show this experimentally? Sure. Do we have to in order to believe it's true? Well, I don't. I trust probability theory and the law of large numbers. -- Rick Block (talk) 03:40, 3 February 2009 (UTC)

It's common to say that you have a 1/17 chance to pick a black marble if you try once, but it's not true. Actually there is no chance directly related to a single specific event. The event is related to (as part of) a group of similar events to which chances are related. These chances are generic and cannot be linked to a unique element of the group. That's why we strictly can only advise a group of players, or a player who will repeat a process. The first spacial similarity focuses on similar players in similar shows, while the timely similarity addresses a certain repeating frequency of the same spacial situation.

What happens if we're in a situation in which we (once) seem to have a chance to pick one out of many, is that we make this event part of a group of (seemingly) similar events, and divide the group in two subgroups: the ones and the manies. Now we want to predict in which subgroup we arrive by event. We have no idea (random) what causes access to one of these subgroups, but we only know that one subgroup is bigger. Is it more likely to arrive in a bigger subgroup, just because it’s bigger? If we try, by process, we will see that most events end up in the bigger group. But it only becomes (generally) true because it becomes real, in a big enough group. Fortunately we don't have to start the whole process ourselves; we can make our event part of a group of events which happened in the past. Or we can make it part of similar single events in similar situations at the same time, assuming that they're all random.

Could we design a test to show this experimentally? Do we have to in order to believe it's true? My answer is: we don't have to do the experiment, but we should indeed do the design! We should describe a theoretical reality to make it credible. One cannot 'once' act random, we really need the bigger picture. Please design your test, as you say you can, and you will see that no forgetting Monty can exist in such an experiment. You will likely replace him with 'another' random event, which will even be repeated, but that's really something else. The situation is presented as random, but it's not. Apart from that, there is no way to define chances in a single specific event. Heptalogos (talk) 21:18, 4 February 2009 (UTC)

Reversion of good faith edits marked as minor

I have seen at least two occasions where good faith edits to this article have been immediately reverted and the reversion marked as a minor edit. This is extremely rude and contrary to WP policy. I understand that editors wish to maintain the quality of this article but this is not the way to do it. Martin Hogbin (talk) 08:34, 3 October 2008 (UTC)

I suppose you refer to edits like this and this (perhaps also this). You're absolutely right. Except, it's maybe a little bit rude to call others extremely rude because they made a slight mistake. (And now I nearly did the same!)--Noe (talk) 11:10, 3 October 2008 (UTC)

I didn't deliberately mark my revert as minor -- looks like Twinkle did that for me, and it appears Huggle does the same. I'll change my settings. I explained my reasons for the revert in the summary line -- the edit was pure OR, something this particular article suffers from a lot. Jonobennett (talk) 12:50, 3 October 2008 (UTC)

Generally, reverting an edit occurs because of vandalism, so all software features dealing with this automatically mark it as a minor edit. This is true of the admin rollback button too. Now, whether these software features should be used to revert good faith edits is a different issue. Mindmatrix 13:38, 3 October 2008 (UTC)

OK, I accept the points made and maybe overstated my case a little, but I first came to this page in response to an RFC which claimed that the current editors were being overly protective of the page. Immediate reversion of somebody's work, with that reversion being marked as minor, is not welcoming to new potential contributors. As I have said before, I believe that the very strict enforcement of no OR might be relaxed a little to the benefit of the article. Martin Hogbin (talk) 16:45, 3 October 2008 (UTC)

Problems with the current Souces of Confusion section

The first two paragraphs give a reasonable overview of what other people have said about the sources of confusion for this and similar problems.

The first sentence in paragraph three is: "A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207)." This gives the impression that exposing information that is already known does affect probabilities, while the truth is that our deeply rooted intuition, in this case, is right on target. The Falk paper gives the same (false)impression but whenever the idea was used the problem turned out to be a mistaken belief that no new information was or would be exposed. In most cases the author pointed out the new information himself but in no case did exposing only known information change a probability.

In this problem the question comes up relative to host intentionally opening a door he knows conceals a goat. The old information was that there was a door, unpicked by the contestant, which hid a goat, and the new information is the location of one of the goats - which has no effect on the probability of the contestants initial choice being right but a great effect on the probability of winning by switching. In the Falk paper referenced (2nd reference) the author does point out that in the variant games, alluded to in paragraph 3, the mistake was in failing to recognize new information was imparted, not that the deeply rooted intuition was wrong. The current section does not mention that and thus solidifies the false impression given before.

Try to imagine the computation of a probability that you would agree would be different if information you knew hadn't been given to you a second time. Yeah, it fairly boggles the mind that anyone would think such a thing is possible. (Just imagine how repeated exposures to same information might give a whole sequence of different answers - it would never be safe to compute a probability!)[This paragraph added in edit]Millbast5 (talk) 02:04, 9 October 2008 (UTC)

The fourth paragraph is all about variant games - those in which the host has a preference for which door to open when there are goats behind both doors the contestant didn't pick. In the Falk paper it is made clear that for that preference to have any effect it must be known to the puzzle solvers - which is not mentioned in the current section. Since no such preference is mentioned in the puzzle, it has no effect whatsoever - that is to say, the whole subject is irrelevant. Clearly no confusion about the puzzle this article is concerned with is going to be eliminated by the fourth paragraph but, since it fails to clearly delineate its subject matter, more confusion can be added.

By the way, if the information were added it would not change the odds of winning by switching, it would only increase the odds in some cases and decrease them an exactly compensating amount in other cases. That is just how irrelevant the whole subject is to the current problem.68.2.55.158 (talk) 01:35, 9 October 2008 (UTC)

I'm not sure why but the preceding edit identifies me by number. In case you're wondering, my name in here is Millbast5.68.2.55.158 (talk) 01:41, 9 October 2008 (UTC) Okay, now I see... it didn't automatically log me in like it did other timesMillbast5 (talk) 01:44, 9 October 2008 (UTC)

I think the point is that if we view the action of opening a door as providing no additional information pertaining to the initial pick (i.e. we already know one of the two unpicked doors is a goat so knowing which does not change the probability of our initial choice being the car) then we're being misled, at least in some circumstances, just as much as we're misled by the "n doors means the probability must be 1/n" intuition. I forget the examples Falk uses for this, but this "no new information" intuition certainly can lead to incorrect conclusions - for example in the (Morgan et al.) case where the host has a preference for one door over another. The main point she (Falk) is making is that people have intuitions about probability, but these intuitions often interfere with the actual solutions and, conversely, solutions appealing to these intuitions are generally inadequate.

The fourth paragraph is not about variant games, but about the nature of the question. The variants are introduced only to show the difference between the two potential questions that might be what is being asked. Many solutions address the unconditional question (ignoring which specific door the host opens, as if the question is raised before the initial pick and before the host opens a door) even though the general statement of the problem clearly (to some) is a conditional question (specifically, with knowledge of which door the host has opened). This particular "confusion" is the main topic of the Morgan et al. and Gillman papers.

If you can clarify these points, please do so. -- Rick Block (talk) 02:58, 9 October 2008 (UTC)

I am more than happy to clarify matters.

Here is an example taken from Falk's article - 'second' refers to our "no-news, no-change" belief, as she calls it. It makes my point very clearly that the error is always in evaluating 'no-news' and never in the intuition/belief itself.

I agree, it is not always obvious what information is gained by a particular action and it is is easy to think that it is none when that is not, in fact, the case.Martin Hogbin (talk) 20:22, 9 October 2008 (UTC)

It is often the case, as in the problem below, that no new information is given about one thing but there is about something else. Someone focussed on the first thing might well jump to the conclusion there is no new information at all. I'm sure that happens with MH, too - host opening a door to disclose a goat reveals no new information regarding the contestants chosen door hence it contains no new information.

Now we just have to bolster your, and others', faith in your intuition that no new information invariably means no change in probabilities!Millbast5 (talk) 22:07, 9 October 2008 (UTC)

>>The second, mentioned in passing, is embedded in Dick’s attempt to induce the warden to tell him the name of one of the other prisoners who will be executed. Dick argues that he already knows that at least one of Tom or Harry is to be executed. Therefore, he maintains, he will receive no information about his own fate by getting one of these two names. If nothing new has been learned, no change in the probabilities should ensue. This is the “no-news, no-change” (or “I’ve known it all along”) belief. The uniformity belief is seldom questioned. It appears to be second nature for many naive, as well as statistically educated solvers. As for the “no-news” argument, it has convinced not only the warden, but most of the readers of the three-prisoner problem as well. It leads to a probability of pardon of l/3 for Dick (the correct answer for this particular problem)...<<

Dick tells the warden he will receive no information about his own fate and he was quite right about that, whether by accident or design. However it is quite clear that the warden did communicate some information and that information can be used to change the calculation of both of the others chances for a pardon - Harry's cut from 1/3 to 0 and Tom's raised from 1/3 to 2/3.

My point is that while we do make mistakes about whether new information has been communicated that in no way indicates that our belief that "no-news, no-change" is not precisely correct. When Falk states that the warden and problem readers were convinced by the "no-news" argument, isn't it clear that they erred in believing that no news was communicated?

When Falk says, "If nothing new has been learned, no change in the probabilities should ensue." she is entirely correct. Since a change in probabilities did ensue that meant something new had been learned and if one goes through the calculation of the new probabilities you can see exactly what that new information is - the identity of one the prisoners to be executed. (Not that it wasn't obvious beforehand.)

There are subtlties in evaluating information but our intuition that "no new information means no change in probabilities" is as solid as a rock.

While it might seem that Falk is indicting our intuition in this regard, on page 207 she correctly points out that the problem lies with the mistaken assumption that nothing new has been learned - she is discussing a different puzzle at that point but the same principle is at work. Without that I might have been concerned that she was as confused as the attributions used in the current "Sources of Confusion" make her seem.

Rick, I will promise you that if you bring forth any purported example where no new information changed probabilities, I will point out the new information. Heck, if you go through the two probability calculations yourself, its hard to believe you won't spot it right off.

Yes, we do have intuitions about probability that lead us astray but bringing up one that never does and treating it like it was a myth or an old wives tale is a terrible way to clear up confusion. Frankly, I think it would be very encouraging to the readers to point out that some of their intuitions are dead right.

I do have to agree with you that conditional and unconditional statements of the puzzle are part of the subject matter of the fourth paragraph. I should have indicted that aspect of it as well. The goal of the section is to reduce confusion about the central problem of the article and dragging in conditional and unconditional probability without a thorough discussion of what they mean is guaranteed to increase confusion - in virtually any context. Perhaps in some remote corner of the article, with warnings posted and a thorough introduction to the concepts one might profitably broach the subject - though, I doubt it.

I stand by what I said regarding the variant games - I assume their introduction was well intentioned but bringing up a variant of the game then proceeding to say and I quote, "In its usual form the problem statement does not specify this detail of the host's behavior, making the answer that switching wins the car with probability 2/3 mathematically unjustified. Many commonly presented solutions address the unconditional probability, ignoring which door the host opens; Morgan et al. call these "false solutions" (1991)." makes it sound like the solutions given in the article are erroneous.

How can that do anything but create more confusion? Solving the variant games was not hard and after doing it, I still had no idea what these statements meant. Exactly what is mathematically unjustified? "false solutions"? I suppose I could somehow dredge up Morgan et al and see what they were talking about but after doing it with Falk and finding that it did not support what "Sources of Confusion" purported, I figured that would be a similar waste of time. And on a subject that is far better left out altogether.Millbast5 (talk) 13:51, 9 October 2008 (UTC)

Of course if no new information has actually been imparted the probabilities cannot change, just as lacking other information about which of N alternatives might be the correct one makes the probability of each 1/N. These are BOTH "true" intuitions but they resonate so powerfully that they are easily misapplied.

The Morgan et al. paper is available online (for a fee) but should also be available in nearly any university library. They argue 1) the problem as generally stated should be interpreted as a conditional problem, and 2) since the problem statement (per vos Savant) doesn't specify the host opens one of two goat doors with equal probability we must treat the host's preference for one door over the other as a variable p. Given these assumptions, the general 2/3 answer is either non-responsive (answering the unconditional question, not the conditional question) or incorrectly assumes a constraint on the host that is not imposed by the problem statement. They argue the "correct" answer to the vos Savant version of the problem (granting that the host must open a door, and won't reveal the car) is that the probability of winning by switching is 1/(1+p), i.e. somewhere between 1/2 and 1 depending on the host's preference for the door that the host opens (in the case where the host has a choice). Relating this back to Falk, this is a case where the "no news" intuition (nothing the host does can affect your initial chance of winning the car, which must remain 1/3) leads to an incorrect answer.

Are there specific changes you're suggesting should be made to the article? -- Rick Block (talk) 14:11, 10 October 2008 (UTC)

Naturally I do have some suggestions for what might reduce the confusion of the readers though I do believe we can do a much better job of providing solutions and try to eliminate confusion that way. The primary source of confusion about this problem is that readers have a very strong intuition that it doesn't matter whether the host purposefully reveals a goat he knows is behind a door or he opens a door at random and by chance reveals a goat. The latter leaves the switch and non-switch strategies equally good - they can reason that out quite well - so how could the former not do the same? Anyway, I will address that source of confusion in a short while.

For the moment I will delve into the issues of paragraph 4.

Recall that the problem posed in "Ask Marilyn" did not state that the contestant picked door number 1, it said "say No. 1" which means that the particular door is not significant in the English speaking world and he's choosing a particular one just for purposes of illustration. Likewise the host opening door No. 3 is purely illustrative. That means the intent of the question is not what is the probability of the contestant winning with different strategies in the case he opened door 1 and the host opened door 3, the question is clearly what are his chances of winning the game with either of two strategies. Not only is that question answerable, it is answerable in case the host has a preference for some doors over others and the answer is the same!

Clearly Morgan et al interpreted the question to be about a particular pair of doors because to fully specify the hosts door opening preferences 3 parameters like p are needed - one for each possible pair of doors he may have to choose between, 1 & 2, 1 & 3, and 2 & 3. Why they choose that interpretation is anybody's guess, but mathematicians do things like that - take a different view and you might have another interesting problem. (I did the same thing you might recall - I chose an interpretation where the puzzle had to solved exactly as stated, without any knowledge or assumptions about the hosts behavior beyond what the puzzle statement supplies. I knew that that was not the intent of the question - for one thing its far too difficult to ask of a general audience. It was too complex to be included in this article, too - it was considered "OR", I believe.)

For illustrative purposes say the contestant picked door 1 and p is the probability the host will pick door 2 to open given the choice between doors 2 and 3. Indeed the probability of winning using the switch strategy with door 2 being opened is 1/(1+p) and the corresponding chance of winning with door 3 being opened is 1/(1+(1-p)). So what are the chances of the switch strategy winning? All we have to is multiply the two probabilities above by the probability of each of the doors being opened and add the two products together - childs play. Door 2 is opened with probability 1/3 + p/3 and door 3 is opened with probability 1/3 + (1-p)/3 thus the chances of winning via the switch strategy is: (1/3 + p/3) * 1/(1+p) + (1/3 + (1-p)/3)*(1/(1+(1-p))) - the algebra is easy and the answer is 2/3, as we knew it would have to be.

Door 1 was just illustrative and since the answer is independent of probability p, the answer will be the same for any other choice by the contestant. So the answer to the problem is that the contestants chances of winning by switching are 2/3 - versus 1/3 by not switching.

I suppose by their peculiar lights it isn't mathematically justified to say the probability of winning by switching is 2/3 when it varies according to which door is opened. Of course, if you do not know what value p has, then you only know one door may be better than the other but not which and by how much. Yet you still know the probability of winning by switching is 2/3 regardless of p - what are they going to say to that? What could they say? "Yeah, but if you don't know p you can't tell what are the chances when Door 3 is opened and thats what the puzzle asked and intended to ask." Frankly, I'd disagree with them if the puzzle had been stated without the writer making it clear the mention of doors was only for illustrative reasons - and so would almost everyone else. I suppose if one of them were contributing here we could have a philosophical-mathematical discussion about how to state puzzles so only one interpretation is possible.

Also note that the puzzle as stated here only asked if switching was better and thats true as long as p is not 0 or 1 and the appropriate door is opened with each value - door 3 when p=0 and door 2 when p=1. In those extremal case not switching is equally as good, not better. If you don't know what p is then you're clearly better off switching since you can never do worse than by not switching and you may do vastly better.

How are the readers going to be helped by going through a mess like this? None of them interpret the problem the way Morgan et al did so one would have to have a very good reason for delving into it. Yes, you could explain that if the contestant did know the value of probability p for the pair doors he did not pick, he'd have a better idea of whether he won or not depending on which door the host opened for those few seconds before the door he settled on was opened to make the probability either 1 or 0. You would also explain that there was no way for the contestant to make use of his knowledge of probability p - except if p was 1 and door 2 was opened or if it was 0 and door 3 opened - then he could not switch and win with the same probability as switching (1/2). And then you could reassure him that the contestants chances of winning by switching were still 2/3, regardless of the value of p.Millbast5 (talk) 13:23, 11 October 2008 (UTC)

Your second to last paragraph above is exactly the same as the conclusion Morgan et al. come to - i.e. if you don't know p you might as well switch since by doing so your chances of winning don't go down but may go up considerably (and, people who a priori decide to switch before seeing which door the host opens win with 2/3 probability). I don't think anyone takes "let's say" as anything other than illustrative - on the other hand, I think the point Morgan et al. make is valid. Their point is that the player is not making a decision at the beginning of the game, before picking a door and before knowing which door the host opens, but after picking the initial door (illustratively, door 1) and after the host has opened a door (illustratively, door 3). At least one of the experimental psychology papers about this problem indicates that this aspect of the problem (putting the player in front of two closed doors and one open door at the point of attempting to evaluate the probability of where the car is), is what makes it resonate so strongly with the "must be 1/N" intuition. Roughly a year ago (in the talk archives) there was an extended discussion here about this, which led to changes to the article including introducing the paragraph you're objecting to. To a large extent, the issue is not what we think but what the published sources say. We could certainly rephrase things for better clarity. Is this what you're suggesting at the very end of your post above? -- Rick Block (talk) 15:27, 11 October 2008 (UTC)

I agree with Millbast5that the issues addressed by the likes of Morgan et al are essentially a diversion from the main question. The current article has too much about 'versions' of the problem but does not provide a convincing resolution of the central paradox. Martin Hogbin (talk) 20:07, 11 October 2008 (UTC)

I'm glad to hear that Morgan et al recognized that short of knowing 'p' took the value 1 or 0, their analysis is entirely without benefit to the contestant and that in no case could he actually gain from it. Of course that means that while the contestant doesn't have to make a decision until the door is opened, he can't do any better than he could by deciding whether to switch or not before the game started. If one looks at their paper in that light perhaps there is a place for a reference to it - in case someone wondered if there wasn't some way putting off the decision might help the contestants chance of winning. However, to appear in the 'Sources of Confusion' section it needs to be enlightening to the puzzle in general, and I see the objections I made originally as still being in full force.

I just had a cool idea - a way that someone who didn't believe switching would improve his chances of winning could hedge, just in case he was wrong. Suppose he thinks door 2 is the door with the car behind it. Instead he picks door 1, intending to switch to door 2 when the host gives him the option to switch. The only way this play could go awry would be if the host opened door 2 to reveal it as a loser - and then he'd have to be glad he didn't pick it and stick with it. Of course, with door 2 out of play he has a real choice about switching. Heck, why not rank all three doors as to his subjective feel of their chances of concealing the prize and he intially picks the worst of the three so he will get to switch to one of his first two choices for sure. He could explain that rationale to his friends and family if they started to tease him about switching from the winning door - and they might even buy it. Or tell them his plan ahead of time and then he probably wouldn't panic under the gun and not switch - lets face it, switching is wussy.

I'm not sure it belongs in the Sources of Confusion section but I've been thinking that someplace in the article we should bring up the idea that people do not like to look indecisive and the switch strategy does give that appearance. Few would fail to switch on that basis if they did believe switching doubled their chances of winning but it probably is part of the reason why everyone who believes the two strategies are equally good elects not to switch.

Sorry, I'm getting off topic here.

Rick, I'm in full agreement that the contestant being faced with two closed doors stimulates the automatic response that they are equally likely to hide the prize. I can't remember if I explained that I believed one reason the urn and balls version of the puzzle is easier to understand is because that image of the contestant is not called to mind. Also it does not appear that the person who knowingly removes the white ball from the urn, after the subject blindly picks a ball, is an adversary while the host is still rather suspicious in that regard.

Published sources should attributed wherever their ideas and results are used but, outside of history and 'related topics', we should not look to refer to them simply because they exist if they do not help us achieve our goals.

I also believe that in an article that is about a popular topic we should have a goal of making the article interesting or even entertaining to read because people will not be enlightened if reading the article is so boring they stop reading. A little lattitude with OR ought to allowed, too, when the subject is just not that serious. Something publishable in a learned journal is not what I mean, of course, but 'novel and creative' enhance an article.

With popular topics, such as this, an article that one has to find the disclosing authority for and then pay thirty bucks to eyeball it, is about as close to useless as you can get - again outside of history and related topics. So if you can't quote enough of it to be helpful you should try to take a course that avoids it altogether - paraphrasing is rife with pitfalls, especially when it is done with an eye to keeping it short. Look what happened with Falk and Morgan et al in the current section - Falk didn't conclude what the section implied and the paraphase of Morgan highlighted their conclusions that made it sound like the puzzle solution was chock full of errors. As I mentioned above the important conclusion from Morgan is that knowledge of the hosts preference for opening doors doesn't affect the chances of the contestant winning, yet that part got left out.

I disagree with the tone of the article (and many of the references) in regard to people's intuition about probability - they are, in fact, generally quite good and we should try to explicitly use them in presenting solutions. Thus they will be nodding their heads and thinking "yes, I do think just that way" instead of feeling like they are being treated as dolts for thinking/believing as they do.Millbast5 (talk) 23:23, 11 October 2008 (UTC)

Isn't the first part of the solution section about as simple as you can get? Three equally likely scenarios, switching wins in two and loses in one, ergo 2/3 chance of winning by switching. The subsequent deeper analysis, using the Morgan et al. reference, addresses a mathematically significant criticism of this solution which is that the solution doesn't utilize the "equal goat door" constraint imposed on the host. Without this constraint, the simple solution may be incorrect making it at least incomplete. You might argue that no one would interpret the problem to mean that the host exhibits a preference for one door over another, but if you don't include this constraint that's exactly what a mathematician would do (find a hole in the argument). If you want the answer to be 2/3 chance of winning by switching, and you want the decision to switch to be made after the host has opened a door, you have to include the "equal goat door" constraint and the solution has to use this constraint. Anything less and you're either asking a different question (like what's the probability of winning if you decide to switch before the host opens a door knowing that the host will open a door), or 2/3 isn't always the right answer. -- Rick Block (talk) 04:30, 26 October 2008 (UTC)

Proposed substitute for the last two paragraphs of Sources of Confusion section

Just for the sake of variety, you get to be the contestant for a while.

There is no question that the most difficult thing to evaluate is the effect of the host intentionally opening a door he knows hides a goat after you've made your initial pick. Everyone can figure out the effect if the host just picks one of the two available doors at random and opens it. A third of the time the action will reveal a car and you'll know immediately that your door and the switch door each hide the car with probability 0, and two thirds of the time the action reveals a goat leaving your door and switch door each harboring the car with probability 1/2. Our intuition is very good about that. If the host jumped ahead of you and opened a door at random before you could pick the outcome is exactly the same. Our intuition is all over that one. Suppose the host didn't wait for you to pick but opened the door he knew hid a goat before you picked. That would be great, you'd win half of the time instead of a third of the time. Also the 'switch' door would hide the car half of the time. Our intuition nails that in milliseconds!

One thing that has been true in every case is that your picked door and the 'switch' door have hid the car with equal probability so it is very tempting to assume the same holds when the host opens the known goat door after you pick and your intuition sends you a strong 'buy' signal on that assumption. Still the smart thing to do is check it out before you buy.

One good way to get a handle on it is to imagine the game being played 9 times all at once. First you make your nine choices. But then you're allowed to send an observer to look behind the doors - she reports that there are nine cars altogether and 3 of them are behind doors that you picked. Now the host goes through all nine games opening a door he knows conceals a goat. The observer does her thing again and reports all the cars are just where they were before - no funny business going on. So there are 18 doors that are closed and three of the cars are behind your 9 of them - that leaves six cars behind the other nine doors. You do the math again to make sure. It doesn't help - there twice as many cars behind the 'switch' doors as behind your doors. Is that darn observer being truthful? What she says makes sense - 3 cars out nines tries is normal and you really didn't expect any more cars to move behind your doors so it sure seems like she is telling the truth. There's no way out of it - the 'switch' doors are twice as likely as yours to harbor a car.

Why does our intuition get this last case wrong? In every case the image we have in our minds is the same - we are standing before two closed doors and one open door, and leaving out the random cases where the open door reveals the car, there is a car behind one of the two closed doors and we don't know which. Generally speaking, when we don't know which door its behind its equally likely to be behind either of the closed doors. It seems only fair - heck it seems right - that it should be equally likely. Even when it is not equally likely but we don't know which door is more likely, its the same as it being equally likely as far as we're concerned. This particular case is the oddball - the doors are not equally likely and we know which one is more likely. It just isn't apparent until we exercise our imagination a bit.

We still can't really fault our intuition because it led us through the 9 games demo just fine, taking it one small step at a time.Millbast5 (talk) 04:44, 13 October 2008 (UTC)

Here is another go it it, where I focus on tracking down the source of confusion

Another possibility for the Sources of Confusion

Comment: This version directly shows "why the probability isn't 1/2" so it replaces the section by that name as well.

The source of confusion is not the puzzle itself. It arises from contemplating what would happen if the host opened one of the 'other' doors at random. The solution in that case is gotten right by virtually everyone: the contestant picks the winning door 1 time in 3, the host opens a randomly chosen door to reveal the car 1 time in 3 and the unpicked and unopened door hides the car 1 time in 3. Now comes the confusion: if you forget about the 1 time in 3 the host reveals the car then the contestant wins 1 time in 2! Of course, if you eliminate half the losing choices for whatever reason you want, the contestant would have winners half the time. Its just simple numbers. It means nothing and means no more if you phrase it this way: when the host reveals a goat by chance the contestant wins half the time. Clearly that is just another way to say you are eliminating half his losing choices. Now the source of confusion leaps into full view - if you attribute the change in frequency of winning to the revelation of a goat instead to the fact that you're disregarding half his losing choices you have the makings of a real misconception.

In the actual problem only goats are ever revealed but if one believes that the revelation of goats causes the contestants chances of winning to increase, it ought to do it even more when more of them are revealed. And so it does, according to the faulty belief - instead of the contestant winning 1 time in 2 when half his losers are removed, his original pick wins 1 time in 2 overall.

This belief gets a boost from another source - the mental image of the contestant having to pick between two closed doors, one of which hides the car. The contestant doesn't know which one, and generally this means that the car is behind either door 1 time in 2 - perfect corroboration! The belief becomes so solid that nothing can shake it. (This is not speculation as a number of the referenced sources demonstrate the absolute conviction of many believers - some of them get irate if one continues to challenge the belief.)

One way to combat the belief is to consider a number of plays of the game simultaneously so as to make the original picks more concrete. Imagine for example, 99 plays of the game in which the contestant picks 33 winners and 66 losers - which is, statistically, the most likely outcome. Now its clear that if no cars or goats are shuttled to and fro behind his chosen doors one can do anything one wants with the 198 other doors and the contestants 33 winners and 66 losers will not be altered. One can open those doors at random to ones hearts content or look behind them and expose goats till the cows come home and the contestants 33 winners and 66 losers sit there undisturbed.

Phrased in frequency terms - the contestant picks the winning door 1 time in 3 and the host purposefully revealing goats does not change that fact. It applies to the imagined contestant choosing between two closed doors, too. True, he doesn't know for certain which door hides the car but he does know his initially chosen door still hides the car 1 time in 3 so, inevitably, it is behind the other door 2 times in 3.

As is apparent at this point, the solution to the game is extremely simple if one just proceeds without doubting that the initial pick remains a winner 1 time in 3 after the host opens a door to reveal a goat. One easy logical step shows the remaining door, the 'switch to' door, hides the car 2 times in 3. This simplicity probably works against it as far as engendering belief - the problem baffles our immediate intuition so it surely it can't be as simple as subtracting 1/3 from 1!Millbast5 (talk) 11:07, 18 October 2008 (UTC)

I think you are quite right in where the confusion lies, but I am not sure that you have found a way of making that clear to the reader.

As you point out at the end, if we take what has been described here as the unconditional case, then there is a a very simple and convincing argument - you have a 1/3 chance of initially choosing the car and you always get the opposite if you swap. With the reasonable assumptions that most people make (host aways offers swap, aways opens a goat door) the same argument 'happens' to apply for the conditional case (since the host action does not change the probability that the player has chosen a car). I therefore suggest that the conditional/ unconditional issue is something of a red herring and should take a secondary place in the article. In the game, as understood by most, the conditional version of the problem has the same odds as the unconditional one. It might have been otherwise, but it was not (as most people understand the problem).Martin Hogbin (talk) 12:27, 18 October 2008 (UTC)

I inserted a new phrase into the middle of the third paragraph for increased emphasis.

I'm not too happy myself with the clarity of the what I wrote. That first paragraph is too long and I don't get to the actual statement of the source of confusion until the end of it. I've also considered starting with the other half of the false argument - the contestant faced with two doors, etc, but I believe the confusion does start with contemplating the effect of opening a door at random. I imagine it varies from person to person where it starts. In any case, if you have any thoughts about how to improve it, I'd be glad to hear them.

You know you don't have to sell me on the idea of eliminating any mention of conditional and unconditional probability from the article. Of course, I'm using the idea myself in talking about the odds under the condition that a goat was revealed but people are quite used to the idea as we phrase things in that way in our everyday speech. "If TMac stays healthy the Rockets have a good shot at the title." We understand perfectly that different conditons change the odds but developing the formalism is out of place in an article like this and using the terms 'conditional' and 'unconditional' without developement is worse. Frankly, I'm not sure at all what you (and others) mean when you use those terms in these discussions. Imagine how someone feels who is a bit shaky when the word probability is used all by itself.

Yes I agree. I was not really making my conditional/non-conditional point to you but to others who might be reading. I think this article suffers from what might be called 'wiki correctness'. In other words it keeps to all the rules but does a bad job. Martin Hogbin (talk) 23:26, 18 October 2008 (UTC)

Monty's Action Does Not Cause The Original Odds To Change.

When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. So it starts out as 1/3 for your door + 2/3 for the remaining doors = 100%. Then he shows a door, but we knew in advance that he was going to show a goat. The odds simply haven't changed following his action. They remain 1/3 for your door + 2/3 for the remaining doors (of which there is now just 1).

That's why on the show, Monty didn't HAVE to show a door (unlike this puzzle), and he could bribe you with cash. Otherwise, there's no show. You always switch.

I'm new to this. Is there a way to get this 'solution' onto the article page?

75.185.188.104 (talk) 17:09, 25 October 2008 (UTC)

Isn't this essentially the same as the explanation in the Combining doors section? -- Rick Block (talk) 17:48, 25 October 2008 (UTC)

It would seem that your solution is not much liked by many of the current editors of this page because it cannot be verified by reliable sources. Martin Hogbin (talk) 22:59, 25 October 2008 (UTC)

Martin - was that directed at me? If so, then I'm confused. I didn't say I didn't like it, but that it's already included in the article (with reliable sources). -- Rick Block (talk) 03:58, 26 October 2008 (UTC)

No. It was directed at all the current editors. If you remember, I first came to this article in response to an RFC which claimed that the current editors were being overly protective. Although I understand why that attitude has some justification, I think that there are possible improvements that could be made that are being resisted.

My central point is this: The Monty Hall problem is particularly notable because, even when people make the 'standard' assumptions, the great majority of them get the wrong answer and it is difficult to convince them of the right one. This article could, and in my opinion should, address that issue better.

The more complex issues of host action/ game rules etc are not part of the core problem. In particular the issue of conditional/unconditional probability only adds to the confusion for many people. It is a fact that, for the 'standard' version of the problem, there is no difference between the two, therefore I put it to you that we do not need to make a major point at the start of the article about something that could have been the case but in fact is not. I have proposed that I very simple, and hopefully convincing, explanation along the lines suggested above should be given a prime position at the start of the article, followed by a section on sources of confusion in the basic case. The more detailed analyses, as already presented, should then follow. I appreciate that this might require a more liberal interpretation of some WP policies but I suggest that it is worthwhile considering this option in order improve the article.

I have tried to be a good Wikipedian by not disrupting this featured article with experimental edits and have set up a draft page for people to attempt to improve the it; so far only one other editor has shown an interest. So, yes, my remark was aimed at getting some reaction but only so that the article might be improved in a way that is acceptable to all. Martin Hogbin (talk) 10:33, 26 October 2008 (UTC)

I agree that the presentation of this article leaves something to be desired. I would think that most users come to this page to settle an arguement, or to enlighten themselves on how they don't 'get' the solution. I know that's why I came here.

So, an elegant, simple solution would serve the readers. I'm a stats and probability geek, but I don't like having to 'prove' something by running simulations (since I don't know who created the simulation, I can't trust it). And, if there is an easier to understand explanation that doesn't require umpteen diagrams, the typical reader would prefer that, as s/he will probably try to explain what s/he read here to other people. 75.185.188.104 (talk) 12:27, 26 October 2008 (UTC)

Isn't the first bit in the Solution section about as simple and elegant as it gets, with a diagram for the more visually minded folks out there? Three clearly equally probable cases, switching wins in two and loses in one. The transitional paragraph about this solution (the one containing the reference to simulations) and the follow-on conditional analysis (with a different diagram that appeals to other sorts of folks) is there to address a significant mathematical criticism (i.e. this solution does not address the problem that's actually posed), raised first by Morgan et al. If we end the Solution section following the first diagram the solution is incomplete as it fails to address the Morgan et al. criticism. IMO, any "simple" solution will inevitably have this same issue. There's an argument that Morgan et al. are splitting hairs over trivial details that most people don't care about, but that's like ignoring negative or imaginary roots of a quadratic equation (x^2 = 4, what's x? most people would say 2, but -2 is also an answer). If the suggestion is to simplify the solution section by eliminating any mention of the conditional analysis I strongly disagree. -- Rick Block (talk) 15:03, 26 October 2008 (UTC)

You refer to "significant mathematical criticism (i.e. this solution does not address the problem that's actually posed),". If such a thing is valid, then you don't have a solution at all. The solution should be, and is, simple. 1/3 + 2/3 = 100%. And Monty's actions do not change this formula. I don't see a mathematical ambiguity.

Glkanter (talk) 16:07, 26 October 2008 (UTC)

This is the same topic discussed in more detail three sections down (about the subtly different question), and indeed Morgan et al. argue the "simple" solution is not a solution at all. -- Rick Block (talk) 16:28, 26 October 2008 (UTC)

Well, I thought we were going to discuss the solution to the Monty Hall Problem. And that the 'Solution' portion of the Wikipedia entry would provide just that. I can see now that there is some other agenda for Monty Hall Problem page. I provided a mathematical proof to a very simple problem. So far, not you, or anybody else has challanged that proof. I already pointed out an error in the VERY FIRST LINE OF THE EXISTING SOLUTION paragraph. Yet you bring up all sorts of 'subtly different questions', 'mathematical inconsisancies', and whatnot. It's your page. Do whatever you want.

Glkanter (talk) 16:42, 26 October 2008 (UTC)

We're off on the wrong foot here. It's not my page, it's openly editable by anyone. In its current form it's the result of the efforts of at least a dozen editors over the span of 7 years. I'm simply one of a number of editors who have contributed to this page for a long time (in my case, since at least June 2005). I'm perfectly happy to discuss changes, and very happy when discussions result in improvements to the article. Please delete the first sentence in the solution section (I'll wait a while and if you don't I will). Thank you for this suggestion. I think nearly anyone would agree it's an improvement (fewer words are generally better). What I mostly meant by the comment from 16:28 above is that the discussion is continuing in the other section (below), so rather than continue here we should continue there. -- Rick Block (talk) 20:48, 26 October 2008 (UTC)

Re: The 'Combining Doors' comment above. Yes, it's similar. imho, these should be a prominant part of the 'Solution', not relegated to 'Aids to Understanding'. Glkanter (talk) 15:12, 26 October 2008 (UTC)

I find the explanation in the combining doors section somewhat abstract. Here's the explanation I've been using with people who have a hard time understanding why switching is the best strategy -- I know this probably shouldn't go on the main page (I'm the only reference I know of) but I wanted to share it.

I'm going to write it in the form of a dialog because that is the shortest form of this explanation and also the form in which it has been used:

"What if I could show you a tricky way to pick two doors instead of just one. Would you agree that you have a 2/3 chance of winning if you could pick two doors?"

"Sure! If I can actually pick two doors I'll have a 2/3 chance of winning."

"OK, tell me (but don't tell Monty) which two doors you want to pick."

"OK, I want to pick doors two and three."

"Great -- now tell Monty you want to pick door number one."

"But I don't want door number one!"

"I know, that's the tricky part. By telling Monty you want door number one you'll get him to open one of the two doors you actually picked -- and the way the rules are setup he'll always pick a door that has a goat. After he opens one of the two doors you actually picked now tell him you're switching that way you'll get to pick both doors two and three."

Am I correct that this would be inappropriate for the main page? Stepheneb (talk) 17:08, 2 January 2009 (UTC)

I would say yes, you are correct that this would be inappropriate for the main page (unless you can find a reliable source for it). It is logically the same as the "Combining doors" explanation. -- Rick Block (talk) 18:53, 2 January 2009 (UTC)

A twist - same choice, different odds?

I follow the reasoning all the way, yet the following twist to the scenario puzzles me. Can someone explain what I'm missing?

1. Contestant chooses door 1, Monty reveals a goat behind door 3 and invites the contestant to revise his choice.
2. Before the contestant makes his final decision known, another contestant is whisked on stage, straight from the dressing room, with no prior knowledge of the first contestant's original choice. This second contestant is presented with the open door 3 (and its goat) and asked to choose where the car is.
3. The two contestants then make their choices, independently of each other. The first contestant, having read this article, switches his choice to door 2. The second contestant tosses a coin on the basis of which he also chooses door 2.

As we know from the article, the first contestant has a 2/3 probability of finding the car. But for the second contestant it's a straight choice between 2 doors, so he has a 1/2 chance of finding the car. So it appears that at the moment the two contestants make their final choices, p(finding the car | choosing door 2) has two different values simultaneously depending on who does the choosing, whereas intuition (mine, anyway) says that the probability should have a unique value, and that it should depend only on the possible permutations of the original distribution of car and goats, and the shared knowledge that door 3 revealed a goat. Help! -- Timberframe (talk) 09:25, 26 October 2008 (UTC)

The probability depends on what the second contestant knows about the situation. If they are simple told,'here are two doors, one hides a car and the other a goat', then they might as well toss a coin. If, on the other hand, they have watched the show on a monitor, or are told, 'this is the door that the contestant first chose, then Monty opened this door to reveal a goat (as he always does)', they would know that the odds are better if they choose the door that the first contestant did not. Martin Hogbin (talk) 10:42, 26 October 2008 (UTC)

Mr. Hogbin is correct. Note that the 2nd player has less likelyhood of picking correctly. 50% vs 67%. This is because he has less knowledge of how the 2 doors came to be. He can only assume randomness, but the first player knows Monty did not act randomly when he opened the door revealing a goat. 75.185.188.104 (talk) 12:17, 26 October 2008 (UTC)

Another way to look at this is to compute the second player's conditional probability of winning the car. At the time the second player is choosing, Door 1 has a 1/3 chance of being the right door while Door 2 has a 2/3 chance. Someone picking randomly between these doors has a (1/2 * 1/3) + (1/2 * 2/3) chance of winning the car. Multiply this out and you get 1/2. In words what this means is that if the second player happens to pick Door 2 he also has a 2/3 chance of winning the car (but this only happens half the time, and the other half he has only a 1/3 chance). If your question is what is the second player's chance of winning if he happens to pick Door 2, the answer is 2/3 (same as the first player). Note that this is true whether or not he watched the show on a monitor. He may well think his chances are 1/2, and they are before he flips the coin, but after he flips the coin and picks a door his chances are either 1/3 or 2/3 (just like player 1) depending on which door he picks. -- Rick Block (talk) 15:24, 26 October 2008 (UTC)

To say that the second player would have a 2/3 chance of winning the car if he picks the door 2 is not, in my opinion, a good way of putting things. It depends on who is making the call. From the second contestant's point of view the chances are 1/2 (if he knows nothing of the history), from the audience's (who have seen what has happened so far) point of view they are 2/3, and from the producer's (who knows the car is actually behind door 1) point of view it is 0. Probability is a state of knowledge.Martin Hogbin (talk) 16:16, 26 October 2008 (UTC)

On the other hand, to say the first player and second player who have picked the same door have different chances of winning seems like an extremely odd way of putting things. The probability does depend on who makes the call, but I think the same perspective should be used in both cases - that being the perspective of a member of the audience. -- Rick Block (talk) 16:35, 26 October 2008 (UTC)

I would say that it is more normal to quote probabilities from the point of view of the person who is making the choice. This does lead to different probabilities for different people doing the same thing but understanding that is basic to understanding probability. For example what are the chances of a player initially picking the car - 1/3. Suppose that we have a different player who somehow knows that car is not behind door 2 - his probability of picking the car is 1/2. Same car, same goats, same doors, but a different answer. Martin Hogbin (talk) 16:47, 26 October 2008 (UTC)

You're right on target. As more knowledge is gained, probabilty increases from 1/2 (random, or no knowlegde), to 2/3 (watched Monty open a door), to 100% (you're Monty). Hopefully, Timberframe's original concern was resolved.

Glkanter (talk) 17:02, 26 October 2008 (UTC)

I'll be more precise. The perspective is generally that of someone who knows all the givens from the problem statement. So, if we're given the unambiguous Monty Hall statement (from the article) and that player 1 had picked Door 1 and the host has opened Door 3 and then player 2 has arrived on the scene and flipped a coin and has selected Door 2, player 2 has a 2/3 chance of winning (not 1/2). This matches the chances player 1 has if switching to Door 2. The probabilities are evaluated with knowledge of all the givens. Different givens result in different probabilities, which is why we'd say Martin's player who (we're given) knows that among the three doors the car is not behind Door 2 has a 1/2 probability of picking the car initially. Same car, same goats, same doors, but different givens so a different answer. -- Rick Block (talk) 17:49, 26 October 2008 (UTC)

I guess the moral of this discussion is to make always make clear what the givens are before stating a probability, which I think none of us actually did. Martin Hogbin (talk) 20:13, 26 October 2008 (UTC)

Thanks guys, got it now. The falacy in my proposition is in the assertion that because there are two closed doors and the car is behind one of them, the second contestant has a 50:50 chance of finding it. This is only true from the second contestant's limited perspective. The conditions of the scenario is posited have in reality already shifted the odds to 2:1. Thanks again -- Timberframe (talk) 19:29, 26 October 2008 (UTC)

BTW

The following is prominant in the article:

"Solution The overall probability of winning by switching is determined by the location of the car...

This result has been verified experimentally using computer and other simulation techniques (see Simulation below)."

This first sentance doesn't mean anything, and in fact is incorrect. The probability is NOT affected by the location of the car. One instance of playing the game is affected, but not the probability.

Further, a valid proof, which I believe my entry is, does not require being "verified experimentally using computer and other simulation techniques". A valid proof stands on it's own.

And lastly, this at the top of the Discussion Page is at a minimum, off-putting:

"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."

As I stated above, valid proofs do not rely on 'experiments'. Yet, there are two references to 'experiments'. Me thinks the lady doth protest too much. 75.185.188.104 (talk) 13:34, 26 October 2008 (UTC)

Are you suggesting deleting the first sentence in the Solution section? If so, that'd be fine with me. The "proof" in the Solution section does not depend on being verified experimentally - that sentence is there in an attempt to further convince the many doubters about the validity of the solution. If you can think of a way to make this more clear, please do so. The banner at the top of this page is there to discourage folks from arguing that the solution is incorrect (we still get a fair number of folks claiming the 2/3 answer is wrong). -- Rick Block (talk) 15:35, 26 October 2008 (UTC)

Yes, being incorrect, the first sentance should be deleted.

Why not move the 'experiment' stuff to 'Aids to Understanding'? A Solution should stand on it's merits. Making reference, in the solution, to another technique weakens it's apparent authority.

In fact, doesn't just having the 'Aids to Understanding' section imply that the proof is lacking clarity? The salient point that I believe is missing, is that the probabilities don't change because of Monty's action. That's why it's 1/3 - 2/3 when you start, and when you finish. Which is the only point of the whole Monty Hall Problem.

Glkanter (talk) 15:57, 26 October 2008 (UTC)

Can I mention again that I have set up this page:user:Martin Hogbin/Monty Hall problem (draft) for those, like myself, who do not find the current solution clear or convincing to develop a better basic solution. The concept is to develop a very clear solution that is verifiable and which can be eventually added to this article. There is also a talk page for discussion on how we might achieve this. Martin Hogbin (talk) 16:58, 26 October 2008 (UTC)

This comes very early in the 'Solution' section:

"A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below)."

I have no idea what this means, and fail to see how it part of a solution to the problem. People come to this page for an explanation. The above paragraph, plus whatever follows it, are not part of the solution to the Monty Hall Problem, and, imho should not be present there. Glkanter (talk) 14:59, 26 October 2008 (UTC)

There are two different questions.

1. If a player decides to switch before the host has opened a door, what is this player's chance of winning? Or, what is the overall probability of winning at the start of the game?
2. If a player decides to switch after the host has opened a door, what is this player's chance of winning? Or, what is the conditional probability of winning given which door the host opens?

The first part of the Solution section addresses the first question. The second question is subtly different, and requires examining only those cases where the host has opened a specific door. The second question can have a different answer than the first question even given the same game setup. This is exactly the same situation as the question raised two sections above about a second player who arrives on the scene after the host has opened a door (substituting before and after this player chooses a door for before and after the host opens a door). Before flipping a coin this second player has a 1/2 chance of winning. After flipping a coin this second player has a 1/3 or 2/3 chance of winning, depending on which door he chooses. Similarly, the first player has a 2/3 chance of winning by switching before the host opens a door, but may have a different chance of winning by switching after the host opens a door depending on exactly how the problem is formulated. In particular, unless the host is constrained to pick between two goat doors with equal probability (in the case the player has initially selected the car), the chance of winning by switching after the host opens a door may be anywhere between 1/2 and 1. An analysis that ignores this constraint and concludes the chance of winning by switching is 2/3 is incorrect, since the same analysis would apply to a different version of the problem (for example, where the host is constrained to always open the lowest numbered door hiding a goat). The typical wording of the problem is arguably asking the second question, not the first. -- Rick Block (talk) 16:07, 26 October 2008 (UTC)

There is no question of the probabilities at the start of the game. It's 1/3 car vs 2/3 goat. That's actually a premise, that everything is random. The only question asked in the problem, which assumes that Monty ALWAYS shows a goat (also a premise), is what are the odds when 2 doors are left? imho, everything else is clutter. I don't know what the word 'conditional' means in the context of this puzzle.

I came to the Wikipedia page because I needed to understand why it was favorable to switch. I've since distilled the reason to a single sentance, 'Monty's Action Doesn't Change the Odds.' As a user, I can assure you that all this other stuff gets in the way of understanding.

Glkanter (talk) 16:21, 26 October 2008 (UTC)

I agree. In the 'standard' version of the problem, it actually makes no difference when the player decides to switch. What is gained by answering a more complicated question that the one most people assume? The 'basic' question is plenty hard enough without trying to make it harder. Martin Hogbin (talk) 16:31, 26 October 2008 (UTC)

The question is why don't Monty's actions change the odds? Asserting this is true because we already know one of the unchosen doors hides a goat is not mathematically sound (and it's actually reasonably difficult to prove this assertion). The standard version of the problem puts the player's decision at the point the host has already opened a door. This makes it a conditional probability question. Any answer that ignores which door the host opens is fundamentally incomplete. -- Rick Block (talk) 17:06, 26 October 2008 (UTC)

Yes, but I am advocating getting the best of both worlds. Have a simple explanation that may not be rigorously complete but that people can understand and follow it with the current explanation that is more rigorous and complete. Martin Hogbin (talk) 17:22, 26 October 2008 (UTC)

This is the exact structure of the existing Solution section. Simple solution first (3 equally likely scenarios, switching wins in 2 of them), followed by a transition to a more rigorous solution. Are we fundamentally quibbling about what the simplest solution is? The one that is presented is meant to be obviously (intuitively) correct and mathematically sound (as far as it goes), but also function as a natural lead-in to the more rigorous solution. Are you arguing that the existing "3 equally likely scenarios" solution is hard to understand? -- Rick Block (talk) 18:14, 26 October 2008 (UTC)

Yes, I suppose I am arguing that simple solution shown is not simple enough. It also has some curious statements such as,'The overall probability of winning by switching is determined by the location of the car'. From whose point of view is that statement made? Martin Hogbin (talk) 19:43, 26 October 2008 (UTC)

Mr. Block, I disagree. There is no 'mathematically sound' problem. If there is, please share it with me. And your statement relating to conditional probability doesn't seem to apply here. The contestant's choice was random, and we don't care which door Monty reveals. You may repeat endlessly your asertion that it is important, but it is not valid. Like I said earlier, you seem to be interested in something other than the clear, concise answer to the Monty Hall Problem. Until you can focus on that, and perhaps show me the flaw with my solution, I'm wasting my time here. As for Monty's actions not affecting the odds, the proof is that the odds do not change once the contestant has chosen a door. His door remains 1/3. Maybe that is a tough concept for the random reader to grasp. That's why it's a paradox! But until you focus on that, and not all this other stuff, you won't be addressing the solution to the Monty Hall Problem at all.

Mr. Hogbin, I don't see how this solution is not 'rigorously complete'. Simple, yes. Incomplete, no.

I am, for the moment, simply accepting the opinion of others regarding rigorous completeness. Martin Hogbin (talk) 19:43, 26 October 2008 (UTC)

Glkanter (talk) 18:02, 26 October 2008 (UTC)

Consider a slightly different problem which is the same in all respects except that the host always opens the lowest numbered door possible. I believe your argument still holds (player's initial pick is 1/3, other two doors are 2/3, we know one is a goat, knowing which one can't affect the initial probability, etc.), leading to the conclusion that the player's chances of winning by switching are 2/3. This is even true if the question is asked before the host has opened a door. The usual version of the problem asks the question after the host opens the door. In this variant, if the player has picked Door 1 and the host has opened Door 3 the correct answer is the chances of winning by switching are 100%. In the unambiguous problem as stated in the article, the host must pick between two goat doors with an equal probability. This constraint, and only this constraint, is the difference between the variant I've suggested and the problem as posed in the article. If the solution does not use this constraint, there's a variant for which the solution produces the wrong result (i.e. the solution is mathematically unsound). -- Rick Block (talk) 18:14, 26 October 2008 (UTC)

Mr. Block, put yourself in the reader's seat. S/he wants to understand the Monty Hall Problem. Not 'a slightly different problem which is the same in all respects except that the host always opens the lowest numbered door possible.' Just the Monty Hall Problem. Now tell me, mathematically, or with symbolic logic, where my solution TO THAT PROBLEM is inaccurate, incomplete, or un-repeatable. That's all I ask. For now.

Glkanter (talk) 18:37, 26 October 2008 (UTC)

(outindent) The question is if it's advantageous to switch after we've picked a door and the host has opened one of the remaining doors. Your solution is the following:

When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. So it starts out as 1/3 for your door + 2/3 for the remaining doors = 100%. Then he shows a door, but we knew in advance that he was going to show a goat. The odds simply haven't changed following his action. They remain 1/3 for your door + 2/3 for the remaining doors (of which there is now just 1).

I've highlighted the problematic parts. These are assertions, not deductions based on what is given in the problem statement. They're true statements, but stating them without justification is equivalent to assuming the solution. We're not given that it makes no difference which door the host opens and we're not given that when the host opens one of the two remaining doors the initial 1/3 odds don't change - this is essentially what we're asked to show. It's like in a geometry proof using that an angle in the diagram is a 90 degree angle because it looks like one without showing why it must be one. If indeed it must be one because of what else is given it's certainly true, but using it without justification is at best incomplete. The simple solution currently in the article (3 equally likely scenarios, switching wins in 2 and loses in 1) doesn't have this problem. Do you find this solution hard to understand? -- Rick Block (talk) 19:21, 26 October 2008 (UTC)

Rick, I take issue with one of your statements: '...true statements, but stating them without justification is equivalent to assuming the solution' is too strong. Stating things without justification is assuming something, but not the solution. Even with the assumption of the statements mentioned, most people would still get the answer wrong, which is the notable feature of all this Martin Hogbin (talk) 19:56, 26 October 2008 (UTC).

Rick, also perhaps you would be kind enough to criticise the solution on my development page under Martin - Very short. Martin Hogbin (talk) 20:26, 26 October 2008 (UTC)

The very short solution (the player has a 2/3 chance of initially picking a goat, and in this case the host must reveal the other goat so this player now has a 2/3 of winning the car by switching) is essentially the "In other words" sentence just before the figure. I don't think it hurts to enumerate the cases, and the diagram definitely helps at least some people so I think this very short solution is fundamentally what's there already. The other advantage of the existing slightly longer version is that it provides a framework to discuss the more rigorous solution. Is the existing solution (up to and including the figure) any harder to understand than this very short solution? -- Rick Block (talk) 21:15, 26 October 2008 (UTC)

Thank you for your response to my question. Again, I respectfully disagree. My statement that it doesn't matter WHICH door Monty shows is true. It's not germain to the puzzle. The odds not changing? That's easy. Monty does not consider the door chosen by the contestant. Therefore, we gain no new knowledge about that selection, and it remains at 1/3. Since the probabilty must equal 100%, the remaining door now carries a 2/3 probability, (100% - 1/3). This is not really a debatable point. It is a bedrock principal of probabilty. The beauty of probability is that you only have to prove a solution one way. You can use my method, or Martin's method. They are BOTH VALID. As are some of the explanations with the diagrams, etc. And yes, the longer the explanation, the harder it is to understand, and explain to others. Why would you NOT want the most elegant solution to lead the paragraph? Any statement, diagram,etc. that is not necessary should be avoided. This is standard proof protocol. When Martin reduces it to 2 statements, that's a thing of beauty. Why interfere with it with extraneous statements?

Glkanter (talk) 21:33, 26 October 2008 (UTC)

Now, let's critique the existing Solution, much like you critiqued my proposed solution.

[Paragraph 1] The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

[Paragraph 2] Tree showing the probability of every possible outcome if the player initially picks Door 1A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).

[Paragraph 3] Referring to the figure above or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138) and considering only the cases where the host opens Door 2, switching loses in a 1/6 case where the player initially picked the car and otherwise wins in a 1/3 case. Similarly if the host opens Door 3 switching wins twice as often as staying, so the conditional probability of winning by switching given either door the host opens is 2/3 — the same as the overall probability. A formal proof of this fact using Bayes' theorem is presented below (see Bayesian analysis).

Paragraph 1 means nothing. It is gibberish.

Paragraph 2 does not respond to the Monty Hall Problem, and does not belong under the Solution heading.

Paragraph3 refers to a special case where Monty opens door 2 only. It too, is not germain to the Monty Hall Problem, which has ALREADY been proven by whatever method leads the discussion.

These three paragraphs of the existing Solution do not add anything to the reader's understanding of the solution, and should be removed. At best they are superfulous, at worst they are confusing, and maybe just wrong.

Glkanter (talk) 22:09, 26 October 2008 (UTC)

And, for the record, there are only 2 scenarios.

1 The contestant chooses a goat, which happens 2/3 of the time. In this case, as Martin so eloquently demonstrated, Monty eliminates the remaing goat leaving the car behind the closed door.

2 The contestant chose the car, in which case Monty eliminates one of the goats. This happens 1/3 of the time.

Therefore, any solution relying on diagrams, etc should have only these 2 scenarios. Anything beyond this convolutes the solution, which, of course should be avoided.

Glkanter (talk) 22:16, 26 October 2008 (UTC)

It doesn't matter which door Monty shows, but only because he's constrained to open one of the two goat doors with equal probability in the case the player initially selects the car. Again, you're asserting this without justifying why this is true. Regarding the odds not changing, are you saying Monty does not consider opening the door chosen by the contestant? This, by itself, is not sufficient to ensure that Monty's actions don't change the initial 1/3 probability. The total probability must be 100%, and before the host opens a door it's surely 1/3 player's door vs. 2/3 for the other two doors (so 2/3 of the players who decide to switch before the host opens a door will win) but the only thing that keeps it that way after the host opens a door is that pesky equal goat door constraint. The best way I know to show that it CAN change is to contrast the problem as stated with a different problem (i.e. the aforementioned "host opens lowest numbered door possible" variant).

With the equal door constraint, if the player picks Door 1 the probability the host opens Door 2 (or Door 3) is 1/2. This 1/2 is not random, but composed of a 1/3 chance that the car is behind the other door plus a (1/2 * 1/3) chance the car is behind Door 1. This fact makes the problem symmetric with regard to which door the host opens. We in effect have two identical subproblems, and it's this that makes it true that it doesn't matter which door the host opens (and prevents the host opening a door from changing the initial 1/3 chance).

In the lowest numbered door variant, the probability that the host opens Door 2 is not the same as the probability that the host opens Door 3. In this variant, the host opens Door 2 if the car is behind either Door 1 or Door 3 (i.e. 2/3 chance), and Door 3 only if the car is behind Door 2 (1/3 chance). The player has a 100% chance of winning if the host opens Door 3, but only a 50% chance of winning if the host opens Door 2. Multiplying out the composite probabilities we get (100% * 1/3) + (50% * 2/3), which totals 2/3 (meaning if you decide to switch before the host opens a door you have a 2/3 chance of winning) but in this variant the chances change to either 1/1 or 1/2 when the host opens a door (depending on which door the host opens).

The ONLY point in bringing up this variant is to demonstrate the flaw in the argument that the host's actions can't change the probability. They most certainly can. The host can't change the fact that 2/3 of players who decide to switch before a door is opened will win, but this doesn't necessarily mean that 2/3 of the players win when the host opens Door 3. This is an important aspect of the Monty Hall problem that, frankly, VERY few people understand (I didn't, until about a year ago - when an anonymous user brought it up on this talk page and provided the initial reference to the Morgan et al. paper and quite persistently argued about it). It is, in a sense, a second order issue (the first order issue being the "is it 1/2 or 2/3" confusion), but treating the problem rigorously, as a conditional probability problem, is not that difficult and leads to a much better understanding. This more rigorous solution is the topic of the paragraphs you've highlighted above. Could these be made more clear? Almost certainly. Should they be removed? I don't think so. -- Rick Block (talk) 00:24, 27 October 2008 (UTC)

Monty choosing the CONTESTANT'S door? Not in the puzzle. Therefore, the rest of your diatribe is meaningless. Equal goat door constraint? Never heard of it.

Look, the opening Wikipedia solution - once you remove the first sentance - proves that the odds are 1/3 vs 2/3. Once that mission is accomplished, the rest is either redundant, or incorrect. Is more than one proof valuable? Sure, but not the malarky that's there now.

Conditional probability? Hogwash!

I'm not familiar with a valid proof that is 'rigorously insufficient'. This isn't a beauty contest, or a political debate. It's math. It's a binary system, either it is, or it isn't a valid proof. The probabilities of those two events equals 100%. Like I said, that's a bedrock principal.

You have failed to demonstrate a weakness in the Wikipedia proof, Martin's proof or my proof. Yet you bring up all these other things. It's a simple puzzle, once you realize that Monty leaves a car on stage 2/3 of the time (thank you Martin). Again, 100% minus 2/3 leaves 1/3 for the contestant's choice.

Glkanter (talk) 01:42, 27 October 2008 (UTC)

Maybe we should take this one step at a time. Do you agree there may be a difference between the probability of winning by switching before the host opens a door and after the host opens a door? Or, at least, that these are different questions? -- Rick Block (talk) 02:09, 27 October 2008 (UTC)

I've read this entire discussion page from top to bottom. Other people have come before me, and pointed out essentially the same things I'm saying. No, thank you, I do NOT want to debate some other puzzle questions with you. It's a simple puzzle. In fact, the puzzle is almost like a Seinfeld episode. Nothing happens! The probabilities DON'T CHANGE! You seem to not understand Probability Theory. Otherwise, you would have acknowledged the validity of the proofs, and therefore the uselessness of those other issues as they relate to the Monty Hall Problem. Despite your valiant efforts at requiring 'a rigorous solution', you allowed a COMPLETELY ERRONEOUS STATEMENT to lead the solutions discussion. And have resisted multiple efforts to improve that section! That erroneous statement is at odds with everything that Probability Theory stands for. At times, you've argued with me over the premises of the Monty Hall Problem. Monty opening the CONTESTANT'S door? Give me a break.

Glkanter (talk) 02:28, 27 October 2008 (UTC)

I'm sorry you're finding this frustrating, but I assure you I understand probability theory. You perhaps missed my post in a section above agreeing that the first sentence in the solution section could (should) be deleted. The question about which door the host is opening was asking for a clarification of a comment you made (I'm well aware the host doesn't open the contestant's door). It is in fact not a simple puzzle, which is one reason why it remains so controversial. I'm very patient, and I'm willing to go through this step by step with you if you'd like. It's sounding like you might not like. That's OK, too. If you are willing, I think the first point is the one I ask above. Are there two possible questions? -- Rick Block (talk) 02:46, 27 October 2008 (UTC)

Well, we could debate your grasp of the science, but that would just get personal. I'd just like the interested reader to be able to see a couple or three simple proofs to help him/her understand the solution. That's really the only reason Wikipedia exists. There is a bunch of stuff under the Solutions heading that interferes with that. I, and people before me, have already demonstrated that. Is it possible to speak with your supervisor, or are you the final word?

Glkanter (talk) 03:13, 27 October 2008 (UTC)

Point by point form your response above...

"It doesn't matter which door Monty shows, but only because he's constrained to open one of the two goat doors with equal probability in the case the player initially selects the car. Again, you're asserting this without justifying why this is true."

False. I simply said "It doesn't matter which door Monty shows'. You added the rest. There's nothing to justify. How does Monty's or the contestant's behaviours change based on the door #s, or which goat he reveals? There's nothing in the puzzle about that.

"Regarding the odds not changing, are you saying Monty does not consider opening the door chosen by the contestant? This, by itself, is not sufficient to ensure that Monty's actions don't change the initial 1/3 probability."

Here's where you challanged the assumptions of the puzzle. Sad.

"The total probability must be 100%, and before the host opens a door it's surely 1/3 player's door vs. 2/3 for the other two doors (so 2/3 of the players who decide to switch before the host opens a door will win) but the only thing that keeps it that way after the host opens a door is that pesky equal goat door constraint."

False. The only thing 'keeping it that way' is that pesky law of probabilty that the outcomes must = 100%.

Glkanter (talk) 03:46, 27 October 2008 (UTC)

I am no more, and no less, an editor than you are - although I am an admin on the en.wikipedia site (admins are merely editors who are viewed by the community as trustworthy enough to have certain capabilities not available to most users, like deleting pages and blocking vandals - but admins distinctly do not work for the Wikipedia Foundation, so talking to my "supervisor" is a curious thought). The "final word" here is the consensus of the editors of this page. If you think there's a conflict you're welcome to follow the steps described at dispute resolution. If you'd like, you're also welcome to make whatever changes to the article you think are appropriate, although since you're apparently a new editor I'd suggest you review the 5 pillars. I'll also point out that this article is a featured article and has gone through two featured article reviews. The "stuff" you're objecting to in the Solution section was added in response to the most recent review, so it's not just me you're arguing with here but essentially the entire set of folks who have an interest in this article.

I'm quite willing to discuss this with you (if you'd like) in a less public forum such as your talk page, or mine, or email. -- Rick Block (talk) 04:30, 27 October 2008 (UTC)

Rick, thanks for your comments on my 'very short' solution. I have to say that Glkanter is not alone in being dissatisfied with the basic explanation in the current article. Myself and at least one other editor have also criticised it. It is also obvious from the reactions of editors just passing through (readers) that the article is not as convincing as it could be. Because of the FA status of this page, I am reluctant to just wade in and start editing it, which is why I set up my development page. The hope was that, on that page and it its associated talk page, we could all (old and new editors) work together to find ways to improve this article without compromising its current status. I would therefore suggest that all persons interested in the quality of this article discuss deficiencies of the current solution and proposals for new ones on that page. In particular, I will answer you questions to me above concerning what I do not like about the current article and I will look forward to your replies there.Martin Hogbin (talk) 09:24, 27 October 2008 (UTC)

My development page was set up to allow discussion rather than edit warring so I do not think that it would be right for old and existing editors to completely ignore my page but then start reverting changes to this page based on a consensus reached by those expressing their views on the development page. We need to move forwards but in the right way. Martin Hogbin (talk) 09:24, 27 October 2008 (UTC)

Proposed Changes

Several editors believe that, although the current article may meet the FA criteria, it still remains unconvincing and also concentrates far to much on variations of the basic problem that are irrelevant to the core problem.

It is therefore proposed that the current solution be replaced with a new version giving simple, intuitive, and verifiable solutions to the basic problem.

Preparation for this new section is taking place on a [[4]] . All interested editors are asked to make their comments on the suitability of this section for inclusion in the current article and to make improvements to the new section. Martin Hogbin (talk) 20:54, 6 November 2008 (UTC)

I couldn't find any discussion on the linked page, but I think the simplest explanation of the solution would be as follows (START):

Sorry, I have now corrected the link [[5]]. Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)

The wording of the puzzle can be misleading because it seems to imply that after the goat is revealed, the player is now picking one of two doors. In reality, when offered a switch, the user is ALWAYS picking between their one door and BOTH of the other doors. Think about it this way. Instead of openening a door with a goat, once the player selects a door, Monty offers the player a choice, they can take whatever is behind their door, or take whatever is behind the two doors they didn't pick. Simple logic tells the player that AT LEAST ONE of those two doors MUST contain a goat, there's only one car. However, the player also knows that two doors are better than one. Since they picked one of three doors, the chance that their door contains the car is only 1/3. The chance that one of the other two doors contains the car is thus 2/3. Whether or not the user switches, we know one of the unpicked doors contains a goat. Revealing the presence of the goat before the switch offers no new information. Monty is not revealing a big secret, everyone knows one of two unpicked doors had to contain a goat. Monty just saved a little suspense and air-time by pre-revealing the goat we knew had to be behind one of the two unchosen doors. Therefore, the odds of winning the car by switching to both of the other doors (before or after the goat is revealed) are always 2/3. The key here is that the user IS switching to BOTH unchosen doors. The one we knew had a goat behind it just happens to already be open.

I think what you are suggesting is pretty much what is on the development page under 'combining doors'.Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)

This solution assumes a couple things:

1. Monty will either randomly or always reveal a goat after the user picks a door. I.e. he knows where the car is.
2. Monty will either randomly or always offer the player the option to switch doors.

Monty randomly offering the trade is equal to always offering the trade, since in this particular instance, the trade was offered. The question becomes moot if Monty didn't offer a trade. If Monty has alterior motives, such as offering a trade only after you already picked the car, then the odds of you winning the car on a trade are 0, and it didn't matter which of the two doors (if either) was opened because once Monty offers the trade you know they both contain goats.

My main point is that not only are the 'assumptions' that you make above the natural ones that most people make, they are also what is clearly stated in the 'Parade' statement of the problem, which is where all the fuss started. It is my contention that the real problem is the one you state above and that this is what the article should concentrate on. Other, essentially academic, possibilities should then be discussed later in the article. Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)

(END) Maybe this isn't any simpler to understand, but this is what I though of as I slowly convinced myself the answer really is 2/3. Hellpimp (talk) 06:41, 7 November 2008 (UTC)

Perhaps you might be able to improve the development article.Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)

Are the doors distinguishable or not? They're numbered, so of course they are. This means there are two questions we can ask.

1) What is the overall chance of winning by switching, averaged across all players?

2) What is the specific chance of winning by switching for only those players selecting a given door (say Door 1) knowing the host has opened a specific door (say Door 3).

The general form of the problem statement asks the second question, not the first one. However, the argument above addresses the first question, not the second. In contrast, the Bayesian analysis section of the article answers the second question. I don't think this is merely an "academic" point. I know the online copy costs money, but I would strongly encourage anyone seriously interested in this problem to read the Morgan et al. paper in the references (it should be available at any university library). If we're attempting to answer the second question, Morgan et al. show the answer is 2/3 chance of winning by switching only if the host is also required to select between two goat doors (if given the chance) with equal probability. Any solution that does not use this constraint (the Bayesian analysis does), is either answering a different question (like the first question, above) or is incomplete (i.e., wrong).

The current structure of the article is to answer the first question first since its solution is easier to grasp, but then acknowledge that the second question is indeed a different question and proceed to answer it. If this structure is not clear, by all means clarify away. IMO, relegating this point to a subsequent "academic" section does our readers an extreme disservice. -- Rick Block (talk) 15:12, 7 November 2008 (UTC)

Although the doors may be distinguishable this is not relevant to the core problem where, to all intents and purposes, they are not. The distinguishably of the doors is another sideline which, in my opinion, should be relegated to an 'academic variations' section of the article.

Regarding the host behaviour on selecting between goat doors, the 'Parade' statement of the problem says,'...the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat...' it does not give us any information as to how the host might choose between two goat doors. In such a case it is not an assumption to take it that this is random but an obligation.

The problem with adding in all these complications is that it makes the article useless at one of its primary purposes which is explaining the solution to readers in a way that convinces them. Martin Hogbin (talk) 18:45, 7 November 2008 (UTC)

And the problem with avoiding these complications is that it makes the article useless at its primary purpose which is summarizing what reliable sources have to say about the Monty Hall problem. We don't avoid things just because they're hard to explain. Where the rubber meets the road here is the references. Two different references that appeared shortly after the vos Savant dust up, Morgan et al. and Gillman, bring up the difference between the two questions that may be asked. Yes, they're both academic papers, but unless there are other references that say these papers are obscuring the "core problem" this notion is simply your opinion. Treatments of the problem in non-academic sources still generally (possibly always) ignore this difference but this does not give Wikipedia an excuse to do so. As far as I know, pretty much all academic sources now include the "equal goat door constraint" in the problem statement. Many do so because it's necessary to justify the Bayesian approach which solves for the probability of winning by switching if the player picks Door 1 and the host opens Door 3. The Bayesian treatments solve for this, rather than simply the probability of winning by switching if the player picks Door 1, because it's what the problem ostensibly asks.

I'm aware that our exchange here has the form of an argument, but I actually don't think we're that far apart. As I've said repeatedly, I'm fine with presenting an "intuitive" solution answering the easier question first. And, indeed, the article currently does this (and whether the current intuitive solution is the most convincing is certainly debatable). However, I'm not fine with the Solution section including only something that one of the main academic references calls a "false solution".

I'll solicit other opinions about this at Wikipedia:WikiProject Mathematics. Most folks there are aware that this problem is a notorious quagmire, so it is somewhat difficult to get them to comment but I'll try. -- Rick Block (talk) 03:44, 8 November 2008 (UTC)

The solutions proposed properly solve the paradox using established Probability Theory. All this other stuff adds no value.

Glkanter (talk) 07:10, 8 November 2008 (UTC)

I got it! Start a new Wikipedia article called 'The Equal Goat Door Constraint Paradox'! Marilyn vos Savant can write a column about it in her general interest Sunday magazine column. People all over the world will marvel at how counter-intuitive the solution is, when all along, after a trip to the nearest university library, Morgan, et al and Bayesian statistics were there to explain it. Glkanter (talk) 16:17, 8 November 2008 (UTC)Glkanter (talk) 16:26, 8 November 2008 (UTC)

I'm responding to Rick's WT:WPM post. Any attempt to make the article clearer is a good idea, but it must be based upon the fact that Wikipedia is an encyclopedia. Its job is to document the knowledge found in reliable sources, not to synthesise explanations in the literature to provide the clearest and most convincing one. This article is not entitled "Explanation of the Monty Hall Problem" and Wikipedia generally doesn't have articles with such titles because they are not encyclopedic and typically involve taking a point of view. The job of this article is not to convince the reader, but to present them with information. That's the difference between an encyclopedia and a textbook: the latter aim to educate the reader, the former are a resource which help the reader to educate themselves.

Wikipedia does not have disclaimers, does not assert viewpoints, and does not contain original research. Geometry guy 19:48, 9 November 2008 (UTC)

The application of WP policies to this problem is full of difficulties. We do not need a cut-and-paste approach to verifiability, neither do we want opinion, but where between these two extremes to we want to be? What is a reputable source? In this subject there have been many sources that would normally be considered reputable that have got the answer wrong. Common sense is no help either. I am not saying it is easy but I believe that we can do better.

Clearly, the current article does not just, 'document the knowledge found in reliable sources' neither should it, in my opinion. What is the point of having an article the contains information if nobody believes it?

Finally, what is the justification for the current article's current concentration on academic variations of the problem. This is not what makes the problem notable. Martin Hogbin (talk) 23:24, 9 November 2008 (UTC)

Between the two extremes we want (from Wikipedia:Featured article criteria) to be factually accurate: claims are verifiable against reliable sources, accurately represent the relevant body of published knowledge, and are supported with specific evidence and external citations.

What constitutes a reputable (reliable) source is defined at Wikipedia:Reliable sources.

Per the quote from the featured article criteria above, the point is to accurately represent the relevant body of published knowledge. What is to be believed about any article is that it does this. References are included specifically so readers can verify that what is said in the article is supported by published sources.

I'm not sure what you mean by the article's current "concentration on academic variations", but (also per the FA criteria) articles should be comprehensive: it neglects no major facts or details. More on this from Wikipedia:Neutral point of view: content must be written from a neutral point of view (NPOV), representing fairly, and as far as possible without bias, all significant views that have been published by reliable sources. -- Rick Block (talk) 15:43, 10 November 2008 (UTC)

Still confused ..

The following introduction is still confused suffering from poor English, a lack of clarity and some misconceptions.

"Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3. Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty. The total probability of winning when switching is thus 2/3."

The object of the problem is not whether the person wins the automobile but how the probability shifts when a door is exposed (with a goat). If the person selects the correct door or not on the first guess is irrelevant to the problem - especially as an introduction.

A breakdown:

(1) Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter.

Is fine

(2) In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

Does not need "in the usual interpretation of the problem" in the introdcution.

(3) Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3.

Should read, "If the player initially chooses the winning door then changing is clearly disadvantageous" I don't even think it is necessary as this is confusing in the introduction.

(4) With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty.

Is badly constructed. It should read, "If the player has initially chosen one of the two losing doors: when one of the losing doors is revealed then switching wins the player the automobile."

However, I would argue that this isn't even needed anyway.

(5) The total probability of winning when switching is thus 2/3.

Is a non-sequitur as the paragraph is referring to two different scenarios (ref to 2 and 3). Both scenarios, incorrect first choice and correct first choice cannot lead to the same probability of winning when switching. --Candy (talk) 22:52, 24 November 2008 (UTC)

So, in summary, you're suggesting a change to perhaps something like:

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3. If the player initially chooses the winning door with probability 1/3, then changing is clearly disadvantageous. If the player has initially chosen one of the two losing doors with probability 2/3, then after the other losing door is revealed switching wins the automobile. The total probability of winning for all players who switch is thus 2/3.

or cutting the explanation after the "In fact" sentence. I'd support cutting the explanation (for reasons discussed at length in previous threads on this talk page this explanation is somewhat less than completely satisfactory). -- Rick Block (talk) 02:54, 25 November 2008 (UTC)

Good faith edits

Reversion of good faith edits should not be marked as minor. Martin Hogbin (talk) 20:27, 11 December 2008 (UTC)

Is there some specific revert you're talking about here? -- Rick Block (talk) 00:49, 12 December 2008 (UTC)

Recent reverts marked as minor:

• Nov. 1 by Belinrahs - reverted humorous comment
• Dec. 3 by Iridescent - reverted vandalism
• Dec. 4 by Paul August - reverted change of 2/3 to 1/3 - sneaky, but possibly good faith
• Dec. 11 by Rick Block - reverted additions - probably good faith - the edit combined an addition to an image caption (that could be good faith) with another addition (that clearly wasn't) - I think Martin missed that bit (I know I did!).--Noe (talk) 12:53, 12 December 2008 (UTC)

Sorry, I did miss that. Martin Hogbin (talk) 20:11, 12 December 2008 (UTC)

External link to ucs.edu page

I've moved the following link which was added to the article here. There's nothing on this page that can't be incorporated directly into the Wikipedia page, per WP:EL that means it is not an appropriate external link.

• The Monty Hall Problem at LEEPS, a lab run by Dr Dan Friedman (a prominent economist at UC Santa Cruz).

-- Rick Block (talk) 14:39, 19 December 2008 (UTC)

--- I believe the above link should be on the main page for the Monty Hall Problem. It cites important academic journals and articles on the issue. We cannot ignore literature that discuss this problem as it was a hot topic of discussion amongst prominent economists and mathematicians. Some of the articles mentioned in the above link do form a basis to the solution that is widely accepted today. - Aadn

From WP:EL#Links normally to be avoided:

1. Any site that does not provide a unique resource beyond what the article would contain if it became a Featured article.

Also from WP:EL (Advertising and conflicts of interest):

You should avoid linking to a website that you own, maintain or represent, even if the guidelines otherwise imply that it should be linked. If the link is to a relevant and informative site that should otherwise be included, please consider mentioning it on the talk page and let neutral and independent Wikipedia editors decide whether to add it.

The first of these definitely applies here and I strongly suspect the second does as well. The LEEPS page itself does not meet Wikipedia's standard for reliable sources, so shouldn't be used as a reference either. You are perfectly welcome to add citations drawn from the LEEPS page to the history section, especially any that are considered major or important. Note that there are dozens (if not hundreds) of published papers on this problem, so clearly most of them will not be mentioned here. -- Rick Block (talk) 05:22, 21 December 2008 (UTC)

The Goat

What if I want the goat and not the car? Should this probability also be addressed, even if it's not part of the original Monty Hall problem? 76.95.124.146 (talk) 15:09, 8 January 2009 (UTC)

This question may be useful in settling the issue of how best to explain the solution. The goats are actually a distraction from the real puzzle, serving much the same purpose in this trick puzzle as "misdirection" serves a magician. If a goat is an anti-prize, worse than neutral, then it may be more difficult to grasp why the revealing of a goat behind a non-chosen door is not useful information in solving this particular puzzle, although it can be useful in solving other puzzles that resemble it. The first step to providing a convincing explanation of the solution may be to remind the solver that one is not actually required to take the goat home if the goat is chosen! (In fact, a goat is equivalent to no prize at all.) This may disappoint 76.95.124.146, but for other solvers, it may be a good first step to understanding the puzzle. Perhaps it is unfortunate if no reliable source has taken this approach.Simple314 (talk) 03:29, 11 January 2009 (UTC)

Addition of simplest explanation

I've undone this edit which adds an introductory paragraph to the Solution section with numerous WP:MOS issues discussing an alternate problem (boxes rather than doors). The addition is essentially the same as the existing "combining doors" section. The text is in the history if anyone wants to talk about it. -- Rick Block (talk) 06:01, 24 January 2009 (UTC)

Minesweeper analogy

I've deleted the addition of the following new section.

Minesweeper analogy

An analogy which might have a more intuitive solution is the computer game Minesweeper with the minor change that the game is played without numbers indicating mines. Suppose one mine (the equivalent of the car) is hidden in a 100x100 grid. The first action by the player is to randomly plant a flag on one of the squares (equivalent of picking a door at the beginning). The player clicks on a square without the mine (this is actually how many versions of minesweeper are programmed, the first click cannot be a mine) clearing the entire grid except for one square and the one flagged by the player (the equivalent of Monty opening the door revealing a goat). For anyone familiar with the game Minesweeper, the intuition here is that the mine is under the unflagged square since the player flagged the square randomly at the beginning of the game. This analogy also removes any possible motivation on Monty's part to trick the player which may be confusing in the classic example.

Although interesting, it's unreferenced and sounds like WP:OR (and requires an understanding of Minesweeper). If anyone can find a reliable source including this analogy we can talk about re-adding it. -- Rick Block (talk) 01:43, 25 January 2009 (UTC)

It's a very good analogy, though, and in fact it's a form of the "increased number of doors" explanation. But it does have a critical flaw: if the player actually manages to flag the mine, all the other squares will get cleared, which would be the equivalent of Monty revealing every single goat. So it actually simulates a very different "Monty" -- one who reveals every goat unless there's one behind your door.--Father Goose (talk) 09:41, 25 January 2009 (UTC)

The description mentions that all the squares except the initially flagged one and one more are cleared (making it not quite like minesweeper). Perhaps a truer version would be you flag one, Monty flags another, and then Monty clicks one he knows isn't the winner. This actually scales down to a 3-square version identical to the MH problem. In any event, if it's not published I don't think we should even consider using it. If it is published we could talk about it, but even then I'm not sure it really adds anything. -- Rick Block (talk) 17:20, 25 January 2009 (UTC)

Still no good: in most cases you'll be left with three unrevealed squares: your flag, the host's flag, and the actual location of the mine. Even in the three door version, it won't work: if both of you pick goats (flag an empty square), when you click on the last remaining square (the mine), the game will move the mine under either your flag or his, presumably randomly. So there's no restricted choice on his part: both his choice and yours are entirely random. It's the "host doesn't know" scenario.--Father Goose (talk) 21:21, 25 January 2009 (UTC)

Wrong explanation

The, what is often called, "simple solution", saying: the player originally has a chance 1/3 of winning, so switching increases this chance to 2/3, is theoretically wrong! This explanation is also offered directly under "Solution" , and although it offers numerically the right answer,it is not correct, as may be seen by comparing it to the statement of the problem right above it. In the problem statement the question is: Should the player switch to Door 2? And in the solution the player is allowed to switch to Door 3 as well. The point is, and I think it is mentioned somewhere in the above discussions, that the posed question can only ask for a conditional probability, given the situation, i.e. Door 1 chosen and Door 3 opened revealing a goat. This is also clear from the equivalence with the Three Prisoners problem and Bertrand's box paradox.Nijdam (talk) 11:26, 3 February 2009 (UTC)

Rather than flat out "wrong", I'd say the simple solution is a correct solution to a slightly different problem. This is the point of the three paragraphs below the large figure in the Solution section of the article. Many (possibly nearly all) people interpret the question to apply to all players unconditionally rather than only those players seeing a specific open door (and a fair number apparently cannot understand that these are actually different questions). These people would claim the wording "say you've picked Door 1 and the host opens Door 3" is saying that we are to assume all specific situations are equally likely, in which case the unconditional and conditional answers must be the same. If you can make this more clear in the article, please do so. -- Rick Block (talk) 14:49, 3 February 2009 (UTC)

This is an other way of putting it. But as the article is so precise in defining the problem, with all guarantees of no possible misinterpretation, the given "simple solution" is no solution to this precise stated problem.Nijdam (talk) 16:04, 3 February 2009 (UTC) —Preceding unsigned comment added by 82.75.67.221 (talk)

I've revised your clarification. Are you OK with the revised version? -- Rick Block (talk) 19:34, 3 February 2009 (UTC)

Any of you recent contributors may not have read this page in its entirety. It's pretty long. So, just to let you know, many, many people are very unsatisfied with the entire article. It's long winded, confusing, and outright wrong in places. But we respect the process, so here we are. Search for my signature, as one example. Glkanter (talk) 03:30, 4 February 2009 (UTC)

Let me see if I've got this right. Glkanter thinks the simple solution (the part above the large figure in the solution section) is ridiculously long winded but conveys the only essential aspect of the solution, and the part below the figure is incomprehensible and wrong and should be deleted. Nijdam thinks the part above the figure is wrong and, if not deleted, should be prefaced with something like "the following is wrong:". How about if you two talk. I'd be happy to try to moderate. -- Rick Block (talk) 05:48, 4 February 2009 (UTC)

I tried to follow the discussion above between the two of you, and I actually got confused, not about the problem or the solution, but about you. I'm not very happy with Rick's revision of my extension to the "simple solution". My extension shows in essence the flaw in this way of reasoning. I showed that the "simple solution" is not a solution to the stated problem. It's like the man, searching in the evening for his lost wallet under a streetlamp where things are well visible, allthough he knows he lost the wallet in a dark side alley. So I invite either one of you to read my extension an comment on it.Nijdam (talk) 10:49, 4 February 2009 (UTC)

I did a little more reading of the above discussion about this topic and I think I find Rick on my side, but Glkanter opposed because of the reasoning: 1/3 + 2/3 = 1, meaning: the initial probablility is 1/3 of picking the car, so whatever happens, as (!?????) this has not changed, the final probability of the alternative should be 2/3. Well as you may guess, I already indicate where it goes wrong. The "initial probability changes" under the assumed condition. Among all possible cases, the ones with the car behind Door 1 and Door 3 openend, only form 1/6th.Nijdam (talk) 11:28, 4 February 2009 (UTC)

Nijdam's version is here. My revision is the next edit in the history. I have two issues with Nijdam's version. 1) It's a little too harsh on the "simple solution". In its current form, this solution doesn't say anything that is actually incorrect it just doesn't precisely answer the question. Also, like it or not, it is what is commonly presented as the "solution". 2) The style doesn't follow Wikipedia style guidelines, in particular see Wikipedia:Manual of Style (mathematics)#Writing style in mathematics. -- Rick Block (talk) 15:02, 4 February 2009 (UTC)

Nijdam, please show me the error in the 1/3 + 2/3 = 1 (100%) solution. I'd also like to know what is unacceptable about Martin's solution, 2/3 of the time I choose a goat, and 100% of those times Monty show me the other goat? Everything else, from Morgan, et al, to Bayesian, to "We have run simulations that prove this" is wasted binary digits. Glkanter (talk) 23:12, 4 February 2009 (UTC)

I thought you were skilled enough to directly understand the problem. First: The "simple solution" is not a solution at all, i.e. not to the stated problem. As Rick puts it: "it just doesn't precisely answer the question"!! As if this is not enough reason to be harsh. 2) I'm really completely stunned about this. Let us translate the situation to an exam. Well, the professor utters: "let's not be too harsh on this "solution". Nothing is really incorrect, it just doesn't answer my question." 3) At least Rick understands it is wrong. 4) Why is it wrong? As I explained in my text, the "solution" offers the possibility that Door 2 is opened, and the statement of the problem excludes this. 5) Now about the specifc question about Martin's reasoning. It has the same flaw: it doesn't answer the posed question. Of course it is the right answer to the right alternative question, which I may present as follows: the player is blind and Monty only tells him: I opened one of the other doors and there is a goat behind it. But that is not the case in de stated problem. There the player sees that Door 3 is opened and shows a goat, and hence the conditions of the problem are accordingly formulated. 6) All this is because after some events has happened, we want to know the conditional probabilities, given these events. Compare it with throwing a fair dice and suppose it shows 6. What is the probability it will show 5? (0 or 1/6???). I hope it becomes clear to you. Sorry, forgot to log in.Nijdam (talk) 23:56, 4 February 2009 (UTC)

In a moderating kind of way, I have two suggestions here. 1) Please comment on content, not on other editors' mental facilities. WP:COOL has some good suggestions. 2) As far as possible, please support arguments with references rather than arguing what you simply know to be the Truth. There are no shortage of references about the Monty Hall problem - any significant point has almost certainly been published (at least once). -- Rick Block (talk) 02:00, 5 February 2009 (UTC)

Following my own advice - to Glkanter, the issue with both your and Martin's solutions is explained in Morgan et al. These are both variants of what Morgan et al call "false solution F1".

Solution F1. If, regardless of the host's actions, 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.

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 siutations here seems to confound many, from whence much of the pedagogic and entertainment value is derived.

Similarly, Gillman 1992 (referenced in the article) criticizes vos Savant's solution:

Marilyn's solution goes like this. The chance is 1/3 that the car is actually at #1, and in that case you lose when you switch. The chance is 2/3 the car is either at #2 (in which case the host perforce opens #3) or at #3 (in which case he perforce opens #2) - and in these cases, the host's revelation of a goat shows you how to switch and win.

This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3. Instead it is still considering the possibility that the car is at #3-whence the host cannot have already opened that door (much less to reveal a goat). In this game-you have to announce before a door has been opened [emphasis in the original] whether you plan to switch.

These are both from academic WP:reliable sources (The American Statistician and American Mathematical Monthly), and the article is (at this point) attempting to present the information contained in these sources.

Is the view that the "simple solution" is sufficient supported by equally reliable sources? If so, which sources and what do they say? -- Rick Block (talk) 04:56, 5 February 2009 (UTC)

Well, don't be offended; take it as a complement instead, because from the discussions above I concluded both of you are well skilled at least in mathematics. Yet you keep asking for arguments, which I gave in the extension of the article and extensively above. Moreover, Rick quotes the same arguments I gave. So what more needs to be said? If the "simple solution" as a solution to the well stated problem, is supported by any source, the conclusion can only be, this source is not reliable. The only necessity is changing he article, as I suggested.82.75.67.221 (talk) 10:01, 5 February 2009 (UTC)

Just to be clear, you're arguing for reinstating this change (right?), which prefaces the simple solution with:

The following "solution" is often stated, but is not the correct solution to the problem, because in the above precise stated problem not only did the player picks Door 1, but is it also known that Door 3 is opened and shows a goat. Let us yet follow the reasoning and see what goes wrong.

and interjects a new paragraph

For the attentive reader it will be clear that the player in this solution is also confronted with the possibility to find Door 2 opened showing a goat and offered to switch to Door 3. This is not allowed in the statement of the problem. However there is a way to work around it. Looking at the pictures we see this is the case in the first and the last picture in the last row. If we restrict the analysis to the other cases (for the probabilistic skilled reader this means conditioning on these events), the (conditonal) probability of winning the car by switching is 1/3 against 1/3+1/6, which is also 2/3.

before the more technically correct solution. The current structure is to start with a commonly presented simple solution being careful to avoid saying anything that is actually incorrect (in particular, qualifying that this solution applies to all players before the host opens a door), and then transition into the more precisely correct solution. I've tried to clarify the transition, as follows:

Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door (Morgan et al. 1991). Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. The difference is whether the analysis, as above, considers all possible scenarios or only the scenarios where the host opens a specific door. The conditional probability may differ from the overall probability depending on the exact formulation of the problem (see Sources of confusion, below).

I think this follows the sources fairly faithfully, leading the reader to the more technically correct solution without making a huge point out of it. I think it is true (I don't have a handy reference, but could perhaps find one) that even though the detailed problem statement is mathematically unambiguous many people misinterpret it as asking about the unconditional probability (for which the simple solution is the correct answer). As I've said, calling the simple solution wrong seems overly harsh - particularly since the unambiguous version is specifically designed to end up with the same answer (I don't have a reference for this either, but it seems fairly obviously true). -- Rick Block (talk) 15:36, 5 February 2009 (UTC)

Well, I find it unacceptable to start with a "solution" that isn't a solution. My suggestion (please improve my English, as I'm foreign):

Proposed Solution section

According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):

• The player originally picked the door hiding the car. The game host must open one of the two remaining doors randomly.
• The car is behind Door 2 and the host must open Door 3.
• The car is behind Door 3 and the host must open Door 2.

This leads to the following situations:

Car hidden behind Door 1		Car hidden behind Door 2	Car hidden behind Door 3
Player initially picks Door 1
Host opens either goat door		Host must open Door 3	Host must open Door 2
This hasn't happened	This happens with "probability" 1/3	This happens with "probability" 2/3	This hasn't happened
	Switching wins in 2 out of 3 cases

Players who choose to switch win if the car is behind Door 2. This is the case in 2 out of 3 cases, so with 2/3 probability.

This means if a large number of players, after choosing Door 1 and being showed a goat behind Door 3, randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

Nijdam (talk) 17:56, 5 February 2009 (UTC)

How about this? (and your English is far better than my Dutch) -- Rick Block (talk) 02:58, 6 February 2009 (UTC)

---

According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):

• The player originally picked the door hiding the car. The game host must open one of the two remaining doors randomly.
• The car is behind Door 2 and the host must open Door 3.
• The car is behind Door 3 and the host must open Door 2.

This leads to the following situations:

Car hidden behind Door 1		Car hidden behind Door 2	Car hidden behind Door 3
Player initially picks Door 1
Host opens either goat door		Host must open Door 3	Host must open Door 2
Switching loses with probability 1/6	Switching loses with probability 1/6	Switching wins with probability 1/3	Switching wins with probability 1/3
Switching loses with probability 1/3		Switching wins with probability 2/3

Looking at the figure above, in cases where the host opens Door 3 (which is the case according to the problem statement) switching loses in a 1/6 case where the player initially picked the car and wins in a 1/3 case where the player initially picked a goat. Players who choose to switch win twice as often as players who stick with the original choice, which is a 2/3 chance of winning by switching.

This means if a large number of players randomly choose whether to stay or switch after initially choosing Door 1 and being shown a goat behind Door 3, then approximately 1/3 of those choosing to stay and 2/3 of those choosing to switch would win the car.

---

We'd probably need to follow this up with a discussion of the typical unconditional solution and why it is not responsive to the question that's asked. I think this structure is probably less accessible than the current structure to someone not mathematically sophisticated (indeed, as Morgan et al. comment, "the distinction between the conditional and unconditional situations here seems to confound many"). Typical Wikipedia style is to start with broadly accessible content and then go deeper (see Wikipedia:Make technical articles accessible), which is the overall intent of the current structure. As I've said the current structure is meant to present an analysis that perhaps doesn't quite address the question that's asked (but is not actually wrong) leading into an technically correct solution. I think this is an article of more interest to laypersons than mathematicians (although mathematicians frequently get this problem wrong as well), so I think anything we can do to make the solution generally accessible is worthwhile. -- Rick Block (talk) 02:58, 6 February 2009 (UTC)

Glkanter's objection

Monty does not pick doors randomly. He always reveals a goat. And it does not matter which goat. I'm still waiting to hear why Martin's solution is not satisfactory.

Glkanter (talk) 23:47, 5 February 2009 (UTC)

Please read the excerpts from Morgan et al. and Gillman, above. Even more directly, Morgan et al. discuss what they call false solution F5:

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

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

We've been through this before, but the easiest way I know of to show there are two different problems, and to figure out which one a given solution is addressing, is to change the assumptions slightly so that the answers to the two problems are different. When the answers are the same it's hard to tell which one a given solution is addressing (is it this 2/3 or that 2/3?). The slight change (also from Morgan et al) is to give the host a preference for one door over the other. To make it the most extreme, let's constrain the host to open the leftmost door if possible (still always showing a goat, but opening the leftmost door unless this would reveal the car). The question remains the same, i.e. if a player initially picks door 1 and the host opens door 3 what is the chance of winning by switching? Martin's solution still applies - the chance of initially picking a goat is 2/3 and the host most show the other goat. So, Martin's solution says the answer to this variant is 2/3 chance of winning by switching (right?). On the other hand, in this variant if the host opens door 3 (which we've said he did) we know for sure the car is behind door 2 so with the given setup (player picks door 1, host opens door 3) the chance of winning by switching is 100%. The essentially "wrongness" of Martin's solution is now making it come out with the wrong numerical answer. It is still the right answer for a different question, specifically what is the chance of winning for a player who decides to switch before seeing which door the host opens. But in both variants, this is not the question we're asking.

I think you don't like analogies but I'll try one anyway. Let's say we're asked what the square of squareroot(4) is. One solution is "squareroot(4) = 2, 2 squared is 4, the answer is 4". This is the right answer, but the solution is technically incorrect because squareroot(4) is either 2 or -2. For this particular numerical problem it doesn't matter if you ignore the negative root, but the solution is clearly wrong. We can expose the flaw by changing the question, for example "what is twice the squareroot(4)". We have a very similar issue here. No one's arguing that the answer (2/3 chance of winning by switching) is incorrect. It is correct. It's just not answering the question that's asked. -- Rick Block (talk) 02:01, 6 February 2009 (UTC)

I'll be honest. I don't understand most of what you guys are talking about. It's a simple puzzle. Do you increase your chances of winning the car by switching? It's been years since I took a Logic course, or a Probability and Statistics course, so I can't cite chapter and verse. But the goal is to present a valid proof with as little complication as possible. So you take shortcuts, which don't affect the proof. Two goats? I only care about the car. Ignore the goats in the proof. 3 doors? There'll only be one 'other one' left when I'm asked to switch. Ignore the door #s in the proof. Conditional probability? Don't know what it means. Don't see how anything is conditional here. Two closed doors when I'm asked if I want to switch, exactly one has a car. No conditions exist.

I do remember something about a requirement for proofs being valid and true. 2/3 of the time the remaining door I didn't choose has a car. Both valid and true.

That's why the real game show didn't work like this. Glkanter (talk) 03:16, 6 February 2009 (UTC)

The essential point is regarding the question that's being asked. Is it

a) "what is the chance of winning for a player who decides to switch before seeing which door the host opens"

or

b) "what is the chance of winning for a player who decides to switch after seeing which door the host opens"

I think most statements of the problem at least implicitly ask the second question (the unambiguous version in the article explicitly asks the second). Martin's solution addresses the first question. Since the numerical answers to these two questions can be different (per the variant mentioned above), a solution for one is clearly not the same as a solution for the other. Simply because a "valid and true" solution for #a ends up with the right numerical answer for #b (which it might for some versions of the problem), doesn't make it a "valid and true" solution for #b.

Putting this yet another way, if there are 3000 people who go on the show and initially pick door #1, roughly 1000 will pick the car and roughly 2000 will pick one of the goats. What Martin's solution is saying is that if all 3000 decide they're going to switch before seeing what door the host opens, roughly 2000 will win the car. However, the problem statement says we know the host opens door 3 so the divisor is not 3000. What is the actual divisor (roughly how many times will the host open door 3) and how do you know this? In how many of these cases will the player have initially picked the car? To answer what the probability is of winning the car by switching if the host opens door 3 these are the numbers you need to know. What does Martin's solution have to say about either of these? -- Rick Block (talk) 04:58, 6 February 2009 (UTC)

Okay Rick, at least I see you completely knows the ins and outs. So what to do? I cannot accept a reasoning that's logically wrong. What about the suggestion I made above, by changing the given "simple solution" in such a way that some possibilities are ruled out as not happened?Nijdam (talk) 11:54, 6 February 2009 (UTC)

There are really only two situations to consider. Not 3, not 6.

2/3 of the time I will choose a goat.

100% of those times Monty reveals the other goat, leaving the car.

=>2/3 of the time I should switch.

1/3 of the time, I will choose the car.

100% of those times Monty reveals one of the goats. It doesn't matter which.

=>1/3 of the time I should not switch.

2/3 is greater than 1/3.

Therefore, by switching, I increase my chances of getting the car when offered the switch.

I earlier said a good proof uses 'shortcuts'. A better phrasing would have been 'eliminates the unessential' from the proof. That is, a goat is a goat, and a door is a door.

Martin's proof, and the one above, address the problem as defined. To deny that is incorrect. [User:Glkanter|Glkanter]] (talk) 13:34, 6 February 2009 (UTC) 63.97.108.114 (talk) 14:19, 6 February 2009 (UTC)Glkanter (talk) 14:22, 6 February 2009 (UTC)

Please reread my responses above. Per Morgan et al., Martin's proof, and your proof above are perfectly correct if the question is "what is the chance of winning for a player who decides to switch before seeing which door the host opens". But, the actual question is "what is the chance of winning for a player who decides to switch after seeing which door the host opens". These questions are different and that they happen to have the same numerical answer in this specific case doesn't mean a solution for one is also a solution for the other. I'm sorry if you don't understand this, but I don't know how to make it more clear. Whether you understand it or not, it is what multiple Wikipedia:Reliable sources say so should be included in the article. -- Rick Block (talk) 15:29, 6 February 2009 (UTC)

Nope. I don't get it at all. My proof above very clearly is the case after showing the goat door. I think this 'before and after' dichotomy is a canard. I've said before, and you just agreed above, that Monty's action doesn't change anything. That is, he provides no new useful knowledge. The proof of this is that the probabilities haven't changed.

It takes exactly one valid proof to solve a problem. That proof has been demonstrated.

Glkanter (talk) 16:19, 6 February 2009 (UTC)

The key here is "valid", which your proof is not according to two papers written by mathematicians published in math journals. What we want in the article is what published papers have to say about this problem. I'd prefer if you understood it, but ultimately whether you do or don't doesn't actually matter. -- Rick Block (talk) 05:40, 7 February 2009 (UTC)

Thank you for the specific response to my question. I'll try to look at those papers. I suppose Morgan, et al is one of those mathemeticians, who's the other?

Do I understand your response to Nijdam correctly? Does "many published sources (possibly all popular expositions of this problem) include only an unconditional solution" mean that the solution I've been advocating is what most other references provide as the only solution?

Would you indulge me? In your own words, tell me which of the eight statements, beginning with '2/3 of the time I will choose a goat' and ending with 'Therefore, by switching, I increase my chances of getting the car when offered the switch.' is either untrue, invalid, or not germain to the question 'Now that Monty has revealed a goat, do you increase your chances of winning a car by switching'?

Thank you. Glkanter (talk) 07:06, 7 February 2009 (UTC)

Right explanation

I'll do one effort to explain the problem. The situation is already conditional on the choice of Door 1 by the player. And we know Door 3 has been opened revealing a goat. Let us assume the game has been played 3000 times. What might have happened?

Player has chosen Door 1
Car hidden behind Door 1		Car hidden behind Door 2	Car hidden behind Door 3
1000 times		1000 times	1000 times
If nothing else is revealed
Player wins the car if he doesn't switch		Player wins the car if he switches
But player gets "information"
Host opens Door 2	Host opens Door 3	Host must open Door 3	Host must open Door 2
500 times	500 times	1000 times	1000 times
This case didn't happen	One of these cases happened		This case didn't happen
	Player wins the car if he doesn't switch	Player wins the car if he switches

Hence only in half the cases did the situation arise in which the player finds himself. So we see it cannot be unconditional, because only a part of all possibilities could have happened. Clear?Nijdam (talk) 18:05, 6 February 2009 (UTC)

Conditional/unconditional

Rick, I thought that it had been agreed that, for the fully defined problem (host alway offers the switch and always opens a goat door), there is no distinction between the conditional and unconditional cases. Martin Hogbin (talk) 22:56, 6 February 2009 (UTC)

In cases of mathematics the right solution is not reached by agreement. I hope I explained clearly enough why the stated problem requires a conditional solution and why this solution differs from the unconditional one (not numerically in the final answer, but that's not the issue).Nijdam (talk) 23:56, 6 February 2009 (UTC)

Martin, no, I don't think there was ever any agreement that in the fully defined problem there is no distinction between the conditional and unconditional cases. The fully defined problem ensures the numerical probabilities are the same, but these remain distinct questions requiring different solutions.

Nijdam - I have said repeatedly I'm OK with starting with an unconditional solution so long as it is clearly identified as an unconditional solution, doesn't actually say anything that's incorrect, and is immediately followed up with a transition to the more accurate conditional solution. My rationale for including an unconditional solution is because many published sources (possibly all popular expositions of this problem) include only an unconditional solution. I think not providing one here would be a significant omission. How to structure the article is an editorial decision, which is reached by consensus. We want the article to be accessible to laypersons, but also correct, which I think sometimes calls for compromises. -- Rick Block (talk) 05:03, 7 February 2009 (UTC)

Well, to be honest, I'm not. I strongly oppose starting with a reasoning that is not a solution to the stated problem. The right solution, as I suggested above, is quite accessible to the layman. If anything has to be said about the "simple solution", it should be as a solution to a (slgihtly, but essentially) different problem. It doesn't seem right to me, to have thoughts like "keep the laymen in the dark, it is no use to explain things properly to them".[Sorry, forgot to login] Nijdam (talk) 12:59, 7 February 2009 (UTC)

Nick, in what way are the conditional and unconditional cases distinct in the case of the fully defined problem? Martin Hogbin (talk) 14:44, 7 February 2009 (UTC)

Conventional Wisdom

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

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

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

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

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