Talk:Monty Hall problem/Arguments/Archive 15

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What if another player picks the other door?

I would know what would happen if another player in the same Monty Hall game had chosen other door (not the one open by the presenter) and given the opportunity to change door also, which would be the probability for him, and the sum of probabilities for both two players?.88.20.162.142 (talk) 03:42, 4 May 2017 (UTC)

Since you don't know which door the host will open until he opens it, the other player only gets to pick a door after the host opens a door. In this case, the other player's door has a 2/3 chance of hiding the car - so this player has a 2/3 chance of winning the car by not switching and a 1/3 chance by switching. The sum of the "not switch" chances are 1/3 + 2/3 = 1 (if neither player switches, one of them will win the car for sure). The sum of the "switch" chances are 2/3 + 1/3 = 1 (if both switch, one of them will win the car for sure). -- Rick Block (talk) 14:59, 4 May 2017 (UTC)

Wrong assumptions: 2/3 win probaility only works if change decision is taken A PRIORI of opening door 3

For the 2/3 win probability to work it is necessary to state a 4th hypothesis, which sorry, NEVER happens in this kind of TV shows:

4) The contester must decide if switch or keep the same door BEFORE Monty opens the door 3.

Of course, in this case he has 2/3 probability of winning if he CHANGES because there are 2/3 probability that the car is behing either door 2 or 3. Then, if he decides A PRIORI that he will change to EITHER door 2 or 3 DEPENDING on what door Monty opens (that will be either door 2 if the car is behind 3 or viceversa if the car is behind door 2) THEN by just saying "I WANT TO CHANGE" he/she is choosing the BEST OF TWO OPTIONS OUT OF TOTAL THREE, and he gets a a 2/3 win probability.

BUT....

IF Monty ALREADY OPENED DOOR 3 and THERE IS A GOAT THERE, THEN, with the GIVEN set of information, he KNOWS that the car is EITHER behind door 1 or 2. And his probability will be 50%. The conditional tree is WRONG because it is valid for the moment BEFORE THE DOOR 3 IS OPENED AND THERE IS A GOAT THERE. When the door 3 is opened AND there is a goat there, then the conditional tree changes completely and all assumptions with door 2 being opened must be set to PROBABILITY ZERO. — Preceding unsigned comment added by 186.23.133.28 (talk • contribs)

Excellent, 186.23; I also deduced that solution, as probability 0+1/2, because the first choice is always a loss (never wins car at 1st choice) which shows goat door. People make the false analogy as 2 chances to win, but first choice only shows goat, as reset 0% chance to win. The issue is false assumptions, as in "Kits, cats, sacks and wives, how many were going to St. Ives?" (not how many were met along the way). -Wikid77 (talk) 06:11, 8 May 2017 (UTC)

@Wikid77: Did you read the rest of this section, in particular my reply just below? -- Rick Block (talk) 15:36, 8 May 2017 (UTC)

Yes, I've read the other replies, and all 24 possible game moves must be counted, as in a truth table to examine all 24 moves and ensure complete coverage of all possibilities. The initial choice of doors, to always show a goat, is a "red herring" to distract from the reality of only 2 choices left, a 50-50 option, and any talk of thirds (1⁄3 or 2⁄3) fails to calculate the probability of winning after the first choice as 0, because the probability of being shown a goat after the first choice is 1. Based on the initial move, as showing the goat, then all remaining probabilities are 0.5 as 50-50 chance of winning the car whether switch to the other door or not. -Wikid77 (talk) 21:58, 11 May 2017 (UTC)

is a "red herring" to distract from the reality of only 2 choices left, a 50-50 option. @Wikid77: You're committing the fallacy of begging the question: assuming the result that you are trying to demonstrate. It is true that at the end there are two choices left, but you are asserting that they are equally probable without proof. You're excluding the possibility that the two choices have different probabilities. The probability of a random experiment depends on how it has been set up, and the previous steps in the Monty Hall problem are relevant; the doors haven't been set up as a "car randomly placed behind two doors with equal probability", so you can't assume that they have equal probability. You can't simply discard the set up of a random experiment and call it a red herring. Diego (talk) 10:26, 12 May 2017 (UTC)

You had it right up to the point where you say "And his probability will be 50%". That is a non sequitur to everything said before. The door not chosen at the beginning of the game has double the probability of the chosen door, because it accumulates the probabilities from the two non-chosen doors (since the door opened by Monty can no longer have a car behing it). When the door is opened, it doesn't change the fact that the non-chosen door is the best of two options out of tree - the only difference is that now we know which those two options the user is getting. Diego (talk) 15:59, 14 November 2016 (UTC)

Exactly - everything you say is correct, except for the last couple of sentences. If you throw all the information you know away, and restart knowing only that the car is behind one of door 1 or door 2, you do end with with probabilities of 50%. But that is NOT what happens. You know the original probabilities were 1/3 for each door. You know the host MUST open door 3 if you've picked door 1 and the car is behind door 2. You can assume the host picks evenly between door 2 and door 3 if the car is behind the door you've picked (door 1). You can and should use this information. Let me fix your last paragraph for you:

If Monty already opened door 3 (deliberately showing a goat there), then with the given set of information, the player knows the car is either behind door 1 or door 2. What we need to determine are the conditional probabilities given the host has opened door 3. For each door, this is the starting probability (1/3 for each door) times the probability the host opens door 3 if the car is behind that door divided by the total probability the host opens door 3. Door 3 is easy. We know the conditional probability the car is behind door 3 is 0 (because we can see a goat there). But working through the computation is pretty easy, too. If the car is behind door 3, the probability the host opens door 3 is 0, so we get a conditional probability of 1/3 * 0 divided by something - so this is going to end up 0. Door 2 and door 1 are a littler harder, but not much. If the car is behind door 2, the host must open door 3 (cannot open door 1), so the composite probability of the car being behind door 2 AND the host opens door 3 is 1/3 * 1 and the conditional probability is this answer divided by something. If the car is behind door 1, the host can open either door 2 or door 3. Assuming the host is indifferent, then the composite probability of the car being behind door 1 AND the host opens door 3 is 1/3 * 1/2 (and the conditional probability is this answer divided by something). The total probability the host opens door 3 is the sum of the composite probabilities, i.e. 1/3 + 1/6 which is 1/2. Now, we can compute the conditional probabilities. For door 1 we get (1/3 * 1/2) / 1/2 which is 1/3. For door 2 we get (1/3 * 1) / 1/2 which is 2/3.

The key here is that if the car is behind door 2 the host MUST open door 3, while if the car is behind door 1 the host can open either door 2 or door 3. This means the chance the car is behind door 1 AND the host opens door 3 is exactly half the chance the car is behind door 2 AND the host opens door 3. These are the only possibilities, so call them X and 2X. We know X+2X must be 1, so X must be 1/3 and 2X is therefore 2/3. -- Rick Block (talk) 16:55, 14 November 2016 (UTC)

Explaining switch/stay is 50% not 2/3

Table 1a: Truth Table of Monty Hall Problem
door with car	1st door try	open door	2nd door try	result	Expected cases out of 900
1	1	2	stay	win	25
1	1	2	switch	lose	25
1	1	3	stay	win	25
1	1	3	switch	lose	25
1	2	3	stay	lose	50
1	2	3	switch	win	50
1	3	2	stay	lose	50
1	3	2	switch	win	50
2	1	3	stay	lose	50
2	1	3	switch	win	50
2	2	1	stay	win	25
2	2	1	switch	lose	25
2	2	3	stay	win	25
2	2	3	switch	lose	25
2	3	1	stay	lose	50
2	3	1	switch	win	50
3	1	2	stay	lose	50
3	1	2	switch	win	50
3	2	1	stay	lose	50
3	2	1	switch	win	50
3	3	1	stay	win	25
3	3	1	switch	lose	25
3	3	2	stay	win	25
3	3	2	switch	lose	25
Totals: 12 stay, 12 switch Wins if switch: 6 switch win in 300 of 900 cases. Wins when stay: 6 stay win in 150 of 900. Odds: 300 to 150, or 2:1, 2⁄3 if switch.
Discussion: That is a ridiculous FAKE TABLE, showing only 24 of 36. Missing 12 are: 6 stayings lose, no staying wins, and 6 switching WIN, no swith loses. --Gerhardvalentin (talk) 19:18, 18 May 2017 (UTC) This is a "truth table" of all possible (true) moves in the game. Perhaps create another table, below, of 36 cases (not including 12 which could never occur where host opens door with car). You will realize only 24 possible gameplays. -Wikid77 (talk) 19:14, 28 May 2017 (UTC)

I have been updating the page to explain the 2/3 paradox as a false analogy to a game where contestant gets 2 tries to win, whereas here the "first try" is always a loss (car not awarded) where host shows a goat instead of player's chosen door. One source mentions probability of showing goat is "1" (and probability of winning car is 0% on first choice). The actual probability of both choices is thus 0+1/2 as 50% regardless of switch/stay at door. However, some readers might need more explanation, so I will add a full truth table (of all possible choices) to calculate the precise answer as 6 switches can win car, and 6 stays can win car, as 6 to 6 or 50% regardless of stay/switch choice of door. Similar brain teasers have been solved this way for over 50 years. See truth table at right.

The mindset of the host should also be explained. In fact, game host Monty Hall typically had only 1 zonk door (or "goat"), where the 2nd door was a medium prize (such as TV+sofa set), against a bigger prize behind 3rd door. However more sources would be needed to document the typical middle prize in the original gameshow. -Wikid77 (talk) 05:54, 8 May 2017, Update: However, the probability of choosing the next door is limited by the first choice, and the truth table column of "900 cases" shows the switched door gains double the wins, compared to stay, as the weighted probability of 300 to 150 wins by always switching (as suggested in comments further below). -Wikid77 (talk) 19:14, 28 May 2017 (UTC)

In the table at right, counting possibilities cannot be used to determine probabilities because the events are not independent. Observe that in the eight cases where door #1 has the car, the first try is door #1 four times. There is no justification concluding that the first try is 50% likely to be correct at the outset, before anything else has happened. The fact that there are more choices of door to open if the first guess was right does not imply that the first guess was more likely to be right. ~ Ningauble (talk) 14:00, 8 May 2017 (UTC)

Well, the truth table at right shows all possibilities, based on the dependent events as the game is played, and counting the various wins will confirm there are 6 wins by switch and 6 wins by stay. Hence, by simulation of the entire game, according to the example given, the results are 50-50 as 50% chance of winning car, whether the contestant chooses to switch the door or not. The initial choice of doors, to always show a goat, is a "red herring" to distract from the reality of only 2 choices left, a 50-50 option, and any talk of thirds (1⁄3 or 2⁄3) fails to calculate the probability of winning after the first choice as 0, because the probability of being shown a goat after the first choice is 1 (the 1st choice is never actually a "1st chance" to win the car). -Wikid77 (talk) 21:47, 11 May 2017 (UTC)

You have failed to demonstrate that all entries in the truth table are equally probable. Your argument is incomplete. Diego (talk) 10:31, 12 May 2017 (UTC)

@Wikid77: In particular, fill in the new column I added to your table above. Think about 900 instances of the game. The car is placed randomly, so it will be behind the each door 300 times. The player picks the initial door randomly, so of the 300 times the car is behind door 1 the player picks door 1 100 times. Of these 100 times, the host can open either door 2 or door 3. Picking randomly which door to open (in this case), means the car is behind door 1 AND the player picks door 1 AND the host opens door 2 50 times. This is the number in the top (and second) row. Please try to fill in the rest. Since your table has both switch and stay (for each case), the sum of the numbers in this column should be 1800 (i.e. each case appears twice). -- Rick Block (talk) 15:31, 12 May 2017 (UTC)

As suggested by User:Rick Block & User:Diego_Moya, I have added the relative probabilities as truth table column "900 cases" (in Table 1a above) to show the weighted probabilities of each gameplay, as switch 300 to 150 stay, proving the 2:1 advantage of always switching doors, tallied among all possible moves. Thanks for feedback. -Wikid77 (talk) 19:14, 28 May 2017 (UTC)

I haven't read your argument in detail, but I agree with your conclusion. The logic tree used by vos Savant depends on there being no new information added during the process. That isn't the case. The host adds new information by revealing a goat, and the logic tree should be restarted at that point. Your original choice then becomes irrelevant, because a) you are allowed to change it, and, most importantly, b) it was never revealed. You are faced with two doors, one of which hides a car, the other a goat, and you have no information as to which is which. Therefore it doesn't matter whether you change or switch. All statements I have read, and, I suspect, computer simulations too, that say differently, rely on the premise that your original choice still has a 1/3 probability of hiding a car. Once the host has opened a door, based on foreknowlege that there was a goat behind, it doesn't.

I believe the first post in the first discussion thread makes the same point in a different way. We're both saying that you have to be careful to eliminate hidden assumptions.21:56, 14 May 2017 (UTC)Nigelrg (talk)Nigelrg (talk) Nigelrg (talk) 01:04, 15 May 2017 (UTC)

That would be true if the host had chosen any one of the three doors to open at random, and it didn't have the car. In that case, the game could be started again, and the two remaining doors would have equal chance.

But that is not how the game is played. The host takes into account which door was chosen by the player in the first step. Even if that door is not opened, it influences which door is opened (because the host must avoid it), so it carries some information to the second step; and this knowledge cannot be discarded. Diego (talk) 16:12, 15 May 2017 (UTC)

Also, computer simulations do not make assumptions about the probabilities of the outcomes. They play the game as stated in the standard assumptions, and the 1/3 vs 2/3 probabilities of lose/win emerge by the law of large numbers, as this is the natural result of the game. Diego (talk) 18:00, 15 May 2017 (UTC)

Thanks for your reply, which I am still considering. I agree that the host's decision generates new information, which it carries to the second step, and I wasn't proposing discarding it. I would use that, and the other facts I mentioned, as a reason for restarting the calculation, using all the new information. I don't believe you can continue to assume that your initial choice still has the same 1/3 probability of being a car. Nigelrg (talk)Nigelrg (talk) 20:04, 15 May 2017 (UTC)

The main point of your reply that I didn't address was the computer simulations. Because the 2 goats are considered identical, there are only 3 initial scenarios: car behind doors 1, 2 or 3. Jumping to step 2, after the host's intervention, the computer has 2 closed doors, behind one of which is a car, and the other a goat. Any other information relates to situations that no longer exist. Probability 50:50. — Preceding unsigned comment added by Nigelrg (talk • contribs) 23:58, 15 May 2017 (UTC)

Any other information relates to situations that no longer exist. Probability 50:50. This is a non-sequitur, an invalid reasoning step. You have two doors in the simulation, but the car was not put behind them with a 50:50 random process. One door was selected by the player from a set of three, the other was selected as the best of the two remaining doors. You can't calculate their probabilities with a uniform distribution, because you won't open one of them at random; you need to calculate them using the conditional probability that the car is behind the door after you switch, or after you stay. Read Rick Block's reply in the next section, which does this. Diego (talk) 08:07, 16 May 2017 (UTC)

In probability theory no single case has any meaning; 'control groups' need to be used

There is a way to explain that several answers -and most logic behind it- can be right, using the 'different control groups' idea. Which is based on the simple fact that in any single case, probability has no meaning. So in probability 'thinking', we always have to use bigger amounts that we also use in statistical evidence, to make it clear that if we follow certain rules (consistent behavior) and repeat it enough times to rule out 'coincidence', the calculated probability is actually exactly that of repeating rules. For that, we first have to decide what control group of consistent behavior the players are in.

Because of the way of questioning: 'if you are the contestant, what should you do?', the contestant has to try to determine what control group he is in, to understand the (behavior) rules of the group, and then decide what to do himself in this single case. Which is of course nonsense, because 1.) his deviation in a single situation has nothing to do with probability calculation, which applies to (endless) repeating, and 2.) his decision about his own behavior automatically creates the same control group, in which all cases follow exactly the same rules. Let me explain by examples, starting with the unexpected.

Assumptions in all cases are the obvious, which is not the problem. The actual problem is in the two questions being asked:
1.) "Do you want to pick door No. 2?"
2.) Is it to your advantage to switch your choice?

Example winning probability = 1/2

Most people think that switching does not change the probability of winning. Let's suppose the contestant is one of those. Let's also suppose that the contestant chooses randomly to switch or not. Now it becomes clear that his control group will answer question 1 randomly with yes or no. Just like the host randomly chooses between two doors that have goats. Rule = random. Now we also know that half of them switches and uses the 2/3 probability, while the other half uses the 1/3, equaling to 1/2. Because we cannot apply probability calculation to a single case, we have to apply these common rules of the whole group (both halves) to this one contestant, because in repetition he is sometimes switching and sometimes not. The correct answer to question 2 is that there is no advantage of switching, because it happens randomly, which is the rule. (And which is very counterintuitive.) The contestants will give the right answer, but not with the right reason.

Please notice that these two questions can be seen as really different questions, or actually the same. When seen as different questions, the first one is about the personal preference of the contestant, while the second one necessarily is a logical one, which can be answered right or wrong. It seems obvious (because of the missing quotes in the second question) that these questions are meant to be one: 'please estimate your advantage and then make your choice to switch or not'. The reason leading to the first answer is a given fact that must apply to the entire control group, while the correct answer to the second question is the result of that reason and thus behavior, which has opposite dependency between both questions as is meant to be.

1/3 < Example winning probability < 1/2

In this situation the contestant still thinks that switching does not matter, but he tends to stick to his initial choice, let's say he does this in 70% of the cases. Chances of winning are (7/10 * 1/3) + (3/10 * 2/3) = 0.4333.. He will answer question 2 with 'no', which is incorrect, because he still switches in 30% of the cases. Given his behavior, the correct answer is 'yes, my winning probability increases from 0.3333 to 0.4333 switching as often as I do', but he is not giving that answer of course. This situation is probably most realistic and also makes it very well clear that a single case or choice does not mean anything in probability; we need to know the consistent processes behind it. In one case he will switch doors, in another he will not, which is not even distributed equally.

Note: the counterintuitive thing about this whole 'one case has no probability' approach, is that it seems that it doesn't matter whatever you choose, having only one chance, which seems to happen a lot in everyday life. Vos Savant helps us to understand: suppose that in 999,999 of 1,000,000 cases you are stupid enough to randomly switch, and only in one case you have a short moment of seeing the light, convincing you to switch. This will not significantly change the odds. However, if this would be a consistent reality (it happens to you once in a million), the only way to calculate it right, is to use the 0.500001 probability of switching. But what if you, when you have this bright moment, simply are aware of that brightness? How could it not be significant? There is only one way to be (probably :)) sure about that, which is your ability to recognize such bright moments, proven by experience. This completely changes the control group of situations, because now you have to add a second group of cases that overall have a high rate of brightness. Which shows us again that only the extent of consistent behavior matters, but also that the odds are really relative and depending on the control group(s) used!

Example winning probability = 2/3

First of all I'd like to emphasize that the key is not in the knowledge, but in the behavior (by knowledge). (Then again it's the knowledge of behavior that enables us to create a realistic control group.) If we check the standard assumptions from the chapter in the article with the same name, we see that the three doors are distinguished as 'originally chosen', 'opened' and 'remaining closed'. Not as numbers. We can safely make the assumption that host nor contestant is consistently paying attention to these numbers. Because they have no reason for that. We only know that the host, when offering the switch, is explicitly using the name of the number of that remaining closed door. It may well be that he is simply reading the number displayed on it, at that moment. It does not mean that any player is aware of the number of the originally chosen door at any time. Moreover, since the possible knowledge of the number of the chosen door does not change the behavior of any player, it does not make sense to use this information in any scenario. Also the information from the puzzle: "You pick a door, say No. 1" may be interpreted as: "this may also be No. 2; we only use the numbers as relative distinction to each other, not as absolute information to recognize one certain door." Because of this, conditional solutions using the specific numbers as absolute information are not only unnecessarily complex; they are even wrong when this interpretation is right. Only if (additional) knowledge changes behavior, it should be used in calculation. We have no clue that this is the case, so the simple solution is most appropriate.

Most important is again the question: is there consistent behavior and what is it? One contestant in one case never shows consistent behavior, whatever he thinks, answers or does. Suppose he knows all about probability theory. He will answer: "Yes, I want to pick door No. 2. If all cases in my control group will always and consistently show the behavior to switch when offered, it will be in the average and relative advantage of this group compared to those groups that show consistent behavior to not switch respectively to randomly switch. There is however no way to know anything about my own chances in this single case, if even such a probability exists." — Preceding unsigned comment added by Heptalogos (talk • contribs) 15:27, 8 January 2018 (UTC)

--Heptalogos (talk) 23:59, 7 January 2018 (UTC)

@Heptalogos: The sentence ' He then says to you, "Do you want to pick door No. 2?" ' is a statement, not a question. The Monty Hall Problem does not require us to answer it. We are asked 'Is it to your advantage to switch your choice?' to which the solution is 'Yes' and the value is 2/3. To illustrate the meaning probability has in any single case, such as this, consider a single fair dice to be thrown once only. The probability it will land a six is 1/6. Freddie Orrell 22:51, 9 January 2018 (UTC)

@Freddie, that's exactly the misconception I mean: there is no way to know the probability for one single throw. We can only make correct statements about probabilities for greater amounts of repetitions of the exact same behavior under the exact same conditions. Statisticians have learned that, and most of them will not deny it, but they simply 'forget' about it sometimes. Because usually it's not an issue. The reason for that is that statisticians always create their own reality in theory as well as in empirical testing. They may however fail to relate a single case in real life to the experiment correctly; nobody will ever know. Look at my first example: the contestant randomly chooses to switch. This will lead to a probability of 1/2 of winning. Because if we test this behavior, the contestant will switch in half of the cases and will not switch in half of the cases. Any single case (switching or not) makes no sense, because that's not his consistent behavior. Stupid statisticians will argue that it is: if he switches, they will use that knowledge and use only 'switching behavior' in their calculations and tests. If he doesn't switch, they will use only 'non switching behavior'. Then the tests of course will prove the theory. :) The whole horrible thing about this, when you think it around one thousand times, is that there is no objective probability at all in any single case. We only obtain meaning or significance from 'law', which is consistency, recognition by repetition. What is the consistent behavior of the contestant? Is there even consistent behavior? Nobody knows, not even the contestant himself: will he react exactly the same in many and all cases? So I read about the assumption that the contestant has to decide what his consistent behavior is, which is one step in the right direction. But still he can decide or 'promise' whatever he wants, that doesn't make it full proof consistent behavior of course. Wrap up: in one single case the contestant can switch, and if he consistently always switches, his chance of winning is 2/3, but if he consistently randomly switches, his chance of winning is 1/2. Actually we can make as many groups and probabilities as we want. To convert a single case to a group, we always have to know the exact rules that apply. Of course we only know that if we turn it around: creating cases by performing rules. Which is what we do in empirical testing. --Heptalogos (talk) 22:47, 10 January 2018 (UTC)

@Heptalogos: You say the probability that a single fair dice thrown once will land a six is unknown. And if I take a fresh, fair coin and toss it once, is the probability it will land heads also unknown? Freddie Orrell 23:34, 10 January 2018 (UTC)

@Freddie, yes it is unknown. Probability in a single case is not even existing as such. There is simply average probability, of control groups. Which is the only thing we can empirically test and therefore prove. The idea that any single case in such a group has it's own probability is just that: an idea. Besides that it cannot be proven, it doesn't even make sense (as ideas sometimes do), because any unique case has the freedom to be influenced by significant forces that do not apply consistently to the whole group. Even if we create cases by performing rules, we should use computing (a program that follows only predefined if-then statements resulting finally in series of 0 and 1) and even then we should rule out any uncommon electric currency that can switch the bit. In empirical testing we usually rule out such deviations, at least afterwards. (If any inconsistency happens, it's a formal excuse to deny the outcome and run the program again.) Then we have probability as an outcome. Which we apply to any case we can reason as being part of our control group. Which is first of all wrong anyway, because it's a single case. And which secondly has a greater possibility to be wrong when the 'natural' single case (control group) has more complexity/diversity than a computing program, like the human individual certainly has, therewith not being in a controlled lab but in a vivid place with lots of stimuli. But most of all, this reasoning is not necessarily the expertise -to put it mildly- of scientists anywhere in the field, publishing articles that dominate Wikipedia. Which becomes very clear by disagreements 'internally'. Wrap up: the outcome of empirical testing is actually still theoretical related to any real life situation. Creating this relation rationally is still theoretical and very tricky. Besides that, it's actually a second probability issue itself. :) --Heptalogos (talk) 10:07, 11 January 2018 (UTC)

@Heptalogos: The unknowns or uncertainties you describe are accounted for in probability, for example by the principles of indifference and propensity. While it is indeed impossible to predict the outcome of a single case, such as a coin or dice, this allows its probability to be determined in the form 1/n. Freddie Orrell 19:25, 11 January 2018 (UTC)

@Freddie, the issue is not about those uncertainties. There are two issues:
1. It is in fact an incorrect statement that one has a probability of 1/2 of throwing head in a single case. We can only state that when repeating enough times, the average probability is 1/2. The advice to 'always switch doors' doesn't make sense to a contestant who is once in a lifetime in a show and has only one opportunity to switch. There is absolutely no way to predict his chances.
2. The question "Is it to your advantage to switch your choice?" can have two correct answers, both yes and no. It can only be answered correctly if you can repeat enough times, fully depending on your consistent behavior. So if you for example decide to consistently switch randomly, it's a very strange question, because it doesn't seem to realize issue 1: in a single case of consistently randomly switching, it is not in your advantage whatever you choose. But it is in your advantage to consistently randomly switch, compared with consistently not switching. --Heptalogos (talk) 21:47, 11 January 2018 (UTC)

@Heptalogos: You are simply repeating the same fallacies over and over - I give up. Best wishes, Freddie Orrell 22:56, 11 January 2018 (UTC)

@Freddie, you are right that I am trying to describe the same issue (it can be reduced to one issue) in many ways. It's because I think it's too hard to understand for most people. I believe it's really not a fallacy, but (indeed) an issue that covers the entire area of probability theory. However, the Monty Hall problem raises this issue by me, for two reasons:
1.) there is not at all any consistent behavior even possible;
2.) the whole discussion has gone to a level that is so exactly and well defined, that it seems ridiculous that the article does not even mention (i.e. that anybody in the professional field did not even publish) the fact that no single case has a probability. I think I will start searching for it, or initiating it. Any help is welcome. --Heptalogos (talk) 13:13, 13 January 2018 (UTC)

Stop over analyzing this

Let's examine the facts:

The original choice can only act against a 2/3 for proportion (of goat/doors) and 1/3 for odds (of finding the car)

The removal of the door changes the proportion of goat/doors from 2/3 to 1/2

If we do a sheer guess, our calculation would tell us that there's 2 doors, 1 guess left, so we'd think it's 1/2

But the trick here is that the (as of yet) not chosen remaining door is not a random door, and therefore a sheer guess isn't possible

Via our initial pick, we disallow our chosen door from being eliminated by the host; our 1/3 choice door stays in the sample space

And based on the parameters of the question, the car cannot be removed by the host; thus, since the car stays available, the total of all odds must equal 1

If the host removes 1/3 and our 1/3 choice stays in the game, then to equal 1 there's 2/3 remaining which must attributed to the as of yet not chosen other door

The question isn't asking us "what is the proportion of remaining goat to doors after the host removes one?" (and FYI, that proportion is indeed 1/2)

Rather, the question is asking us "Is it better to switch?"

Thus, in the final analysis, there are two numbers which apply: The final goat/doors ratio; which is 1/2 and odds attained if one switches; which are 2/3

98.118.62.140 (talk) 14:39, 3 February 2018 (UTC)

Another way to think about this

If the aim were to find 1 of the 2 goats, instead of the car, but all other rules were the same, then what? The host must remove a goat - he cannot remove the car. Thus, the other door which remains must contain what? Either a goat or a car. If it contains a goat, you are already on a car. If it contains a car, you are already on a goat. When you first picked a door, you picked one door from a 2/3 chance of getting a goat. If you stay, staying with your first choice will win you a goat 2 times for every 3 times you play. But, since the host must remove a goat, if you choose again, you will only get a goat 1 time for every three switches. Why? Because 2 out of 3 times you are already on a goat after your first choice. Thus, you only can possibly switch to a goat 1 out of 3 times. Thus, finding the car is the opposite of this: 1 out of 3 first choices get you the car. And if you stay with that choice, you can't then put yourself into the remaining 2 out of 3 pool; because you are stuck at 1 out of 3. When you switch, you leave the 1 of 3 pool of choices and you switch to the 2 of 3 pool of choices. And the 2nd choice is indeed two of three; 3 doors, one chosen by you, one removed by the host (no car) and one which you leave behind when you switch. If you switch, in only 1 time out of 3 plays will you accidentally leave the car behind, because in only 1 out of 3 plays, will you have actually picked the car to begin with. The odds of switching improve over the original odds because the removal of the door by the host enriches the prevalence ratio of cars to the available remaining choices. There initially is 1 car in a pool of 3 doors, then there is 1 car in a pool of 2 doors, which means it looks like this: 1/3 (car prevalence of 1st choice) ÷ 1/2 (car prevalence of 2nd choice) = 2/3 if switch. Of course, this last sentence looks like Ma & Pa Kettle math [1] Xerton (talk) 18:26, 4 February 2018 (UTC)

---

Reductio ad Absurdum of the "vos Savant" Solution

Your first choice had a 1/3 probability of being a car, as did the other two doors. If you continue to say that it has a 1/3 probability, after the host's intervention, then you must say that the remaining unopened door still has a 1/3 probability of being a car, which is clearly absurd.

What has happened is that the host has introduced new information by opening a door, so the original logic tree must be modified or restarted from scratch. Nigelrg (talk) 21:14, 15 May 2017 (UTC)

Indeed, once the host has opened another door to reveal a goat, then the only chance to win the car is a 50-50 choice to stay or switch to other door, as probability 0.5 either way. Conversely, some people claim Monty Hall would show the car if behind the first door chosen; otherwise the car would be behind the other door, as 100% chance of winning car by switching door (not merely 2⁄3). Of course, such a game would be absurd; hence, the only sensible game would include the chance of the car behind first door chosen, as again 50% chance of win, whether stay or switch. There is no other sensible conclusion. -Wikid77 (talk) 03:27, 16 May 2017 (UTC)

@Nigelrg: You are exactly correct. In fact, numerous mathematicians have described vos Savant's reasoning as somewhat less than complete (see https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions). Before the host opens one of the doors, the probability the car is behind each door is 1/3. After the host opens one, the probabilities must now be reevaluated. We're certain the probability the car is behind the door the host opens is 0, but what about the other two?

@Wikid77: Wikid77 - please pay attention to this.

If the player picks door 1, and the car is behind door 2, then the host MUST open door 3, so the composite probability the car is behind door 2 AND the host opens door 3 is 1/3 * 1, i.e. 1/3.

If the player picks door 1, and the car is behind door 1, then the host can open either door 2 or door 3. Let's say the host doesn't care which door he opens in this case (e.g. he flips a coin to decide). This makes the composite probability the car is behind door 1 AND the host opens door 3 equal to 1/3 * 1/2, i.e. 1/6.

If the host opens door 3, these are the only two possibilities. To express these as conditional probabilities, we divide each by their sum. 1/3 + 1/6 = 1/2, so the conditional probability (given that the host has opened door 3) that the car is behind door 1 is 1/6 / 1/2 = 1/3. And the conditional probability (given that the host has opened door 3) that the car is behind door 2 is 1/3 / 1/2 = 2/3.

As it turns out, the probability the car is behind door 1 doesn't change. But we really can't just assume that. We should compute the conditional probabilities. As one source puts it "The host can always open a door revealing a goat and the probability that the car is behind the initially chosen door does not change, but it is not because of the former that the latter is true." -- Rick Block (talk) 04:09, 16 May 2017 (UTC)

@ Rick Block. I believe that your analysis only holds true if the player has decided whether to stay or switch before the host opens the door. The first post of the first discussion thread makes this point, and the main article addresses it in some depth, as stated in the unidentified post above: https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions. No such stipulation is made in the description of the game, so we should assume that the player makes his decision after the host opens the door. I prefer to use Occam's Razor, and cut out unnecessary detail. Because the player has no information about the location of the car, other than the fact that it's behind one of two closed doors, he/she has a simple 50:50 choice.Nigelrg (talk) 02:43, 18 May 2017 (UTC)

@Nigelrg: First, the analysis I presented above is not "mine", but it is what the most reliable sources in the field of probability use. It most definitely does NOT hold true only if the player decides whether to stay or switch before the host opens the door. It evaluates the probabilities explicitly AFTER the host opens a door. Yes - you don't know for sure which of the two remaining doors the car is behind, but you have "partial" information. This partial information (the fact that the host MUST open door 3 if the car is behind door 2, but has a 50/50 choice which door to open if the car is behind door 1) is what lets you conclude that the chance of winning the car by switching (AFTER the host opens a door and there are only two doors remaining) is 2/3.

Lets take this a bit further. Say we have a deck of 52 cards, and the ace of spades "wins". If I (the "host") shuffle the cards, give you one (face down), and now (looking at the rest) discard all but one making sure I'm not discarding the ace of spades is it now 50/50 that you have the ace of spades? Or is it 1/52 that you have it and 51/52 that I have it? There are only two choices - you have it or I have it. You don't know for sure which one it is. In case it's not obvious the analogy to the MHP is this - instead of putting a car behind one of 3 doors we're putting the ace of spades among 52 cards. Instead of the player picking a door, we're dealing a random card to you. Instead of the host opening one "losing" door resulting in only two doors being left - the host host is discarding 50 losing cards resulting in only two cards being left. Please try this (really, no kidding) - say 20 times and let us know how it turns out (how many times do you end up with the ace of spades vs. how many times does the host end up with it). The notion that "you don't know for sure" and "there are only two choices" necessarily leads to the chances being 50/50 is simply incorrect. This is the entire point of the Monty Hall problem. It goes against a deeply held belief. The bottom line is that humans suck at conditional probability (or, perhaps, schools suck at teaching elementary probability concepts). -- Rick Block (talk) 06:49, 18 May 2017 (UTC)

I'll think over your interesting post later. I had some further thoughts myself, but your post may have negated them. Re: your last sentence, it's not just schools in general. Your views are opposed by post-graduates in math from some of the world's better universities. They're not infallible, of course, but it's a bit like climate change. When a large body of experts say one thing, there's a high probability that they're right.Nigelrg (talk) 18:05, 18 May 2017 (UTC)

I'm dealing with your post piecemeal :-) The deck of cards example is only relevant to one option in the Monty Hall problem - the case in which the player has pre-decided to stick (once you've picked a card, you can't change it). Therefore I don't think it's relevant to the problem as a whole≈Nigelrg (talk) 20:05, 18 May 2017 (UTC)

That's not right. In the formulation of the card game aboive, it does not preclude that the player could still switch cards with the dealer, turning around the probabilities of having the ace of spades. This would make the game equivalent to a Monty Hall game with 52 doors, where the host opens 50 doors which don't have the car, and the player can decide to switch or not after that. Diego (talk) 23:06, 18 May 2017 (UTC)

Re: "partial" information. I think you've neglected the situation when the car is behind door 3, so the host must open door 2. Therefore the host has 2 possible actions when the player's door hides the car, and only 1 possibility when it doesn't. Either way, the player gets new information. I came back to this post months later and corrected it. I made a typo/brain fade when I originally wrote it.I hope my correction didn't change any responses.≈Nigelrg (talk) 21:00, 18 May 2017 (UTC)Nigelrg (talk) 06:07, 11 November 2017 (UTC)

My views are NOT oppposed by post graduates in math (well, not any in probability or statistics) anywhere in the world. This is a classic problem - it appears in many elementary probability textbooks (with the exact answers I'm giving you). So, yes, it's a little like climate change in that the science is settled. But this is math, so not only is the science settled it's actually proven.

@Nigelrg:I thought you wanted to talk about a case where the host has already opened a door and there are only two possibilities for where the car is - for example, the player picked door 1 and then the host opened door 3. In this case the car is manifestly not behind door 3. I'm not 'neglecting" the situation where the car is behind door 3 - we're explicitly talking about only a subset of cases where the host has opened door 3, which means the car is not there.

Here's yet another way to think about it. Imagine 300 shows where the player has initially picked door 1. We'd expect the car to be behind each door about 100 times (right?). So, now the host opens door 2 or door 3. If we want to think about only the shows where the host has opened door 3 we're not talking about 300 shows anymore - but only some subset of the entire 300. Can you answer the following questions (thinking about 300 shows where the player has picked door 1 and the car is behind each door 100 times)? -- Rick Block (talk) 14:58, 19 May 2017 (UTC)

In how many of the shows where the car is behind door 1 does the host open door 2? ______

In how many of the shows where the car is behind door 1 does the host open door 3? ______

In how many of the shows where the car is behind door 2 does the host open door 2? ______

In how many of the shows where the car is behind door 2 does the host open door 3? ______

In how many of the shows where the car is behind door 3 does the host open door 2? ______

In how many of the shows where the car is behind door 3 does the host open door 3? ______

In how many shows overall does the host open door 2? ______

In how many shows overall does the host open door 3? ______

In how many shows where the host opens door 3 is the car behind door 1? ______

In how many shows where the host opens door 3 is the car behind door 2? ______

If you pick door 1 and the host then opens door 3, are you more or less likely to win the car if you switch?

This is a good example, Rick. Answer: It is completely impossible that you ever can be less likely to win the car if you switch.
This is valid not only in that (given?) case if you pick door 1 and the host then opens door 3 (in order to show a goat), but this is valid in any case, regardless which door you may pick, and regardless which other door the host may be opening (in order to show a goat).

Once more: You never are (nor can be) "less likely" by switching. This is valid in any situation given.

Your chance to have picked the car by luck is 1/3, so on average the risk to lose the car by switching is still 1/3, but
your chance to have picked one of the two goats (=wrong guess scenario) is 2/3, so on average the chance to win the car by switching doors is 2/3. And the host's opening of a door (to show you the SECOND goat) does not alter the scenario you're actually fixed in.
The chance to have picked the car by luck (lucky guess scenario) is only 1/3, and the chance that you are fixed in the wrong guess scenario, having picked a goat (thereafter the SECOND goat has already been shown to you !) consequently is even 2/3.
So in 2/3 of all cases you will win the car by swithing doors.

As to the host's behavior, he will never give you any hint on the scenario you're actually (unchangingly) fixed in (Henze, Mladinow et al). --Gerhardvalentin (talk) 12:20, 20 May 2017 (UTC)

Nigelrg seems to think that the analysis of the situation AFTER the host opens a door is that there are two doors and we don't know where the car is, so therefore (??) the chances are 50/50. The point is that the structure of the imaginary show (the real show wasn't actually run like this) gives us some information. -- Rick Block (talk) 15:50, 20 May 2017 (UTC)

Some readers believe in miracles, may be. Or they forget about the significant progress of the advancement made in course of the imaginary "show". But reality never will. --Gerhardvalentin (talk) 17:02, 20 May 2017 (UTC)

This post is very useful to settle the topic. — Preceding unsigned comment added by 190.199.242.101 (talk) 23:15, 16 August 2017 (UTC)

Thank you. IP190.199.242.101, for your compliment. --Gerhardvalentin (talk) 16:08, 25 April 2018 (UTC)

Bottom line is that the idealised Monty's behaviour is precisely equivalent to his telling you, "If you have picked the wrong door, then the prize is HERE". And you had 2 chances in 3 of being wrong. What more needs saying? Fredd169 (talk) 13:12, 23 December 2017 (UTC)

There are ONLY 2 choices left: "Impossible"/Confusing to have a present x/3 statistic.

The fact is, when there are only 2 doors left to choose from, there are then ONLY 2 choices left; and it is therefore impossible to have an x/3 statistic among those at that point in time because you no longer have a choice of 3 but only of 2. Thus, the way some of the article is written is incredibly confusing for the 9/10 people who notice this clear and indisputable fact: There are only 2 choices left.

Seeing that even a famed mathematician thought similarly makes it clear that if you are going to use x/3 when you have left only 2 choices, then you'd better be crystal clear about the logic of application, which from reading much of the article isn't that convincing or clear; because using an x/3 stat when there are only 2 choices left makes little sense to most of us.

Also, placing that remaining x/2 likelihood back in time when the INITIAL chance was 1/3, and continuing to use x/3 makes little sense to most of us.

Saying anything new about x/3 when there are only 2 choices left makes little sense to most; so Why be so confusing with the math? we think. The fact is, the REMAINING chance has only 2 choices possible, and so therefore the remaining statistic MUST be in reference to that: It must be stated as x/2.

Hardly a clear justification was made in my reading of the most of the article for using x/3 at all, at the point where there were only 2 choices left. It is that justification that needs to be made, and clearly applied, which would help the hapless reader understand at least some of the gobbledygook; because otherwise, it continues to seem nonsensical. And after all, isn't this conundrum in particular one that needs to be crystal clear to the general reader who was mystified in the first place when reading the "Ask Marylin" article? Misty MH (talk) 23:55, 9 March 2018 (UTC)

I hand you a die with 2 white sides and 4 black. You only have 2 choices. Roll it, what do you think your odds are if you choose white? — Preceding unsigned comment added by Nijdam (talk • contribs) 11:17, 10 March 2018 (UTC)

The formula ---> (#favorable events / #possible events) is applicable only if you have the same information about the options. Remember the division distribute equal amounts. For example, if I have a cake and two persons and I want to give each one the same amount, then I should give 1/2 to each. But certainly that's not the only way to distribute it. I could give 4/5 to one and 1/5 to the other, or give all the cake only to one, etc.

The probability is a measure about how much information we have. The confusion is to think that the options must be equally likely in any case. In this game, yes, we have two options, but do we have reasons to distribute the probability equitably? No. The contestant's door was chosen randomly from three, meaning that it has 1/3 probability to be the correct. On the other hand, the host knows the positions and must leave the car hidden (because he must reveal a goat door), so always the contestant failed at first, the other door the host leaves closed is the correct one, and we know this is more likely to have happened than the opposite.

In the second selection, we don't have a random door vs another random door. Basically, you have an option selected by someone who hits the correct 1/3 of the time and another chosen by someone who chooses the correct in the other 2/3 cases. Which one do you prefer? The important thing here is nor how many options do you have, but that they were chosen by two different people, one with more knowledge than the other. The host's closed door tends to be the correct with more frequency than the contestant's selection. — Preceding unsigned comment added by 190.36.105.224 (talk) 05:28, 13 March 2018 (UTC)

@Misty MH: I guess Leonard Mlodinow, in his book The Drunkard's Walk, shows it best, in underlining the neglected, but obviously constricted role of the host in the MHP. His role leads to conditional probabilities of surprising 1/3 : 2/3. About Mlodinow, Just have a look there.
It's hard for anyone to grasp that the host can act "randomly" only in 1/3 of all cases. He can act randomly only in 1/3, only in case that the guest (by luck) selected the door with the car (i.e. only if the guest actually is in the "lucky guess scenario").
Only in this special case (1/3) the guest should stay and never switch.

But in 2/3 of cases the guest will be in the "wrong guess scenario" (2/3), having selected one of the two wrong doors (two out of three). Because in that "guest's wrong guess scenario", the host's two doors having the car and a goat resp. a goat and the car, the host nevermore can act randomly, because he never shows the car but the second goat only. So his selection which door to open, in openening always the door with the goat, leads to conditional probabilities. He keeps the secret car undisclosed behind his still closed second door. In that 2/3 switching wins the car for sure.

So in 2/3 of all cases, as a "conditional probability", switching doors helps the guest to get the car.

The contestant, in opening a door, does not know the scenario he actually is in. He only knows that - by the host's constraint to never showing the car - by switching he will win the car with double (conditional) probability. Admittedly, on the long run only.

I agree with you, this fact that in 2/3 the host will never act randomly leads to a conditional probability of 2/3 to win the car by switching is hard to tell, using a few words only. --Gerhardvalentin (talk) 15:15, 22 April 2018 (UTC)

I read this as a criticism of the article, not the math. What Misty MH is saying is that the article is not clear, because it immediately delves into the "simple" solutions which more or less ignore the conundrum most people encounter which is that at the point of the decision there are only 2 doors involved, not 3. In particular, we know vos Savant's explanation from her original column is unconvincing based on the reader reaction she got. -- Rick Block (talk) 14:53, 23 April 2018 (UTC)

Yes, you are right. As to my understanding, Misty MH says that the article should clearly show that any assemblage of both unselected (host's) two doors is more likely to contain the prize. Exactly they have probability of 2/3, compared to any originally selected door of only 1/3.

And it's a fact that the host AVOIDS to show the car, but in any case he shows a goat. Solely in the contestant's lucky guess scenario, the host actually having two goats, he can act randomly. But if the CONTESTANT SELECTED ONE OF THE TWO GOATS (with twice chance) the host avoids to show his car, the "strict condition" is that he will show the SECOND GOAT only. So the 2/3 chance to win by switching doors is a typically "conditional probability". It's that easy.

And the famous mathematician Henze says that strictly speaking "math" is unnecessary to solve the MHP. And also Misty MH says: Why be so confusing with the math?

I say once more: math is rather unnecessary to fully understand vos Savant's MHP. That means that the whole ado about the host's biased behaviour belongs to a completely different article, you can name it "Index of lessons on conditional probability theory, based on the famous MHP"

And Marilyn vos Savant says that only in case that you already do have additional information of a factual existing preference of the host, and its extension, it might be sensual to do maths. But this never will be the case in the MHP, so any of similar considerations do never address the MHP of vos Savant, that definitely excludes any "additional (hidden?) information". No hidden hints are given. Such assumptions have to live their own life completely outside of vos Savant's famous MHP paradox.
(Though mathematically fully correct, M.et al. have disqualified themselves, in inventing ungiven additional assumptions, far outside of vos Savant's MHP, similar to "if wishes were horses, beggars would ride" and similar to "if you know that the biased host is paralyzed, then ...").

Any of such unbased considerations, only basing on ungiven and unproven assumptions (suitable for lessons in probability theory only), are obfuscating for the reader and prevent understanding the core of the famous paradox:

"Two still closed doors, why does the door offered have double chance, compared to the door originally selected?"
Answer: because there is'nt a new simple probability of 50:50, but because there meanwhile emerged a conditional probability of 1/3 : 2/3.

See theoretical physicist Leonard Mlodinow, who worked together with Steven Hawking, his diagram (1/3 : 2/3 on the long run) here:
The article, for years and years, has been confusing. It is time to detoxify it from disconcerting, misleading and truthless "conditional bias" math garbage that clearly belongs to a different article.--Gerhardvalentin (talk) 09:55, 25 April 2018 (UTC)

@Rick Block EXACTLY! You said it much better than I did! :) — Preceding unsigned comment added by Misty MH (talk • contribs)

With a small change to the game, you could be told that you could either have what's in the door you chose, or take what you want from the other two doors - in effect Monty did just that by showing you the door not to bother with - you'd rather swap to the 2/3, right? — Preceding unsigned comment added by Gomez2002 (talk • contribs) 16:07, 19 July 2019 (UTC)

Where did the article go?

Where did the article go? I clicked on the "Article" tab, and there was nothing. Misty MH (talk) 09:12, 21 March 2019 (UTC)

Weird. I found it, I guess. It's like this is an orphaned Talk page. Is "/Arguments" a normal additional page to a regular Talk page? The article is here https://en.wikipedia.org/wiki/Monty_Hall_problem Misty MH (talk) 09:14, 21 March 2019 (UTC)

@Misty MH: No, this isn't normal. Subpages for talk pages are normal for archives (which will present the same "missing" article tab; mainspace articles can't have subpages either), but not for actual discussion. This is an unusual case because of the nature of the article. See also Talk:0.999.../Arguments. –Deacon Vorbis (carbon • videos) 16:13, 17 October 2019 (UTC)

Law of large numbers proves this (not disproves as originally written)

The below information is accurate so leaving it for prosperity whereas my code was not. However, as a cautionary tale: The code I had originally written was tossing out every attempt where the host picked either the winning door or the contestant's instead of just picking the other door. This is more likely when the contestant picked a nonwinning door. This incorrectly shifted the expected results of the contestant winning to 50%. Now to the original argument.

We have 3 doors.

1 with a car behind it and 2 with goats - Assumed randomly assigned

Contestant randomly selects door - Also assumed random

We have to assume that the host cannot pick the door with the car AND cannot pick the contestant's door but other than that they randomly pick an available goat door (ultimately meaningless as it'll always come down to prize door and a goat door)

So at this stage we have 4 possible states...

1. Contestant picked the right door. Host picks goat door A. Contestant chose correctly

2. Contestant picked the right door. Host picks goat door B. Contestant chose correctly.

3. Contestant picked goat door A. Host must pick goat door B. Contestant chose incorrectly.

4. Contestant picked goat door B. Host must pick goat door A. Contestant chose incorrectly.

Host opens a goat door...

At this point the host HAS opened a door and it's known to be a goat and the problem has completely changed. We're left with the new states of...

1. Contestant picked the right door. If they stay they win but if they swap they lose.

2. Contestant picked the goat door. If they stay they lose but if they swap they win.

This can be programmatically shown by this pseudo-code.

Randomly pick 1 of 3 doors to have the prize.

Randomly pick 1 of 3 doors to be the contestant's pick.

Randomly pick 1 of 3 doors to have the host pick a door to throw out. If it's the prize door or contestant's pick just randomly pick again until it's not.

If the prize door is the contestant's pick we add 1 to the win column. Else we add one to the lose column.

Run it for as long as you want, assuming 1 million+ times to be at least semiaccurate, then then unsurprisingly find out that the tally is approximately 50/50. It cannot be argued against because we did the entire Monty Hall Scenario. We have a random prize door. We had a contestant pick a random door. We had the host pick a random goat door. We then compare if the contestant already picked the right door or not. If they always swap doors or not doesn't impact their chances of winning in the slightest.

If we don't do exactly what the pseudo-code does we don't fit the description of the Monty Hall Problem but rather a situation where we're doing math for before the host opens a goat door but after the host has announced the door has a goat behind it. — Preceding unsigned comment added by 184.166.97.144 (talk) 03:07, 28 June 2020 (UTC)

Nope, it's definitely 2/3 for switching and 1/3 for staying. If you implemented your description of the scenario correctly, a simulation should verify that. See the main article for a more detailed explanation. –Deacon Vorbis (carbon • videos) 03:26, 28 June 2020 (UTC)

Ran the simulation. Worked exactly as I said it did. 50/50 — Preceding unsigned comment added by 184.166.97.144 (talk) 03:38, 28 June 2020 (UTC)

Please indent your replies and sign your posts with 4 tildes; see Help:Talk for more info; thanks.

Then you made a mistake in implementing it; there are references in the article that describe doing exactly this and that it supports the analysis showing a 2/3 win chance for switching. –Deacon Vorbis (carbon • videos) 03:49, 28 June 2020 (UTC)

I gave exactly the code I used. Random numbers for everything. If you see a problem with the pseudo-code point out exactly where I made an incorrect assumption.184.166.97.144 (talk) 06:52, 28 June 2020 (UTC)

No, you gave a very very rough description of an algorithm, certainly not rising to the level of pseudocode, let alone actual code. What you described seemed more or less in line with the usual version of the problem, so if you got a different result, you must have erred in its implementation. Without seeing that, it's impossible to say how. –Deacon Vorbis (carbon • videos) 12:45, 28 June 2020 (UTC)

3 random ints [0,2]. They stand in for prize door, chosen door, and the door the host shows to have a goat. Ensure that the host hasn't picked either the prize door or the contestant's door but rather a free goat door, and then just a comparison if the prize door is the chosen door. It's that easy.184.166.97.144 (talk) 19:19, 28 June 2020 (UTC)

But if you're simply checking to see if the player's choice matches the car as you say, then you're simply checking to see if those two random numbers were equal (the value of the host pick is irrelevant), which is 1/3, just as it should be. Again, if you're getting 1/2, then you're implementing something that doesn't match what you've described, because what you've described plainly would result in a 1/3 win chance. –Deacon Vorbis (carbon • videos) 23:18, 28 June 2020 (UTC)

Found the mistake. It didn't help that the the error shifted it to ~50% at large numbers which exactly fit my misconception but it did. I apologize for the inconvenience.184.166.97.144 (talk) 01:00, 30 June 2020 (UTC)

No worries; it happens. –Deacon Vorbis (carbon • videos) 02:27, 30 June 2020 (UTC)

using the same logic with 2 players

I can't figure out what’s wrong with this line of reasoning: You choose door #1, host reveals door #3. A viewer watching the show at home had picked door #2. If your odds to win by switching are 2/3, then the odds for the viewer to “win” are also 2/3 by switching. So each of you will win 2/3rds of the time by trading doors. How is that possible? 71.162.113.226 (talk) 16:31, 19 September 2020 (UTC)

The host is guaranteed to open a door that the contestant in the show didn't choose. This doesn't apply to the viewer at home. It's possible that the viewer chooses door 2 and then the host opens door 2. So it's trivial that a lot of the time the viewer is better off changing their choice. The information the host is giving is the same: out of the two doors that the contestant didn't choose one doesn't have the prize behind it. The viewer can use this information to see that the unchosen, unopened door is more likely to have the prize. MartinPoulter (talk) 14:43, 20 September 2020 (UTC)

What’s wrong with that line of reasoning is: the viewer has 2/3 chance of “winning” the car by NOT switching (from door #2 to door #1). The information they and the contestant both have about those doors is the same. The viewer is observing the contestant’s probability set, not generating one of their own. Freddie Orrell (talk) 19:32, 28 September 2020 (UTC)

You said: "It's possible that the viewer chooses door 2 and then the host opens door 2." With all due respect, I'm not talking about that case. I'm only talking about the case where it is in the interest of both the contestant and the viewer to switch. The contestant would be better off with the viewer's door and vice versa, as each will win 2/3 of the time by switching. 71.162.113.226 (talk)

What's your basis for "2/3 of the time" if you're only talking about a specific case? The probabilities are completely different if you get to specify which door wins. Put another way, if you want to say there is a certain probability of winning when you explicitly exclude the case where the viewer chooses the door that will be opened by the host, you need to provide a calculation of that probability. MartinPoulter (talk) 13:12, 23 September 2020 (UTC)

For both the contestant and the viewer, staying wins 1/3 of the time, switching wins 2/3 of the time. This is the logic of the “correct” solution. Obviously, if the viewer is eliminated by Monty’s door opening, then the viewer’s chance of winning is 0. But when both contestant and viewer are still in it after the door is opened, each has a 2/3 chance of winning by switching, since their original 1/3 chance of winning by staying hasn’t changed. 71.162.113.226 (talk) 18:44, 24 September 2020 (UTC)

'their original 1/3 chance of winning by staying hasn’t changed' you're making a false assumption there. When the host acts, he adds information about doors unchosen by the TV contestant. The contestant should then switch to the door he has more information about. In your example the at home player has learned information about a door he has already chosen, so he needs to stay with what he's picked. There is no contradiction. - MrOllie (talk) 18:50, 24 September 2020 (UTC)