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Latest comment: 8 years ago118 comments9 people in discussion
OK, thanks, moved to here. Quoting our exchange so far from the MHP page:
"I suggest the word 'always' is removed from the Standard Assumptions. If taken to mean 'every time the MHP is stated' it is redundant, and if 'every time the game show plays the three-door game' it is incorrect. In the citation 1991a, vos Savant uses 'always' only in the context of the six games that exhaust all the MHP possibilities. If the host acts non-randomly, the MHP odds apply in those games where he reveals a non-chosen goat even if in other games he reveals the car, the chosen door or nothing.Freddie Orrell (talk) 06:34, 11 October 2015 (UTC)
I disagree, per the sources that are cited in this section. And it is not correct that "the MHP odds apply in those games where he reveals a non-chosen goat even if in other games he ...". See, for example, the variant Monty Hall himself apparently created on the spot during a 1991 interview with John Tierney from the NY Times (cited in the article). In this "Monty from Hell" variant, switching always loses. If you'd like to discuss this I suggest we move to the /Arguments page. -- Rick Block (talk) 16:16, 11 October 2015 (UTC)"
I belive variant host behaviors based on a motive to either help or hinder the contestant, have no bearing on the MHP solution. In the absence of any being stated in the MHP wording we may assume there are none. Alternatively, if we nevertheless suspect a motive there is no indication of its nature; there being an equal and opposite possible helpful variant for every possible hindering one, the uninformed contestant is randomly affected in equal measure so their decision is unchanged. Regarding the word 'always', the MHP solution turns on the host acting non-randomly, not consistently. Thus if the circumstances arise of a non-chosen goat being deliberately revealed and a switch offered, it's 2/3 by switching; the host could have revealed the car and said 'So long, sucker'. The latter might be termed a variant host behavior, but is more properly a different scenario to the MHP. In a series of games on the show there could legitimately be some variant scenarios and some MHPs, but the host need not 'always' act the same. I don't think the cited sources use 'always' in quite the way implied by the Standard Assumptions quoted.Freddie Orrell (talk) 18:26, 11 October 2015 (UTC)
Discussion with Rick
The question is what do you mean by "probability" and from whose perspective? You are distinguishing deliberately revealing a goat (implying the host has not opened one of doors 2 and 3 at random and has just happened to reveal a goat) from other host behaviors. Why? How is the host randomly opening a door and revealing a goat different from the player's perspective? To some extent I agree that from the player's perspective, if nothing is said about the rules governing the host's behavior, perhaps the "best guess" is 2/3 chance of winning by switching but this is a distinctly Bayesian view of probability. To make the number of times switching wins (given that the player has picked door 1 and the host has opened door 3) approach 2/3 over a large number of samples (a frequentist view of probability), the host must not only open a door non-randomly (deliberately show a goat), but the host must also always make the offer to switch, and must randomly pick which door to open if the player has initially picked the car. Without these constraints, the frequentist probability (the fraction of times switching wins over a large number of samples) can approach anything from 0 to 1.
As to what the specific sources that are cited in this section of the article say, have you read them? For example, Krauss and Wang provide what they call a mathematically explicit version that includes all of the "standard" assumptions (they don't use the word "always" but instead make the same conditions rules governing what the host can do - which amounts to the same thing). Most of the early discussion following vos Savant's initial article was about whether the host always opens a door showing a goat and always makes the offer to switch. In her later columns she clarified these points. Selvin responded to letters about his initial problem with the same conditions (adding that if the player initially picks the car, the host picks from among the remaining two choices randomly).
These are points that have been previously argued here extensively. My advice is that if you want to pursue this, then find sources supporting your view. -- Rick Block (talk) 04:27, 12 October 2015 (UTC)
Imagine you are asked the MHP question. Can you not answer it accurately without inquiring 'Did the host also reveal a goat in yesterday's show?' And what if he did not?Freddie Orrell (talk) 19:27, 12 October 2015 (UTC)
No, you cannot answer it accurately without knowing more, or assuming more, about the host. I don't know how much math you know, but conditional probability is the relevant area of probability theory. In particular, assuming you've picked door 1 then the probability the car is behind door 2 given the host has opened door 3 is
P(car behind door 2) * P(host opens door 3 given the car is behind door 2) / P(host opens door 3)
What the standard assumptions do is make
P(car behind door 2) = 1/3 [pretty much everyone assumes this]
P(host opens door 3) = 1/2 [the host is equally likely to open door 2 or door 3 - this requires some thought]
P(host opens door 3 given the car is behind door 2) = 1 [the host MUST open door 3 if the car is behind door 2 and you've picked door 1]
So (with these assumptions) the probability the car is behind door 2 given the host has opened door 3 is (1/3 * 1) / 1/2 = 2/3, which is the "normal" solution. Without these assumptions or with different assumptions, we can make the probability the car is behind door 2 anything between 0 and 1. For example, in the "Monty from Hell" variant I referred you to, P(host opens door 3 given the car is behind door 2) is 0 (if the car is behind door 2 Monty doesn't open a door and doesn't give you a chance to switch), which makes the probability the car is behind door 2 given the host opens door 3 equal to 0 (i.e. you lose if you switch).
There are other approaches using, for example, Bayesian probability - but if you are ending up with a 2/3 chance of winning by switching you're effectively making the "standard" assumptions (one way or another). -- Rick Block (talk) 22:39, 12 October 2015 (UTC)
Thanks for that answer, during which I was already typing. So firstly, to respond specifically to your last-but-one post above:
I mean the probability of choosing a goat or the car from the contestant's perspective.
I am distinguishing deliberately revealing a goat from other host behaviors because the contestant can infer the former (by the information that he knows what's behind the doors being superfluous if he is random) while the latter cannot be inferred from the information given.
Randomly revealing a goat would look no different, but the contestant has the information to infer it is not random on this occasion.
The Bayesian may calculate (not guess) 2/3, as the set of all possible unstated hindering variant behaviors and the set of helpful ones are equal in likelihood and opposite in value, thus neutral to the solution. Although I think the absence of any positive information means the assumption of 'no motive' is a safe one.
The frequentist answering the MHP question (ie 'In the circumstances stated in this scenario, is it to your advantage to switch?') would only count those games where the MHP circumstances apply, and their solution would approach 2/3.
As to the specific sources, making the same conditions is not the same as using the word 'always' - the precise conditions might only occur once in a series of similar games, or even once in all time, with the result on that occasion still 2/3.
I should emphasise that while variant scenarios may occur without affecting the MHP when it occurs, if variant host behavior can be discerned by the contestant a game meeting all the other conditions cannot be the MHP; as vos Savant put it "Anything else is a different question".
Regarding your most recent post, I get basic conditional probability though am a novice in math and stats. The first time the MHP ever occurred the probability was 2/3. If it never occurred again, it was still 2/3. Now every time those conditions come together it's 2/3. If the host revealed the car on yesterday's show but you don't know why, it's 2/3 if the MHP comes together today. I am not disagreeing with the assumptions, or that they 'must' accompany the MHP, just with the word 'always'.Freddie Orrell (talk) 23:33, 12 October 2015 (UTC)
You say above Randomly revealing a goat would look no different, but the contestant has the information to infer it is not random on this occasion. What information? If you're attempting to address this strictly from the contestant's point of view, I think you have to include "host randomly opened a door and happened to reveal a goat" in your analysis - as well as "host only opens a door and makes the offer to switch if the contestant originally picks the car". If the player knows nothing about the host's behavior, it boils down to a choice between two alternatives, i.e. 50/50 (this is vos Savant's "little green woman" example). Even the game theoretic approach (you're trying to win the car, the host is trying to keep you from winning the car - the strategy of picking a random door and then switching wins the car 2/3 of the time) assumes the host always opens a door revealing a goat, and always makes the offer to switch. Are you saying if these conditions are not present, then it's not the MHP? If so, that's essentially what adding "always" is saying (alternatively, what adding "always" is saying is that if these are not restrictions on the host's behavior, the probability of winning by switching may not be 2/3).
And what is the difference between saying the assumptions "must" accompany the MHP as opposed to expressing these as something the host "always" does?
It sounds like you're saying if you don't know what rules the host is following then the probability of winning by switching is 2/3. It's not. It's 1/2. It's 2/3 only if 1) the host must open a door showing a goat (i.e. the host always does this), 2) the host must offer the switch (i.e. always does this), 3) the host must pick which of two losing doors to open randomly (always does this, too). If you don't know #1 and #2 are true, then your chances of winning by switching (from the player's perspective) are 50/50. If you don't know #3 is true (but you do know #1 and #2 are true), then a strategy of always switching wins with probability 2/3 - but the chance of winning if you pick door 1 and see the host open door 3 might be anything between 1/2 and 1 (another way to say this is that the Bayesian odds of winning by switching from the player's perspective in this case are 2/3).
Everything I've said here is in numerous sources which I could provide if you'd like (i.e. I'm not just making this up - it's not "my opinion" but rather what the sources say). -- Rick Block (talk) 06:25, 13 October 2015 (UTC)
I'm sorry, I have evidently not been very clear; I thought I was saying that as the contestant is informed that 'the host ... knows what's behind the doors' they may infer he is acting deliberately and not randomly. This is because the information that he knows would be superfluous if he was acting randomly. It is of course possible that he knows and yet flips a coin to decide what to do, but one might expect more than half that information to be given were that the case. So that's how the contestant comes by the knowledge that in the circumstances of the MHP, the host is revealing the door deliberately.Freddie Orrell(talk) 22:07, 13 October 2015 (UTC)
Got it. How about whether the host is only making this offer because you've originally picked the car? This is the "Monty from Hell" scenario which makes the odds of winning the car by switching 0 (you always lose if you switch). The 2/3 chance of winning by switching depends on the host being required to open a door showing a goat and making the offer to switch. If these aren't required, the odds are not 2/3. If you see this only once, and the rules are not clarified, your chances of winning by switching are indeterminant (could be anything, so 50/50). -- Rick Block (talk) 00:12, 14 October 2015 (UTC)
The set of all possible hindering variant behaviors (eg 'Monty from Hell') and the set of helpful ones (eg 'Angelic Monty') are equal in likelihood and opposite in value, thus neutral to the solution in the absence of any information as to whether any, or if so which, variant is taking place. The assumption of 'no motive' is a safe one, in a single isolated game or any series of games where such information is absent.Freddie Orrell (talk) 06:23, 14 October 2015 (UTC)
Is this your own analysis, or something you've seen published somewhere? -- Rick Block (talk) 15:12, 14 October 2015 (UTC)
No, I'm afraid I have not seen it anywhere else. It is the result of me coming across the MHP over a year ago and buzzing around various online forums with great interest; after I 'got' the MHP I still fell prey to gambler's fallacy among other misconceptions but am trying to learn to develop a more systematic approach using common logic rather than math, in an effort to be more rigorous in my thinking.Freddie Orrell (talk) 20:03, 14 October 2015 (UTC) I have submitted my analysis for publication as I believe it challenges the Standard Assumptions given here.Freddie Orrell (talk) 10:24, 15 October 2015 (UTC)
The formal math approach for the type of reasoning you're doing here is Bayesian analysis which provides a framework to reason about probabilities given various levels of uncertainty. For example, one of the missing assumptions in vos Savant's statement of the problem is that if the player initially picks the door hiding the car the host opens one of the other doors randomly. If the host does not do this the probability of winning by switching (even if the host always opens a door and always makes the offer to switch) is between 1/2 and 1 - but "averaged" across all possible host behaviors the probability is 2/3. This "averaging" is the result of a Bayesian analysis where this specific aspect of the host's behavior is left as a random variable.
If you do not make any of the standard assumptions (in particular, the assumption that the host always makes the offer to switch) the range of host behaviors results in a probability of winning by switching from 0 to 1. You can do a Bayesian analysis of this situation and come up with an "average" probability, but you need to identify what random variable(s) you're considering. For example, one might be the probability of whether the host makes the offer to switch if you've initially picked the car. Another might be the probability of whether the host makes the offer to switch if you've initially picked a goat. Once you've done this, you can compute a probability "averaged" across all these possible host behaviors. Your statement that the "no motive" assumption is neutral and that helping or hindering behaviors are equally likely is an informal argument that could be formalized with a Bayesian approach. I haven't seen anyone attempt this with none of the standard assumptions, but my guess is it ends up with a probability of 1/2, not 2/3. -- Rick Block (talk) 16:00, 15 October 2015 (UTC)
I'm not sure the method of reasoning, whether Bayesian, frequentist or whatever, is material to the assumptions. The MHP might be paraphrased as "If these particular conditions are met, what is the probability?". Your approach appears to be to answer the question "If these conditions are always met, what is the probability?" while mine is to ask "On the occasions when these conditions arise, what is the probability?".Freddie Orrell (talk) 10:29, 17 October 2015 (UTC)
If by "my approach" you mean the Bayesian approach, what it actually does is allow you to compute a probability given any level of uncertainty. If you want to answer "on the occasions when these conditions arise, what is the probability" you still need to understand what you're certain about vs. what you have no clue about (completely uncertain). Are you saying you're certain about only the items in vos Savant's description of the problem (i.e. you're on a game show, there are 3 doors with 1 car and two goats, you've picked door 1, the host knows what is behind the doors, the host has opened door 3 showing a goat, the host has offered you the chance to switch to door 2) but know nothing else? There's a very nice paper by Puza et al [1] that offers a completely general solution to the MHP expressing the probability the car is behind door 2 in terms of 6 component probabilities:
(1) The probability the car is hidden behind door 2
(2) The probability the player picks door 1 given the car is behind door 2
(3) The probability the host opens door 3 given the car is behind door 2 and the player picks door 1
(4) The probability the car is hidden behind door 1
(5) The probability the player picks door 1 given the car is behind door 1
(6) The probability the host opens door 3 given the car is behind door 1 and the player picks door 1
The answer is (1)x(2)x(3) / [ (4)x(5)x(6) + (1)x(2)x(3) ]. With the standard assumptions this ends up being ( 1/3 x 1/3 x 1) / [ (1/3 x 1/3 x 1/2) + (1/3 x 1/3 x 1) ] which is 2/3. If you know nothing about one or more of these, you can still compute an answer using a Bayesian analysis. This is what I said above I haven't seen anyone attempt. For example, Puza et al say "[if we know none of these] your chance of winning the car by switching cannot be determined and could be anything from 0 to 1". This is not quite true, since you CAN use Bayesian techniques to compute an expected probability given complete uncertainty about all of these. I haven't worked through this computation, but (again) my guess is it ends up 1/2. -- Rick Block (talk) 16:58, 17 October 2015 (UTC)
Thanks for the reference; I have previewed the first page free online. Math and stats not being my field, your summary of it is most welcome. I really meant our approach to the conditions, before making any calculations - perhaps I should have paraphrased the MHP as "If these particular conditions are met, and no others, what is the probability?". To allow 'and no others' requires an approach to what we're certain about (what we are told unambiguously, eg one car and two goats; the host opens a goat door) and what we're uncertain about (what we have no information about, and what is ambiguous). As I see it, we have to start somewhere and reject unrealistic possibilities (eg the stagehands switching the contents to bug Monty) otherwise a yes/no answer is impossible and surely the questioner is expecting one by asking 'Is it to your advantage ...'. Since we cannot calculate a yes/no answer with incomplete information, why wouldn't the questioner include all the information we are expected to use? This approach suggests that complete uncertainty actually implies certainty - in the absence of any information, we need not consider the stagehands' trickery as if it is not mentioned we can be certain it is not relevant (by not happening or by being effectively random). That leaves what is ambiguous but necessary to permit calculation, which is where we need to make assumptions. Perhaps the first assumption should be that a car is more advantageous than a goat.Freddie Orrell (talk) 18:47, 17 October 2015 (UTC)
Yes. Everyone assumes the car is more advantageous than a goat. Then, to compute the probability of winning by switching you need the six items listed above. The "random" values for each would be 1/3, 1/3, 1/2, 1/3, 1/3, and 1/2 which makes the probability the car is behind door 2 end up 1/2. This is not quite the same as the Bayesian analysis I've been mentioning, but I believe it ends up with the same answer (with significantly less computation). -- Rick Block (talk) 21:44, 17 October 2015 (UTC)
Discussion with Martin
Freddie, you say 'If the host acts non-randomly, the MHP odds apply in those games where he reveals a non-chosen goat even if in other games he reveals the car, the chosen door or nothing'. If I understand you correctly, that statement is incorrect. It is esential for the given solutions to be correct that the host does not have the option of revealing the car. If we know that the host has chosen a door randomly and on this occasion just happens to have chosen a door hiding a goat, there is no advantage in switching. That is one of the counterintuitive things about the MHP. Martin Hogbin (talk) 15:20, 13 October 2015 (UTC)
May I please ask for your response to the following question (admittedly much less thought-provoking than Parade's version):
'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. 2, which has a car. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?'Freddie Orrell (talk) 19:47, 13 October 2015 (UTC)
Switch my choice to what? Martin Hogbin (talk) 21:57, 13 October 2015 (UTC)
Well, if you like, answer for both cases.Freddie Orrell (talk) 22:22, 13 October 2015 (UTC)
OK I will switch to door 2. Martin Hogbin (talk) 23:04, 13 October 2015 (UTC)
Then you have 100% chance of the car. Now this question: '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, and who on yesterday's show opened a door which had a car, today 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?'Freddie Orrell (talk) 06:23, 14 October 2015 (UTC)
Can you first tell me some things about the setup and rules of the game.
I have seen at least one previous game and know where the car was placed that time. Do I assume that the car is placed in a way that is unknown to me (for example randomly) each time the game is played?
In the previous game, was the player allowed to switch to the door that the host opened to reveal the car? If so is this option always available?
Do I know how the host chooses which door to open? Martin Hogbin (talk) 13:42, 15 October 2015 (UTC)
Do I always get offered the swap?
'Can you first tell me some things about the setup and rules of the game.' I'm afraid not; you only have the information in the question to go on.Freddie Orrell (talk) 15:17, 15 October 2015 (UTC)
Then the answer is then very hard to calculate; to most intents and purposes it is indeterminate. From my state of knowledge about game shows in general I would be completely baffled. It would be a strange game indeed in which the host was allowed to reveal the star prize and the player was then allowed to chose the revealed prize. In such a strange show, maybe the car is always placed behind door 2 or maybe it is placed randomly, who knows. Maybe the host usually offers the swap when the player has originally chosen the car to try and save the production company money, on the other hand maybe he wants to increase the audience by making the player win as often as possible. Oddly enough, a pidgeon might have a good chance of making the correct decision if it was allowed to watch enough shows.
As a Bayesian it is theoretically possible to chose probabilities for all the unknowns facts and then do a complex calculation to get an answer but it would not be simple and I have no interest in trying such a thing. Martin Hogbin (talk) 15:45, 15 October 2015 (UTC)
'Suppose you're on a game show in which nobody involved is following any strategy to either help or hinder the contestant, 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 and decides what to do deliberately rather than at random, 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?'Freddie Orrell (talk) 22:45, 16 October 2015 (UTC)
It might be easier if you explained exactly what you are getting at rather than asking a series of ill defined questions.
I assume that you are asking me to answer your question from the Bayesian (I am happy to use a frequentist approach if you prefer but let us stick to one or the other) perspective of a contestant on the show. To answer that question, we need to agree on exactly what the state of knowledge of the contestant is. Without that agreement the question itself is ill defined and no solution is possible.
So, do we agree on the following:
1 The player believes that the car has initially been placed behind each door with equal probability.
2 The player believes that the swap is always offered.
3 The player has no knowledge of the host's possible preference for any particular door or goat.
What does the player believe the rules to be concerning which door the host opens? Is the player's understanding of the rules that the host must always reveal a goat or does the player think that the host may reveal the car. If the host is allowed to reveal the car, on what basis does he do that?
After a door has been opened, what possible doors does the player believe that he can switch to. Specifically, does the player believe that he may be offered the option of switching to the door that the host has opened.
Without agreeing exactly what the player knows it is not possible to answwer the question. Martin Hogbin (talk) 10:59, 17 October 2015 (UTC)
What I am getting at is explained in my original suggestion above. May I invite you to answer using whatever means you did for the original MHP; my question is identical save for the addition of two items of information.Freddie Orrell (talk) 11:12, 17 October 2015 (UTC)
To answer the original MHP it is necessary to make some assumptions about the rules of the game and what the player knows about those rules and the game. From what vos Savant says about the many letters she received, we know that most people make the same assumptions. These are what are called the standard assumptions in the article. What assumptions would you like me to make in answering your question? Martin Hogbin (talk) 12:18, 17 October 2015 (UTC)
Perhaps my question can be answered using the information given in the question, without any assumptions being necessary.Freddie Orrell (talk) 12:36, 17 October 2015 (UTC)
We already know that is not so. As I said above maybe the contestant knows that the car is always put behind door two. Nearly everybody assumes that is not the case, but without that assumption, the question has a different answer. Martin Hogbin (talk) 13:02, 17 October 2015 (UTC)
If the contestant knows the car is always put behind door two, why don't they just choose door 2 and not switch? Sounds like a sure-fire winning strategy.Freddie Orrell (talk) 13:34, 17 October 2015 (UTC)
Yes, of course it would be with that particular assumption. On the other hand maybe the contestant knows that the car is never put behind door two. With that assumption switching would be an advantage.
I am really finding it hard to understand what point you are trying to make. In any probability problem it is essential to know exactly what the problem is before giving an answer. It may be that some of the assumtions do not matter but in this case some certainly do. To put it another way, if you change the question you might change the answer. Martin Hogbin (talk) 13:57, 17 October 2015 (UTC)
If it cheers you up, I agree that it is true that, given just the problem stated by you, using my knowledege of probability, game shows, and life in general that I would probably switch. We know that there are circumstances where there is a clear advantage to switching (for example MHP with standard assumptions) and there are not many realistic scenarios where it would be a disadvantage to switch, so overall, switching is probably the best strategy. If that is your point then fine, it is a valid observation. However, it has as more to do with real world experience than it does with probability it it is not the reason that the MHP is so infamous. The MHP, when understood as most people naturally do, is a simple brain teaser that most people get wrong, not an exercise in intuitive Bayesian estimation. Martin Hogbin (talk) 15:48, 17 October 2015 (UTC)
The idea was that the information given in the question (and no more) might be sufficient to calculate the solution. If we have to account for everything else possible, say whether the host's breakfast put him in a bad mood so he skews the game, surely we would never be able to answer anything.Freddie Orrell (talk) 16:03, 17 October 2015 (UTC)
I do not think that there is much more that I can say. With no clear question there can be no clear answer. Martin Hogbin (talk) 21:34, 17 October 2015 (UTC)Agreed. With no limit to the possible influencing factors to be weighed, there can be no clear answer.Freddie Orrell (talk) 22:02, 17 October 2015 (UTC)
In real-world cases it is possible to come to some conclusions, using Bayesian methods, based on real-world knowledge and experience. If, for example, I were going on a real game show, I could use all my knowldge from all sources to decide on the best strategy. What the host had for breakfast could be important. Another person may have different knowledge and experience and and may therefore come to a different conclusion.
In the case of mathematical problems, set for students of probability, the aim is to define the problem precisely, so that there is only one correct answer. A combination of clear language, mathematical notation, and accepted conventions is used to acheive this.
In the case of vos Savant's version of the MHP she was writing a puzzle for a general interest magazine. She had in mind a clear and unambiguous problem but unfortunately she did not express that problem as clearly as she might. Despite that fact, we do know, from her own words, exactly what she intended the problem to be and that that is how the vast majority of the population did, in fact, understand the problem. It is that version, and the surrounding controversy, that this article is about. There are many possible variations and interpretations that are also briefly mentioned.
It is my guess, based on experience here, that most of the different interpretations of the problem arise from people who get the answer to the question wrong and then look for ways to defend their answer. It is a great pity that vS did not describe the problem more clearly, but she was writing a puzzle for a magazine, not a paper for a mathematical journal. Discussion could then have centred on why so many people get this simple problem wrong rather than on ways to get different answers. You are welcome to cary on considering your own personal interpretation of what the actual problem was, or was intended to be, or should have been but that is not relevant to the WP article and I doubt that you will discover any new methods or principles in the field of probabilty by doing so.
I am always interetsted in discussing probability with people but probably that should be done in user space unless it relates to the MHP. Martin Hogbin (talk) 09:36, 18 October 2015 (UTC)
The assumptions
The standard assumptions are those that most people make when they answer the question. They are based on people's knowledge of game shows in general and what people thought that vos Savant's precise question was intended to be. They cannot be challenged using mathematics or logic. First you decide, on whatever basis you choose, what the exact question is, then mathematics gives you the unequivocal answer. Martin Hogbin (talk) 16:32, 15 October 2015 (UTC)
FourthKind, I have asked three simple questions below to see where we agree and where we disagree. If you will not answer those then I can not help any further. Martin Hogbin (talk) 15:12, 15 February 2016 (UTC)
Copied from talk by SPACKlick (talk) 12:07, 10 February 2016 (UTC)
Note: a previous version of the "Fourthkind resolution" section was originally placed here, but that user removed it from here and placed it elsewhere several times during various rewrites. Diego (talk) 22:32, 2 March 2016 (UTC)
The game clearly begins at when the second question is asked. You have a choice between two options which are different so the strategy is about how to choose which door to pick.
When you make that choice the probability isn't 50-50 it's 1:2 or 33-66. It's not about "reading tells" or any of the fluff, it's a simple mathematical fact that given that one of the three options was chosen and a losing option was removed from the other two, switching from the chosen option is twice as likely to win.
The first "pick" simply arbitrarily choses one of the doors which cannot be opened to reveal a losing prize. It doesn't matter which door is picked but it does matter that both contestant and host know which door cannot be revealed.
Where you say the show doesn't begin till after the second question is answered your conflating the probablistic cascade beginning with the cascade being determined.
Your roulette example is not analogous because the croupier knowing which bets are placed doesn't affect the outcome at the end but the host knowing which door is picked does affect the outcome.
Consider the following. If you pick a door with a goat behind it on choice 1, the host doesn't have any choice of which door to reveal. If you pick the door with the car behind it the host does have a choice which door to reveal because they're both goats. The chances of winning by switching at the end truly are 66.6% SPACKlick (talk) 12:07, 10 February 2016 (UTC)
Do you agree with the following statements?
Players who always switch will win 2/3 of the time.
Players who never switch will win 1/3 of the time.
Players who randomly switch (that is to say switch 1/2 the time) will win 1/2 the time. Martin Hogbin (talk) 13:19, 15 February 2016 (UTC)
Fourthkind you are assuming at Q2 that the two options are equiprobable to be a car, they're not. Because of the process whereby the host chooses the door to open before Q2 it is twice as likely that the reference door is a goat than a car, that's the entire point of the paradox. SPACKlick (talk) 10:23, 23 February 2016 (UTC)
FourthKind, user SPACKlick is right. Probability Theory says that you can only calculate probability as winning outcomes divided by the possible outcomes when all options are equiprobable, but here the options are not equiprobable so you cannot apply that formula.
The sample space of MHP is 2, but the probability of winning is not 1/2 because the process Q1 has made one outcome more likely than the other. This has nothing to do with how the term "sample space" is defined, but with the way you arrive to the values in that sample space.
In probability you always have to ask "how was this sample space created?", as small changes in the creation process can lead to huge changes in the probabilities. Diego (talk) 12:08, 23 February 2016 (UTC)
SPACElick thanks for responding to my Final Resolution
You state: winning outcomes divided by the possible outcomes
In MHP it is 1/2. That is my point. Before answering Q2 there is 1 winning outcome of 2 possible outcomes, thus 1/2. not 3 possible outcomes. So 1/2 NOT 1/3. Do you believe in MHP there are 3 possible outcomes at Q2? Think again. With all due respect you mixed TGP with MHP a common flaw in reasoning in MHP.
FourthKind FourthKind (talk) 14:02, 23 February 2016 (UTC)
I didn't say "winning outcomes divided by possible outcomes" that was Diego Moya and his point is valid. That maths only applies if all the outcomes are equally likely. In MHP there are two outcomes for each choice. The door you choose hides a car or hides a goat, however those two options are not equally likely. It's like throwing a dart at a board, if the board is in two sections divided down the middle the odds of hitting either section is 1/2 but in this case one door is more likely than another to contain the car so it's like a board divided unequally. The door you originally chose has a car behind it only 1/3 of the time.
So I ask again, what do you think is different about the MHP from the TGP that means these odds are different between the two. SPACKlick (talk) 14:10, 23 February 2016 (UTC)
Please respond
If you want to be taken seriously you need to respond to other editors comments. Please could you answer these three simple questions. A simple 'yes' or 'no' for each will do. That will help me see wher you are coming from.
Do you agree with the following statements?
Players who always switch will win 2/3 of the time.
Players who never switch will win 1/3 of the time.
Players who randomly switch (that is to say switch 1/2 the time) will win 1/2 the time.
FourthKind, could you please elaborate on how you think the MHP differs from the other problems. SPACKlick (talk) 12:53, 23 February 2016 (UTC)
FourthKind I understand you to be saying that it is incorrect that in the MHP players who always switch will win 2/3 of the time and that players who never switch will win 1/3 of the time.
You are just plain wrong. Can I ask you if you think this simple card game is equivalent to the MHP.
A simple card game
Take the queen of hearts and two spot cards, say the 2 and 3 of clubs. A person, who I will call the host, shuffles the cards an puts them randomly, in three positions, which we will call position 1, position 2, and position 3, face down on the table. A person, who we will call the player, has no information as to which card is where. His aim is to pick the queen. He chooses a card and puts his finger on it but does not look at it. The host then always looks at both the remaining cards, without letting the player see the faces. He then must always turn over a spot card, which he does. The host may not chose the card that has been turned over (obviously he would not want to anyway) but is always allowed to chose between his original choice and the remaining, unfaced, card on the table.
The player has decided in advance that he is always going to swap and take the unseen card remaining on the table. How many times out of three, on average, would you expect the player to get the queen? (Please do not respond until you are absolutely sure what the rules of my game are. If you have any doubts at all please ask before responding. Do not mention any options or ifs or buts, just ask me to clarify before you answer). Martin Hogbin (talk) 13:14, 23 February 2016 (UTC)
FourthKind
Excellent! Now we are getting somewhere (I presume you meant 2/3 if you always swap). Could we go through my game step by step and you tell me where the difference lies. I have some idea from what you say above but to avoid unnecesary argument let us go through it one step at a time.
Take the queen of hearts and two spot cards, say the 2 and 3 of clubs. A person, who I will call the host, shuffles the cards an puts them randomly, in three positions, which we will call position 1, position 2, and position 3, face down on the table.
Does that match the placement of the car and the goats behind three doors in the MHP? Martin Hogbin (talk) 14:30, 23 February 2016 (UTC)
FourthKind thanks again Martin. My bad, a typo I clarified. Thanks for pointing it out and assuming correctly. I can't answer any more today but will respond to any questions. Thanks FourthKind FourthKind (talk) 20:20, 23 February 2016 (UTC)
Hi Martin. The game I called TGP is what you would probably call the 3 card game exactly. This model has a few flaws built in when trying to compare it to MHP or the Monty Hall Show. Here are a few
TGP or 3-card-game is a game which essentially shows if the contestant choose the 2/3 set they will win 2/3 times, choose 1/3 set and they will win 1/3 times. The "game" essentially begins with a sample space of 3 then locks you into a lopsided decision proving essentially itself alone and then comparing it to MHP.
TGP "tries" to explain TM because it is not at precise as TM which precisely shows the 1/3-2/3 set model. TGP has flaws in the model itself let alone trying to compare it to MHP. 3-card-game has no room for "random" (flip a coin) at Q2. The real choice in TGP which results in consequences is Q1 and essentially locks the Contestant with the reference door as 1 of 2 “choices” later at Q2. Not so in MHP. The answer to Q2 is where real consequences will occur in MHP.
TGP has flaws when trying to compare it to MHP. These are a couple. The short-run fluctuations are results of a few factors: shuffling 3 cards equally randomly, choosing equally randomly, short run variances, etc. The TGP or 3-card model are the most flawed models of all. So much so, I don't spend too much space in the Final Resolution explanation.
Thanks again I hope this helps.
FourthKind FourthKind (talk) 11:40, 24 February 2016 (UTC)
FourthKind, I am trying to go through this step by step so that I can understand where your thinking differs from that of the rest of the world. You say that my game is not exactly the same as the MHP and I want to know exactly where it differs.
So that we can all talk the same language, can you confirm that the first step shown above namely, 'Take the queen of hearts and two spot cards, say the 2 and 3 of clubs. A person, who I will call the host, shuffles the cards an puts them randomly, in three positions, which we will call position 1, position 2, and position 3, face down on the table', exactly matches your understanding of the placement of the car and the goats behind three doors in the MHP. Without going through the two problems step-by step, I fear we will make no progress. Martin Hogbin (talk) 11:48, 24 February 2016 (UTC)
A simple Monty Hall Problem
FourthKind, in your "silent 4th assumption" you say that the Monty Hall Problem (MHP) requires the sample set to be "equally balanced" between several runs.
But what if the Monty Hall Problem is played only once? (say, if instead of a regular quiz show, it was a single-event show played during the SuperBowl). In that case, "all runs" is the same as "a single run". What would be the probability of winning by switching doors, in that on-off single instance of the game, if all three standard assumptions are made?
In that experiment, the idea of a sample set balanced between runs is meaningless, as there are no several runs. There is only one car, and one player. Yet the way to calculate the probability of winning is the same one explained in the Wikipedia article:
Suppose that the car was behind door "A" (which door is "A" in this argument doesn't matter; because of symmetry, the following argument runs the same whether "A" is 1, 2 or 3). Picking at random at Q1, the player will chose a door with a goat behind it (either door "B" or "C") with probability 2/3; only with probability 1/3 will chose door "A".
In the first case, switching doors will win the game (because the other goat has been discarded in Q2), and we have seen there are two ways this thing can happen (i.e. two outcomes out of three), with the player having first chosen either door "B" or "C".
Only in the second case, switching doors will lose the game, because the player would abandon the door with the car and chose the remaining goat. As we have seen, this only happens in one outcome out of three (when the player had chosen "A" at first), and has probability 1/3.
Do you agree with this analysis of the probabilities of this one-off "Super Bowl Monty Hall Problem" (SBMHP)? If not, why not? Diego (talk) 10:56, 24 February 2016 (UTC)
Hi Diego, I just saw your addition and only have time for your first question;
I changed it to Sample Space a typo I had Sample Set
In the Monty Hall show I addressed the issue above.
A single run is fraught with issues. I don't defend the multiple run issue with the models proposed TM or TGP or MHP, although it can be challenged. I don't know how many Monty Hall Shows aired. The TM and TGP and MHP have some serious issues with trying to defend themselves in a one time experiment (run). I am not defending that side of the argument.
My initial answer of 1/2 is the same for a single episode of the Monty Hall Show or an infinite number of Monty Hall Shows.
Thanks for your insight, I will have to answer the other issue you brought up later.
FourthKind FourthKind (talk) 12:35, 24 February 2016 (UTC)
what are the issues with a single run? the probably of winning if you stick is 1/3, if you switch its 2/3 - as Diego explained above.Jonpatterns (talk) 12:47, 24 February 2016 (UTC)
Well, I have introduced this simple version because I think it will help you understand the "repeated" version better. As I have shown, the "Super Bowl Monty Hall Problem" is a well-defined experiment, and the argument above is proof that the player in that single event would have more chances by switching than by staying; so your intuition about the probability of winning being 1/2 is wrong for that problem. Take your time to understand the argument and see why it's true, by counting the number of options. I don't expect a reply today. Although this doesn't affect the argument, it may help you if I say that the car in that fictional event was behind door 3, because Monty has 3 children (so the car's placement was not chosen at random, and "A"=3).
I think the problem you have in understanding the Monty Hall Problem is because you're trying to separate the Q1 step and the Q2 step as if they were unrelated, but they are not.
If the question was stated as the mini-problem "There are two doors, one has a car behind it and the other has a goat, both placed at random with equal chances. Choose one, and then decide whether you want to switch to the other one", then you'd be right that the probability of winning is 1/2 either way; but that is NOT the Monty Hall Problem.
In the Monty problem, the fact that you first choose among three doors (not two) and that Monty *chooses* which door to open in order to discard one option, does transfer some information between steps Q1 and Q2, and therefore the car and goat have not been "placed at random with equal chances" behind the two remaining doors; so Q2 is not the same as the mini-problem, and does not have the same probabilities. Diego (talk) 12:56, 24 February 2016 (UTC)
Hi FourthKind, I'm trying to compare the SBMHP with the Monty Hall Show, because the reasons that make it more likely to win if you switch are the same in both problems. I think what you call the MHP is the same thing that I have called the "mini-problem" above, which is different from the Monty Hall Show. Correct me if I'm wrong, as I don't fully understand your terminology. Diego (talk) 13:35, 24 February 2016 (UTC)
Hi good to know people are still questioning. I have a question for you. It is not a trick question. If someone, say a world renowned mathematician, told you that the Hosts 2/3 side was was weighted but not precisely by how much and the Host exposed 10 cars and 15 goats only 2 doors remain out of 27 doors at Q2 what is the probability of you picking a car from the remaining 2 doors?
Excuse me? I don't think I understand the question. I don't know what "the Hosts 2/3 side was weighted" means; and if 15 cars and 10 goats have been exposed, doesn't that mean that there are at least 25 doors? Diego (talk) 22:26, 24 February 2016 (UTC)
Yes 27 doors total. Sorry. Weighted the set of 26 doors remain before Q2. 26 unknown then 25 exposed then 2 unknown then Q2. Your pick from Q1 & Hosts door.2 doors remain. What would be the probability of winning with MHP rules?
FourthKind FourthKind (talk) 20:06, 25 February 2016 (UTC)
Ok, I see. The problem here is I don't think your definition of the MHP, TM and TGP are well-defined models; they have ambiguous or incomplete information, and therefore I find it impossible to reason about them correctly. To define a probability problem, you need to identify what your random variables are, and what are the probabilities of each outcome in the sample set. But in your models there are no mention at all of how the doors relate to the outcomes, which is essential to the problem.
Therefore I can't understand what the differences between MHP and the other problems are. I cannot answer any of the doubts that I have:
What does it mean for TM to have a sample space of (goat, goat, car)?
What does it mean for the outcome in TM to be "goat"? Is it the first goat or the second goat? What probability has each of these outcomes?
What door did the player open if the outcome is "goat" in the TM or the TGP?
How is that different from having an outcome of "goat" in the MHP model?
How mane goats exist behind doors in the MHP model, one or two? How many doors?
I cannot answer any of those questions in your model, and therefore the model is not precise enough for me. I can't use your defined terms and be sure about what I'm saying; which means that I can't understand the model from the incomplete definition that you provided. At the very least, you'd have to define in each of those three models how the outcomes "goat" and "car" relate to "door 1", "door 2" and "door 3", to know what you had in your head when you defined them. Please, provide a more detailed definition of the models, or I won't be able to further discuss your ideas. Diego (talk) 10:23, 26 February 2016 (UTC)
To make it easy, these are the random variables that I'm thinking about:
Variable "Position" is a random event that has three possible values: { (car, goat, goat); (goat, car, goat); (goat, goat, car) }
Variable "First pick" is a random event that has three possible values: { door 1; door 2; door 3 }
Variable "Door which Monty opens" is a random event that has three possible values (which are dependent of the values of "Position" and "First pick": { door 1; door 2; door 3 }
Variable "Final decision" is a random event that has two possible values: { switch; stay }.
Alternatively, you can represent a variable "Final pick" with values { door 1; door 2; door 3 }, with the restriction that "Final decision" != "Door which Monty opens".
Variable "What the player gets" is a random event that has two possible values: { goat, car }. Its value is fully defined by the value of "Position" and "Final pick"; although it can also be described in terms of "Position", "First pick", "Door which Monty opens" and "Final decision". A "win" event happens if and only if "What the player gets"="car".
Could you explain the relation between MHP, TM and TGP with these variables that describe the various steps in the original problem? This is the level of detail that I need to understand your model. Diego (talk) 11:10, 26 February 2016 (UTC)
Rick Block gives it a try
I agree with FourthKind that there is a flaw in the "usual" explanations of the MHP, and it is roughly what he has identified. These explanations change the problem from "given you've originally picked door 1 (or whatever door you've originally picked) and have seen the host open door 3 (or whatever door the host actually opens) at which point there are only two doors involved, should you switch" to "what is the best constant strategy, staying with your original choice or switching (or, alternatively, should you decide to switch before seeing which door the host opens)". The latter question (and usual explanations), if you've originally picked door 1, considers the possibility of switching to either door 2 or door 3 - which is clearly not what's going on after the host has opened door 3. The folks who use these explanations, in particular vos Savant in her original Parade magazine columns, are strongly criticized in various mathematical journals (see https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions). The essential problem here is that even though the "usual" arguments do not apply to the situation most people understand is posed by the problem, they DO come up with the correct numeric answer (!).
I've used this analogy before. It's as if someone has asked "what is the value of NN for N=2?" and being given the answer "4. Because 2+2=4". 4 is the right answer. 2+2 does equal 4. And the fact that 2+2 equals 4 is an essential part of the answer. But there's a little more to it. -- Rick Block (talk) 16:47, 25 February 2016 (UTC)
Ok, I understand the subtle difference between both those definitions, and you're right that the "three standard assumptions" are not enough to calculate a definite answer. However I don't agree that repeating the experiment and having a "usual" strategy is required, which is what FourthKind stated; the problem would still be well defined and have a 2/3 probability even with a single run. The essence to have a fully defined problem is knowing enough information about how the host will behave, so that you can calculate the prior and conditional probabilities of each outcome. You don't need a repeating problem (and thus a "constant strategy"); if you have full information, there can be a winning strategy even if the game is played once.
In this case and if I'm not mistaken, adding the assumption that "if there are two goats, Monty chooses which one to open at random with equal probability" is enough to make both questions equal (i.e. choosing to switch before or after the door is open would be irrelevant).
So yes, you need to add some more assumptions to fully define the problem, but no, the "silent 4th assumption" presented by FourthKind is not a requirement to have a 2/3 winning probability; in fact, you can have that same probability without having the car equally distributed among the 3 possible positions. Even if the car is always placed after door 1, the probability of winning by switching will still be 2/3 if the player is not aware of that fact and chooses the first door at random. Diego (talk) 09:53, 26 February 2016 (UTC)
I completely agree with you. Even further, by using a Bayesian approach if you lack any information about how the host chooses which door to open if his choice is between two goats you also come up with the 2/3 win by switching result. However, first things first. Can we get FourthKind to fill in the blanks below? -- Rick Block (talk) 15:45, 26 February 2016 (UTC)
@FourthKind: Can you think about 300 games where the player has initially picked door 1 (we're immediately after your Q1)? Of these 300 games how many times would you expect the car to be behind:
door 1? ____
door 2? ____
door 3? ____
(the sum of these must be 300)
Now the host must open a door (and this door cannot be door 1). How many times in each of the cases above does the host open door 2, and how many times does the host open door 3?
car behind door 1, host opens door 2? ____
car behind door 1, host opens door 3? ____
car behind door 2, host opens door 2? ____
car behind door 2, host opens door 3? ____
car behind door 3, host opens door 2? ____
car behind door 3, host opens door 3? ____
(the sum of these must be 300)
Of the original 300 shows, how many times are you expecting the host to open
door 2? ____
door 3? ____
(the sum of these must be 300)
Given the host opens door 3 (i.e. looking at only those cases), the probability the car is behind door 1 is your answer for "car behind door 1, host opens door 3" divided by the total number of times the host opens door 3. Similarly, the probability the car is behind door 2 is your answer for "car behind door 2, host opens door 3" divided by the door 3 total.
What numbers do you come up with? Please show your work. -- Rick Block (talk) 16:50, 24 February 2016 (UTC)
FourthKind Puzzle
Hi good to know people are still questioning. I have a question for anyone. It is not a trick question. Mathematicians might get this one. If you can answer it, you will be able to solve this 40 year old mystery.
If someone, say a world renowned mathematician, told you that the Hosts 2/3 side was was weighted but not precisely by how much and the Host exposed 15 cars and 10 goats only 2 doors remain out of 27 doors at Q2 what is the probability of you picking a car from the remaining 2 doors?
FourthKind FourthKind (talk) 18:52, 28 February 2016 (UTC)
FourthKind, your first step is wrong (where you say: "Reasoning: In MHP and the Monty Hall Show at Q2 the contestant is left with 2 equal weighted possible outcomes “choices”). It is not true that the two outcomes (finding a goat if you switch, vs finding a goat if you stay) are equally weighted in the Monty Hall Show. This invalidates your whole argument. Whatever it is what you call MHP, it is not equivalent to the game that is played in the Monty Hall Show.
If you play the MHP, it will only show a 50% win rate if you ignore the Q1 step and don't perform the pick-door-and-reveal-goat process, and start the problem with just two doors. But if you skip that first step, you are not playing the Monty Hall Show game. Diego (talk) 15:01, 26 February 2016 (UTC)
FourthKind - can you please fill in the blanks in the section above this one? After doing that we can talk about your model, but I'd really like you to work through the 300 show example first. Note that I agree with you that the popular explanations do not explain what the probabilities are after the host has opened a door (at your Q2 decision point). -- Rick Block (talk) 15:51, 26 February 2016 (UTC)
Can we get back to this after you fill in the blanks above? -- Rick Block (talk) 17:00, 26 February 2016 (UTC)
I figured a method of explanation. Rick, I'll show it shortly.
Once again, can you please fill in the blanks above? -- Rick Block (talk) 19:19, 28 February 2016 (UTC)
VSM changes the question. I agree with that. Her model addresses the changed question, which is what is the probability of winning for a strategy of staying with your original choice or switching. It sounds like you agree with this, and you agree that the probability of winning for a "stay" strategy is 1/3 while it is 2/3 for a "switch" strategy. Are we in agreement so far? -- Rick Block (talk) 01:33, 29 February 2016 (UTC)
You agreed above that always switching wins with probability 2/3 while always staying wins with probability 1/3. Now can we talk about 300 shows? I think you've agreed that out of 300 shows if all of the players pick door 1 and then none of them switch, about 100 will win the car. I think this means you're expecting the car to be behind door 1 about 100 times out of the 300 games, and I think it's not unreasonable to think you think the car will also be behind door 2 and door 3 about 100 times (each). This is the vos Savant model of the problem. Are we still on the same page? -- Rick Block (talk) 05:42, 29 February 2016 (UTC)
Thanks Rick for getting back. I have an easier explanation I am working on and it will answer the questions and a lot more. I will answer this shortly.
FourthKind FourthKind (talk) 10:38, 29 February 2016 (UTC)
Rather than write more, can you please listen a bit? I really think following through what happens with 300 shows will help you understand. At Q1, the player has picked a door (let's say it's door #1). Out of 300 shows we have the car behind each door 100 times.
Now, the host opens a door. He can't open door #1, so he opens either door #2 or door #3. If the car is behind door #2, he MUST open door #3 (and if the car is behind door #3 he MUST open door #2). On average, what happens if the car is behind door #1? -- Rick Block (talk) 19:17, 29 February 2016 (UTC)
You're wrong at the same place you're always wrong, which is that you assert (incorrectly) that after the host has opened a door (say #3) the remaining 2 doors have an equal chance of hiding the car. I can show you exactly where you go wrong if you'd like. -- Rick Block (talk) 20:27, 29 February 2016 (UTC)
Do you have a deck of cards? If so, please try this experiment. Think of the ace of spades as the car. Shuffle. Deal one card to the "player" (this represents the player's choice of door at Q1, but out of 52 "doors" rather than just 3). Look at the remaining 51 cards and discard 50 of them that are not the ace of spades. Now there are two cards left - the player's original choice and one more. Should you switch, or is this a 50/50 choice? Please actually do this (say 20 times) and record whether staying or switching wins. This shouldn't take you more than 5 minutes. Please report your results. -- Rick Block (talk) 22:29, 29 February 2016 (UTC)
FourthKind, I agree with Rick Block that you should follow the scientific method and try out the experiment to test your solution. Your solution explains differences between the theoretical model aFourthKind (talk) 02:02, 2 March 2016 (UTC)nd what happens in reality, but you have not examined reality yet, and therefore your solution is incomplete.
If your reasoning is correct, when playing the card game that Rick proposed it should make no difference whether to change cards every time or to select the final card at random. Have you tested your hypothesis to see if it holds in reality?
You can play the game through this online simulator, although I think it's better if you try it yourself with a deck of cards instead of trusting the software in that simulator.Diego (talk) 11:02, 1 March 2016 (UTC)
Thanks so much for your insight, it's good to know people still question on this page and it is even available via Wikipedia.
The simulator, I believe is also a "program" experiment of the flawed vos Savant model VSM. That is the whole point of what I stated. In essence you then would be asking me to have my idea tested with the very thing my idea is discrediting. A dichotomy for sure there. Not sure it would give me anything new, but thanks for the idea. I would like to thank you for sticking with me and not knocking me off. I am completely new to Wikipedia and excited to contribute. I still don't know my way around and sometimes place my ideas on the wrong page. When I realize this, I immediately remove it. If I don't realize it you guys take care of it for me without throwing me completely out of Wikipedia. For this I want to thank you. Any other thoughts? I have a job and I will answer you at the first opportunity, as I found this Monty Hall Problem fascinating.
This will be my final thought, as I don't think I can add much more to help you. It seems clear that you understand the reasons why the model offers a 2/3 probability of winning, and why an uninformed choice offers a 1/2 probability. However, I think you have a different definition of the problem than the one the mathematicians are using, and this is the reason why you are reaching a different solution.
The Monty Hall Problem, as stated in the Wikipedia article, allows the player to make two decisions:
The first decision is taken during the Q1 step, where the player has a choice among three doors. Then Monty opens a door which was not chosen by the player and does not contain the car.
The second decision is taken during the Q2 step, where the player can switch among two doors; one of them contains a goat, and the other contains a car; but one was picked by the player, a fact that was used to discard the reference door, and the other was not. The fact that the user picked a door in Q1 can not be discarded in the Monty Hall Show game; without that fact, the decision "switch or stay" (which is the question that Monty asks) is meaningless.
Now, you are redefining the second decision as if the player was given a choice among the two remaining doors, but that is not what happens in the game. The options that the player has in Q2 are {switch, stay}, but you said that the options in Q2 are {door 1, door 2} (which could have the outcomes {goat, car}). Indeed, the game that you described in the FourthKind Resolution is the "little green woman" game, which is the following:
During the Q1 step, the player has a choice to pick one among three doors. Then Monty opens a door which was not chosen by the player and does not contain the car.
Suddenly, an UFO appears and a little green woman steps down. The woman does not know which door was first picked by the player.
Monty tells the green woman that one door contains a valuable item, and asks her to choose one of the doors to open it.
Therefore, the decision taken during the Q2 step in this game is one with a choice among two doors; any door has the same chances to be selected by the little green woman, who knows nothing about step Q1, the picked door and the reference door.
The game that you have explained is in your solution is the "little green woman" game, not the Monty Hall Show game. You have unfairly redefined the term "switch" to mean a "choice" between {first door, second door}, by changing the game that is being played.
The VSM is an accurate model of the Monty Hall Problem, where a player decides to switch or stay; in reality, if you play the Monty Hall Problem you win with 2/3 of games if you decide to switch.
Meanwhile, the FourthKind Resolution is an accurate model of the "little green woman" game; in reality, if you play the "little green woman" game you win 1/2 of games if you choose a door to open at random.
FourthKind's flaw of reasoning: MOVING THE GOALPOSTS: is a type of informal fallacy where one problem is stated, but someone changes the problem in the middle of the argument. From Wikipedia: https://en.wikipedia.org/wiki/Moving_the_goalposts.
During 40 years, mathematicians have solved the Monty Hall Problem where the player is asked to switch doors, by correctly comparing it to the accurate VSM where the decision is "switch or stay" from a door with 2/3 chances of having a goat. FourthKind has unfairly redefined the term "switch" to mean "choice" in the Q2 step, moving the goal of playing the Monty Hall Problem game to the goal of playing the "little green woman" game, where the decision is "open left door or right door" by tossing a coin at random with 1/2 probability.
I hope this is enough for you to understand the error that is contained in the FourthKind Resolution argument. Diego (talk) 13:53, 1 March 2016 (UTC)
Thanks for the insight. When I say random choice. Just "toss a coin", in your mind, and tell the host the result by naming it "switch" or "stay". I thought that might be obvious so I left it alone. Thank again Moya, honestly you guys keep me going. I hope this helps.
If you toss a coin to decide whether to switch or stay, you are not calculating the probability of "winning if you switch" (which is what mathematicians say is 2/3), but the probability of "winning if you toss a coin AND switch if it's tails" (which mathematicians agree is 1/2). I.e. you are moving the goalposts. Diego (talk) 14:16, 1 March 2016 (UTC)
I completely agree with Diego here. FourthKind - can you please try the experiment with the cards? You've spent way more time arguing your point than it would take to try this. And, BTW, your whole approach here constitutes wp:Original research, which means none of it can be added to a Wikipedia article (even if you were correct, which you aren't). Any content added to Wikipedia articles must be based on wp:reliable sources. This particular page is not an article, and is here to help people who have trouble understanding the solution to the MHP. The 2/3:1/3 solution is correct. If you disagree, you are not understanding something. Rather than continue to argue that you're right (and all the mathematicians in the world are wrong) - which at some point will exceed the patience of everyone here - it would be best if you actually listen to what some of us are saying. The experiment should help. Thinking about what happens in 300 shows should help. Repeating yourself over and over as if you're saying something new does not help. And, please don't take this too harshly. As the article says, Paul Erdős (one of the most prolific mathematicians ever) remained unconvinced until being shown a simulation. -- Rick Block (talk) 15:41, 1 March 2016 (UTC)
Just to be clear, we're talking about the probabilities in the situation depicted here, where the player has originally chosen door 1 and then the host has opened door 3. This is your Q2 - and the car is certainly behind one of door 1 or door 2. However, the probabilities are not equal. If you decide to stay at this point you will win the car with probability 1/3. If you decide to switch at this point you will win with probability 2/3. Vos Savant (and many others) confuse the issue by talking about what happens if you decide to stay or switch effectively at Q1 (before the host opens a door). It turns out the answer is the same, but the thought process to arrive at the answer is somewhat different and the usual solutions do not directly address the situation at hand. -- Rick Block (talk) 18:28, 1 March 2016 (UTC)
Hi guys. The issue is exactly what you are exactly pointing out. If you take a hard choice at Q1 like in VSM, you will get your 1/3-2/3 just like you think. One small problem with that thinking. No one makes a hard choice at Q1 except in VSM. NOT in MHP. NO statistical choice until Q2. NO HARD choice in the Monty Hall Show. It is not a math flaw as much as it is a reasoning flaw, literally to begin with. Think again.
Your choice at Q1 affects what happens. And it affects what happens in a way that results in the probabilities the car is behind door 1 and door 2 being different. Have you tried the experiment with the cards? Please do. -- Rick Block (talk) 19:28, 1 March 2016 (UTCThanks for sticking with me. In MHS/MHP you get a door exposed before ANY real pick. At Q2 AFTER seeing a goat. Not in VSM. Contestant picks a hard choice BEFORE swing a goat. Not very fair at all. That is why the win rate is impossible in a real life show. This is why.
@FourthKind: please come back to this conversation after you've done the card experiment described by Rick. It's not in anyone's interest for you to ignore the suggestion. Cheers, MartinPoulter (talk) 19:59, 1 March 2016 (UTC)
I understand what you're saying and agree vos Savant's model relates only to Q1. But the decision at Q1 does affect the probabilities at Q2. Have you done the card experiment? If you just keep insisting you're right, one of two things will happen. Everyone will simply stop responding to you, or actions will be taken against you preventing you from editing. This is a seriously unintuitive problem. Everyone understands that. But insisting you're right without listening to what others have to say is considered disruptive and won't be tolerated indefinitely. -- Rick Block (talk) 20:26, 1 March 2016 (UTC)
I would like to support what Rick has said above. You seem to be endlessly repeating your own personal theory without listening or responding to anyone else. You have rather rudeley ignored my request above to state the exact difference between my proposed card game and the MHP. You also seem to be making up your own non-standard terminology, which does not help anyone understand what you are talking about. Ther is limit to how far people will go here to accommodate your misunderstandings of this problem. Martin Hogbin (talk) 08:51, 2 March 2016 (UTC)
Martin, I think this may be a case of WP:HONEYTRAP. (I'm an optimist, otherwise I'd think it's an elaborate WP:DENY situation). I'm trying to be careful and not bite the newbie. Diego (talk) 15:53, 2 March 2016 (UTC)
You may well be right but what is clear is that this conversation is not leading to an improvement to the article or to the world's understanding of the MHP or to FourthKind's education. Martin Hogbin (talk) 14:16, 3 March 2016 (UTC)
Speak for yourself... ;-) This conversation inspired me to write a clarification in the lede for a point that confuses many people and its explanation was buried in a section. Diego (talk) 14:31, 3 March 2016 (UTC)
Thanks again, I respectfully agree with the card game. It wins if certain conditions are met. We agree to that point. I think we simply disagree on the conditions of the results. I believe that is the disagreement. I have great respect for knowledge and discussing different ideas. I do not mean any disrespect toward the moderators. You have my history to show that I don't. I hope the CAPS didn't offend anyone. I thought it would help in the discussion. I am typing on my phone to address concerns and am having a hard time with quotation marks. Please don't take offense for me using caps it is easier for me. I think we are on the same side regarding solving the puzzle.
Thanks again I really appreciate you questioning me. It helps everyone,I think.
You agree with the card game "if certain conditions are met". Have you actually done the experiment (take 52 cards, deal one, discard 50, leaving only 2, ...)? What were the results? What "certain conditions" are you referring to? -- Rick Block (talk) 23:21, 1 March 2016 (UTC)
FourthKind, you ask: " Why would one take VSM for an explanation of why a contestant has an advantage at Q2 when Q2 contains 2 equal weighted members of the 2 Sample Space at that moment in time, at Q2 over any other reason than itself." There are two reasons:
1) The VSM answers the question that defines the Monty Hall Problem, and the 1/2 equal weighted probability is an answer to a different question, not to the Monty Hall Problem.
2) It is not true that "Q2 contains 2 equal weighted members of the 2 Sample Space at that moment in time"; because of the way that the two doors have been selected from a set of three doors, the members of the 2 Sample Space are not equally weighted.
Explanation of the first reason:
It is not true that the solution to the Monty Hall Problem is a false dilemma. The formulation of the problem never assumes that there are only two strategies (always switch/never switch), this part of your reasoning is a strawman argument that you set up in order to discredit it, but it is not a flaw in the original solution of the VSM model.
The question that defines the Monty Hall Problem is, "is it to your advantage to switch your choice?", which is the same as "Is it better to always switch than any other possible strategy?" This does not imply that "never switch" is the only other possible strategy, that is something that you just made up. It is perfectly possible to compare "always switch" with "switch when a random toss coin says so".
In particular, if you compare those options, the problem is equivalent to "Is the probability of always switching higher than the probability of switching after a toss coin?", which means "Is 2/3 > 1/2?". As the first value is indeed higher than the second, the answer to the question formulated in the problem is "yes". I can't possibly fathom what reason has led you to believe that the "always switching" strategy should not be considered as part of the problem, when it is the question that defines the problem itself; nor why you insist that "toss a coin" is the only strategy that should be used, when the problem explicitly asks about a different strategy.
Of course if you change the initial conditions, you can reach different conclusions; but then you are not solving the same problem, nor answering the same question. From your reasoning it does not follow that "is it to your advantage to always switch your choice?" should be answered with "no", because along the way you have changed to a different question.
Explanation of the second reason:
You ask: "What would it need to overcome this 1/2 equal weighted probability? If one believes they are unequal weighted members, is it because VSM tells them so?" The probability is not equal in real life, and the difference is not because the VSM model says so; it is because in reality, the process to place a car behind two doors is not "place a car behind two doors at random with equal chance", but "place a car behind three doors at random with equal chance, and open a door that contains a goat and was not selected by the player". Both processes end up with two doors with one goat and one car, but they arrive to that sample space through different steps.
In probability it's very important to take into account the process used to create the values in the sample space: you can't assume that two sample spaces are equal merely because they contain the same values. The process that you use to create the sample space affects its probability. You are assuming that the sample space in Q2 is equal to a sample space that is created from scratch with the two values having equal probability, but that assumption is wrong. This is the core of your mistake, really. Every time you modify your explanation, you are making this same mistake. There are times when you can calculate the probability of a sample space with (number of "win" values / total number of values), but not always. You need to learn to recognize them or you will always make this mistake.
You ask "What event could have possibly happened between the acceptance of a “reference door” (begin) to making a “choice” at Q2 (end) that could have taken place of any statistical significance to make the 2 members of the Sample Space somehow unequal?" "One should ask: Has any event of statistical importance occurred?": The answer is yes: between Q1 and Q2, Monty has opened a door with enough care to avoid revealing the car, and this has statistical significance.
You say "because 1 of the equal weighted members of the 3 equal weighted members (exposed goat-door) is removed to bring the Sample Space now to 2 equal members of the Sample Space of 2. Then because 1 of the equal weighted members of the 3 equal weighted members (exposed goat-door) is removed to bring the Sample Space now to 2 equal members of the Sample Space of 2."
You would be right if the "reference door" was opened at random, without taking into account the position of the car nor the door picked at Q1 (the "Monty Fall" problem); but that is not what happen between Q1 and Q2. (Remember, the process you use to create a sample space affects its probabilities).
What happens in the show is that the reference door is carefully selected to create a sample space with two unequal values (one value is the door that the user picked at Q1, with 1/3 probability of containing a car; the other value is the door not opened by Monty, with 2/3). Again, this is not because the VSM says so, but because Monty opened the door in reality using a particular criteria.
In summary:
The VSM is a model that is a correct solution to the Monty Hall Problem.
FourthKind Resolution is a model that is a correct solution to the "Monty Fall Problem".
Thanks Diego, but I respectfully disagree. In this case the host has 2 goats to choose from either one doesn't matter. That is why in this case it is unnessary.
I respectfully disagree in reality Ther were always 3 balanced members. Walk it thru without thinking VSM. It's VSM that sets the unequal members right from the start and doesn’t look back. In MHS it is 3 equal until Q2 them 2 equal.
Thanks again Diego I think this helps a lot to talk with you.
Well, you're wrong. In the two cases where the player didn't pick the car, one of the goats is behind the door that the player picked at Q1 and the other goat is not, so they are not equal and it matters which one Monty is revealing.
And you are wrong about the VSM. In MHS, in step Q2 the two values are not equal probability. The probabilities in Q2 are equal only in the "Monty Fall problem", where the host opens a door at random, but not in the MHS, where the host selects the door to open with a non-random process. Again, the process you follow to define the space state changes the probabilities. This is basic probability theory. Diego (talk) 15:36, 2 March 2016 (UTC)
Diego (talk) 15:27, 2 March 2016 (UTC)
Rick Block's last attempt
I'll give this one more try. If FourthKind refuses to respond I will not respond further. -- Rick Block (talk) 16:28, 2 March 2016 (UTC)
@FourthKind: Can you think about 300 games where the player has initially picked door 1 (we're immediately after your Q1)? Can you please fill in these blanks? I believe what you're saying is that vos Savant's model only talks about the first table, and I agree with this. If you fill out the second table (adds some columns to the first table), then we can talk about the probabilities after the host has opened door 3 (or door 2).
car behind door 1
?
car behind door 2
?
car behind door 3
?
Total
300
Now the host must open a door (and this door cannot be door 1). How many times in each of the cases above does the host open door 2, and how many times does the host open door 3? At Q2 one (and only one) of these has happened - i.e. the host has either opened door 2 or door 3.
car behind door 1
?
and then host opens: door 2
?
door 3
?
car behind door 2
?
and then host opens: door 2
?
door 3
?
car behind door 3
?
and then host opens: door 2
?
door 3
?
Total
300
total times host opens: door 2
?
door 3
?
Hi Rick, can we try this. Would you be kind enough to fill these yourself and I know I'll understand your point.
Please at least try to fill these in by yourself. Trust me, it will be worth the effort. -- Rick Block (talk) 20:50, 2 March 2016 (UTC)
Rick, if we are still on speaking terms I would greatly appreciate it if you would be so kind as to respond just one question. Do you believe the reference door Q1 answer has staticalsignificance for probability? Very greatful if you do.Thanks. This is not a joke honest. I am trying to see your reasoning. Sincerely. You have helped me a lot and I appreciate it.
With all respect, you seem to have a weak understanding of probability theory, as many of your expressions don't make much sense or at best have loose meanings. A probability is a property of a random event, which is a measurement of the actual value of a random variable. For that reason, to calculate a probability you have to carefully define your variables and events. You should strive to make your sentences more precise and use mathematical terms.
In the Monty Hall Problem, the event of a "win" is the probability that the player finds a car behind the door when she opens it (i.e. that the variable "price behind the door that the players open" is "car"). Therefore, strictly speaking it doesn't make sense to talk about the probability of winning at the step Q1, because at that point the player has not opened any door and that event has not happened.
The event that can be measured at Q1 is the event that there is a car hidden behind the door first picked by the player (which the player did not open at that point). That event has a probability of 1/3 at that point, and it's also 1/3 at Q2, because it is not changed by the action of the host to open a door if he carefully avoids opening that first one. Diego (talk) 22:00, 2 March 2016 (UTC)
Wow! So many misunderstandings, so little time to answer them!
It is not true that the player makes only one choice. There are two decisions made during the game (I'm avoiding the word "choice" or "preference", as it doesn't matter); and both decisions affect the outcome at the end of the game. The first decision does not make you win or lose, but it affects which options are available for the second decision:
The first choice is what door should be picked at Q1. It affects the door that the host will open: if the player had picked a different door, there is a possibility that the host would later open a different door. Therefore, the final outcome of the game is affected by the value of this first decision.
The second decision is made at Q2, whether to stay with the door picked at Q1 or change to the other door. Note that the shape of this decision depends on what happened during step 1. If the player had picked a different door at Q1, it's likely that there would be a different set of two doors to choose at Q2.
It seems that you are confused by the wording in the Vos Savant "Ask Marilyn" column. That is because the column is not a full formal proof, and it makes some inferences that are not stated explicitly. For example, when it says "The first door has a 1/3 chance of winning", we understand that she means "the player would win if she stays with the first door until the end of the game", but it is not written.
You would benefit from reading a full mathematical specification of the Monty Hall Problem, which would eliminate all ambiguities. I've searched for one to show it to you, but I haven't found any; the best I've found is this webpage that explains the problem in more detail, including the relation between the first decision and the second decision (although it uses a Kenmore washer as the prize instead of a car). Or you can try this page and play the game for yourself, and see how many times you can win in reality with each strategy. Then compare it with the Monty Does Not Know game (that works like the "Monty Fall" game that I talked about earlier). Diego (talk) 13:47, 3 March 2016 (UTC)
@FourthKind: If you would take the trouble to fill in the table above, I think things will become more more clear for you. Until then, this is my last comment. -- Rick Block (talk) 16:04, 3 March 2016 (UTC)
@Diego Moya: The page you've referenced (this webpage) contains the same sloppy reasoning as vos Savant. There is no foundation presented to conclude that the prior probability of the prize being behind either door 2 or door 3 (before the host opens a door) is the same as the posterior probability of the prize being behind one of these doors given the host opens the other one (and, without further clarifications, this may not even be true). It's basically saying that because we know P(car behind door 2) + P(car behind door 3) = 2/3, we can conclude that P(car behind door 2|host opens door 3) + P(car behind door 3|host opens door 3) must also be 2/3 (??), and therefore, since we know P(car behind door 3|host opens door 3) is 0, P(car behind door 2|host opens door 3) must be 2/3. Comparing P(car behind door 2) + P(car behind door 3) with P(car behind door 2|host opens door 3) + P(car behind door 3|host opens door 3) is comparing apples and oranges. With rational assumptions, it indeed turns out that P(car behind door 2|host opens door 3) is 2/3, but it is distinctly not because P(car behind door 2) + P(car behind door 3) = 2/3.
A much better reference would be the discussion of the MHP in the Grinstead and Snell textbook, available online at https://math.dartmouth.edu/~prob/prob/prob.pdf. The discussion starts at the bottom of page 136. -- Rick Block (talk) 18:24, 3 March 2016 (UTC)
FourthKind: Random and Conditional Probability relating to VSM vs. Reality (The Monty Hall Show)
Latest comment: 8 years ago1 comment1 person in discussion
FourthKind: Random and Conditional Probability relating to VSM vs. Reality (The Monty Hall Show)
Conditional Probability in Vos Savant Model uses probability theory and displays it’s strength with a 2/3 win rate. This is a fact I believe within it's own model. In reality/Monty Hall Show, an event other than a goat-door being exposed can occur. On paper, in computer models, in games, with a finger on a choice, “always-stay” or “always-switch” results are based on an extremely important idea. That idea, is that the only a single significant event takes place in VSM/Games/Explanations between Q1 and Q2 is that a “goat-door” is exposed. This is not a correct assumption when comparing the VSM vs. reality. Reality or the Monty Hall Show is a dynamic situation. VSM is a static model. Events can and do occur between Q1 and Q2, between minutes on the Show. On paper or in a computer model (milliseconds between Q1 an Q2) or in games, no important events are able to occur, just chose at the beginning of the experiment, game, model and the “event conditions” (of 1) have been set before the game begins. The “conditions”, part of Conditional Probability, have only 1 event predicted. The exposing of the “goat-door”. If another condition is introduced it is not addressed in any of the models, or games. Conditional Probability works both ways, and effects reality. Can the models handle a “random” (2nd) event between Q1 and Q2? These models assume only 1 event. In Probability Theory it is known that in Conditional Probability, models based on unbalanced members of the Sample Space in a short run (1 show) can be affected wildly, with probability moving from 0-1.
A “random “choice” is an event between Q1 and Q2 on the Monty Hall Show. In fact it was relished by viewers, not knowing what would occur. In essence, the showmanship or excitement factor of the Show, without it, not much of a Show. The Models listed above, do not address this. A “random choice” is an event, and a possible event and an important possible event in a dynamic reality setting. What if the contestant “tossed a coin” and chose “randomly”. This event alone throws off all those models. What happens now? The models will decay to 1/2 win rate the more “random” is introduced within the games, models, they are not allowed. No 2nd event allowed. Now 2 events occurred in The Monty Hall Show or reality, just before Q2 answer, the moment just before the utterance leaves the contestant’s lips, and either the word “stay” or “switch” is conveyed to the Host by the contestant. 2 events, not 1 as formulated within the models.
Event #1: The Host altered reality when exposing the “goat-door”.
Event #2: The Contestant altered reality when they “tossed a mental coin”.
The Contestant, or the Little Green Lady, currently, in reality, now have a 1/2 probability of choosing the “car-door”. Not 1/3 as fixed in the Models or Games. This is why this seems counter intuitive to most people, the 2nd event is not considered in the Models. This is never explained , at least anywhere I have found. This is also the reason (the event) why it is possible to alter reality if there was one involved in the Million Door Model. The 2nd event between Q1 and Q2. Flip a coin at Q2, and the win rate is now 1/2, reality just got altered 2 times within the time space between Q1 and Q2, not just 1 alteration with the 999,998 doors being exposed by the Host. In some of the models there is almost no time elapsed, thus little can happen and a “random” event is not allowed.
This is why the VSM Model should not be compared as a Model to predict probability in the Monty Hall Show. No “random” or possible 2nd event allowed within the VSM. How can a static model with 1 event between Q1 an Q2 be predictive of a dynamic show? What if the Contestant hears, from the left side of the stage goat bleating coming from door #1 the “reference door” on the left side of the stage and Monty revealed a goat in door #3 on the right side of the stage and the goat is still walking off stage at door #3. An event has just occurred to change the strategy. This is an example of the problem with trying to predict outcomes with event information still available to the contestant but not the model/game. Dynamic vs. Static can be extremely problematic via Use of Inference, Bayes' Theorem. Dynamics of reality, probable events, changes outcomes.
The absence of “random” within VSM, alone is enough to rethink the answer to the Paradox Question applying VSM.
Latest comment: 8 years ago6 comments4 people in discussion
FourthKind "preference" or "choice" part 1
For simplicity I decided to refer to the Q1 answer in VSM and reality (The Monty Hall Show) with different words for comparison of VSM vs. reality (The Monty Hall Show).
I use the term "preference" when referring to Q1 answer in reality (The Monty Hall Show)
I use the term "choice" when referring to Q1 answer in the VSM (vos Savant Model)
One main issue I observed is the concept of the terms used in each being exchanged equally when comparing, VSM vs. Reality (The Monty Hall Show). How these terms are defined and how they are applied when comparing VSM vs. Reality. A word or term can change but the concepts should be similar when comparing VSM vs. Reality. I have noticed similar words like "decision", "pick", "choice", "reference door", "Q1 answer", "Q2 answer". These terms are used to compare VSM to Reality and and also change through time within both VSM and Reality. I have issue with precisely how those terms/words are exchanged between VSM vs. Reality. Is this not fair to insure VSM is an accurate comparison, that is: "Does the Model fit/explain Reality, accurately"?
Game Show Problem
(This material in this article was originally published in PARADE magazine in 1990 and 1991.)
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
Craig F. Whitaker
Columbia, Maryland
Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?
My concern is not one of mostly math but one of mostly reasoning. I believe there is an issue with that statement made above and the reasoning that followed.
When I read Marilyn’s statement: “The first door has a 1/3 chance of winning”
The way it appears, is that “choice” seems to be interpreted by Marilyn as that of the contestant, and that choice has an event attached to it. VSM (vos Savant Model) attached the event of winning, not just containing a goat. Is this correct? This “choice” at that moment in reality had a probability of 0 for win/loss by choosing it as a reference door. It is the fuel VSM runs on. It is a model for strategic advantage in a show in this case. A strategy which would give the contestant the win probability of 2/3 vs. a random choice at Q2 with a win probability of 1/2. Win probability of 2/3 VSM vs. 1/2 reality (Monty Hall Show).
There is a 0 (zero) probability in reality, meaning impossible for an outcome of win or loss takes place at Q1. There is no outcome at Q1 in reality (The Monty Hall Show). This moment, at Q1, in VSM, sets the advantage of 2/3 win rate and 1/3 loss rate right then and there within the model, VSM. The snapshot of time is locked and frozen to predict probability of outcomes for the hopeless contestant in VSM, Finger-On-The-Pea, 3-Door Game, and Million Door Game. These are examples/games outside of the reality of the Monty Hall Show. Fun games, and wonderful examples, and a brilliant way to explain Conditional Probability to average folks. The VSM model/games have been helping regular folks and students for decades, helping explain probability theory, including myself. I think the model is a fantastic teaching tool.
I have great respect for the VSM (Model) in explaining this visually.
What I am honestly concerned with is how this Model was used to explain what happens on the Monty Hall Show. I question it’s use of that Model in this particular case, the Monty Hall Problem or the Monty Hall Show. Monty Hall himself had issues with the Monty Hall Show compared with VSM.
A Hard Issue
There is a 0 (zero) probability in reality (The Monty Hall Show), meaning impossible for an outcome of win or loss takes place at Q1. Doesn’t this curious conundrum bother anyone else? I am sure it has for decades considering all the discussion on the MHP page in Wikipedia: https://en.wikipedia.org/wiki/Monty_Hall_problem, and with the public at large years ago, considering the news articles at the time.
I have an issue with the public applying the VSM as it relates to The Monty Hall Show, and expecting it to prove an advantage win rate exceeding 50% in a one-time event on the Monty Hall Show. That is, applying VSM and thus answering a paradoxical Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) from the Monty Hall Problem, Wikipedia page: https://en.wikipedia.org/wiki/Monty_Hall_problem. The issue I have is with the application of VSM and it’s resulting accuracy relating to answering that one single question. Fun nevertheless, The Monty Hall Paradox Question.
I question, how a reference door, at Q1 in reality, becomes a “choice” in VSM before Q2 answer in either VSM or reality (MHS)? In reality this “preference” (decision yet to be made) becomes a “choice” from the contestant’s lips in reality only after the goat-door is revealed, and Q2 is answered. That moment answering Q2. In VSM (the Model) the “preference” is the keystone of that Model.
Isn’t this the case here where the Model is trying to fit into reality? A key term, a “preference door” is adjusted to a “choice” from whomever and squeezed into a Model which appears to be a fantastic strategy (with a probability 2/3 win rate) for the MHS, but has conditions. I am questioning these conditions and how important they are to the resulting belief that the VSM (Model) can overcome these conditions and thus be “The Model” for strategy “playing” the Monty Hall Show.
"preference" or "choice"?
The condition necessary for VSM to function as an advantage is “locking” the contestant into 1 of 2 choice options at Q1. VSM claims it isn’t if you want a higher win rate. The rest will be history for VSM.
This is essentially what happens in VSM. Make a “choice” (1/3 win probability at Q1), select a door, answer to Q1. Flip that selection “choice” for the other member of the Sample Space of 2 within VSM and run the Model. It shows an advantage of 2/3 win rate.
The problem with that it is not an accurate comparison to reality, The Monty Hall Show. The Monty Hall Show designed the Show for excitement and thrills so the producers had an “out” clause in the Deal in 2 ways.
1: The nature of the show gives the opportunity to firm up their preference (with no outcome even possible, probability of “0” here), at Q1, with an actual answer, carrying the weight of probability at Q2 (outcome is possible). This preference in reality at Q1 was meant to create excitement to see whether or not the contestant would actually stay or switch or if confused or even not confused the contestant was always inherently afforded an opportunity to choose “random” and stay or switch with that “random” reasoning, essentially no reasoning, random, without reason. A method of reasoning “random” choosing is not allowed in VSM.
2: The idea that a contestant can choose randomly at Q2. This is not possible either within VSM. There is no “random” choice option in VSM. VSM is a strategy model it doesn't need to have it, after all it is to show how to win, not how to lose less, it purports. This random option is available and used quite often in reality/The Monty Hall Show.
3: The missing "random", an insignificant element?
Again, isn't this the model trying to fit reality, with a key term changed and concept changed? There is no "out clause" in VSM when comparing it's conditions with reality's (MHS) conditions. This bothers me as a possible Model comparing it with The Monty Hall Show or MHP.
I have always argued this previously, but not so well.
@Fourthkind:, you're not engaging in covnersation, you're ignoring questions and comments and you're repeatedly posting, microediting and removing essays of repetetive preaching, please either engage in discussion or this whole thread will end up being deleted as disruptive. The process from Q1 to Q2 affects the probability at Q2. The random strategy is one of many strategies considered for answering Q2. It turns out the best strategy isn't random, or always stick or switch if the host opens door 3 or anything else. The best strategy within the game is always switch.
Your attempt to distinguish the monty hall problem from the model of the problem is unhelpful. The model is the show as described in the problem. Below are all the relevant elements, you tell me which of them you think the model is missing or are not in the show.
1) There are three doors
2) There is a prize behind one of the doors
3) The host knows which door holds the prize
4) The contestant first chooses a door
5) The host then always reveals a door which the contestant has not chosen intentionally avoiding revealing a prize
6) The host then always asks the contestant whether they would like the prize behind their original door or the other door.
7) The contestant then chooses which door they want the prize from
8) The question of the Monty Hall problem is what strategy can the contestant use to maximise their win percentage.
The answer to that question is that always choosing the other remaining door has the greatest chance of winning at 2/3 chance of winning. Rick has tried to help you understand the probabilities above but you have steadfastly refused to answer. Please don't just repost an essay again specifically engage with the points raised. SPACKlick (talk) 20:31, 6 March 2016 (UTC)
@FourthKind: When other editors say that you are ignoring questions and comments and editing disruptively, that's mainly because you are not following the conventions and guidelines for participating in talk pages. In particular you are expected to follow the following steps:
Reply to a comment or group of comments by placing your own post right below the comments that you want to answer.
Put colons ":" in front of each new line to indent your post, putting one more colon than the comment above it.
You are expected to not make changes to your posts after someone else has already replied to them, except in very limited circumstances. In particular, it is problematic that you are deleting whole versions of your essay and placing a new version at a different place in the page. If you intend to rewrite it many times, I suggest that you move it to a subpage in your user space such as User:FourthKind/Fourthkind resolution of the Monty Hall Problem, and link to that page from here.
You are expected to reply to the questions that other editors have asked you, even if briefly, before introducing a whole set of new ideas.
Mr.Diego and SPACKlick I am saddened and sorry you are upset with me not answering your questions, but honestly I have been working on an idea I would like to convey to the discussion.
The edits were only my verbiage. I only thought my explanations were not concise enough and wanted to clear them up.
I was unaware of editing one's statement was disruptive and will not edit myself so much in the future. Thanks for your help all along. I will study edits in WP also. I honestly have learned a lot about the VSM and Monty Hall Problem.
I appreciate it and have found a way to express my input within the discussion. This, one of my final statements will explain everything to you. I hope you read it and let others read it and will limit my additions to a minimal. Also I will answer any questions now as I think I have also narrowed down your questions with this next explanation. "Random and Conditional Probability"
FourthKind, my mistake. You only need to put the colon in front of lines after a newline, when you start a new paragraph, not for every line in the text edit box. I've fixed your post for you. Diego (talk) 20:43, 7 March 2016 (UTC)
Pages can be created in a users 'userspace'. A page could be created for the essay on your Monty Hall solution, and edited as much as you like. The essay could then be linked to in discussions on this page. That way the conversation would flow better and you'd have your solution in one place. To a create a page, go to your userpage and type / 'name of page' in the url bar and press enter. For example, here is a page called 'new page' in my userspace User:Jonpatterns/new page, more info here Wikipedia:User pages.Jonpatterns (talk) 20:46, 7 March 2016 (UTC)
Anyone, shouldn't a Valid Strategy (VSM) be Defined vs. Valid Definition of Reality?
Latest comment: 8 years ago22 comments4 people in discussion
Shouldn’t reality be defined in MHP/MHS at the point in time just before the contestant’s decision of Q2? At that point, just before answering Q2 reality should be defined via the perspective of the contestant. At that real moment in time (reality) is when the setting the members of the sample space should be defined, from the perspective of the contestant in reality (MHP/MS) which, then should be 2. In reality, the sample space should be set at 2.
VSM takes the point of view, in the model, that there are 3 members of the sample space. My argument is VSM is unfairly assessing reality to be 3 members for the “strategy” to prove 1/3 vs. 2/3 within the model. That is why some (me also) argue that the model sets an unfair definition of reality, because VSM assumes because of the “preference door” (Q1 answer) is a choice” and at that moment 3 members are set. Shouldn’t it be 2 members in reality, just before Q2. I am saying it never was 3 members, it should be 2 in reality. Probabilities should be predicted from that point in time. Then at 2 members take in the information that transpired up to the moment before Q2 not from Q1 as assumed in VSM.
Q1 was never a “choice”.
If one is to take any model seriously as a strategy in reality, shouldn’t that model assess reality accurately? Shouldn’t the conditions the model is trying to strategize FROM be accurate, then prove an advantage from that point forward. It seems to me the VSM is strategizing from the “reference door” in reality. Remember that “reference door” in reality has a probability of 0. Why make this VSM’s starting point?
Why would you limit reality to a single point in time, when in reality a lot of actions have happened before that point that have influence over its outcome?
What you are doing is proposing a new model (let's call it the FourthKind's model, or FM) where two doors appear fully formed out from nothing, with a price and a goat randomly placed behind them with equal probabilities. But in reality, in a game of the Monty Hall show, the chances that the price is behind a particular door are not equal for both remaining doors, because the process of finding those two doors from a set of three has discarded some options, and made one outcome more probable than the other. You say that the "reference door" has a probability of 0, but the Monty Hall Problem doesn't ask about that door, it ask about one of the other doors (the one that the player didn't pick at Q1, let's call it "switch door"); precisely because the reference door is open and does not have a price, the "switch door" has a probability of 2/3.
It is important to make models about the problems; in fact that is what we always do - you cannot reason about reality directly, because the reality of the problem is not inside your brain, where all the arguments happen; you can only build models and see how well they predict your future measurements of reality. For that reason, it is important that you actually play the game and see which one of the models (the VSM or the FM) provides better predictions. This is why I and all the other editors insist that you play the game in reality and see if it behaves like you think it should, or not.
Models should be made as simple as possible, but not simpler (as Einstein said). The handicap in your model is that it is simplifying too much, discarding important details that are relevant to the problem at hand, and thus the FM model does not properly represent the reality of a full Monty Hall's show played by the Monty Hall's Problem rules. It is OK that your model contains only two doors with two values behind them, discarding the rest of details of reality; but it is not OK to then assume that those two values have equal probability; that part of your model does not match the reality of the Monty Hall Problem game. Diego (talk) 11:37, 9 March 2016 (UTC)
Thanks so much Diego, by the way is that how I should properly refer to you? Please let me know.
One should employ Conditional Probability. Just exactly when. I am saying when is just before Q2 not after Q1: as in VSM. It should be employed just before Q2 when the sample space is set at 2.
Here is an example why:
1) the host asks you to pick a reference door from 1 Million Doors
2) the contestant picks a door of reference (say door #555,555)
3) the host reveals 999,998 doors, leaving the reference door and another door (say door #222,234). Both doors (door #555,555 and door #222,234) remain closed.
4) the contestant now has a sample space of 2 set. 2 doors. Door #555,555 (reference door) and door #222,234 the door the host left closed.
5) the contestant now has 3 options to choose one of 2 decisions.
3 options of reasoning to choose 1 of 2 doors
A) the contestant stays with the reference door #555,555. The contestant reasons this has a probability of winning is 1/1,000,000
B) the contestant can choose randomly, toss a mental coin in their head and the probability of wining is 1/2
C) the contestant can, with the power of Conditional Probability, choose door #222,234 ( the unopened door the host left) and the probability of winning is 999.999/1,000,000.
Take option “C” the probability of winning is 999,999/1,000,000
Conditional Probability shows this. The difference is Conditional Probability should be employed at Q2, not at Q1. That’s a big difference. Employ CP after the 999,998 doors are opened. Not at Q1 when no doors were opened. That also allows a “random” possibility, not a good avenue to follow, in this case ( 1/2 vs. 1/1,000,000), but a option of reasoning or avenue none the less. Random” avenue of choice actually exposes the flaw of reasoning in VSM and all the examples. It forces one to have a finger on a shell. No “random” allowed here. Stick with setting the sample space at 3 within the model, not the sample space of 2 within reality.
You can call me Diego, that's my fist name. Ok, first thing, let's have a clarification of terminology. When you say "reference door" you're talking about the door selected by the player at Q1? I though you were talking about the one that Monty opens. Sorry if that may have confused you in my previous comments.
Second, I think this version of your problem description is much clearer than the previous one, looks more accurate. You are right that the conditional probability of winning should be calculated at Q2, it doesn't make sense to calculate it at Q1. (Although I'm not sure why do you believe that the VSM calculates it at Q1? That's not how I interpret the VSM myself; I think it is the "conditional probability of winning at Q2, given that at Q1 the player has picked this door").
The question stated by the Monty Hall problem, if applied to this model would be: "what of the three options (A, B, C) has the highest conditional probability of winning?" The answer is clearly C, which has a probability of winning of 999,999/1,000,000; much higher than B, with 1/2. This latest model of yours is an accurate representation of the Monty Hall Problem with 1.000.000 doors, congratulations. Diego (talk) 13:09, 9 March 2016 (UTC)
Thanks Diego for looking at this. This has been a concern of mine since I got here. Why "always" anything? One needs "always" in VSM to avoid "random". The VSM chooses the snapshot of time at essentially a weak point (probability of winning 1/3) then flips the weak decision(?) for a stronger one "always-switch" and then sets the problem as only 2 options to choose, "always-stay" or "always-switch" of course "switch". This is not the case in MHP/MHS. At Q1 in reality it is not a possible member of the sample space until just before Q2, then it is 2 not set at 3 at Q1 in VSM. Remember no probability even exists between Q1 & Q2 as the 2 members are not decided yet. In reality the probability of the reference door winning is 0. That's impossible, because the Show does not end there. "Choosing" it has no consequence of either winning or losing. One can't win or lose yet. Not yet. So 0 probability. The left over door from the host and the reference door picked by the contestant in the beginning. This is why everyone gets so confused.
FourthKind, the "always" part comes directly from the question asked in the problem. The question "Is it advantageous to switch?", if you repeat the game, is equivalent to "Is it advantageous to always switch?" (as compared to other strategies, like switching some times and staying some others). That's part of the problem definition. You need the "always" because, if you change it, you are solving a different problem, not the Monty Hall Problem. For example, if you toss a coin to decide to switch, you will not always switch, only when it is tails; so you can't use the coin toss to answer the question about always switching. All your concerns about calculating probabilities at Q1 are distracting you, as the probability of "winning at Q1" is meaningless (winning can only happen after a door is opened). Diego (talk) 14:01, 9 March 2016 (UTC)
Hi Diego, thanks for challenging me again.
I thought the Paradox Question is
From the front page of Wikipedia----"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?" (End quote here)
If this is not it? If not, what is the MHP Paradox Question? Thanks again.
Yes, that is the question. "Switch your choice" means "always switch your choice", not "sometimes switch your choice" (which is what would happen if you toss a coin as the strategy for deciding).
Or alternatively you can think about the question in Game Theory terms, and interpret it as "how often should you switch your choice to maximize your chances of winning?". If you solve the problem, the solution is "always" (because "always switching" has a higher probability of winning at Q2 than "tossing a coin to switch or not"). Both interpretations are equivalent in this problem. Diego (talk) 15:50, 9 March 2016 (UTC)
I disagree with Diego there. The question is about this one case. Whether switching or not gives you the best chance of winning based on the information you have. The only way to calculate that is to consider what happens in the general repeated case. SPACKlick (talk) 15:55, 9 March 2016 (UTC)
Can you clarify? Do you mean that they are not equivalent because the first is about one run, and the second is about the repeated experiment? Diego (talk) 16:06, 9 March 2016 (UTC)
Would not the FourthKind Model (FKM) then be more accurate in defining reality, a moment before Q2 is asked, then all information is assessed. Isn't this a better time to consider all information and decision options, one of them being guess. Maybe the FKM is a better model for MHP/MHS which does incorporate "random", which is possible in reality but not VSM.
The FourthKind Model is an accurate model of reality if you drop the requirement that both doors have the same probability of containing a car. You *can* represent the reality of the MHP in a model with two members in the sample space (i.e. two doors and two objects behind them), if that's how you like to represent it. But then the probability of winning in the reference door is an inverse function on the number of doors that Monty discarded in previous steps, and the probability in the "switch" door is a direct function on that value (the exact formula is given here). Diego (talk) 16:28, 9 March 2016 (UTC)
I am saying isn't it better to wait until a moment before Q2 to assess all information, then employ CP to help the contestant make the use of all their decision possibilities? That is what I propose.
With sincerely all due respect, isn't this a case of trying to fit the model (VSM) into reality, instead of building a model that fits (describes reality) defines reality more accurately?
Pretty much everyone agrees that you should assess at that point, once Q2 is asked which way to answer it, the issue is you assume as there are tow options the odds are 50/50 but that's not the case, the reference door is twice as likely to contain a goat as the other door. SPACKlick (talk) 16:51, 9 March 2016 (UTC)
SPACKlick, thanks for joining. I am not saying both doors before Q2 have equal probability. I am saying a moment before Q2 access all the options, then if the contestant decides that Conditional Probability is the best of the 3 options with 2 possible doors (outcomes) in reality (stated above in my argument), THEN employ CP and then "switch". I am not disregarding CP. It works. Just employ it as a possible decision route, then use it a moment before the answer of Q2. If one wants to. They might have heard a goat bleating from the door CP said to switch to..etc
Has anyone read this section of the article? FourthKind is saying essentially the same thing, in his own inexpert sort of way. -- Rick Block (talk) 17:18, 9 March 2016 (UTC)
Rick with all due respect, this is not my argument. Excerpted from above: "The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open."
Just before Q2 we DO know. This is not correct. This is not my argument. Honestly, employ Conditional Probability only if the contestant decides to just before answering Q2.
Rick, rethink for a second,who hinders the contestant? The VSM model does, not reality.FourthKind (talk) 17:52, 9 March 2016 (UTC)
This is the exact point this section of the article is making. VSM addresses a slightly different problem (that has no Q2). If you want to know whether to switch at Q2, you should evaluate the conditional probabilities at that point (not at Q1). If you're making a different point, how is your point different? -- Rick Block (talk) 17:57, 9 March 2016 (UTC)
Wouldn't a better model be a model that defines reality (conditions) with more information at Q2 as it's starting point?
Latest comment: 8 years ago16 comments4 people in discussion
Wouldn’t a model which would define reality (events between Q1 and Q2 and the conditions right at Q2), the moment before answering Q2, be a superior model from one which presets events/conditions at Q2 (VSM)? Conditional Probability” would also be a strategic option to employ, or not, as would also a "random" capability to employ or not within the model. VSM assumes itself to be the better possible strategy vs. random, so why consider “random” within the model? Is it the best possible choice, to “always-switch” (VSM)? If it is really the "best" model. How can it claim to be the best strategic model to choose (answer MHS Q2), when the strategy is essentially chosen at Q1. Events on a show are possible between Q1 and Q2 (other than goat-door exposed)? VSM considers itself to be a better strategy than "random". Is it for a single show?
That depends on conditions or events that could occur between Q1 and Q2. On what happens in a single MHS (real-time conditions at Q2 in reality). Remember, the MHS is a one time event and as someone once said “Is it to your advantage to switch your choice?". Notice the word “your”. The you is the contestant and winning a car. Not in a repeat simulation of VSM. (Monty Hall Show). Probability Theory with the use of Conditional Probability can show, with multiple runs, the win/loss probability with more and more certainty. But wouldn’t a wiser idea to determine a strategic choice (case here without employing Conditional Probability) be more “certain” with information (possible events between Q1 and Q2) available and considered at Q2 not at Q1? I am not saying don't employ Conditional Probability just if/when to consider employing it's use. Essentially after observing the goat-door being opened by the Host before Q2? This just makes more sense. What is holding back the contestant from employing such a strategy model vs. reality? Could it be the VSM model itself? If it is maybe it’s time to tweak the model. Say with a new model. Maybe a VSM Dynamic Model? Or a FourthKind Dynamic Model (FKDM)
What I am saying is wait closer to Q2 when the assertion from VSM at Q1 when this assertion is closer to being a fact, just before answering Q2 in reality. Wait until all information is assessed to the last possible second to make the best strategic decision for the new strategy to surpass the real opponent, 1/2. The challenge would then become for everyone on Wikipedia (MHP) page to surpass this new strategy.
Now will you fill out the table about 300 games from above? -- Rick Block (talk) 16:12, 10 March 2016 (UTC)
I agree with Rick Block here. To answer your question about which model is the best one, you need to understand how each model is representing data; and the best way to understand that is through an example. If you fill out Rick's table, it will be easier for all of us to talk about the relative merits of each model. Diego (talk) 16:25, 10 March 2016 (UTC)
Rick thanks so much. I get it, more than you might think, 1/3-2/3. Richard Gill 2011 made a flaw in reasoning. He said "in words, the information which door is opened by the host (door 2 or door 3?) reveals no information at all about whether or not the car is behind door 1," (end quote here). He overlooked: It's not the "removing" of the door #3 that leaves information, it's the information gained by what he "leaves alone". Door#2 and door #1 he thinks he CAN'T TOUCH door#1 so door #2 gains the lost 1/3. Thus now 1/3 vs. 2/3. (Door#2).
EVERYONE has the rules wrong. Q1 is not the choice, but it is the probability of that choice. VSM works, but not why or I should say where. At Q2 not at Q1.
Rick, hi, I thought the other day you knew what I meant about, "we both know what this means." I will expect great grief over saying such a thing, but it's true. Are you looking at the above? Thanks.
FourthKind, with all respect, I believe you have a mental block around the question that happens at the step Q1 of the show (although this impression could be my misunderstanding of what you are trying to say). That first decision may not be "the choice" in the problem with respect to being a mathematical variable that directly defines the outcome of "winning the game", but it certainly is "a choice" in the common English meaning of the word: during the game, the player is asked to make a decision of which door should be the reference door, and to announce it aloud; and that decision influences the probabilities of events during the rest of the game.
No matter at what time in the game you put the focus of your model, the model needs to take into account that decision and how it affects the final choice at Q2. Different models may place focus at different times; but any model that doesn't account for the decision at Q1 may be wrong; and models that properly predict what happens in reality will be right (even if they are clumsy for not placing the focus only at Q2 like you suggest).
For what is worth, I agree that we could try to include in the article a model of the problem that focus only on the decision taken between the final two doors, as it may help people who think like FourthKind. Although we need to find a reliable source that has published such a model (and making it clear that including it doesn't mean that other models are wrong). Diego (talk) 12:34, 11 March 2016 (UTC)
P.S. I like your insight that 'It's not the "removing" of the door #3 that leaves information, it's the information gained by what he "leaves alone".' That explains intuitively why, in the "Monty Fall" problem (where a door is opened at random and the "switch" door has not been selected with care), the probabilities are not weighted toward the "switch" door. Diego (talk) 12:44, 11 March 2016 (UTC)
Hi Diego and Rick, thanks so much for your help. Your sharp eyes are my best challenge. I agree with you almost completely. Except for the one thing. Please remember that I am not being disrespectful whatsoever I am just stating a case. Here it is:
From the eyes of the contestant on the show (defining reality).
Reality is the contestant comes on stage, Host asks to choose the reference door, or choice.
The contestant sees the Host open a “goat-door”.
The contestant has 2 options: guess (1/2) or choose to employ Conditional Probability and choose a probability 2/3 (maybe). The contestant employs Conditional Probability and “switches” from the reference door or choice. What rule did the contestant break? He broke a VSM rule (not keeping the reference door a “choice” that VSM depends on to work at Q1) that is the rule the contestant broke. Not a rule in reality. Remember keeping Finger on Shell, this is to lock in at Q1 the reference shell as the “choice” in question. No room for “random”. Just play by VSM rules.
Did the contestant fool the Host into revealing a “goat-door” other than the contestant’s choice/pick/reference door/etc. under false pretenses?
What rule did the contestant break within reality at Q2 by employing Conditional Probability right then and there only, at Q2?
It’s VSM's problem to explain why the contestant has their hands tied, or finger stuck on a shell isn’t it? It breaks VSM rules, not reality’s rules.
All other CP models can use CP, no problem with that. Still taking Q1 into account and working off that. I’m not saying Q1 isn’t fair to use. Not at all. No problem. Q2 would be a good spot. Same reality at any model using CP.
VSM doesn’t like it (CP) there (at Q2), it (CP) needs to be employed at Q1. If so that’s VSM’s problem isn’t it? That is why VSM shows a loss rate of 2/3. It “cherry picked” the weakest possible moment (Q1 at 1/3) in it’s own model and flipped it (2/3) to explain CP. I think it’s a great teaching tool. It’s a teaching tool squeezed into a game show and now has issues. Don’t you think?
I'm not saying any of the models are "wrong" using Conditional Probability. They can fairly use Q1 and work off of it. I agree Q1 is paramout in setting up the problem to begin with. That's how it's strategic advantage in reality is 2/3(maybe). It works great. VSM has it employed CP at a weaker spot, at Q1. And thus has issues with "random" and others issues everyone will show. I am not defending it (VSM) as the most accurate model to compare to reality. I have always argued "within the model" remember, haven't I?
Uh, could you please clarify where in the VSM is it stated that there's a rule for 'not keeping the reference door a “choice”'? I don't understand what you mean by that sentence, but if I can read it in the original, it will be clearer to me. Diego (talk) 14:02, 11 March 2016 (UTC)
Hi Diego, I am using that figuratively. The "reference door/choice/pick, etc." are just interchangeable terms. VSM uses the "reference door" to work ahead of a Q2 decision so it calls it a "choice", no problem. To work from. VSM needs the "reference door" to be the "choice" (1/3). Fair enough. No problem with Conditional Probability use whatsoever. Isn't that term "reference door/pick/choice, etc." necessary to select as a pivotal point for VSM. That's ok too. I'm just pointing out that focus point is figured out at Q1. Why not wait and with the exact same logic use it at Q2? The contestant, answering will make it THE "choice". And everything works out. VSM assumes that the "reference door" is THE "choice" from Q1. It's essentially a "choice-not-yet-made" in reality. By forcing (Finger on the Shell) that as "THE choice" (within VSM) it calculates off that point in time. Unwisely I think. It's still the same math, MHP/MHS problem, with a slight reality check. That THE choice" won't be made until answering Q2 in reality. It was essentially answered at Q1 in VSM. Just the timing at Q1 it needs to call the reference door "THE choice". It is necessary for timing. It essentially removes "time" between Q1 & Q2. Doesn't it? Just moves it forward. Same thing except stuff can happen in reality between Q1 & Q2. Like facing a 1/2 "random" choice just before Q2. By jumping time it doesn't allow for "random" just itself as the 2 choices, "always-switch" or "always-stay" subtle, very subtle but jumping over time within reality.
Time is why many are fooled. Time between Q1 & Q2. In a model, there is not much of a time lapse, milliseconds. Humans can feel time. This is why some get fooled into thinking Q1 and Q2, there is not much difference. They didn't "see" it "could" matter "when" to employ CP. So why take a good model and fool with it? VSM thinking. Between Q1 and Q2 is everything in reality. A lot can happen (events) in reality, not so in a model. People jumped their reasoning over Q1 to Q2. Keeping the ideas (VSM ideas) of Q1 and that movement in into Q2. That's why I say 2/3 maybe? I haven't done the math. People will now and we can "maybe" get an even more accurate model of a win rate of 2/3 (long run) vs. 1/2. Can't we?
On paper that time lapse doesn't even exist. So why think (CP placement wise only) Q1 is any different from Q2? in terms of CP. It isn't in reality. But it is important in terms (important conditions) of VSM to keep it (CP) there at Q1, before even seeing the goat-door exposed Q2 sees the goat-door exposed but it's still the same CP math, except just before Q2 not at Q1, just timing of CP. Almost same results except it takes in consideration of knowledge between Q1 & Q2. Given no other events of consequence, the results could be 2/3 maybe. This is my "maybe". What if a bleating goat out of the door the contestant is supposed to switch" to. Bail out of CP and "stay" or if unsure at all choose "random", if no unusual events, stay the course with CP and "switch".
1) It employs Conditional Probability in reality too soon. On paper it looks fine. Why have a strategic model that ignores events between Q1 & Q2 in reality? I think because it assumes no events of consequence will occur. Dosen't it make sense in reality for strategy to hold your "choice" until the last possible second when the Host is admonishing you for not answering Q2. Then choose your strategy. I am surprised Game Theorists haven't caught this idea years ago.
2) No "random". No "out-plan". No 1/2 to fall back on if in doubt not to "switch". Otherwise stay the course of "conditional Probability", "switch" then and only then at Q2, when ALL information and NO other information can squeeze in between what you have witnessed (Q1 up to Q2 answer) and your decision "THE choice".
Latest comment: 8 years ago8 comments2 people in discussion
Let's think about 300 games where the player has initially picked door #1. If this bothers you because it ignores games where the player picks other doors, then we could start with 900 games where the player's initial pick is random. Out of these 900 we'll expect there to be about 300 where the player picks door #1. Either way I want to trace through what happens in 300 games where the player initially picks door #1.
At Q1 (VMS model), we expect the car to be behind door #1 (the initially chosen door) about 100 times. And behind door #2 and door #3 also 100 times (each). VSM says if you stay with your initial pick (door #1) you'll win about 100 times out of the 300 games we're thinking about, while if you switch (to whichever door the host does not open, which might be either door #2 or door #3) you'll win whenever the car is behind either of these doors - so, 200 times out of the 300 games. So, you're better off deciding at Q1 that you'll switch. This is where the VSM model ends.
But now the host opens a door which cannot be door #1, and this definitely changes things. For example, we can't be talking about the same 300 games any more because in 100 of those the car is behind door #2 and in another 100 the car is behind door #3, so whichever door the host opens we're obviously no longer talking about 300 games! VSM clearly no longer applies.
If the car is behind door #1, how many times does the host open door #2 and how many times does the host open door #3?
If the car is behind door #2, how many times does the host open door #2 and how many times does the host open door #3?
If the car is behind door #3, how many times does the host open door #2 and how many times does the host open door #3?
In ??? games out of 300 where the player initially picked door #1, the host opens door #3 and asks if the player now wants to switch to door #2. Should you switch?
In the problem as stated, we're told the host opened door #3 and now (at Q2) we're asked if we want to stay or switch (while we're standing in front of only two closed doors, per the image to the right). The VSM model doesn't directly help us make this decision. It considers all 300 games, not only those where the host has opened door #3. It compares deciding to stay or switch at Q1, not at Q2 (!!??). To determine the probability of winning after we've seen the host open door #3, we need to know how many times the host opens door #3 - which is the sum of some of the answers to the above questions - and then how many times the car is behind door #1 and door #2 out of only those games. We won't be talking about all 300 games any more, but only some subset of those.
How many? And, in only those, how many times is the car behind door #1 vs. door #2?
By filling out the following table, we can see what happens.
Rick thanks for reponding. The math for Conditional Probability is absolutely correct. We both agree there my chart is your chart and math also. Math differences are not the issue here. Same idea, same CP, same math except. The contestant should decide when and where to EMPLOY the strategy in reality. I think at Q2 when the contestant has seen more than just a goat-door exposed. VSM assumes that's all that matters, just answer the question Q2. The contestant is facing 1/2 vs. 2/3 (possibly). I say it should be now that the contestant decides to EMPLOY CP. Didn't VSM employ it at Q1. That is why VSM faced 1/3 vs. 2/3 at Q1 AND ALSO AT Q2. And in reality the contestant faces 1/2 ("random") vs. 2/3 at Q2. At Q1 what did the contestant really face in reality? The contestant faced a 1/3 probability of winning vs.what? This is still unknown yet completely for them yet. Until After the goat door revealed then, before answering Q2 they face 2/3 vs 1/2. Timing CP, not forced at Q1 to follow through. Who says the contestant has VSM'S only 2 answers 1/3 or 2/3? VSM does not reality.
I don't have any idea what you're talking about (I'm sorry, but it sounds like gibberish to me). Can you please just fill in the table? I don't want an essay. I want 6 numbers (actually 8, but 2 are just sums). -- Rick Block (talk) 20:27, 11 March 2016 (UTC)
Thank you. So, now, we're at Q2. We've seen the host open door #3. There are only two doors that are not open (door #1 and door #2). The only way we can get here is from our starting point of 300 games where the player initially picked door #1. The numbers you've put in the table above say that we're now (at Q2) talking about only 150 games. These are the ones where the host has opened door #3. So, now, thinking about ONLY these games (the ones where the host HAS ALREADY opened door #3 and we're looking at only two closed doors), what are the chances of winning if you stay with door #1 vs. winning if you switch to door #2? Hint: the answer can be figured out from the table above. -- Rick Block (talk) 21:19, 11 March 2016 (UTC)
Has your head exploded? Or are you starting to see that at Q2, with only two doors to choose from, the odds are not 50/50? BTW - I completely agree that vos Savant's model does not explain what's going on at Q2 (at least not without some more explanation). -- Rick Block (talk) 03:10, 12 March 2016 (UTC)
Rick, thanks for asking. No my head didn't explode. I am defining reality (what's going on at Q2), not VSM model at that point, Q2. I am stating that from the contestant's perspective (not employing Conditional Probability Theory) from the contestant's perspective, in reality, the contestant would be facing a decision (1/2 vs.1/2), not VSM perspective decision (1/3 vs. 2/3) now that there are 2 members of the sample space (reduced by 1 via goat-door revealed). This is a valid perspective of the average contestant on the show (in reality) isn't it?
Rick, with all due respect. There are a minimum of 3 perspectives to consider when discussing the MHP. I think this will clear something up.
1) Reality (show) perspective
Rick’s question: “what are the chances of winning if you stay with door #1 vs. winning if you switch to door #2? Hint: the answer can be figured out from the table above.”
My answer: Rick, thanks for asking. No my head didn't explode. I am defining reality (what's going on at Q2), not VSM model at that point, Q2. I am stating that from the contestant's perspective (not employing Conditional Probability Theory) from the contestant's perspective, in reality, the contestant would be facing a decision (1/2 vs.1/2), not VSM perspective decision (1/3 vs. 2/3) now that there are 2 members of the sample space (reduced by 1 via goat-door revealed). This is a valid perspective of the average contestant on the show (in reality) isn't it?
It’s my mistake, I should have taken your Hint: “the answer can be figured out from the table above.” I should have answered CP perspective I made the error of answering in the “reality” perspective as the only 2 perspectives with which to answer your question is either VSM or CP. You are correct in that it doesn't make any sense to answer in "reality" perspective.
2) vos Savant Model (VSM) perspective
3) Conditional Probability (CP) perspective
Question: You asked me to fill in the probabilities of the chart you purposed to me here in this perspective.
Answer: I filled in the correct probable assumptions.
Rick’s question: “what are the chances of winning if you stay with door #1 vs. winning if you switch to door #2? Hint: the answer can be figured out from the table above.” Here I am assuming you are asking me a question from the CP perspective. It’s my mistake, I should have taken your Hint: “the answer can be figured out from the table above.” I should have answered CP perspective I made the error of answering in the “reality” perspective as the only 2 perspectives with which to answer your question is either VSM or CP. You are correct in that it doesn't make any sense to answer in "reality" perspective.
You are now jumping through time from Q1 to Q2 in all 3 perspectives. VSM an CP perspectives are almost exactly alike as VSM employs Conditional Probability.
I have a dilemma here. Should I answer VSM perspective or CP perspective or Reality perspective? I answered from “reality” perspective.
So I answer:
I am defining reality (what's going on at Q2), not VSM model at that point, Q2. I am stating that from the contestant's perspective (not employing Conditional Probability Theory) from the contestant's perspective, in reality, the contestant would be facing a decision (1/2 vs.1/2), not VSM perspective decision (1/3 vs. 2/3) now that there are 2 members of the sample space (reduced by 1 via goat-door revealed). This is a valid perspective of the average contestant on the show (in reality) isn't it?
This is a reality perspective answer. I believe now you were asking me a question from simultaneously VSM and CP perspectives. Thus confusing you to ask “if my head exploded” a fair question, I suppose.
The issue with this particular problem, the Monty Hall Problem, is that discussions jump from a question in one perspective and then replied with the answer in another perspective. I jumped perspectives, sorry for that. If we discuss any further argument I will ask “from which perspective” of you in the future as this will make our discussions a lot clearer now.
ALL that being said, and that's a lot of said.
Question: “what are the chances of winning if you stay with door #1 vs. winning if you switch to door #2?”
The chart is a VSM chart employing CP (Conditional Probability Theory is an theory and thus has no games, Q2, or doors), so I will answer this question VSM perspective.
Answer:
1/3 probability of winning by “staying” with door#1
2/3 probability of winning by “switching” with door#2
This time I took your hint.
OK, we both agree on the VSM outcomes. Thanks for explaining that. Very sorry Rick. I do not want to be difficult whatsoever, I just want to be very accurate. I believe in the future, I don't need to explain the above again. That's it.
Can VSM theory show or prove an advantage over “random” 1/2 in reality, in a single experiment (show)?
Latest comment: 8 years ago5 comments2 people in discussion
vos Savant Model (VSM) is an elegant, efficient model designed to prove a specific idea. This model a is proof and learning tool of the merits of conditional probability thus how to gain advantage over a probability of winning 1/3. That is, when faced with a 1/3 (1st member of the sample space, “always-stay”) possibility of winning in a model with 3 members of a sample space, and a 2nd member of the sample space is removed, why it is advantageous to select ("always-switch") the original 3rd member of the sample space. This model proves it is advantageous "always-switch", winning 2/3 strategy vs. an "always-stay", winning 1/3 strategy.
Conditions set within VSM prove it is superior vs. an opponent of 1/3 (an initial “choice” of one-in-three) within VSM and in reality.
I believe the single most important question should be: Can VSM theory show or prove an advantage over “random” 1/2 in reality, in a single experiment (show)?
Do any renowned mathematicians believe that VSM cannot accomplish this? Is this even in question anymore or is it accepted as fact that VSM accomplishes this?
The answer is yes, because of probability theory not "VSM theory". Imagine a 6-sided die that has two sides with a number "1" and four sides with a number "2". If you roll this die the chances that it comes up with a "1" are 2/6 while the chances of it coming up with a "2" are 4/6. There are only two possibilities, but these possibilities do not have equal probability. The MHP is the same way. Two choices at Q2, but unequal probabilities. -- Rick Block (talk) 18:43, 13 March 2016 (UTC)
I found one mathemetician on the Monty Hall Problem page. Richard Gill 2011. I believe Mr. Gill might have made a flaw in reasoning in his argument.
Mr. Gill stated: "in words, the information which door is opened by the host (door 2 or door 3?) reveals no information at all about whether or not the car is behind door 1," (end quote here).
I believe that he overlooked one concept.
It's not the "removing" of the door #3 by the Host that leaves behind information, it's the information gained by what the Host "leaves alone" (door#2), which is 1/3 probability that now can be essentially added to door#2. Door#1 CAN'T BE TOUCHED so then door#2 gains the lost 1/3 probability (now over-weighting the 2 members of sample space of 2). So now door#1 (probability 1/3) vs. door#2 (probability 2/3).
What Professor Gill says is accurate. There's an article about him here, see Richard D. Gill. With all due respect, what are your credentials? --
Thank you as always Rick, I will read the full article. My credentials are not, most would say, equivalent to Professor Gill's credentials, so I will not press my viewpoint. I can though challenge his viewpoint, as I did. But that challenge will end here. I have to gain respect here, at MHP.
Could phrasing of Standard Assumption #1 be causing most of the confusion in MHP?
Latest comment: 8 years ago3 comments2 people in discussion
Could phrasing of Standard Assumption #1 be causing most of the confusion in MHP? I think so. Here is why.
Clear any doubt for the 3 implied reasons “why” a Standard Assumption #1
Pick a door
Not the contestant’s door
WHY not the contestant’s door (off-limits) within the game, not simply because it doesn’t make any sense not to.
MHP Standard Assumption #1: “The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).”
The example below uses the answer of Q1 to be door #1, as the contestant’s pick or “tentative choice”.
I am referring as to WHY the host chooses ANY door other than the contestant’s “tentative door” (door #1). This may sound absurd here but please bear with me and I think you might see why. I reason this might be why the 1/2 “camp” or side of the MHP hold the “tentative door” (door #1) with not as much respect as VSM (1/3-2/3) “camp” or side.
From VSM perspective MHP Standard Assumption #1 might have these 3 implications
1) the host must open a door
2) a door that was not picked by the contestant (door #1), so door #2 or door #3
3) the reason WHY “The host must always open a door that was not picked by the contestant” is “because the show ends there” reason and the “off-limits” reason.
The reason WHY “The host must open a door that was not picked by the contestant” is because that “tentative door” (door #1), will with time (at Q2) essentially becomes then the “un-tentative” if you will, or the “always-stay” member of the sample space of 2 decision at Q2. Currently in time (Q1), before the host opens ANY other door, that “tentative door” carries the probability of 1/3 attached to that “tentative door” (door #1/Q1 answer). This essentially means the contestant’s initial pick, at Q1, is currently a main element set to run VSM as the “always-stay” member. The “always-stay” member is (before the host removes one of the 3 members via exposing either door #1 or door #2) one of 3 key members in the sample space of 3 BEFORE the host opens either door #2 or door #3. The “tentative choice” (“always-stay” member) to predict VSM outcomes after Q2, which it does by employing Conditional Probability. 1/3 vs.2/3.
From a reality perspective MHP Standard Assumption #1 might have these 3 implications
1) the host must open a door
2) a door that was not picked by the contestant (door #1), so door #2 or door #3
3) the reason WHY “The host must always open a door that was not picked by the contestant” is “because the show ends there” reason.
The reason WHY the host must open a door that was not picked by the contestant is “because the show ends there” reason. Weighting probabilities will take place at Q2 when the sample space is set at 2. Yes, currently the probability of winning is 1/3 with door #1 but the contestant does not have to make a final decision here anyway, at Q1. At Q2 the probability of winning could be 1/2.
Notice here, I believe the perspective of the 1/2 proponents at Q1 understand this “tentative door” (door #1 in this case) is a “stay” or “switch” term to explain their final choice to the host at Q2. The “choice-yet-to-be-made”.
Before Q2, WHY the host chooses to expose a specific door is everything. I am not referring to how the host chooses between the 2 remaining doors (door #2 or door #3).
Does the host choose another door besides the “tentative door” (door #1) because it is “off-limits” reason or “because the show ends there” reason or because of both reasons? Yes, certainly if “always open another door, one that was not picked by the contestant”, otherwise the show ends there if the host opens the “tentative door” (door #1). But it is not apparent WHY as currently phrased. Open to maybe interpretation by 1/2 camp and VSM camp. I believe because of the specific manner the Standard Assumption #1 is phrased.
If Standard Assumption #1 were rephrased to: “The host must always open another door, one that was not picked by the contestant”. This phrasing would make crystal clear the 2 reasons WHY to choose only door #2 or door #3 AND leave door #1 alone because it is “off-limits” and also “because the show ends there” reason.
This would constrain the host from opening the contestant’s door pick and explain it is clearly “off-limits”, not simply, singularly “because the show ends there”. This would keep the “Finger on a Shell” so to speak. WHY the host chooses any door would become crystal clear. The 1/2 camp actually never wants to “KEEP their finger on the shell” at Q1. They will decide at Q2 when to “PUT their finger on the shell”.
WHY to keep door #1 pick “off-limits” reason not “because the show ends there” reason to the host when “opening a door not picked by the contestant”. The reason “WHY” this single idea, means absolutely everything. VSM, CP, 3-Card Game, goat-in-door drawings, Million Door Example, Finger-On-A-Shell, the MHP on Wikipedia, endless debate of 2/3 win rate with “always-switch” vs. 1/2 win rate with “random”.
The phrasing could be: "The host must always open another, a door that was not picked by the contestant”.
I believe this could clear things up with many arguments and confusion and rule/assumption interpretation. Maybe I am totally wrong, but for some it will be at least worth looking at to compare 2 perspectives. The VSM camp and the 1/2 camp. I have been in both camps.
FourthKind, if that wording for Standard Assumption #1 helps you understand the puzzle, go with it. I wouldn't change it in the article (for me, the current wording "a door that was not picked by the contestant" already implies "another door", so I find your wording a bit redundant for my tastes), but you can think about it that way. Diego (talk) 12:30, 14 March 2016 (UTC)
Diego. Thanks you are completely correct. My point is not that it is a shaky interpretation in itself. But that SA #1 to 1/2 camp might not see the subtle reason WHY the host chooses another door (missing the "WHY" part) the part that is important to the idea of 3 members of the sample space at Q1. This would reveal both the 2 reasons + WHY (essentially 3 reasons now) a SA #1, not just for the 2 reasons for SA #1 (another door + for the show not to end there). I added a small bit at the top to clear this idea up. Thanks
A Taste Test: A Matter of Taste or a Matter of Proof?
Latest comment: 8 years ago1 comment1 person in discussion
FourthKind: A Taste Test: A Matter of Taste or a Matter of Proof?
Notice the 3 Standard Assumptions all pertain to the host’s actions after Q1. What happened at Q1 with the contestant? This has to be surmised. It is implied the contestant already picked a “random or not random” door. This is the “neutralization” time for the contestant, as explained by Nalebuff (1987).
“Vos Savant asks for a decision, not a chance. And the chance aspects of how the car is hidden and how an unchosen door is opened are unknown. From this point of view, one has to remember that the player has two opportunities to make choices: first of all, which door to choose initially; and secondly, whether or not to switch. Since he does not know how the car is hidden nor how the host makes choices, he may be able to make use of his first choice opportunity, as it were to neutralize the actions of the team running the quiz show, including the host.”(end quote here)
Isn’t it important to make clear within the rules of the show not to alter the contestant's “neutralization” (Q1 answer)? This answer of Q1 by the contestant is only surmised (without proof) by Standard Assumption #1, with the statement: “The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).”
WHY the host opens “a door not picked by the contestant”. Because the host has 2 choices, because the show would end, or because the “neutralization” should not be touched? Better yet, all 3 would be answered if rephrased. It has been assumed “because the show would end there and can’t move on from that point”. This is a great reason for the argument that “the host needs to choose another door”. This is obvious and correct for 2 reasons, but SA #1 shouldn’t stop there and leave out the 3rd reason. The not so obvious 3rd reason is WHY another door other than the contestant’s door? SA #1 needs to add the PROOF for the 3rd reason. No surmising would be necessary, no implying would be necessary of SA #1. The surmising lacks proof pertaining to the 3rd REASON why the host opens another door. That 3rd reason is to protect the “neutralization” of choice by the contestant after answering Q1.
The 3 Standard Assumptions begin after Q1 is answered. These pertain to the host’s limits and actions. What about the contestant’s limits/actions before Q1? Are these “protected” via Standard Assumption #1? Or not? It is not clear from SA #1 that the contestant’s “neutralization” is protected. Rephrasing would be the proof it is protected and off limits, or “finger on the shell” so to speak. Or doesn’t the “neutralization” NEED to be protected anyway? If not there is no contestant “neutralization” that even took place at Q1. Should this “neutralization” be implied or surmised or “protected with proof” within the rules. If not protected wouldn’t this be incredibly unfair to the contestant and ALSO problem considerations involving strategies, thus solutions to the MHS/MHP?
If SA #1 were rephrased: “The host must always open another door, one that was not picked by the contestant”. This would be the proof for protection of that “neutralization” not just implied or surmised. This is extremely important to the entire problem isn’t it? After all the contestant only gets 2 decisions to make and 1 of them, the contestant’s answer to Q1 is skipped over only to be implied or surmised via SA #1. It’s protection via the Standard Assumption #1 is not proven. Does this sound rational?
SA #1 should make clear that this “neutralization” is protected, not only “because the show would end there and can’t move on from that point” reason, or the “host has to choose another door reason”. Make it crystal clear here at SA #1 that it is all 3 reasons. After all the 3 Standard Assumptions, make no mention of the contestant, and the contestant's actions must be “surmised” via the 3. Doesn’t the contestant’s “choice” answering Q1 deserve a clear protection?
Is this phrasing merely a matter of taste? Here is a taste test.
If it were phrased this way (above phrasing), doesn’t this make the Monty Hall Show just another, 3-Card Game example, Finger on a Shell example, Million Door example, Conditional Probability example, or even Monty Hall problem on Wikipedia? Yes, I believe the Monty Hall Show would become precisely another example. Let the VSM/Conditional Probability strategy decide right then and there , at Q1.
By not rephrasing SA #1, if SA #1 obviously means the “same anyway” implied/surmised logic from VSM proponents, aren’t VSM proponents kidding themselves? I think this rephrasing would give VSM proponents a “bad taste” in their mouths, that the MHS is reduced to exactly another VSM/Conditional Probability example. This is precisely what would happen if the SA #1 was phrased this way, just another “finger on the door” example. This would and should leave a “bad taste” in everyone’s mouth.
So why leave the phrasing alone if it obviously means the same thing? Why leave it just like it is currently (Standard Assumption #1 of MHP)?
“Repeated plays also make it clearer why switching is the better strategy. After the player picks his card, it is already determined whether switching will win the round for the player. If this is not convincing, the simulation can be done with the entire deck. (Gardner 1959b; Adams 1990). In this variant the car card goes to the host 51 times out of 52, and stays with the host no matter how many non-car cards are discarded.” This is why: “After the player picks his card, it is already determined whether switching will win the round for the player.”
So again: Why leave the phrasing alone if it obviously means the same thing? Why leave Standard Assumption #1 just like it is?
Should the MHS/MHP rules IMPLY "placing the finger on the shell" so to speak, or PROVE the contestant’s protection “placing their finger on the shell" (answering Q1) so to speak?
VSM proponents, MHP proponents, many people do not want to reduce the Monty Hall Show or the Monty Hall problem to another Conditional Probability exercise. It would leave a “bad taste” in the mouths of many. I believe the verbiage would reduce it to an exercise and many think SA #1 should be left as it is, ambiguously (exercise or The Monty Hall Show/Monty Hall problem) neglected.
Latest comment: 8 years ago3 comments3 people in discussion
FourthKind: Down the Rabbit Hole
The Monty Hall Problem is a Conditional Probability exercise employing the vos Savant Model (VSM) to “maximize” a winning strategy.
The Monty Hall Show was a TV show in reality. Vos Savant Model strategy, using Conditional Probability Theory, was adopted by the public to obtain a winning probability of 2/3, or at the least exceeding a “random” 1/2 strategy.
The Monty Hall Problem on Wikipedia, goat-in-door-drawings, 3-Card Game, Finger-on-a-Shell, Million Door Example, are all games/exercises/examples of Conditional Probability Theory problems solved via VSM. Great games/exercises/examples with the same strategy employed by the vos Savant Model to explain CPT. The Monty Hall Problem on Wikipedia, with respected references, explains the winning probability to be 2/3. The page describes in detail, a strategy employing Conditional Probability via VSM. The page also twists the concept of the game show with various versions also with detailed explanations and these explanations can add even more twists. To go as far down the rabbit hole as the public wants. Also to “maximize” (possibly improve over 2/3?) winning strategy. A Quantum Version plays with this idea.
The Monty Hall Problem on Wikipedia: https://en.wikipedia.org/wiki/Monty_Hall_problem
“The behavior of the host is key to the 2/3 solution. Ambiguities in the "Parade" version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution (vos Savant 1990a) printed alongside Whitaker's question implies and both Selvin (1975a) and vos Savant (1991a) explicitly define the role of the host as follows:
1.The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).
2.The host must always open a door to reveal a goat and never the car.
3.The host must always offer the chance to switch between the originally chosen door and the remaining closed door.” (end quote)
Proposed Rules which define both the host and contestant could be as follows for the Monty Hall Show, which arguably may not be similar to Ms. vos Savant rules. They define clearly the roles of the contestant and host.
1: The contestant chooses a door from 3 doors, only 1 of which contains the prize.
2: The host always exposes a prize-less door from the other 2 doors.
3: The host always offers the contestant to choose between the 2 unexposed doors.
These proposed 3 rules clearly place the contestant’s “finger on the Door” if you will. The contestant made their 1st decision, choosing an initial door at Q1. The contestant’s 2nd decision to “switch” or “stay” is at Q2. The rest then becomes a Conditional Probability exercise with a 2/3 probability of winning excluding the 2 issues stated below.
The Monty Hall Show and VSM have 2 issues/concepts that are not addressed when employing VSM for a winning MHS strategy.
1) How the initial door is chosen by the contestant. In other words the contestant’s “strategy” in answering Q1. The contestant can use any strategy imaginable for this choice. This Q1 “choice” by the contestant has been stated as a “neutralization” choice. This is one difference between virtual (Monty Hall Problem) and reality (The Monty Hall Show).
VSM has no consideration for any “neutralization” of the host or producers biases as stated above. There are hosts or producers, no biases, in VSM. A model which has precise initial door distribution before Q1, perfect rotation of the contestant door choice distribution at Q2 complete with all possible combination of outcomes. Here lies one discussion area between VSM and reality, The Monty Hall Show.
2) The time interval between Q1 and Q2. A LOT can happen in reality in the few minutes between Q1 and Q2 on the Monty Hall Show. No model has been developed to predict some, any, or ALL the events between Q1 and Q2 to date.
The VSM ignores (incomplete defining of reality here) time between Q1 and Q2. It can ignore it. It’s a strategic model and do what it wants. If we adopt that model, that’s our problem, isn’t it? A model that does not and can’t address the time gap between Q1 and Q2. No time consideration. Here lies a great area of contention. It is a fantastic Conditional Probability teaching tool squeezed into a game show strategy with no consideration of time between Q1 and Q2. An example is “random”. A perfectly plausible strategy in reality, but with no consideration in VSM. Another is information gained by the contestant between Q1 and Q2. The sound of a “bleating-goat” from one of the doors, the host “tells” and so on. All considered “variables”, which are not and cannot be addressed within VSM. The model’s strategy is powerful enough (2/3 probability of winning) to simply ignore/overcome the variables. Wikipedians didn’t ignore the variables, and these have not been overcome in reality thus endless discussion and argument.
In a model or game there is virtually no time lapse or need for time consideration. Nothing to alter probabilities can even happen between Q1 and Q2 in a model, game or exercise. In the models/games/explanations time is ignored. Should it be in reality?
In 1992 Monty Hall was interviewed by the New York Times. Monty stated:
"[After I opened a door with a goat], they'd think the odds on their door had now gone up to 1 in 2, so they hated to give up the door no matter how much money I offered...The higher I got, the more [they] thought the car was behind [the other door]. I wanted to con [them] into switching there. That's the kind of thing I can do when I'm in control of the game. You may think you have probability going for you when you follow the answer in her column, but there's the psychological factor to consider." (end quote)
VSM strategy ignores these 2 differences. They don't need to be addressed in VSM. The public adopted VSM. This is the center of most discussion. The incomplete (my belief) role definitions rules add more discussion. The math within VSM has been decided/proven years ago by many great mathematicians.
A specific example of Conditional Probability Theory has been explained employing the vos Savant Model. VSM outcomes (2/3 win rate) were proof needed to employ this strategy as being the de facto strategy to obtain similar outcomes (2/3 win rate) in MHP/MHS. The 2 issues/concepts stated above are painfully missing in VSM. And the pain is apparent in discussions.
“In an attempt to clarify her answer she proposed a shell game (Gardner 1982) to illustrate: "You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? (end quote)
Motivated by a 2/3 winning strategy, the public by adopting the VSM model with it’s issues and deficiencies, didn’t we all “put our finger on a shell”, agreed?
PS: Great thanks to Diego and Rick and Martin for putting up with me as I descended the rabbit hole. Probably (2/3 probability) my last "lecture". Thanks again all of you.
Yes, we all put out finger on a shell. Because that's the rules of the game. Don't confuse the game here with the game show "Let's Make a deal" the two are not similar and are hardly comparable. 176.248.76.162 (talk) 16:29, 17 March 2016 (UTC)
FourthKind, I think you are still missing an important detail: the Monty Hall Problem in its VSM formulation, which is the topic of the article, does not attempt to describe the whole reality of the Monty Hall Show spectacle.
The article already explains this indirectly, when it says that "The problem is actually an extrapolation from the game show", and that Monty in the show didn't play following the rules stated in the problem.
You had the right idea when you described the VSM as a good probability teaching tool; the reason why it works well as such is because that is its intended purpose. It is not fair to blame that model for not being able to represent everything that may happen in reality, as that was not its goal. The goal of the model is to build a quiz, a puzzle that illustrates how the rules of probability often produce "paradoxical" non-intuitive results. Mathematicians focus in that version of the problem because that paradoxical property of the model makes it interesting.
As I said before, if you try to build a model for events and rules in the game other than those stated in the "VSM-finger on a shell-3 card-million doors" puzzle, you are solving a different problem, which most of us will not find as interesting as the VSM one (or at least just interesting in different ways). Diego (talk) 14:40, 18 March 2016 (UTC)
Food For Thought
Latest comment: 8 years ago10 comments6 people in discussion
Assume we have 2 contestants.
Contestant 1 is brought onto the stage, and it is explained to them that there are 3 closed doors, and behind two of them are goats, behind the other is a Car.
They are then asked to select a door at random, and advise what they think there chances of winning the car would be. They would reply 1 in 3 chance.
They are then asked, if they were allowed to switch to both of the other two doors, what would their chances be then, and would they be prepared to change their selection. They would reply, 2 in 3, and yes, they would be happy to switch.
They are then asked, if you were going to open both the other doors, would it matter which order they were opened. They would reply, no it wouldn't. And then asked, since there has to be at least one goat behind those two doors, if I open that one first, before opening the 2nd one, will that change your odds?. And they would reply, it would not.
Without having opened any doors, get them to confirm what they think the chances are of their original selection having the car, and they would say 1 in 3.
Put a blindfold on them, then open one of the other two doors, in particular the one that has a goat behind it.
Bring contestant 2 onto the stage, who has not seen any of the earlier actions. It is explained to them that there are two closed doors, behind one is a goat, the other a car.
They are then asked to select a door at random, and advise what they think there chances of winning the car would be. They would reply 1 in 2 chance.
They are then asked, if they were allowed to switch to the other closed door, what would their chances be for it to be the car, and would they be prepared to change their selection. They would reply, 1 in 2, and that it would make no difference to switch.
They are then asked to confirm, that they are saying it will not matter which door they open, they have a 1 in 2 chance with each of them. Yes, that is correct they would say.
The host then points to the door that contestant one selected, and contestant one is asked to confirm what they thought their chances were for their original selection, and they would say 1 in 3. Contestant 2 is then asked about the same door, and asked to confirm what they thought their chances were of having the car, they would say 1 in 2.
They are both talking about the same door, one is saying 1 in 3, the other 1 in 2. They are both correct.
How can that be?
Note that if the game was played multiple times, contestant 2 would win the car 50% of the time, and they would end up selecting the same door as contestant 1 50% of the time. However contestant 1 would only win the car 33% of the time.
Whilst contestant 2 is pointing to the door that contestant 1 selected, and saying their chances would be 50%, if they were encouraged to switch to that door when they hadn't picked it, their actual winning chances would drop to 33%.
So contestant 2 is saying 50% for both doors, but always selecting the door of contestant 1 reduces their chance to 33%.
How can that be? — Preceding unsigned comment added by 121.222.9.68 (talk) 05:16, 7 April 2016 (UTC)
>They are both talking about the same door, one is saying 1 in 3, the other 1 in 2. They are both correct. How can that be?
This can be because they are using different processes to choose their respective doors, and there are times when the door they pick is not the same; it's therefore perfectly reasonable that they can have different probabilities.
> Note that if the game was played multiple times, contestant 2 would win the car 50% of the time, and they would end up selecting the same door as contestant 1 50% of the time. However contestant 1 would only win the car 33% of the time. [...] How can this be?
This happens because, those times when contestant 2 chooses the other door that contestant 1 did not pick, she is twice as likely to win than when she picks the same door. Thus contestant 2 averages to 50% of wins, but contestant 1 does not.
> always selecting the door of contestant 1 reduces their chance to 33%. How can this be?
And this is because, if contestant door always selects the same door, she is no longer randomly picking a door at 50%, but (unknowingly to her) she picks a door which was selected through a process with a lower probability of containing a car.
Actually, if contestant 2 repeatedly played the game and was forced to always choose the same door as contestant 1, she would win only 33% of games; so, following the scientific method she should assume that there's something funny going on in the way that contestant 1 picked her first door; contestant 2 could learn this way that contestant 1 did not pick between doors with a random 50% choice.
The part where contestant says "it will not matter which door they open, they have a 1 in 2 chance with each of them" is correct if and only if you also assume "as long as you choose between them randomly with 1/2 probability, and don't use information provided by contestant 1 to choose the door". Diego (talk) 17:19, 7 April 2016 (UTC)
Hi, I think a concept is still ignored here. The part where the contestant "would reply 1 in 3 chance". They would be wrong. Does the game end there with 3 doors? Does the host open the other 2 doors right then and there with no Q2? No. The concept is they don't choose between winning and losing at Q1. At Q2 they have 2 doors to choose from, win or lose, right then and there in the game. The probability was never 1 in 3 unless you use Conditional Probability and try hard to lose repeatedly. Again, only in the model. Only when the finger is on the shell and the contestant can't choose randomly when 2 shells remain because the rules keep one's finger on the shell at a moment in time with a probability 1/3 if the game ended then. Only when the player has 2 options of 1/3 or 2/3 in the 3-Card Game, only when 1 million doors remain is the probability 1/1,000,000. These models/games all narrow down to 2 possible shells, 2 possible cards, 2 possible doors, with win/lose probability of 1/2. Not less. The models show less than a 1/2 probability because they predict so. That's it. At Q1 winning or losing probability is 0 in reality. That means impossible. In the model the winning probability is 1/3. So be it, it's a model. And it is impossible to win or lose at Q1 because the game goes on to Q2 where a real choice is narrowed down to 2 "choices" at Q2 later in the show. Not a reference door, not a finger on a shell, when the probability is 1 in 3 at that moment but the game is never over at that moment at Q1. Not repeated runs of the experiment/game with a cherry picked moment in time from the model. In reality it is impossible with 2 choices to lose 2/3 times. How can a choice between 2 doors in 1 Million doors have a loss rate of 999,999/1,000,000 with 2 doors left? The model told everyone so. That's why. Not in reality.
"Does the game end there with 3 doors? Does the host open the other 2 doors right then and there with no Q2?" In this version of the game, as stated by the IP editor, for contestant 1, the answer is yes. The first contestant is never given a Q2 option to change their first decision in the steps defined above. Diego (talk) 16:40, 11 April 2016 (UTC)
The answer is because probabilities are not absolute but relative to information. An individual coin toss is, say, 100% chance heads and 0% chance tails because the physics is such that it will land heads. To someone who doesn't know enough about the physics but just coin tosses in general can only calculate the probability to 50%/50%. Someone who knows that coins land more often the same way as they were tossed from and that the coin was tails up to begin with would calculate 48%/50%. Someone who knew the person tossing the coin was a skilled manipulator aiming for heads might calculate the probabilities as, say, 95%/5%. These are all correct for the given information. In the monty hall problem the host knows that switching is either 100% to win or 0% to win. The contestant doesn't have as much information so can calculate switching is 66% to win. Someone brought on stage at the end has even less information and so can not calculate more accurately than 50% to win. Information is the missing component here. SPACKlick (talk) 23:02, 9 April 2016 (UTC)
@Fourthkind
Re your comment:
I think a concept is still ignored here. The part where the contestant "would reply 1 in 3 chance". They would be wrong.
Sorry, YOU are wrong. I presented a modified Monty Hall problem, where the contestant is asked, after having selected a door, what they think their chances are. It is their chances right there, at the time they are asked the question. What happens after that is not relevant to the question they are asked, when they are asked.
The subsequent discussion was to highlight that the Monty Hall problem is really a choice about one door or two. If the contestant was asked, would you think you had a better chance if I allowed you to pick two doors instead of one, they would say yes. We all would. The original Monty Hall problem asks the question is such a way that it disguises this fact.
Also, a 1 in 3 chance is not the same as the number of outcomes. There are only two outcomes, win or lose, assuming one car only, regardless of the number of doors. 1 in 3 chance means what it implies, that they would win once in every three times they played.
Re your comment:
In reality it is impossible with 2 choices to lose 2/3 times. How can a choice between 2 doors in 1 Million doors have a loss rate of 999,999/1,000,000 with 2 doors left? The model told everyone so. That's why. Not in reality.
Well, how come reality proves YOU wrong. Yes, after having picked a door, they get a choice, stay with the door they picked, or select the unopened door of the other two. Two options, two possible outcomes. But as noted, we could have three or more options, still only two outcomes, so don't confuse outcomes with chances. In reality, as the tests have proven, play the game 100 times, stay with your original choice, and 33 times, give or take a few, you will win the car. Switch your choice, your other option, and 66 times, give or take, and you win.
Win or lose is an outcome. It is not a choice. Two doors is a choice, but doesn't mean you have the same chance with each. Reality proves you win only 1 time in 3 with your original selection.
Also, if the host said, we are going to play this game 3 times, after you pick a door, once I will put the car behind that door, the other two times I will put the car behind one of the other two doors. Then they open one of the remaining two doors that doesn't have the car. What say your chances of winning that car with your original selection. 1 in 3. Fact. And your chances of winning by changing selection. Two times out of every 3. Fact. — Preceding unsigned comment added by 124.177.113.188 (talk) 08:54, 10 April 2016 (UTC)
Hello User talk, with all due respect, I believe you still miss the concept of 2/3 loss probability or win probability 1/3. And here is why. Can you recreate a 2/3 loss rate or 1/3 win rate in reality without a knowledge of Conditional Probability? The answer is no.
You stated: "Win or lose is an outcome. It is not a choice." You just made the choice an outcome. An outcome with a 1/3 win rate. An outcome with false loss probability of 2/3 in reality.
You just said: "Sorry, YOU are wrong. I presented a modified Monty Hall problem, where the contestant is asked, after having selected a door, what they think their chances are. It is their chances right there, at the time they are asked the question. What happens after that is not relevant to the question they are asked, when they are asked." (end quote)
The truth is it is relevant to the question "what happens after that". They win or lose after that, not then, when asked the question. It is everything. Their perception of the probability of winning is incorrect at 1/3. What happens later at Q2, with a 1/2 probability of losing is the true loss probability. Unless, as I stated way above, they use CP and try hard to repeatedly lose.
Your original statement: "what they think there chances of winning the car would be. They would reply 1 in 3 chance." (end quote)
If the contestant is aware of CP, then why would they say such a thing about losing 2/3 times at a 1/2 win/loss probability rate at Q2 in reality. Who are you trying to impress with a loss rate of 2/3 at a 1/2 probability at Q2? The model told you so. A teaching model to impress children, stretched and squeezed into a game show, then denied as an accurate comparison (see above comment to me) to the game show "Let's Make a Deal". The win rate with CP can exceed 1/2 win rate in reality but not the loss rate of 2/3. This is not accurate in reality. Only if you are forced to "keep your finger on the shell" will you achieve a 2/3 loss rate. No contestant is forced to do such a thing. Are they? Did the model tell them so? It is like a religion. I wrote about this in detail, but it it archived by now.
You stated: "Can you recreate a 2/3 loss rate or 1/3 win rate in reality without a knowledge of Conditional Probability? The answer is no."
You're wrong. As the IP said, if you ask "Do you want to pick one door at random, or two doors at random?", the contestant can guess that the probability of winning is higher by keeping two doors instead of one, without using conditional probability. Same as if you offer "do you want me to give you for free one lottery ticket, or ten thousand lottery tickets?" Of course the latter has more chances of winning than the former.
Deigo, do you realize what you are saying here? You "proved" me wrong with 2 questions. A simple question. How can an experiment with 2 probable outcomes have any other than a close 1/2 win or lose outcome?
That is possible because not all random experiments have a uniform distribution; there are many other common probability distributions. My previous post was a proof by counterexample, showing an instance of an experiment with two outcomes with different probabilities. Another example is the unfair coin, so called because its both outcomes (heads and tails) do not both have 1/2 probability. Your main error in all this discussion always has been thinking that, because one process has two outcomes, each one necessarily must happen half the time; but probability doesn't work that way. You've started from that fixed position and reasoned your way back to the problem, finding the "green woman" version as the only way compatible with your pre-existing conclusion; but your reasoning only works for some cases, and fails for the most common formulation of the problem. Diego (talk) 21:42, 13 April 2016 (UTC)
"What happens later at Q2, with a 1/2 probability of losing is the true loss probability." Again wrong. You're assuming that the contestant at Q2 chooses without having any information of what happened before, but that is not what happens in the several versions of the game as stated.
In particular, the game posted by the IP at this section does NOT have a choice for contestant 1 at Q2; in that game, the first contestant is deliberately forced to stick with the first choice. In that case, it's impossible to change the first 1/3 probability to anything else; in that game, contestant 1 will win only on in three games. There is nothing that contestant can do to improve their chances, as in that game the only choice of contestant 1 is one door among three.
I have a simpler version of the game:
There is a contestant. Someone explains her that there are 3 closed doors, and behind two of them are goats, behind the other is a Car.
The contestant is asked to choose two doors and discard one; she can get the best one of whatever prices are revealed when those two doors are opened.
Just after the contestant picks those two doors, she is asked if she wants to discard both and switch to the remaining door.
Should she accept that final offer? She is choosing among two options (getting the better price in the set of two doors, or the only price in the set of one door). Do both options have 1/2 probability of winning a car? Does the player need to know conditional probability for making the choice? Diego (talk) 13:27, 11 April 2016 (UTC)
Re Quote above: ....Only when the finger is on the shell and the contestant can't choose randomly when 2 shells remain because the rules keep one's finger on the shell at a moment in time with a probability 1/3 if the game ended then.
So you agree then.
In the modified Food for thought game, contestant one is never given the option to switch. There is no question 2. Their chance, when asked the question, is 1 in 3. When another door is opened to reveal a goat, and always a goat, it does not change the chance of that original selection, because, as you put it, the finger is on the shell, and they can't choose randomly between the two doors.
Contestant 2 is allowed to randomly choose between the two doors, and their chance is 1 in 2.
In the original Monty Hall problem, there is no winning, or losing, the game doesn't end. After the door is opened to reveal the goat, the contestant might choose from a number of strategies to give themselves the best chance of winning. But the problem isn't about winning, it's about answering the question. Their strategies
1: Stick with their original selection
2: Change to the other selection
3: Make a new selection randomly
4: Bribe the host, offering to share the winnings
There could be other strategies as well, but these, as well as strategy 4 above, are off into fairyland.
Strategy 3 would provide a 1 in 2 chance of winning. But here's the catch. The contestant can't choose randomly when 2 doors remain. The original problem doesn't explicitly exclude them from doing this, however they are required to answer a question, and they are only give a choice between strategy 1 and 2. They do not get to randomly make a new selection. Stay with the original choice or switch. Their original choice remains a 1 in 3 chance. You have acknowledged that. Randomly selecting a door would give them a 1 in 2 chance, but that option is not available to them, they are required to answer the question. Stay with their original selected door, 1 in 3 chance, or switch.
When the door is opened to reveal the goat, and the question is asked. do you want to stay with the original selection or switch, the contestant could elect to randomly select a door, say with a coin toss, and if this new selection matches the original, they would advise the host they would stay, if different, they would switch. Their winning chances would now be 1 in 2. Is that the best strategy though? Why choose a 1 in 2 chance if there is an option for an even better chance than 1 in 2?
Real world, if you found yourself on the stage, with the host, and with the standard 3 assumptions, i.e about the goat always being revealed etc, which of the following 3 strategies would you employ, in order to give yourself the best chance to win?
1: Stick with their original selection
2: Change to the other selection
3: Make a new selection randomly
Re: your comment: Only if you are forced to "keep your finger on the shell" will you achieve a 2/3 loss rate. No contestant is forced to do such a thing. Are they?
Exactly. As you clearly state, if they are forced to stay with their original selection, they will achieve a 2/3 loss rate. Or 1/3 win rate. Same thing. So why do it? If sticking with the original selection gives only 1/3 win chance, then doesn't that tell the contestant not to do it. That is, don't stay with your original selection. So what then, switch or random?
If one stays with the original selection, whether forced or voluntary, there is only 1/3 chance. Play the game 3 times, win only once. Play the game 300 times, win 100 times. So what chance does that leave with switching? Or with random? Play the game 300 times, how many times will one win by staying, switching or random?
I've Finished Eating
Latest comment: 8 years ago2 comments2 people in discussion
Hi Diego, you still avoided the exact question posed. How can an experiment with 2 probable outcomes have anything other than a close 1/2 win or lose outcome without the knowledge of conditional probability from the players? You gave an answer for probability distribution and unfair coin with biased probabilities.
Choose randomly with 2 possible outcomes and get a close 1/2 win or lose ratio. That is a fact you don't seem to acknowledge. Why is this so difficult to swallow? Did the model tell you so? Remember the model has no room for random. Just always-stay or always-switch and proof will follow. Always choose randomly and no advantage. Try it. It works. Let me know. Alternate answers at Q2 and you will get a 1/2 win/loss ratio. It's proof Diego. Whether the "correct" answer is on one side or the other side of the "choice" the dumb, random will find it 1/2 times. You are not correct. Outcomes not probabilities. I said "outcomes" and you responded with 2 examples of variances in "probabilities". Probability distribution and unfair coin. What does a biased-weighted coin have to do with outcomes in the Monty Hall Show in a one time event? I think the concept of "outcome" is missed here by you. Outcomes can be counted physically and thus proven in the real world.
Even if someone did have ANY coin and ANY distribution in a one-time event isn't it random anyway because the player chooses which is win or lose from heads or tails, which is random in itself because of no bias/knowledge used in guessing in a one-time event, with 2 equal probable outcomes created in this circumstance by the guesser. A guess is a guess in a one-time event with 2 probable outcomes. No one can prove why someone won or lost in a single event with 2 outcomes. Probabilities yes, outcomes no. The proof is in repetition of any experiment is science. Proof by repetition of a single event of 2 outcomes, no. Unfair distribution is neutralized at Q1 in Let's Make a Deal in reality, Ms.Vos Savant said so herself in an interview which you also seem to ignore.
The game played on Wikipedia is the "Monty Hall Problem", which has no neutralization available as explained above to me by Talk: "Yes, we all put out finger on a shell. Because that's the rules of the game. Don't confuse the game here with the game show "Let's Make a deal" the two are not similar and are hardly comparable. 176.248.76.162 (talk) 16:29, 17 March 2016 (UTC)" end quote.
As for all your other comments, I will ignore here as I can tell by your attempts of destruction to every one of my comments that something is getting to you so I will relieve you of future pain and suffering.
Thanks for the comments anyway, at least someone is watching here.
FourthKind, I'm saddened that you thought my attempts to enlighten you were attacks to your comments, when they were intended to help you. You have a severe misunderstanding of what the other editors have been talking about, and we have tried to explain you our position, but obviously there has been an insurmountable communication problem. We all recognized long ago that, if you play the Monty Hall Problem with a strategy of deciding the final door at Q2 with a fair coin, the probability of winning is 1/2; what you don't understand is that this result is not a solution of the question stated by the Monty Hall Problem, which in the end asks which is the best possible strategy.
Now, I'll address your main points.
you still avoided the exact question posed. [...] You gave an answer for probability distribution and unfair coin with biased probabilities.
I did not avoid the exact question. My answer about biased probabilities is the answer to how a random process with two outcomes (head and tails) can have different probabilities for each outcome (which is the definition of biased coin). A biased coin is an instance of a random process which has two outcomes, and where each outcome does not have 1/2 probability of happening. Or, in the case of my modified version of the game where Monty never reveals a goat and therefore the player chooses between "one door" or "two doors", there's no room to apply conditional probability as there never is a Q2 question; yet choosing the "two doors" option is twice as likely to produce a "win" outcome.
Choose randomly with 2 possible outcomes and get a close 1/2 win or lose ratio. That is a fact you don't seem to acknowledge
I didn't acknowledge it because the meaning of those words, as a mathematician would understand them, is not a universal fact. There are random processes that behave that way, such as playing the Monty Hall Problem by throwing a fair coin; but there are random processes which don't behave that way, like playing the Monty Hall Problem and always switching, or playing the Monty Hall Problem using a biased coin; and that is a fact that for some reason you don't acknowledge.
Remember the model has no room for random. Just always-stay or always-switch and proof will follow. Always choose randomly and no advantage. Try it. It works. Let me know. Alternate answers at Q2 and you will get a 1/2 win/loss ratio. It's proof Diego.
And I have acknowledged that if you play that way, you'll get that 1/2 ratio; I never denied that. What I deny is that throwing a coin is the best strategy, because there are other better ones.
Now, your use of the term "random" may be the source of your misunderstanding. When people talk about random processes, it means that the outcome happens at random, not that the decisions of the players are taken at random. In the Monty Hall problem, the random event refers to the position of the car behind one door, not to the player's choice, which is a decision (and may be decided by any strategy, random or not). In other words, you have an essential misunderstanding of what the other editors have been talking about when we used the word "random".
The "always switch" strategy does not make the game suddenly not random, because the placement of the car behind each door is still a random process. And the distribution of the car placement behind both remaining doors after Q2 is *not* a 1/2 distribution, which is what the model describes, but also what happens in reality if you follow the "always switch" strategy.
You can try *this* strategy in the real world, and it will work out by making you win much more often - not because a model says so, but because the deterministic strategy produces that 2/3 distribution in the random part of the game (the placement of the car with respect to the chosen door).
What does a biased-weighted coin have to do with outcomes in the Monty Hall Show in a one time event?
For a start, if you play the Monty Hall Problem with a biased coin that says "switch" 95% times, you won't win 1/2 times, but much more often. So it's not true that choosing between two doors at random at Q2 will always produce 1/2 probability of winning; it depends of which kind of "random" you use (you can select the doors with a random 1/2 process, or with a different random process; but you only accept the first strategy as "reality").
Even one-time events can have probabilities defined, in this case by assessing how many of the possible outcomes of the experiment produce a "win"; each outcome will have a probability calculated from the problem's full logical possibilities. A frequentist statistician would tell you that, of the three possible outcomes of the random placement of the car, two of them produce a "win" if you play the game by switching, and one of them produces a "lose". A Bayesian statistician would tell you to assign equal probabilities to each outcome if you don't know additional knowledge (such as deciding between two doors if you don't know how anything about how the car has been placed behind them), but to refine the conditional probability if you have additional information (such as knowing how the host has discarded one door at Q1).
In that context, playing than one-off event with a strategy of throwing a fair coin will have a calculated probability of 1/2 for winning that single game, deciding to switch will have a calculated probability of 2/3, and throwing a 95% biased coin will have a calculated probability of slightly below 2/3 of winning the single game. This is not calculated from repeated experiments but by the combined possible outcomes of the random car placement and random (1/2, 95% or 100%) selection of "switch" or "stay".
Don't confuse the game here with the game show "Let's Make a deal" the two are not similar and are hardly comparable.
As stated in the article, the real show wasn't comparable either to the version where the player chooses, because the host not always offered the players the possibility to switch, nor would always open a door containing a goat; we have always been talking about different versions of a mathematical problem, all of which can be played in reality as games, but none of which are comparable to the original show.
It's up to you accept these comments as advice to make you understand how random processes are studied; I hope some time in the future this conversation helps you learn how information is used at different steps of a process involving random events. Diego (talk) 13:46, 15 April 2016 (UTC)
Fess up
Latest comment: 8 years ago4 comments4 people in discussion
I have observed thoroughly what's going here. My issue has never been with the math of conditional probability. As you can see I have passed all you or your editors attempts to discredit my knowledge of the math issue when tested. Logic questions of specific rules and definitions have consistently been answered and directed toward me as though it is my lack of understanding the math. It is a lack of understanding your logic, not the math.
What is happening here is consistently avoiding logic challenges and questions replied with math answers. Shaky rules and problem definitions answered mathematically with great detail how I am not comprehending it's complexity. It is because of certain logic issues of the rules and definition of the exact problem, as I have always stated, the math would THEN be an issue.
Apparently here, one can distance themselves from "Let's Make a Deal" when necessary and attach themselves to "Monty Hall" problem when convenient.
Because of the shaky rules, the math, as I have pointed out then becomes a problem. Ms. Savant herself won’t defend her model under certain conditions (rules). On her website she makes no connection, and specifically does not mention the words “Let’s Make a Deal” or “Monty Hall”. Why is this? She knows better. Yet the game show, hosted by Monty Hall is used the very name for a problem(?) on Wikipedia. What problem is that? Exactly what problem would that be to “maximize” game strategy for math which has withstood challenges from 40 years ago? Diego you said a purpose of the Wikipedia page is to “maximize” strategy of the game. What new “maximization” of game strategy has occurred in 40 years? Does anyone EVER expect to exceed a 2/3 win rate on a specific conditional probability problem as the "Monty Hall problem", whatever that is?
The article and discussions are about a SPECIFIC logic problem which is called "The Monty Hall Problem". This problem is only loosely based on he game show "Let's Make a Deal". The reason people distance from the show is that the show is in no way relevant to the problem. Even when discussing the problem you have said a choice between two options cannot have a 2:1 probability and must be 50/50. Are you now changing that position? SPACKlick (talk) 12:05, 20 April 2016 (UTC)
Indeed. The problem includes what we know about the setup. If all we know is that there are two doors behind one of which is a goat (or nothing) and behind the other the car, the chances of picking the car are 50/50. But in the Monty Hall problem we know more than that, and what we know affects the chances of where the car is.
For example, if we knew the host rolls a 6-sided die and puts the car behind door #1 if the roll is 1 or 2 and puts the car behind door #2 if the roll is 3,4,5, or 6 we know (because of what we know about the setup) that the car has a 1/3 chance of being behind door #1 and a 2/3 chance of being behind door #2. In this case do you want door #1 or door #2? There are only two doors. Does it matter? If you do this (rolling a die, etc) 150 times, you'll observe the car is behind door #1 about 50 times and behind door #2 about 100 times.
The setup in the MHP has exactly the same result. If you randomly place the car 300 times, the player initially picks door #1, and then the host opens one of door #2 or door #3 showing a goat (randomly picking which of these to open if the car is behind door #1), you'll be left with about 150 instances where the car is behind door #1 about 50 times and behind door #2 about 100 times. In some thread above, FourthKind agreed with this. That means at Q2, when there are only two doors, the chance the car is behind the door the player initially picked (door #1 in the concrete example cited in the the problem statement) is 1/3 while the chance the car is behind the other door (door #2 in the example) is 2/3.
Two doors, but unequal chances. -- Rick Block (talk) 15:19, 20 April 2016 (UTC)
FourthKind, you have stated that you understand the logic behind the 2/3 probability in the model that describes the Monty Hall Problem, and your uneasiness is with the way we defend that model as an accurate representation of the rules stated in the article, over the strategy of choosing a door at random - which, if I understand you right, you think is a superior representation.
You are right that the Monty Hall Problem only represent a specific situation that might happen during the "Let's Make a Deal" show, and not the full game. I think what you miss is that your preferred scenario where a player chooses at random amongst two doors is itself a model, and it doesn't represent the real show either. As explained in the article, the real host of the show didn't always open a door, and didn't always offer an option to switch doors; therefore, neither the "2/3 scenario" nor the "1/2 scenario" are accurate descriptions of the show, only a subset of it. We choose to focus on these scenarios because of their interesting mathematical properties, not because they best describe the original game.
In that respect, you were right when you said that the model is chosen as a tool to teach probability; and not when you said that the 1/2 random choice represented Reality. There is no model that represents everything that could happen in the TV show because the rules in the show were not well defined, and they where changed on a whim of the producers and the host; any mathematical study will necessarily be a simplification of whatever may happen in reality. Savant's Monty Hall Problem is inspired by the "Let's Make a Deal" show, not a model of it; that's the reason why it doesn't even try to a include a complete set of rules that describes everything that may happen. The "always switch" strategy is the one that maximizes the changes of winning under the limited set of rules described as the "standard assumptions" in the article.
I now think that I have misunderstood your concerns during the whole conversation. If this is so, I apologize. I first thought you were saying that, if you played the game following the rules as stated by Savant, you wouldn't win 2/3 of times even if you followed the "always switch" strategy. Now I realize that your concern the whole time is regarding how Savant's model is (loosely) related to the show. I hope my above explanation clarifies this for you. Diego (talk) 22:23, 20 April 2016 (UTC)
it's just this simple....
Latest comment: 8 years ago4 comments4 people in discussion
since which of the 3 door will be the winning door is initially set up at random, any unopened door has the same probability of being the winner than any other unopened door…no matter what happens, one unopened door cannot become more probable than another...
thus, if 3 doors are unopened, probability for each is 1/3...if 2 doors are unopened, 1/2...24.59.126.91 (talk) 19:22, 11 May 2016 (UTC)
Why do you think that, no matter what happens, the probability of unopened doors cannot change? If we open one door and it contains a car, do the other unopened doors still have 1/3 probability of having the only car? Diego (talk) 23:20, 11 May 2016 (UTC)
Diego, the previous poster was NOT arguing that the probability of an unopened cannot change, rather, that the probability of two unopened doors is the same. You will note that they changed the probability from 1/3 to 1/2 after one door was opened, so they are saying the probability for both unopened door changes. Of course, they miss entirely the point, that it is really a choice between 1 door or 2 when the option to switch is offered. — Preceding unsigned comment added by 123.211.46.30 (talk) 07:55, 13 May 2016 (UTC)
You can change the problem by moving the decision point to before the host opens a door (and, after deciding whether to stay or switch the host then opens a door), and doing this does indeed make it a choice between one door or two doors. As typically stated, at the point the option to switch is offered, door #3 is already open - so the choice is actually between only two unopened doors. However, due to the process the host uses to open a door (including some assumptions typically not stated) the (conditional) probability the car is behind the remaining two doors is not the same. -- Rick Block (talk) 15:32, 13 May 2016 (UTC)
Counterargument
Latest comment: 7 years ago3 comments3 people in discussion
This problem is a case that shows that dogmatic mathematics sometimes are not related to reality merely by ignoring a fact of that reality and by sinning against one of its own rules.
There is one premisse in the consideration of the problem that is completely ignored in the pro-switching advice. In the beginning it is indeed true that there is only a 1/3 chance that the car is behind the chosen door or in the 100-door situation a 1/100 chance. However, that original situation is abandoned as soon as one door is opened; at that moment it does not matter whether it has existed because it is not continued but completely replaced by something else with its own situational considerations. Consequently, it is faulty arguing that it is allowed to apply the initial rules to a completely different situation, and therefore then the switching advice has no basis. The new situation is a case on its own right of a choice between two objects, not at all related to the initial one.
In mathematical terms the flaw is, that a different rule in the same mathematics does not allow to unconditionally transfer the value of one element of a matrix to one other one, something that is sinned against in the 2/3 arguing. The one-dimensional three-element matrix is replaced by a two-element one while the 3-element considerations are not, but are bent to fit for continuation by the no-no of assigning the addition of two element values to one defined element. The sum of three elements initially is 1, but if one of those elements is eliminated, the sum does not stay the same but is reduced with the value of that element. If I want to maintain the same sum, the value of the eliminated element must be shared by both remaining ones, bringing both up to 1/2. It is not allowed to transport the whole value of the eliminated element to just one selected other one.
To show this flaw in this kind of logic, let us turn the situation around and start with a two-door situation; there then is a 1/2 chance. We add now 1 empty door and apply the same faulty logic as before. This would thus lead to the chance of the chosen door being reduced to 1/3. Not true? Indeed not, because the considerations of the initial matrix situation can not be transported to the new one. The only difference between the two cases is that, because of the procedure being reversed by adding a NULL-element, in the decreasing doors problem the final situation is the ruling one (decreasing + replacement) while in the increasing doors one it is the initial one (increasing + non-replacement). The procedure itself is immaterial and switching doors does not have any effect in any of the two.
That a math rule can be invalidated by one of its co-rules is so nicely shown in the well-known x2-x2=x2-x2; x(x-x)=(x+x)(x-x); x=x+x -> x=2x.
What happens before can evidently have an impact on the probability of what comes after, for instance at the beginning you accept there is a 1/3 chance. This is determined by the fact that the doors are previously arranged with 1 car and 2 goats. If each door was given a prize from a pool of options the those odds may change dependent on how it was selected.
It is clear from practical tests of the monty hall problem that the way the door to open is chosen has an impact on the proportion of cases where the originally chosen door and the remaining door contain the prize. The maths, which is accurate, demonstrates how this proportion is determined. SPACKlick (talk) 22:35, 3 August 2016 (UTC)
(edit conflict) I have no idea what you mean by "a different rule in the same mathematics does not allow to unconditionally transfer the value of one element of a matrix to one other one". The rest of your comment is so vague that you don't make any points I can refute. I'm not sure what you mean by "co-rules" either. The second statement in the last line of your comment does not imply the third or fourth ones, which are false for all real x. If you don't believe that the probability of winning when you switch doors is 2/3, try it yourself. Do a simulation and see how often you win when you do and when you don't switch doors. KSFTC 22:39, 3 August 2016 (UTC)
The probability is 1/2. Where is the trick?
Latest comment: 7 years ago3 comments3 people in discussion
The chance of winning that the player has changing the first choice is 1/2 and not 2/3; the solutions that lead to the value 2/3 contain a trick that misleads the reader.
The game has two phases:
1) the initial choice between the 3 doors,
2) the opening of a door with the goat and the possibility to change the initial choice.
The second phase make the first one completely irrelevant and the trick contained in the solutions consists to continuing to consider it.
More specifically the opening of a door (2nd phase) redefine completely the game as a game with odds 50-50 (the player chooses between the 2 doors left closed). The initial selection (1st phase) become only one of the possible choice between the two doors left.
The trick contained in the solutions is to hide the transformation of the game due to the 2nd phase and continuing to assume the odds of 1/3 of the initial choice even though the transformation nullified the effects of the first phase.
In particular, what doesn’t work in the table with the 3 rows corresponding to the 3 initial choices and the 6 possible results of the game?
The opening of the door with a goat (es. goat A) from the host eliminates all the branchs where the host shows the other goat (goat B) and the two results that remain show one in two chances of winning whether the player change the initial choice or not. Alessmaga (talk) 12:54, 10 August 2016 (UTC)
The issue is that there are two different probability distributions. The 50/50 one is: given two doors, one of which has a car, if I pick one, what is the chance that I pick the one with the car? That distribution has four possible outcomes, and in two of them I pick the door with the car.
To see how the 2/3 figure is obtained, suppose that I always switch. Then we get this table of all the possible outcomes (as you mentioned):
Car in Door 1
Car in Door 2
Car in Door 3
First choose door 1 then switch
Lose
Win
Win
First choose door 2 then switch
Win
Lose
Win
First choose door 3 then switch
Win
Win
Lose
As you can see, there are 6 scenarios in which the switching strategy wins, and 3 in which it loses. When people talk about a 2/3 probability of winning after a switch, that collection of 9 possible outcomes is what they are referring to. — Carl (CBM · talk) 14:00, 10 August 2016 (UTC)
@Alessmaga:@CBM: There's more to this little problem than meets the eye. As Carl says, if you consider always switching you can easily see that if you do this your probability of winning the car is 2/3. But this moves the decision point to before the host opens a door rather than after. If you consider the situation after the host opens a door (say door #3), the odds at that point are also 1/3 if you stay and 2/3 if you switch. Let's assume you initially pick door #1 and the host opens door #3. The probability the car is behind door #2 AND the host opens door #3 is 1/3 times 1 (the host must open door #3 in the case the car is behind door #2) while the probability the car is behind door #1 AND the host opens door #3 is 1/3 times 1/2 (in the case the car is behind door #1 the host opens either door #2 or door #3 - assuming the host is indifferent means there's a 1/2 chance the host opens either remaining door). These are the only possibilities - so if you switch to door #2 you win with probability 1/3 * 1 (= 1/3), while if you stay with door #1 you win with probability 1/3 * 1/2 (= 1/6). Scaling these up as conditional probabilities yields the surprising 1/3:2/3 result. -- Rick Block (talk) 15:44, 15 August 2016 (UTC)
You do not obtain a 2/3 chance of winning by switching, you only increase your chance from 1/3 to 1/2 , I have posted my computer simulation and flowchart at the bottom of this page. My solution shows independently obtained results, thus process based and not logic based so help locate any flaws in the logical premise. So in summary, Switching gives you an advantage of 1/6 more than the 1/3 initial chance without a switch.
The Monty Hall problem can be explained very very simply without the need for complicated mathematics or theories and the like.
Latest comment: 7 years ago1 comment1 person in discussion
Indeed, it can be explained in plain English to the layman as simply as this:
There is one car and two goats behind the three doors. So, when the contestant chooses a door (door one in this instant) he has a two out of three chance of choosing a goat. [Point 1]
Monty Hall knows where the car is, and obviously he cannot open the door concealing it, or the game would be over - hence he must choose a door with a goat behind it. So when he chooses a door (door three in this example) it [always] has to have a goat behind it 100 per cent of the time.[Point 2]
Now simply combine the facts in Point 1 and Point 2 above and you have the solution: ie.
Two thirds of the time x 100 percent of the time door two has the car behind it.
Or put another way:
Two thirds of the time the car will [always] be behind door two.
So, the odds are in favour of swapping doors as two thirds of the time the car will be there.
It really is as simple as that.
Postscript.
Turning the above on its head we get the following:
If the car is behind door one, the chance of this happening is one out of three.[Point 3 say]
Monty can now choose either of the remaining doors two and three as they both have a goat behind them. The same principal as stated in Point 2 above still applies, however, ie he must choose a door with a goat behind it 100 per cent of the time.
So combining Point 3 and Point 2 above, you find that only one third of the time x 100 per cent of the time he is worse off if he swaps to door 2.
Conclusion:
Two thirds of the time he is better off swapping to door two; and one third of the time he is worse off by swapping to door two - and the thing that tips the balance away from a 50:50 split between sticking with his original choice or swapping, is brought about by the fact that Monty knows where the car is, and then shows his hand by revealing where it is not.
(Dave Richards, Denby Village, Derbyshire, England, 2016)— Preceding unsigned comment added by Dave Richards (talk • contribs) 21:07, 23 August 2016 (UTC)
Example door numbers should be removed from the quoted letter at the beginning of the page as they just confuse you
Latest comment: 7 years ago7 comments7 people in discussion
Example door numbers should be removed from the quoted letter at the beginning of the page as they just confuse you. If we are specifying that the player choose door 1 and the host opened 3 and it contained a goat, there is no advantage in switching. both switching and stay give a 50% chance of winning (because if the host opened door n3, it means the car could not have been there in the first place, in this case, so only two possibilities are left). You should just explain that the player picks a door and the host opens another one with a goat and he ask the player if he wants to switch to the remaining door. It is the sum of all the cases together that gives you an advantage by switching. — Preceding unsigned comment added by 2001:B07:644C:A33:1188:5791:7107:1D12 (talk) 10:28, 28 August 2016 (UTC)
That's not true, Given that you picked door 1 and the host opened door 3 it is still twice as likely the car is behind Door 2 as Door 1. SPACKlick (talk) 11:43, 28 August 2016 (UTC)
Comment from original author of the talk:
Yes the probability of winning is still twice as likely if you switch, but this is only true if you assume their choices happened naturally and before.
What i'm saying is that it is hard for a reader to understand this immediately; the reader could unwittingly not think that door 1 and 3 had been voluntarily chosen, therefore starting to think about possibilities from there on, ignoring that the current situation is just one of the outcomes caused by different choices.
That is the whole point of the Monty Hall problem, that is why people do not think that changing door increases the probability of winning, but this page should be about understanding the problem. — Preceding unsigned comment added by 2001:B07:644C:A33:2002:B345:7843:2FE6 (talk) 22:48, 30 August 2016 (UTC)
What do you mean by "this is only true if you assume their choices happened naturally and before"? And, to your original point, removing the door numbers from the problem statement would be changing a quote. The whole point of the Monty Hall problem is to force the reader to imagine standing in front of two unopened doors and at that point deciding whether to switch or not - and how we got there does indeed matter. Without this, the problem might as well be "what is your chance of picking which of 3 doors is hiding the car"? The point of interest is after the host opens a losing door, precisely because there are then only two unopened doors. -- Rick Block (talk) 16:10, 31 August 2016 (UTC)
it means that a readers could initially not think that the answer to the problem lies in the fact that the current outcome is a result of a series of different determined choices, he might instead just think that the car could be in his door (1) and the other available one (2), so the probability for the car to be in both doors is 50%.
Without explicitly specifying door n1 and n3, you underline that choices can be made and the answer lies there; in this way, the reader might start reasoning directly from there and get to the solution quickly.
If removing words from the quote is not allowed, why don't just explain the problem directly?
Also, i'd like to specify that in some other languages the problem is stated like i'm suggesting. — Preceding unsigned comment added by 2001:B07:644C:A33:3C12:5330:1BE0:504C (talk) 20:14, 31 August 2016 (UTC)
"The host, who knows what's behind the doors, opens another door" seems perfectly clear about the sequence of events and the intentionality of host's action. It is not made less clear by labeling the door as "say No. 3". Labeling the doors makes it easier to discuss them. ~ Ningauble (talk) 16:53, 3 September 2016 (UTC)
It does make it less clear simply because you don't chooce yourself, thus it is less likely that your thoughts are going to go in the right direction, which is that the solution lies in analysis all the possible choiches.
I'm the original author of this talk and i'm going to stop it here as i'm battling alone against everybody. — Preceding unsigned comment added by 78.13.147.90 (talk) 20:35, 3 September 2016 (UTC)
flawed logic?
Latest comment: 7 years ago2 comments2 people in discussion
Surely there are 4 possibilities and not 3
if any door is selected by contestant, the host has 4 possible selections(door to open)
1. if contestant selects correct door(say door1), host could select door2
2. if contestant selects correct door(say door1), host could select door3
3. if contestant selects incorrect door(say door1, car behind door2), host must select door 3
4. if contestant selects incorrect door(say door1, car behind door3), host must select door 2
hence in case 1,2 switching will lose, case 3,4 switching will win, hence 50% not 67% as common logic would suggest.
By always opening a goat door(and never the contestants or car door), this new information does not favour either of the remaining doors, both remaining doors probability would increase from 33% to 50% — Preceding unsigned comment added by 211.27.69.72 (talk) 03:35, 27 November 2016 (UTC) g.bardwell
The issue with the logic you've proposed is that those 4 options aren't all equally likely. The contestant is twice as likely to pick a wrong door first as a right door so the 3rd and 4th options are twice as likely as the 1st and second.
To break it down you have (A) the scenario where car is behind door 1 (1/3), (B) where it is behind door 2 (1/3) and (C) where it is behind door 3 (1/3). If the contestant in (A) picks door 1 50% of the time the host opens door 2 (1/6 overall) and 50% they open door 3 (1/6). In (B) if the contestant picks door 1, 100% of the time the host opens door 3 (1/3). In (C) if the contestant picks door 1 then 100% of the time the host opens door 2 (1/3).
In those scenarios you can see that if you've picked door one and door 3 was opened it's twice as likely you're in C as in A. Hence you should switch SPACKlick (talk) 05:24, 27 November 2016 (UTC)
Proposed Solution and Computer Simulation to show evidence
Latest comment: 7 years ago4 comments2 people in discussion
After having an epiphany at 2:00AM and writing computer models and flow charts all night, I have determined that the page https://en.wikipedia.org/wiki/Monty_Hall_problem has several flaws. Mainly that according to the page you should have a 2/3rd chance of winning if you switch, and a 1/3 chance of winning if you don't. I using deductive reasoning and computer models have come to the conclusion that if you don't switch you have a 1/3 chance of winning, and if you do switch you have a 1/2 chance of winning.
https://github.com/McClainJ/Monty_Hall_Simulationhttp://i.imgur.com/PoJzAoG.jpg <- Yes, I know the flow chart is not the prettiest thing, but the simulation doesn't seem to have any errors so I am forming this as my theory. This simulation does the exact process described by the problem, it does not perform a mathematical simulation but does perform a procedural one. Therefor the results are provably accurate. I ran each simulation 5 times for a total of 50,000 cycles for both switch only and never switch.— Preceding unsigned comment added by Electroninja (talk • contribs) 02:06, 9 February 2017 (UTC)
I haven't reviewed your code, but you're clearly doing something wrong. If you have 1/3 chance of winning if you don't switch and 1/2 chance of winning if you do switch where is the other 1/6? It is perhaps not entirely coincidental that 1/2 + 1/6 = 2/3.
Looking at your flowchart, in the "switch" case you show two doors left where the car might be. Didn't the host open one of these doors? In particular, if you have initially picked a goat (with probability 2/3), isn't the only option available when you switch the car? The flowchart needs to be a little more complicated, perhaps somewhat more like the tree diagram in this section of the article. -- Rick Block (talk) 08:09, 9 February 2017 (UTC)
In regards to the flow chart, I agree it is not the best way to organize it, the top two on the right are not a representation of the top two in the middle, they are meant to represent the 2 unopened remaining doors.
Wait, you didn't review the code, but assumed I did something wrong, do you see the problem with that?
Just check the code, it is not even 130 lines of code, and I guarantee that it plays the game perfectly. (just change the to_switch value when you run the code to true or false for the various simulations.)
And I think the main difference between your analysis and mine, is that you see the choice as a single linear chain, where it is realistically 2 separate decisions, not from which door to pick. but whether to enter into a separate decision. If you don't switch , then you never entered into the second decision, which explains the 1/3rd win rate, but if you do switch then you are entering into a scenario when you have chosen 1 of 2 results, which gives you the 1/2 chance of winning. Both of our lines of logic seem to follow deductive reasoning, and since there are separate results, something is going on. — Preceding unsigned comment added by Electroninja (talk • contribs)
I've had a quick shufti at the code, and admittedly, I'm a beginner with C languages. in enact_Switch When the car is behind Door 2 and the choice door is 2 you have the code Doors[(2 - rand()%2 )].goat_Revealed = true; Which opens the door 1 or 2, it should be Doors[(1 - rand()%2 )].goat_Revealed = true; to open 1 or 0. I think this will account for most of the discrepancy in your switching data.
I've included a VB coding of the game and the results from a hundred million runs that I just ran.
VB simulation
Option Explicit
'SwitchWin(33,334,712) SwitchLose(16,678,126) WinIfSwitch(0.666523104)
'StayWin (16,658,911) StayLose (33,328,252) WinIfStay (0.333263782)
Sub MontyHall()
'run counter
Dim run As Long
'car and choice and revealed placement
Dim car As Byte
Dim choice As Byte
Dim revealed As Byte
'totals
Dim switchwin As Long
Dim switchlose As Long
Dim staywin As Long
Dim staylose As Long
For run = 0 To 1000000
'Choose a door for the car
car = Int(Rnd * 3)
'Contestant chooses a door
choice = Int(Rnd * 3)
'Host opens door
Select Case car
'Where car is 0
Case 0
Select Case choice
Case 0
revealed = Int(Rnd * 2) + 1
Case 1
revealed = 2
Case 2
revealed = 1
End Select
'Where car is 1
Case 1
Select Case choice
Case 0
revealed = 2
Case 1
If Int(Rnd * 2) = 1 Then
revealed = 0
Else
revealed = 2
End If
Case 2
revealed = 0
End Select
'Where car is 2
Case 2
Select Case choice
Case 0
revealed = 1
Case 1
revealed = 0
Case 2
revealed = Int(Rnd * 2)
End Select
End Select
'Randomise switch; 0 stay 1 switch
If Int(Rnd * 2) = 0 Then
'If you stay then you win if you're on the car
If car = choice Then
staywin = staywin + 1
Else
staylose = staylose + 1
End If
Else
'switching is going to the door which is neither the choice nor the revealed
'all sum to 3 so that door is 3-choice-revealed
choice = 3 - choice - revealed
If choice = car Then
switchwin = switchwin + 1
Else
switchlose = switchlose + 1
End If
End If
Next run
Cells(1, 1) = switchwin
Cells(1, 2) = switchlose
Cells(1, 3) = switchwin / (switchwin + switchlose)
Cells(2, 1) = staywin
Cells(2, 2) = staylose
Cells(2, 3) = staywin / (staywin + staylose)
End Sub
^thank you so much for helping me fix that small typo, I ran the simulation again 5 times after the fix, and the results are the same still 50/50 if you switch, 1/3 if you don't. I have updated github with the more accurate code.
I fear I do not know enough about VB to analyze your code, most of it seems to not make sense. I will have to look up operators for VB and see what I am missing. is * your modulus operator? Your code may not have copied correctly. — Preceding unsigned comment added by Electroninja (talk • contribs) 05:22, 10 February 2017 (UTC)
* is the multiplication operator. To roughly explain the code in English:
Define Variables, start a for loop for 100,000,000 runs of the game
Rnd is a random number between 0 and 1. Int is the integer part of it. So Int(rnd * 3) is a random number between 0 and 2
This is used to pick a random car and choice
Select Case picks based on the chosen variable. So it's like your reveal goat function. Where both choice and car are the same it randomly selects one of the other two doors otherwise it picks the only available door and so sets revealed to that door.
It then randomly selects whether to switch or stay. If it switches then it uses the property that the sum of all three doors is 3, the door you can switch to is the one that isn't the chosen door or the revealed door so 3-choice-revealed = remaining door. (say choice was 1, and revealed was 0, 3-1-0 = the door you switch to.
It then checks if the final chosen door is the same as the car. If it is, it adds one to the win, if not it adds one to the loss
The final section of code just prints the results to an excel spreadsheet.
Also, I've spotted a second flaw with your code Where you have
You're switching from door 2 to door 1 if door 1 is open. I've run your code with the correction and it doesn't produce 50/50 any more. It produces a clear 2/3. — Preceding unsigned comment added by SPACKlick (talk • contribs)
Thank you so much for helping me find this dude. Finally, the universe makes sense again. I am gonna make the edits and publish this a Monty hall simulation. Link is posted here in C++ to help those coding simulate this problem. This is what peer review is for, finding code errors that occurred when writing code at 3:00am, and restoring balance to the universe.
