Talk:Monty Hall problem

"The Paradox" section

In order to help the reader to clearly distinguish the standard paradox's premise and quite other problems, I propose to say in section "The Paradox" as follows:

"The correct answer, that swapping doors doubles the chance on the prize because players who swap have 2/3 chance of winning the car whereas players who stick only have 1/3 chance, is based on the premise that the host knows which door hides the car and, after the player's initial selection, intentionally reveals a goat from behind one of the two remaining unselected doors. If the player by luck selected the door with the car (chance to be in that lucky-guess-scenario irrevocably is 1/3 only), then both remaining doors hide goats and the host may choose either door at random, and switching doors loses. But if the player initially selected one of the two doors each hiding a goat (irrevocably 2/3 chance to be in such wrong-guess-scenario), then the host's choice is no longer at random, as he has no options but to show his only goat, the second goat, and is forced to offer his still closed door that hides the prize. So in both wrong-guess-scenarios, switching doors wins the car for sure (Mlodinow 2008).

Consider a completely different problem that contradicts the assumptions of the standard paradox: the host, instead of intentionally revealing a goat, reveals one of the remaining two doors at random. If the host opens a door at random and reveals a goat simply by chance, then the odds are reduced from the standard paradox's 2:1 in favour of switching to 1:1 only. The latter odds track the common intuitive wrong answer because half of the potential winning cases are wasted when the host accidentally reveals the car, and by that discards such plain winning event (Rosenthal2005a) (Rosenthal, 2005b)."

Would be great if you could help to say that crystal clear in plain English. Gerhardvalentin (talk) 21:48, 23 November 2013 (UTC)

Two points of clarification:

1. I have placed the above post under a new heading because it appears to be a separate issue, unrelated to use of the word "ludicrous" in the lede section, which was the subject of the thread above.
2. Much of this language already appears in the abovementioned "The Paradox" section. It would be helpful to state specifically what you propose to change, and why you think it would be better.

(Cf. a similar proposal a few months ago.) Thanks. ~ Ningauble (talk) 14:54, 24 November 2013 (UTC)

?? – Clarification: Please read the wording of my above proposal, that's exactly what I propose to change. IMO this is better for the reason I also called above:

• In order to help the reader to clearly distinguish the standard paradox's premise and quite other problems.

And there are lots of reasons, see the sources:

1. E.g. with one host and two players that simultaneously both do select different doors, say #1 and #2, and the host opens #3, the only unselected door. And there is quite a lot of other aberrant "problems". Since, you should preclude and obviate.

2. In first selecting a door, the player knows that in "exactly 1-out-of-3", he inevitably will be in the lucky guess scenario. And in "exactly 2-out-of-3" he inevitably will be in one of the two wrong guess scenarios. This rate of 1/3 : 2/3 can never be changed nor updated for the actual game being, this rate remains immutable. Please note that this is not only "an overall rate", no, this immutably is the actual rate of any single game. So we should preclude and obviate. Consider that even M et al's host can never change that rate, he just could be taken to give further advice as to the actual scenario the player actually is in, e.g. in disclosing that the player actually is in one "wrong guess scenario" for sure, and thus the "conditional probability" to win by switching approximates 1. But no-one, neither M et al's host, can change the rate of 1/3 : 2/3. And it is wise to help the reader to be aware of this fact, just from the beginning. No OR, see Mlodinow who has been tagged as careless. Once more: This ratio of 1/3 : 2/3 is invariable, for any single actual game.

3. And it is wise to help the reader to see just from the beginning that, in both wrong guess scenarios, the host never is "eqully likely", but is forced to offer his door with the car behind. Yes, "is forced". Despite I constantly heard that this was OR.

So please read again what I wrote above: In order to help the reader to clearly distinguish the standard paradox's premise and quite other problems.

I ask you to help to improve that important section, it should be as crystal clear as possible and, just from the beginning, help to avoid any possible ambiguity. Yes, we should preclude and obviate. Regards, Gerhardvalentin (talk) 22:22, 24 November 2013 (UTC)

I think that the present text in the article is perfectly adequate. It makes exactly the same points as you want to make, and it makes them more clearly (it is much shorter, and it is written in more or less decent English). Richard Gill (talk) 16:07, 26 November 2013 (UTC)

This probably seems so obvious to me. It is entirely due to how the host is forced to act in only one case:If you pick door 1, and it is door 2, the host is forced to pick 3, therefore you should always switch to door2 to take exploit the advantage of this..

(12/11/2013) — Preceding unsigned comment added by 99.232.160.130 (talk) 19:46, 11 December 2013 (UTC)

Yes, exactly. For this famous paradox it helps to say that there are exactly three possible scenarios: the one and only "lucky guess scenario", where the contestant by luck first selected the door with the prize behind (in only 1 out of 3 cases) and switching doors will lose the car. In this scenario both unselected doors hide goats and the host may randomly open any of them.

But in both "wrong guess scenarios", where the guest first selected one of those two goats, the host obviously is forced to show the other goat, and to offer a switch to his door that definitely hides the car.

And please consider that in the beginning, the contestant knows exactly that his first choice has 1/3 chance only. And as to that famous paradox, after the host has shown a goat and offered a switch, the contestant still knows that his first choice unchanged has only 1/3 chance, so switching doors doubles his chance to win the car. Gerhardvalentin (talk) 20:48, 14 December 2013 (UTC)

And all this is written in the article! Readers who read carefully can find everything they need. It does not need to be repeated again and again in different words. Wikipedia is not a text book, it is not a pedagogical guide, it's an encyclopedia. The purpose of wikipedia is not to teach but to help people find information. Richard Gill (talk) 20:52, 17 December 2013 (UTC)

Simpler more intuitive explanation

The contestant chooses a losing door in two thirds of all games played. The winning door is chosen in only one third of all games. Therefore, in only one third of all games is the winning door and the chosen door the same. Since the host never opens the chosen door nor the winning door, it is only in these one third of all games that the host has the option of opening either of the two remaining doors. Whereas, in two thirds of all games, the chosen door and the winning door are different doors leaving the host only one door to open. In these two thirds of all games, the door which the host leaves unopened is the winning door. In only one third of all games played, does switching lead to the wrong door. In two thirds of all games, switching leads to the winning door.

If others agree this is a more intuitive explanation which leads readers to the correct conclusion more quickly, please comment and replace the paragraph below which appears as the fifth paragraph in the article with the paragraph above.

"Contestants who switch have a 2/3 chance of winning the car, while contestants who stick have only a 1/3 chance. One way to see this is to notice that, 2/3 of the time, the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. "Switching" only fails to give the car when the player picks the "right" door (the door hiding the car) to begin with. But, of course, that will only happen 1/3 of the time."

--75.199.38.85 (talk) 14:09, 20 December 2013 (UTC) --Jerrykeywest (talk) 14:16, 20 December 2013 (UTC)

• Keep existing. I like the existing version; its topic sentence gives good guidance. The proffered replacement does not. Glrx (talk) 00:08, 21 December 2013 (UTC)
• I too prefer the existing wording. Someone tried rewriting it somewhat like the above but only made it less clear and obvious, so I've restored the original wording.--JohnBlackburne^words_deeds 23:59, 21 December 2013 (UTC)

Sample space question

In this problem, is it correct to say that the sample space is this:

1. At 1st choice; 1 car, 2 goats, 3 doors.

2. At 2nd, choice; 1 car, 1 goat, 2 doors.

In other words, does the Sample Space shrink when the host removes a door from the available choices?

Tweedledee2011 (talk) 19:05, 22 December 2013 (UTC)

NO; read again or consider remedial mathematics. Cheers Crusoe8181 (talk) 09:40, 24 December 2013 (UTC)

Not a very helpful or polite comment. Martin Hogbin (talk) 00:26, 26 December 2013 (UTC)

I'm not sure you understand my question. A sample space is the set of all possible outcomes of an experiment. When a door is removed by the host, it ceases to be part of the sample space, because it's no longer choose-able and if it's no longer choose-able, it's not part of the second experiment. When that happens, you are left with 2 doors, 1 goat and 1 car. Therefore, if door #3 is removed, the remaining possibilities are D1G D1C D2G D2C. The sample space at the point of the second choice is two possible positions for either the goat or car. The first experiment results in us getting an unknown outcome from an event. We choose a subset of the sample space (a door), but we don't yet look behind it. Then comes another experiment. This is when we make another choice, and switch (or stay). This event results in a knowable outcome because as part of this event, we open the door. Suffice it to say, this question is dishonest because the first event does not come to fruition - the outcome does not become known as a result of the event. Rather, the outcome of the removal of the door is pre-determined by a prior event - the prior inspection of the door contents by the host to verify that only a goat door will be removed. What this problem actually does is inject a player into a partially completed series of experiments, and then illegitimately allows the player to calculate a choice based on the prior knowledge of the host - which was acquired externally to the experiments conducted by the player. In other words, the "winning" method requires reliance on information acquired outside of the boundaries of the player's experiments. Not only that, but by removing a door, that door can no longer be chosen, so it's not possible that the removed door has the car. As a result, to include any math related to the state of the removed door, is to calculate odds based on information which depends on a set that no longer exists - the set of three doors. At the point of the second choice, the removed door is no longer part of the set which available to choose from, so it's not part of the sample space. Hence the sample space of the second choice must only be D1G D1C D2G D2C. That's two doors, and the possibility that either goat or car may be behind either of them. The odds of the second choice, if calculated using only the possibilities available to the second event, are 50/50. Tweedledee2011 (talk) 20:37, 24 December 2013 (UTC)

Please read again section "The paradox". Your reasoning does not address the famous paradox, where the host opens one of the two REMAINING doors in order to show a goat, AFTER the guest already has made his first decision – with a chance on the car of only 1/3 (in that 1/3 staying wins), but with a chance on a goat 2/3 (in that 2/3 switching wins).

You refer to another variant (3 doors, originally 1/3:1/3:1/3) where a goat had been removed BEFORE the guest afterwards makes his first decision out of only 2 doors (1/2:1/2). In your aberrant variant, chances of sticking:switching indeed are 1/2:1/2. Please read the introduction. Gerhardvalentin (talk) 16:38, 25 December 2013 (UTC)

Perhaps I did not make clear what I am saying. Allow me to simplify: At the point in the game, where the door is removed, the removed door ceases to be part of the sample space of the game. That is what I am saying. Why do I say this? Because of the very definition of sample space which is "A sample space is the set of all possible outcomes of an experiment". This game is not one single experiment. It is a series of experiments. The knowledge acquired about the true door contents, is acquired by the host prior the beginning of the game, that's experiment #1 - this is when he familiarizing himself with all the doors/contents. The first choice of a door by a player, that's experiment #2. But experiment #2, unless the player sticks with that door, never comes to fruition, because it's not opened. After the player makes choice number one, the host removes a door - but that removal is NOT an experiment. The information required by the host so as to make a removal of a goat-door was acquired in experiment #1, prior to the start of the game. What I am claiming is that the goat-door removal by the host, reduces the size of the sample space because the player is NOT allowed to choose the removed door, therefore the removed door DOES NOT hold any possibility of having the car (or goat) for the player. To be part of the sample space, a door must have a bona fide possibility for both yes and no. The sample space at the player's first choice is D1G D1C D2G D2C D3G D3C in however a pattern you want to arrange it, provided that you limit the car to one unit and fill the other two with goats. Those are the total possibilities in the starting sample space (set of possibilities). However, when a door is removed, if the rules of the game are followed, the removed door CAN NOT have a C or G possibility for the player's next choice. Therefore, since the removed door is not available to the player, it's not possible that it's part of the game's sample space. The only doors which can be counted as part of the sample space of the game, are doors which have both POSSBILITIES to the player - which requires that they be available to be chosen. Any door which is not available to be chosen, fails the predicate definition of Sample Space and is therefore excluded from the game's sample space at the moment it fails the test. It's not the fact that removed door #3 has a goat which makes it fail. Rather, it's the fact that it's not choose-able which makes it fail. It fails the definition of possible, therefore it's not part of the player's sample space after it's removed. And to be extra clear, let's be sure to call each of the player's chocies what they are: Events. And since an event is a subset of a sample space, then if at the point of the second choice, the 3rd door is not part of the set from which the choice can be made; it's not, and can't be part of the event which is the second choice. My contention is that the removed door, because it's not available to the player to choose, is not part of the player's sample space. In other words, the size of the sample space available to the player changes when the door is removed. This is what I am saying. Tweedledee2011 (talk) 19:40, 25 December 2013 (UTC)

Tweddledee, I think you are possibly misunderstanding the concept of a sample space. A sample space is, 'the set of all possible outcomes or results of that experiment'. Follow the link to see what is meant by 'outcomes'. Martin Hogbin (talk) 00:31, 26 December 2013 (UTC)

Martin - the wiki article about 'outcomes' which you directed me to, says the following:

'In probability theory, an outcome is a possible result of an experiment'.

If that is an accurate statement, then it proves my point, which is this: The total possible outcomes from the second choice is less than the total possible outcomes of the first choice, because after the first choice and prior to the second choice, two of the possible outcomes are removed from the game. The removed outcomes are D3G and D3C. If D3 is removed from the game, it's not possible for the player to choose it, so anything it contains is not a possible outcome of the second choice. The player can not conduct an experiment on the 3rd door - because it's removed. If the player deduces information about the third door, that information was not arrived at as a consequence of the player's experiments. Rather, it's only if the player knows that the host will not remove the car door, that the player acquires information about the third door. That information, again, was not arrived at from the player's experiments. Rather, as I said above, it precedes the player's actions in the game. For this trick question to work, it must rely upon information which was developed before the start of the game, by the host's experiment (see above). Again, let's return to my question: Does the removal of the door reduce the size of the sample space for the player's 2nd choice? I say it does, and if it does, then playing this game honestly would require disregarding any information known or deduced about the contents of door #3. This is why if a new person comes in and chooses, without any information, there is no advantage to switching. The advantage to switching lies in the knowledge about the 3rd door. But to be intellectually honest, one has to admit that it's illicit to use that knowledge if door #3 is removed from the game - because door #3 is no longer on the playing field - and it's no longer part of the player's sample space. Tweedledee2011 (talk) 04:03, 26 December 2013 (UTC)

This illustrates sample space size reduction from door removal

Tweedledee2011 (talk) 07:03, 26 December 2013 (UTC)

You are correct that the host opening a door reduces the sample space. Looking at this in more detail (assuming the player picks door 1) the sample space before the host opens a door is:

D1=C, D2=G, D3=G, HostOpens=D2

D1=C, D2=G, D3=G, HostOpens=D3

D1=G, D2=C, D3=G, HostOpens=D3

D1=G, D2=G, D3=C, HostOpens=D2

The probabilities of these are 1/6, 1/6, 1/3, 1/3 (in the case the car is behind door 1 the host can open either door 2 or door 3, assuming this is an equal choice the probability of either of these is 1/2 the probability the car is behind door 1). If the host then opens door 3 the remaining possibilities are

D1=C, D2=G, D3=G, HostOpens=D3

D1=G, D2=C, D3=G, HostOpens=D3

with original probabilities 1/6 and 1/3. Scaling these up as conditional probabilities (given the host has opened door 3) yields the typically stated 1/3 chance of winning if you stay (with door 1 given the host opens door 3) and 2/3 chance of winning by switching (to door 2 given the host opens door 3).

There being N possible outcomes does not mean the probability of each occurring must be 1/N. -- Rick Block (talk) 22:06, 29 December 2013 (UTC)

Here's my concern: The Monty Hall problem is a deceitful pastiche which does not teach either reasoning or probability very well. If one takes all the information into account, then yes, switching is the best thing. However, if just prior to choice #2, you ask the question "What are the odds of the car being behind either door?", the answer is 50/50. The odds of finding the car takes into account the information derived from the removed door. But the distribution of cars to doors at the point of choice #2 is clearly 1 in 2 - as my sample space diagram shows. The smugness with which the experts explain the "switch is better" option hides the fact that there are multiple sets of data involved here: The original sample space and the smaller, "2nd choice" sample space. This is why, if a new player enters at choice #2, his odds are 50/50 - because he has no access to the information derived from the larger sample space. Tweedledee2011 (talk) 18:32, 30 December 2013 (UTC)

In your question (just prior to choice #2) what do you mean by "what are the odds of the car being behind either door?" If you mean what is the limit of the fraction of times the car will be behind door 1 (or door 2) if you repeat this setup numerous times (I'll assert this is what most people think of as "probability"), the odds are 1/3 for door 1 and 2/3 for door 2. If you mean what is the conditional probability the car is behind door 1 (or door 2) given the player has initially picked door 1 and the host has opened door 3 (knowing it will reveal a goat and choosing door 3 randomly if the car is behind door 1), then the odds are also 1/3 for door 1 and 2/3 for door 2. If you mean what are the odds of selecting the car by random choice (as a new player who enters at this moment must), the odds are 50/50 (since there are two doors). However, this latter meaning is not what most people think of as "probability". The reason the random choice ends up 50/50 is because a random choice between any two alternatives always results in a 50/50 success rate. In particular, if the probability the car is behind door 1 is 1/3 and the probability it is behind door 2 is 2/3, then the overall success of a random choice is (1/2)*(1/3) + (1/2)*(2/3), which is 1/2! This is not saying anything about the "probability" (in either of the other two meanings I've suggested) of the car being behind each door - just that the chance of picking the car with a random choice is 1/2.

Lets say you are at a university where you know 90% of the students are male and 10% female. If you run into a random student what are the odds this student is male? Is it 50/50 because there are only two choices (male or female)? If you encounter 100 students and flip a coin to guess male or female (lets say heads means your guess is male), you'll be correct about 50% of the time. However, I think most people would say the odds are 9/10 male and 1/10 female (of the 100 students you encounter, about 90 will be male and about 10 will be female). The Monty Hall problem is similar. At choice #2 the car really is twice as likely to be behind door #2 than door #1! Try the card experiment - see for yourself. Take the ace of spades and 2 red 2s. Shuffle. Deal 3 cards. Look at the 2nd and 3rd (the player picked #1). Turn over a red 2 (flipping a coin to decide which one to turn over if both are red 2s). If you happen to have turned over the 3rd card (host opened door 3), make a note of which card is the ace (#1 or #2). Do this until you have about 30 samples (more is better). You'll see the ace will be card #1 about 1/3 of the time and #2 about 2/3 of the time (as you do more and more of these experiments the proportion will become more and more exact). Note that it is important always to look at both cards and actually flip a coin to pick which one to turn over if both are 2s (lets say heads means you flip over #2). If you just look at #3 and flip it over if it's a 2, the experiment won't work (you're not following the proper rules). You have to both flip over a card you know is a 2, and pick evenly between them if both are 2s. If you do this experiment, please report back what you find. -- Rick Block (talk) 21:16, 30 December 2013 (UTC)

Ahem. Paradoctor (talk) 22:01, 30 December 2013 (UTC)

The problem with this question, is how the vernacular trips people up. The odds of finding it are not the same for the original player as they would be if a new player came in just prior to choice #2. This is because a 2nd player coming it at that point would not have the information gleaned from the larger sample space which existed prior the door removal. And, even though if this series of events were not interrupted, then yes, switching would be better, even for a new player, the new player could not know that - because he lacks the information. What I am saying is that this is a trick question and this article is weak, because it omits a proper explanation of "sample space" and "experiment". If each choice is a separate experiment, then the reduction of the size of the sample space must be taken into account. This is so because in the real world, when calculating probabilities, there is no perfect knowledge of the sample space. For that reason, this question (and article) deludes people into misapplying the parameters of a true probability event. For something to be part of a sample space, it has to hold the possibility of being true - but the removed door fails this test. By restricting the removed door to only goat doors, it's not possible to say that the removed door is part of the sample space. And if it's not part of the sample space, then making calculations based on knowledge of it, is outside the scope of the second experiment. That is, if the 2nd experiment (2nd choice) is discrete. So then, perhaps that's the real question: From a standpoint of calculating the event of the 2nd choice, is the 2nd choice a discrete choice or is it part of a series of choices? If it's a discrete choice, then it can't use the information gleaned by the door removal as that act pre-dates the experiment which is the event of the second choice. By definition, an "event" is a subset of a "sample space" and at the point of the 2nd choice, the removed door - and all its possibilities are no longer part of the sample space. And as such, it's axiomatic that the removed door - and all its possibilities - must be excluded from the 2nd choice. This question isn't about the math, this question is about how to honestly explain what logic-rules should be used to apply the math, and how, at which stage in the process. Tweedledee2011 (talk) 23:25, 30 December 2013 (UTC)

Since this discussion is not apparently about the article but rather the math behind the problem, it is probably more appropriate for the /Arguments subpage. We could move it there if anyone here is unduly bothered. -- Rick Block (talk) 22:29, 30 December 2013 (UTC)

I'm not happy with the suggestion that I am "unduly" concerned.

If the discussion is on topic, it belongs here. If it's not, it doesn't belong anywhere on Wikipedia, per WP:NOTFORUM. Paradoctor (talk) 02:31, 31 December 2013 (UTC)

Please do not move this discussion. It's needed here to clarify whether or not the change in the sample space should be fully explained in the article. Tweedledee2011 (talk) 23:09, 30 December 2013 (UTC)

If we're talking about the article, and not the math behind the problem, then the first question has to be what references are you talking about? If there are no references supporting your point of view, then the discussion is over. If there are references, then the question becomes how (or possibly even whether) to include what they say into the article. -- Rick Block (talk) 23:54, 30 December 2013 (UTC)

I have linked to event and sample space - are those enough? Tweedledee2011 (talk) 00:05, 31 December 2013 (UTC)

No. You need reliable sources discussing sample space in the context of the Monty Hall problem, otherwise this is WP:OR. Paradoctor (talk) 01:27, 31 December 2013 (UTC)

I don't understand you saying that. It's axiomatic that sample space is an essential aspect of a probability calculation, so to omit any reference to sample space in an article about the most well-known probability brain teaser seems needlessly obtuse to me. Tweedledee2011 (talk) 01:39, 31 December 2013 (UTC)

I agree with Paradoctor. A general discussion of events and sample spaces as applied to the Monty Hall problem would be original research (OR) without a specific reference. Since there are hundreds of published articles and multiple books specifically about and covering every conceivable aspect of the Monty Hall problem, virtually any approach (or any thought about it) has already been published - and if it hasn't been published it's OR (pretty much by definition).

Another approach for you here would be to suggest a specific change - exactly what are you suggesting be changed and in exactly what section of the article? For example, are you suggesting expanding the section "The little green woman"? Perhaps I'm misunderstanding, but it sounds like you're suggesting adding something that says the probability is 50/50 after the host has opened door 3 (when the player is deciding to switch from door 1 to door 2) because there are only two possible locations for the car. There aren't any reliable sources that say this (it's not correct), so if this is what you're looking for I think you're going to be disappointed. If this is indeed what you're thinking I'd be happy to discuss the math with you on the /Arguments subpage if you'd like. Rick Block (talk) 01:47, 31 December 2013 (UTC)

No what I am saying is that this article, by omitting any explanation of 'experiment', 'sample space' or 'event' leaves novice readers at a disadvantage. This article, as a matter of editorial choice, ought to explain those concepts. As it stands, even some of the questions I asked here have not been answered. For example, if the sample space shrinks upon door removal, how can one honestly say that the 2nd choice can calculate in the information which the removed door provided? In other words, a door choice is an "event" and an event is a subset of a sample space. But the event of the 2nd choice is not a subset of the original sample space. Rather, it's only a subset of the reduced sample space. And the reduced subset, I contend, is a wholly distinct sample space from the original one, not a subset of it. The removal of the door from the original sample space destroys that sample space because by definition, a sample space is all the possible outcomes of an experiment (or trial), but when you open a door, you eliminate that door as a source of possible outcomes for the 2nd choice. This is why I'm saying the article doesn't explain things well. It might very well be a distinction without a difference, but I still think it should be explained how we are still able to calculate the odds, even though the sample space is modified. We are able to do so, essentially, by thinking "outside the box" at the point of choice #2. That's because it's literally true that the information derived from the removed door of the original sample space is from a transient superset which no longer exists (the original sample space). Tweedledee2011 (talk) 02:20, 31 December 2013 (UTC)

Please use proper WP:indentation.

"the reduced subset, I contend, is a wholly distinct sample space from the original one" (my emphasis) That is the problem. What you contend is not encyclopedic. Is there a reliable source saying this? Paradoctor (talk) 02:37, 31 December 2013 (UTC)

Paradoctor: I don't understand your point. Either you agree or disagree. If you agree, then help me bring my contention into line by finding a good source. The absence of a source is why it's on this page - to be discussed. But discussing it means just that - discuss it - not reject it out of hand merely because it's not sourced yet. If it's not a distinct sample space, then why not? And if is, then help me find a good citation. Tweedledee2011 (talk) 02:49, 31 December 2013 (UTC)

Quote from talk page guidelines: "Use indentation as shown in Help:Using talk pages#Indentation (or, more specifically, Wikipedia:Indentation)". Please note that a guideline is "a generally accepted standard that editors should attempt to follow".

"bring my contention into line by finding a good source" WP:BURDEN. Furthermore, I don't think there is such a source.

"Either you agree or disagree." My opinion on this matter is just as uninteresting as yours here. We report what the sources say, not our own musings. Paradoctor (talk) 03:41, 31 December 2013 (UTC)

(responding to Tweedledee) A full decision tree enumerating all events that might happen is presented in the Grinstead and Snell book (an elementary probability textbook in the article's references, available online). Although presented there as a decision tree, this tree is for all intents and purposes the same thing as a sample space. A similar tree (from both Carlton and Chun, also in the references) showing only the events following "player initially picks door 1" is presented in the article in the "Conditional probability by direct calculation" section. If you're having trouble understanding the text in this section or simply think it's unclear, it's fair to ask here for the text to be clarified. I hope you don't take offense, but it seems like you're having trouble understanding the basic concepts of sample spaces and how they change in response to events. This sort of discussion belongs on the /Arguments subpage as this article is not a tutorial on basic probability theory. And, perhaps a more direct response, the article does not mention sample spaces because that's not how most references approach the problem. -- Rick Block (talk) 06:09, 31 December 2013 (UTC)

(responding to Rick Block) I think you are mistaken that a sample space changes in response to an event. Think about the search for the SS Central America. The search method they used took into account not only the probabilities of the wreck being in various quadrants in the search grid, but also the likelihood of actually being able to successfully locate it in the grid, even if the correct grid was searched. Suffice it to say, unless and until the wreck was found, then in regards to any particular quadrant, unless the searchers were certain (after inspection) that the quadrant was empty, then it wasn't eliminated from the sample space. But what is it that enables quadrant elimination? Inspection, that's what. However, in this game it's not an inspection by the player which eliminates the door. Rather, it's the prior knowledge by the host which eliminates the door. The actual opening of the door is a perfunctory action to prove that removal is valid (i.e.; has no car). Suffice it to say, it was not the action of the host's inspection of the door which eliminated it. Rather, it was the action of the host to deem the door eliminated which does that. An "event" is not an "action" in this context of this vernacular. Rather, an "event" is a subset of a sample space. But the subsets are entirely arbitrary and do not eliminate themselves from further consideration. Rather, it is our confidence in the accuracy of the information yielded by the experiment (look and see) that leads us to eliminate things from further consideration. However, in the game, it's not the player who eliminates the door, it's the host. And if the player is not aware of the fact that he can use the derived information about what the removed door does not contain (the car) to best formulate his next choice, then he's not able to understand that his results can improve by switching. So without beating a dead horse, I still think that this article could benefit from explaining the changing composition of the sample space and how a wise player can, even if the samples space shrinks, use the information from the previously large sample space (and removed door). In other words, because this article is written by those well versed in the topic, it overlooks that novices might like to see information about sample space in this article. Tweedledee2011 (talk) 07:38, 31 December 2013 (UTC)

Four users have told you that you're wrong. Two have told you that this discussion doesn't belong here. You have been offered to continue the discussion at /Arguments. Further edits in violation of WP:TPG will be deleted. How much farther you escalate this is up to you now. Paradoctor (talk) 07:57, 31 December 2013 (UTC)

Tweedledee, although I do not agree with some other editors about absolutely everything, we all agree that, in the standard interpretation of the problem, the probability of winning by switching is 2/3 and not 1/2. This page is for improving the article but you have not found any valid reason to make changes to the article so further discussion here is pointless.

If you want to discuss the underlying theory of this problem you are welcome to continue this discussion at the /Arguments page, where I and others will be happy to discuss more general issues of probability with you. Martin Hogbin (talk) 09:56, 31 December 2013 (UTC)

2 small corrections?

After rereading the current article I noticed 2 things that appear to be incorrect/incorrectly sourced.

• In the lead there is currently the following line: "The argument relies on assumptions, explicit in extended solution descriptions given by Selvin (1975a) and by vos Savant (1991a)". However as I understand Selvin is giving the extended solution description, in particular the explicit mentioning of the assumptions, in his 2nd letter only. Hence the source for that sentence should be (1975b) rather than (1975a) or at least (1975a, 1975b).
• The Recent discussion section lists several publications that have referenced the WP article on MHP. It states: "Nowadays, Wikipedia is frequently cited as a source for Monty Hall problem (it is cited, for instance, by Ruma Falk, Rosenhouse 2009, Gill 2012, Gnedin 2012).". The only publication by Falk given in the References section however is an article from 1992, which certainly doesn't reference WP (only being around since 2001). So if there is another (later) publication by her which indeed references WP, then it should be added to References section otherwise she should be removed from that enumeration.

--Kmhkmh (talk) 09:36, 31 December 2013 (UTC)

 Done Just out of curiosity: Is there a specific reason you didn't do it yourself? Paradoctor (talk) 16:56, 31 December 2013 (UTC)