Monty Hall problem
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Vos Savant's response was that the contestant should switch to the other door. Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance.
The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A key insight is that, under these standard conditions, there is more information about doors 2 and 3 that was not available at the beginning of the game, when the door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities.
Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong. Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy. Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result.
The problem is a paradox of the veridical type, because the correct result (you should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand's box paradox.
- 1 The paradox
- 2 Standard assumptions
- 3 Simple solutions
- 4 Vos Savant and the media furor
- 5 Confusion and criticism
- 6 Solutions using conditional probability and other solutions
- 7 Variants
- 8 History
- 9 See also
- 10 References
- 11 External links
Steve Selvin wrote a letter to the American Statistician in 1975 describing a problem loosely based on the game show Let's Make a Deal,  dubbing it the "Monty Hall problem" in a subsequent letter. The problem is mathematically equivalent to the Three Prisoners Problem described in Martin Gardner's "Mathematical Games" column in Scientific American in 1959  and the Three Shells Problem described in Gardner's book "Aha Gotcha".
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? 
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  printed alongside Whitaker's question implies and both  and  explicitly define the role of the host as follows:
- The host must always open a door that was not picked by the contestant.
- The host must always open a door to reveal a goat and never the car.
- The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
When any of these assumptions is varied, it can change the probability of winning by switching doors as detailed in the section below. It is also typically presumed that the car is initially hidden behind a random door and that if the player initially picks the car, then the host's choice of which goat-hiding door to open is random. Some authors, independently or inclusively, assume the player's initial choice is random as well.
The solution presented by  in Parade shows the three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking door 1 in each case:
Behind door 1 Behind door 2 Behind door 3 Result if staying at door #1 Result if switching to the door offered Car Goat Goat Wins Car Wins Goat Goat Car Goat Wins Goat Wins Car Goat Goat Car Wins Goat Wins Car
A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three.
An intuitive explanation is that if the contestant picks a goat (2 of 3 doors) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (1 of 3 doors) the contestant will not win the car by switching.) The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.
Another way to understand the solution is to consider the two original unchosen doors together. As Cecil Adams puts it, "Monty is saying in effect: you can keep your one door or you can have the other two doors". The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from their original chosen door to the set of both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.
As Keith Devlin says, "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.' "
Vos Savant suggests that the solution will be more intuitive with 1,000,000 doors rather than 3.  In this case there are 999,999 doors with goats behind them and one door with a prize. After the player picks a door the host opens all but 1 of the remaining doors. On average, in 999,999 times out of 1,000,000, the remaining door will contain the prize. Intuitively, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one initially. Stibel et al. proposed working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options.) They report that when increasing the number of options to over 7 choices (7 doors) people tend to switch more often; however most contestants still incorrectly judge the probability of success at 50/50.
The answer to the Monty Hall Problem. That is answering the question: Is it to your advantage to switch your choice? The Answer is “no”. Here is why. Monty asks the game show contestant a grand total of 2 questions in the game.
1st question: What door do you choose ("pick")?
2nd question: Do you want to “switch” your initial door “pick”(the answer to the 1st question) to another door?
I purpose that the “game” does not even begin for the contestant (and also the game show producers) until the 2nd question is answered by the contestant. Precisely then the probabilities (for each show) become fixed and also clear: 1 in 2 or a 50/50 chance of winning or losing. A chance to drive around in a new basic model, red American auto. The banter before this moment is game show fluff (sorry Monty) for the viewing audience intently observing the host for any “tells”. After all, even the viewing audience knows it’s more exciting before the final “pick”(2nd question), it’ NOW win or lose. The viewer excitement is believing they could probably win because they watched many shows before and can “read” many of Monty’s “tells” and after all it’s now is a 50/50 chance to drive home in style.
The 1st “pick” is a cosmetic one for the show to set up the real drama of the 2nd “pick” later. It is literally inconsequential to any probable outcome (winning or losing) other than setting up the real “pick” which is: The contestant’s answer to the 2nd question (the first and only real pick). The contestant NEVER wins or loses on the first pick because the game ALWAYS continues despite ANY pick of the contestant (the game is NEVER over after the 1st question). There is no win or loss even possible at that moment(contestants can still move their bet later via the answer to the 2nd question).
The Standard Assumptions:
1. The host must always open a door that was not picked by the contestant.
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.
Mathematical proof following the “Standard Assumptions” for the problem:
The Contestant’s probability for winning(driving themselves home in style) after answering the host’s 1st question: 0%
The Contestant’s probability for losing(catching a ride with a relative or friend) after answering the host’s 1st question: 0%
The Game Show Producers probability for winning(not giving away the red American auto) after answering the host’s 1st question: 0%
The Game Show Producers probability for losing(finishing the paperwork on the red American auto) after answering the host’s 1st question: 0%
The Contestant’s probability for winning(driving themselves home in style) after answering the host’s 2nd question: 50%
The Contestant’s probability for losing(catching a ride with a relative or friend) after answering the host’s 2nd question: 50%
The Game Show Producers probability for winning(not giving away the red American auto) after answering the host’s 2nd question: 50%
The Game Show Producers probability for losing(finishing the paperwork on the red American auto) after answering the host’s 2nd question: 50%
The game truly begins only AFTER the contestant answers the 2nd question. Only after that event, probabilities take their place as the first truly real “pick” is now fixed (all bets are in) for probable outcomes.
A similar example: Take for instance an American roulette wheel at a Casino in Reno, Nevada. There are 38 numbers on the wheel. The croupier spins the ball around the wheel (Monty calling a contestant to come up to the stage). Hopeful gamblers gather around a table placing their bets on various numbers on a table representing each of the possible 38 numbers on the roulette wheel (Monty asking the contestant the 1st question). Say an unclever Probability Theorist is in the group of gamblers and places a stack of gambling chips on number “1” on the table, then quickly slides their bet to number “2” on the table all the while the roulette wheel is still spinning and the ball slowing down around the wheel. The Theorist again moves their bet to number “3” on the table, then number “4”, then “5” then “6” and so on until the Theorist finally rests their bet on number “19” all the while the croupier in amazement is observing the gambler. The Theorist is now content in the fact they have culled the probability toward their favor. After all they now better than 50/50 chance to win. They have already “picked” 19 numbers out of 38. The croupier sees the ball is slowing down and waves his arm over the table: a universal gesture for the gamblers to stop placing bets, as the croupier knows the ball is succumbing to gravity and will momentarily drop in one of 38 cups on the wheel. The ball drops in one of the 38 cups (the answer to Monty’s 2nd question). This is when the game begins. Real consequences will occur, win or lose (drive home with mom or with the convertible top down in your new red American auto). A single bet on one of 38 choices. Answering a simple question from Monty: Do you want to stay or switch (with one goat exposed it’s now a “real” 50/50 probability)? The moving of bets around the table before the ball dropped was literally inconsiquential to the probabilities of the outcome or the percieved advantage by the unclever Probability Theorist. There was 1 in 38 possibilities before the spin ever took place. Monty is ALWAYS going to show you a goat(removing one of the 3 possibilities). The probabilities were fixed from the before the ball was spun or before Monty even called you up to the stage. I’ll bet the croupier knew before sliding the roulette ball that it would drop in one of the 38 cups on the wheel.
Change your pick if you want, but it’s still 50/50. I hope my contribution can help clear up this decades old controversy.
Contributed by FourthKind 2/10/2016.
Vos Savant and the media furor
Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. She received thousands of letters from her readers—the vast majority of which, including many from readers with PhDs, disagreed with her answer. During 1990–1991 three more of her columns in Parade were devoted to the paradox. Numerous examples of letters from readers of Vos Savant's columns are presented and discussed in The Monty Hall Dilemma: A Cognitive Illusion Par Excellence.
In an attempt to clarify her answer she proposed a shell game 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? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger." She also proposed a similar simulation with three playing cards.
Vos Savant commented that though some confusion was caused by some readers not realizing that they were supposed to assume that the host must always reveal a goat, almost all of her numerous correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer ("switch") was wrong.
Confusion and criticism
Sources of confusion
When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter. Out of 228 subjects in one study, only 13% chose to switch. In her book The Power of Logical Thinking, quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer". Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans.
Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show,  and do not fully specify the host's behavior or that the car's location is randomly selected . Krauss and Wang  conjecture that people make the standard assumptions even if they are not explicitly stated.
Although these issues are mathematically significant, even when controlling for these factors nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter. This "equal probability" assumption is a deeply rooted intuition. People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not. Indeed, if a player believes that sticking and switching are equally successful and therefore equally often decides to switch as to stay, they will win 50% of the time, reinforcing their original belief. Missing the unequal chances of those two doors, and in not considering that (1/3+2/3) / 2 gives a chance of 50%, similar to "the little green woman" example.
The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as: 1) the endowment effect; people tend to overvalue the winning probability of the already chosen – already "owned" – door; 2) the status quo bias; people prefer to stick with the choice of door they have already made; 3) the errors of omission vs. errors of commission effect; all else considered equal, people prefer any errors they are responsible for to have occurred through 'omission' of taking action, rather than through having taken an explicit action that later becomes known to have been erroneous. Experimental evidence confirms that these are plausible explanations which do not depend on probability intuition.
Solutions using conditional probability and other solutions
The simple solutions above show that a player with a strategy of switching wins the car with overall probability 2/3, i.e., without taking account of which door was opened by the host. In contrast most sources in the field of probability calculate the conditional probabilities that the car is behind door 1 and door 2 are 1/3 and 2/3 given the contestant initially picks door 1 and the host opens door 3. The solutions in this section consider just those cases in which the player picked door 1 and the host opened door 3.
Refining the simple solution
If we assume the host opens a door at random, when given a choice, then which door the host opens gives us no information at all as to whether or not the car is behind door 1. In the simple solutions, we already observed that the probability that the car is behind door 1, the door initially chosen by the player, is initially 1/3. Moreover, the host is certainly going to open a (different) door, so opening a door (which door unspecified) does not change this. 1/3 must be the average probability that the car is behind door 1 given the host picked door 2 and given the host picked door 3 because these are the only two possibilities. But these two probabilities are the same. Therefore, they are both equal to 1/3. This shows that the chance that the car is behind door 1 given that the player initially chose this door and given that the host opened door 3 is 1/3, and it follows that the chance that the car is behind door 2 given the player initially chose door 1 and the host opened door 3 is 2/3. The analysis also shows that the overall success rate of 2/3, achieved by always switching, cannot be improved, and underlines what already may well have been intuitively obvious: the choice facing the player is that between the door initially chosen, and the other door left closed by the host, the specific numbers on these doors are irrelevant.
Conditional probability by direct calculation
By definition, the conditional probability of winning by switching given the contestant initially picks door 1 and the host opens door 3 is the probability for the event "car is behind door 2 and host opens door 3" divided by the probability for "host opens door 3". These probabilities can be determined referring to the conditional probability table below, or to an equivalent decision tree as shown to the right. The conditional probability of winning by switching is (1/3)/(1/3 + 1/6), which is 2/3.
The conditional probability table below shows how 300 cases, in all of which the player initially chooses door 1, would be split up, on average, according to the location of the car and the choice of door to open by the host.
Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem. Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent.
Initially, the car is equally likely behind any of the three doors: the odds on door 1, door 2, and door 3 are 1:1:1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is by definition the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given the host opened door 3, the probability the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.
Richard Gill analyzes the likelihood for the host to open door 3 as follows. Given the car is not behind door 1, it is equally likely that it is behind door 2 or 3. Therefore, the chance that the host opens door 3 is 50%. Given the car is behind door 1 the chance that the host opens door 3 is also 50%, because when the host has a choice, either choice is equally likely. Therefore, whether or not the car is behind door 1, the chance the host opens door 3 is 50%. The information "host opens door 3" contributes a Bayes factor or likelihood ratio of 1 : 1, on whether or not the car is behind door 1. Initially, the odds against door 1 hiding the car were 2 : 1. Therefore, the posterior odds against door 1 hiding the car remain the same as the prior odds, 2 : 1.
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, and this is precisely what is alleged to be intuitively obvious by supporters of simple solutions, or using the idioms of mathematical proofs, "obviously true, by symmetry".
Consider the events C1, C2 and C3 indicating the car is behind respectively door 1,2 or 3. All these 3 events have probability 1/3.
The player initially choosing door 1 is described by the event X1. As the first choice of the player is independent of the position of the car, also the conditional probabilities are P(Ci|X1)=1/3. The host opening door 3 is described by H3. For this event it holds:
Then, if the player initially selects door 1, and the host opens door 3, the conditional probability of winning by switching is
Strategic dominance solution
The Monty Hall problem is also much studied in the literature on game theory and decision theory, and also some popular solutions correspond to this point of view. 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.
A strategy of contestant involves two actions: the initial choice of a door and the decision to switch (or to stick) which may depend on both the door initially chosen and the door to which the host offers switching. For instance, one contestant's strategy is "choose door 1, then switch to door 2 when offered, and do not switch to door 3 when offered". Twelve such deterministic strategies of the contestant exist.
Elementary comparison of contestant's strategies shows that for every strategy A there is another strategy B "pick a door then switch no matter what happens" which dominates it. No matter how the car is hidden and no matter which rule the host uses when he has a choice between two goats, if A wins the car then B also does. For example, strategy A "pick door 1 then always stick with it" is dominated by the strategy B "pick door 2 then always switch after the host reveals a door": A wins when door 1 conceals the car, while B wins when one of the doors 1 and 3 conceals the car. Similarly, strategy A "pick door 1 then switch to door 2 (if offered), but do not switch to door 3 (if offered)" is dominated by strategy B "pick door 3 then always switch".
Dominance is a strong reason to seek for a solution among always-switching strategies, under fairly general assumptions on the environment in which the contestant is making decisions. In particular, if the car is hidden by means of some randomization device – like tossing symmetric or asymmetric three-sided die – the dominance implies that a strategy maximizing the probability of winning the car will be among three always-switching strategies, namely it will be the strategy which initially picks the least likely door then switches no matter which door to switch is offered by the host.
Strategic dominance links the Monty Hall problem to the game theory. In the zero-sum game setting, discarding the non-switching strategies reduces the game to the following simple variant: the host (or the TV-team) decides on the door to hide the car, and the contestant chooses two doors (i.e., the two doors remaining after the player's first, nominal, choice). The contestant wins (and her opponent loses) if the car is behind one of the two doors she chose.
Solutions by simulation
A simple way to demonstrate that a switching strategy really does win two out of three times with the standard assumptions is to simulate the game with playing cards. Three cards from an ordinary deck are used to represent the three doors; one 'special' card represents the door with the car and two other cards represent the goat doors.
The simulation can be repeated several times to simulate multiple rounds of the game. The player picks one of the three cards, then, looking at the remaining two cards the 'host' discards a goat card. If the card remaining in the host's hand is the car card, this is recorded as a switching win; if the host is holding a goat card, the round is recorded as a staying win. As this experiment is repeated over several rounds, the observed win rate for each strategy is likely to approximate its theoretical win probability.
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. 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.
Computer simulation using the Monte Carlo method is also possible. The following Python code defines and runs a function
montyhall() that allows the user to select a strategy and test its efficacy in winning a number of trial games.
import random def montyhall(strategy='swap', trials=1000): """Simulate the Monty Hall Problem using the Monte Carlo method. Arguments: strategy (string): Player strategy to 'swap' or 'stick' (default 'swap') trials (int): Number of trials for simulation (default 1000)""" # Create 3 mystery doors and initialise the wins counter doors = range(0,3) wins = 0 for trial in range(trials): # Randomly assign a winning door winning_door = random.choice(doors) # Player initially selects a random door player_door = random.choice(doors) # Host opens a random door, but not the player door or winning door host_doors = [door for door in doors if door not in [player_door, winning_door]] host_door = random.choice(host_doors) # Player decides whether or not to swap their current door if strategy == 'swap': player_door = [door for door in doors if door not in [player_door, host_door]] # Host opens the player's final door choice # If the player wins, increment the win counter if player_door == winning_door: wins += 1 # After running all specified trials, output the ratio of wins to trials winrate = float(wins) / float(trials) print(strategy, trials, wins, winrate) # Test the function with the two different strategies. # Notice how the win rate varies with the strategy chosen! montyhall('swap', 10000) montyhall('stick', 10000)
Criticism of the simple solutions
As already remarked, most sources in the field of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) given the contestant initially picks door 1 and the host opens door 3; various ways to derive and understand this result were given in the previous subsections. Among these sources are several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but ... shaky" , or do not "address the problem posed", or are "incomplete", or are "unconvincing and misleading", or are (most bluntly) "false". Some say that these solutions answer a slightly different question – one phrasing is "you have to announce before a door has been opened whether you plan to switch", emphasis in the original).
The simple solutions show in various ways that a contestant who is determined to switch will win the car with probability 2/3, and hence that switching is the winning strategy, if the player has to choose in advance between "always switching", and "always staying". However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given the player has picked door 1 and the host has opened door 3. As one source says, "the distinction between [these questions] seems to confound many". This fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values. For example, assume the contestant knows that Monty does not pick the second door randomly among all legal alternatives but instead, when given an opportunity to pick between two losing doors, Monty will open the one on the right. In this situation the following two questions have different answers:
- What is the probability of winning the car by always switching?
- What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?
The answer to the first question is 2/3, as is correctly shown by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1/2. This is because Monty's preference for rightmost doors means he opens door 3 if the car is behind door 1 (which it is originally with probability 1/3) or if the car is behind door 2 (also originally with probability 1/3). For this variation, the two questions yield different answers. However, as long as the initial probability the car is behind each door is 1/3, it is never to the contestant's disadvantage to switch, as the conditional probability of winning by switching is always at least 1/2.
Four university professors published an article (Morgan et al., 1991) in The American Statistician claiming vos Savant gave the correct advice but the wrong argument. They believed the question asked for the chance of the car behind door 2 given the player's initial pick for door 1 and the opened door 3, and they showed this chance was anything between 1/2 and 1 depending on the host's decision process given the choice. Only when the decision is completely randomized is the chance 2/3.
In an invited comment  and in subsequent letters to the editor, Morgan et al. were supported by some writers, criticized by others; in each case a response by Morgan et al. is published alongside the letter or comment in The American Statistician. In particular, vos Savant defended herself vigorously. Morgan et al. complained in their response to vos Savant (1991c) that vos Savant still had not actually responded to their own main point. Later in their response to Hogbin and Nijdam (2011) they did agree that it was natural to suppose that the host chooses a door to open completely at random, when he does have a choice, and hence that the conditional probability of winning by switching (i.e., conditional given the situation the player is in when he has to make his choice) has the same value, 2/3, as the unconditional probability of winning by switching (i.e., averaged over all possible situations). This equality was already emphasized by Bell (1992) who suggested that Morgan et al.'s mathematically involved solution would only appeal to statisticians, whereas the equivalence of the conditional and unconditional solutions in the case of symmetry was intuitively obvious.
There is disagreement in the literature regarding whether vos Savant's formulation of the problem, as presented in Parade magazine, is asking the first or second question, and whether this difference is significant. Behrends concludes that "One must consider the matter with care to see that both analyses are correct"; which is not to say that they are the same. One analysis for one question, another analysis for the other question. Several discussants of the paper by,  whose contributions were published alongside the original paper, strongly criticized the authors for altering vos Savant's wording and misinterpreting her intention. One discussant (William Bell) considered it a matter of taste whether or not one explicitly mentions that (under the standard conditions), which door is opened by the host is independent of whether or not one should want to switch.
Among the simple solutions, the "combined doors solution" comes closest to a conditional solution, as we saw in the discussion of approaches using the concept of odds and Bayes theorem. It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities. But knowing the host can open one of the two unchosen doors to show a goat does not mean that opening a specific door would not affect the probability that the car is behind the initially chosen door. 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. If the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation) this probability indeed remains unchanged, but if the host can choose non-randomly between such doors then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat and (in the standard interpretation of the problem) 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. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially chosen appear persuasive, but the assertion is simply untrue unless each of the host's two choices are equally likely, if he has a choice. The assertion therefore needs to be justified; without justification being given, the solution is at best incomplete. The answer can be correct but the reasoning used to justify it is defective.
Some of the confusion in the literature undoubtedly arises because the writers are using different concepts of probability, in particular, Bayesian versus frequentist probability. For the Bayesian, probability represents knowledge. For us and for the player, the car is initially equally likely to be behind each of the three doors because we know absolutely nothing about how the organizers of the show decided where to place it. For us and for the player, the host is equally likely to make either choice (when he has one) because we know absolutely nothing about how he makes his choice. These "equally likely" probability assignments are determined by symmetries in the problem. The same symmetry can be used to argue in advance that specific door numbers are irrelevant, as we saw above.
A common variant of the problem, assumed by several academic authors as the canonical problem, does not make the simplifying assumption that the host must uniformly choose the door to open, but instead that he uses some other strategy. The confusion as to which formalization is authoritative has led to considerable acrimony, particularly because this variant makes proofs more involved without altering the optimality of the always-switch strategy for the player. In this variant, the player can have different probabilities of winning depending on the observed choice of the host, but in any case the probability of winning by switching is at least 1/2 (and can be as high as 1), while the overall probability of winning by switching is still exactly 2/3. The variants are sometimes presented in succession in textbooks and articles intended to teach the basics of probability theory and game theory. A considerable number of other generalizations have also been studied.
Other host behaviors
The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. However, vos Savant made it clear in her second follow-up column that the intended host's behavior could only be what led to the 2/3 probability she gave as her original answer. "Anything else is a different question".  "Virtually all of my critics understood the intended scenario. I personally read nearly three thousand letters (out of the many additional thousands that arrived) and found nearly every one insisting simply that because two options remained (or an equivalent error), the chances were even. Very few raised questions about ambiguity, and the letters actually published in the column were not among those few." The answer follows if the car is placed randomly behind any door, the host must open a door revealing a goat regardless of the player's initial choice and, if two doors are available, chooses which one to open randomly. The table below shows a variety of other possible host behaviors and the impact on the success of switching.
Determining the player's best strategy within a given set of other rules the host must follow is the type of problem studied in game theory. For example, if the host is not required to make the offer to switch the player may suspect the host is malicious and makes the offers more often if the player has initially selected the car. In general, the answer to this sort of question depends on the specific assumptions made about the host's behavior, and might range from "ignore the host completely" to "toss a coin and switch if it comes up heads"; see the last row of the table below.
Morgan et al.  and Gillman  both show a more general solution where the car is (uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the statement of the problem in Parade despite the author's disclaimers. Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q, having a value between 0 and 1. If the host picks randomly q would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. If the player picks door 1 and the host's preference for door 3 is q, then the probability the host opens door 3 and the car is behind door 2 is 1/3 while the probability the host opens door 3 and the car is behind door 1 is (1/3)q. These are the only cases where the host opens door 3, so the conditional probability of winning by switching given the host opens door 3 is (1/3)/(1/3 + (1/3)q) which simplifies to 1/(1+q). Since q can vary between 0 and 1 this conditional probability can vary between 1/2 and 1. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. However neither source suggests the player knows what the value of q is so the player cannot attribute a probability other than the 2/3 that vos Savant assumed was implicit.
|Possible host behaviors in unspecified problem|
|The host acts as noted in the specific version of the problem.||Switching wins the car two-thirds of the time.
(Specific case of the generalized form below with p=q=½)
|The host always reveals a goat and always offers a switch. If he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q=1−p.  ||If the host opens the rightmost door, switching wins with probability 1/(1+q).|
|"Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door. ||Switching always yields a goat.|
|"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly.||Switching always wins the car.|
|"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.  ||Switching wins the car half of the time.|
|The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. He offers the option to switch only when the player's choice happens to differ from his.||Switching wins the car half of the time.|
|The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise. ||Switching wins 1/2 the time at the Nash equilibrium.|
|Four-stage two-player game-theoretic. The player is playing against the show organizers (TV station) which includes the host. First stage: organizers choose a door (choice kept secret from player). Second stage: player makes a preliminary choice of door. Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium.||Minimax solution (Nash equilibrium: car is first hidden uniformly at random and host later chooses uniform random door to open without revealing the car and different from player's door; player first chooses uniform random door and later always switches to other closed door. With his strategy, the player has a win-chance of at least 2/3, however the TV station plays; with the TV station's strategy, the TV station will lose with probability at most 2/3, however the player plays. The fact that these two strategies match (at least 2/3, at most 2/3) proves that they form the minimax solution.|
|As previous, but now host has option not to open a door at all.||Minimax solution (Nash equilibrium: car is first hidden uniformly at random and host later never opens a door; player first chooses a door uniformly at random and later never switches. Player's strategy guarantees a win-chance of at least 1/3. TV station's strategy guarantees a lose-chance of at most 1/3.|
|Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car hasn't been revealed.||Switching wins the car half of the time.|
D. L. Ferguson suggests an N-door generalization of the original problem in which the host opens p losing doors and then offers the player the opportunity to switch; in this variant switching wins with probability (N−1)/[N(N−p−1)]. If the host opens even a single door, the player is better off switching, but, if the host opens only one door, the advantage approaches zero as N grows large. At the other extreme, if the host opens all but one losing door the advantage increases as N grows large (the probability of winning by switching approaches 1 as N grows very large).
A quantum version of the paradox illustrates some points about the relation between classical or non-quantum information and quantum information, as encoded in the states of quantum mechanical systems. The formulation is loosely based on quantum game theory. The three doors are replaced by a quantum system allowing three alternatives; opening a door and looking behind it is translated as making a particular measurement. The rules can be stated in this language, and once again the choice for the player is to stick with the initial choice, or change to another "orthogonal" option. The latter strategy turns out to double the chances, just as in the classical case. However, if the show host has not randomized the position of the prize in a fully quantum mechanical way, the player can do even better, and can sometimes even win the prize with certainty.
The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889 in his Calcul des probabilités. In this puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold. As in the Monty Hall problem the intuitive answer is 1/2, but the probability is actually 2/3.
The Three Prisoners problem, published in Martin Gardner's Mathematical Games column in Scientific American in 1959, is equivalent to the Monty Hall problem. This problem involves three condemned prisoners, a random one of whom has been secretly chosen to be pardoned. One of the prisoners begs the warden to tell him the name of one of the others to be executed, arguing that this reveals no information about his own fate but increases his chances of being pardoned from 1/3 to 1/2. The warden obliges, (secretly) flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. The question is whether knowing the warden's answer changes the prisoner's chances of being pardoned. This problem is equivalent to the Monty Hall problem; the prisoner asking the question still has a 1/3 chance of being pardoned but his unnamed colleague has a 2/3 chance.
Steve Selvin posed the Monty Hall problem in a pair of letters to the American Statistician in 1975. The first letter presented the problem in a version close to its presentation in Parade 15 years later. The second appears to be the first use of the term "Monty Hall problem". The problem is actually an extrapolation from the game show. Monty Hall did open a wrong door to build excitement, but offered a known lesser prize – such as $100 cash – rather than a choice to switch doors. As Monty Hall wrote to Selvin:
And if you ever get on my show, the rules hold fast for you – no trading boxes after the selection.— (Hall 1975)
A version of the problem very similar to the one that appeared three years later in Parade was published in 1987 in the Puzzles section of The Journal of Economic Perspectives. Nalebuff, as later writers in mathematical economics, sees the problem as a simple and amusing exercise in game theory.
Phillip Martin's article in a 1989 issue of Bridge Today magazine titled "The Monty Hall Trap" presented Selvin's problem as an example of what Martin calls the probability trap of treating non-random information as if it were random, and relates this to concepts in the game of bridge.
A restated version of Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990.  Though vos Savant gave the correct answer that switching would win two-thirds of the time, she estimates the magazine received 10,000 letters including close to 1,000 signed by PhDs, many on letterheads of mathematics and science departments, declaring that her solution was wrong.  Due to the overwhelming response, Parade published an unprecedented four columns on the problem.  As a result of the publicity the problem earned the alternative name Marilyn and the Goats.
In November 1990, an equally contentious discussion of vos Savant's article took place in Cecil Adams's column The Straight Dope. Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. After a reader wrote in to correct the mathematics of Adams's analysis, Adams agreed that mathematically, he had been wrong, but said that the Parade version left critical constraints unstated, and without those constraints, the chances of winning by switching were not necessarily 2/3. Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two.
The Parade column and its response received considerable attention in the press, including a front page story in the New York Times in which Monty Hall himself was interviewed.  Hall appeared to understand the problem, giving the reporter a demonstration with car keys and explaining how actual game play on Let's Make a Deal differed from the rules of the puzzle. In the article, Hall pointed out that because he had control over the way the game progressed, playing on the psychology of the contestant, the theoretical solution did not apply to the show's actual gameplay.
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|The Wikibook Algorithm Implementation has a page on the topic of: Monty Hall problem simulation|
|Wikimedia Commons has media related to Monty Hall problem.|
- The Game Show Problem–the original question and responses on Marilyn vos Savant's web site
- University of California San Diego, Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game
- Monty Hall at DMOZ
- "Stick or switch? Probability and the Monty Hall problem", BBC News Magazine, 11 September 2013 (video). Mathematician Marcus du Sautoy explains the Monty Hall paradox.