# User:SPACKlick/Monty Hall Problem(draft)

In search of a new car, the player picks a door, say 1. The game host then opens one of the other doors, say 3, to reveal a goat and offers to let the player pick door 2 instead of door 1.

The Monty Hall problem is a brain teaser, in the form of a probability puzzle (Gruber, Krauss and others), 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 (Selvin 1975a), (Selvin 1975b). 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 (vos Savant 1990a):

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. (vos Savant 1990a) 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.

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 (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999).

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.

## Standard assumptions

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 (vos Savant 1990a) solution printed alongside Whitaker's question implies and both Selvin (1975a) and vos Savant (1991a) explicitly define the role of the host as follows

1. the host must always open a door that was not picked by the contestant (Mueser and Granberg 1999),
2. the host must always open a door to reveal a goat and never the car (Adams 1990)
3. 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 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 picked the car, then the host's choice of goat-hiding door to open is completely random. (Krauss and Wang, 2003:9) Some authors, independently or inclusively, assume the player's initial choice is completely random as well. Selvin (1975a)

## Solutions

### Simple Solutions

The solution presented by vos Savant (1990b) 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: 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.

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 Car Goat
Goat Car Goat Goat Car
Goat Goat Car Goat Car
Car has a 1/3 chance of being behind the player's pick and a 2/3 chance of being behind one of the other two doors
The host opens a door, the odds for the two sets don't change but the odds move to 0 for the open door and 2/3 closed door.

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 (Carlton 2005, concluding remarks). 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. (Adams 1990; Devlin 2003, 2005; Williams 2004; Stibel et al., 2008) As Cecil Adams puts it (Adams 1990), "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 switch from their original chosen door to both remaining doors. The switch in this case clearly gives the player a 2/3 probability of choosing the car.

As Keith Devlin says (Devlin 2003), "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. (vos Savant 1990a) 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 out of 1,000,000 times that remaining door will contain the prize. Intuitively, the player should ask how likely is it, that given a million doors, that they managed to pick the right one initially. Stibel et al. (2008) 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 still incorrectly judge the probability of success at 50/50.

### Solutions using conditional probability

The simple solutions above show that a player with a strategy of switching wins the car with overall probability 2/3 (Grinstead and Snell 2006:137–138 Carlton 2005). 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 (Selvin (1975b), Morgan et al. 1991, Chun 1991, Gillman 1992, Carlton 2005, Grinstead and Snell 2006:137–138, Lucas et al. 2009). The solutions in this section consider just those cases in which the player picked door 1 and the host opened door 3.

#### Conditional probability

Tree showing the probability of every possible outcome if the player initially picks Door 1

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 the car is behind door 2 and the host opens door 3 divided by the probability the 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 (Chun 1991; Carlton 2005; Grinstead and Snell 2006:137–138). The conditional probability of winning by switching is (1/3)/(1/3 + 1/6), which is 2/3 (Selvin 1975b).

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.

Car hidden behind Door 3
(on average, 100 cases out of 300)
Car hidden behind Door 1
(on average, 100 cases out of 300)
Car hidden behind Door 2
(on average, 100 cases out of 300)
Player initially picks Door 1, 300 repetitions
Host must open Door 2 (100 cases) Host randomly opens Door 2
(on average, 50 cases)
Host randomly opens Door 3
(on average, 50 cases)
Host must open Door 3 (100 cases)
Probability 1/3
(100 out of 300)
Probability 1/6
(50 out of 300)
Probability 1/6
(50 out of 300)
Probability 1/3
(100 out of 300)
Switching wins Switching loses Switching loses Switching wins
On those occasions when the host opens Door 2,
switching wins twice as often as staying (100 cases versus 50)
On those occasions when the host opens Door 3,
switching wins twice as often as staying (100 cases versus 50)

#### Bayes' theorem

Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem; among them Gill, 2002 and Henze, 1997. Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent (Rosenthal, 2005a), (Rosenthal, 2005b).

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 as door 1.

Richard Gill (2011 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" (Bell 1992).

#### Direct calculation

Consider the events ${\displaystyle C1,C2}$ and ${\displaystyle C3}$ indicating the car is behind respectively door 1,2 or 3. All these 3 events have probability 1/3. .

The player picking door 1 is described by the event ${\displaystyle X1}$. As the first choice of the player is independent of the position of the car, also the conditional probabilities are ${\displaystyle P(Ci|X1)=P_{1}(Ci)=1/3}$. For ease of notation the conditional probability given X1 is denoted by ${\displaystyle P_{1}}$

The host opening door 3 is described by ${\displaystyle H3}$. For this event it holds:

${\displaystyle P_{1}(H3|C1)=1/2}$
${\displaystyle P_{1}(H3|C2)=1}$
${\displaystyle P_{1}(H3|C3)=0}$

Then, if the player initially selects door 1, and the host opens door 3, the conditional probability of winning by switching is

${\displaystyle P_{1}(C2|H3)={\frac {P_{1}(H3|C2)P_{1}(C2)}{P_{1}(H3)}}}$

${\displaystyle ={\frac {P_{1}(H3|C2)P_{1}(C2)}{P_{1}(H3|C1)P_{1}(C1)+P_{1}(H3|C2)P_{1}(C2)+P_{1}(H3|C3)P_{1}(C3)}}}$

${\displaystyle ={\frac {P_{1}(H3|C2)}{P_{1}(H3|C1)+P_{1}(H3|C2)+P_{1}(H3|C3)}}={\frac {1}{1/2+1+0}}={\frac {2}{3}}}$

### Strategic dominance solution

Going back to Nalebuff (1987), 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.

Following Gill, 2011 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 (Gnedin, 2011). No matter how the car is hidden and no matter which rule the host uses when there is 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.

Thus, if the car is hidden by means of some non-uniform randomization device, 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 of Gill, 2011, discarding the nonswitching 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 if the car is behind one of the two doors they chose.

### Solutions by Simulation

Simulation of 30 outcomes of the Monty Hall problem

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 (Gardner 1959b; vos Savant 1996, p. 8). 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 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. (Gardner 1959b; Adams 1990). In this variant the car card goes to the host 51 times out of 52, and stays with the host no matter how many non-car cards are discarded.

## History

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 (Barbeau 1993). 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 (1959a, 1959b), 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 (Selvin 1975a), (Selvin 1975b). 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.

"You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I'll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical illiteracy in this country, and we don't need the world's highest IQ propagating more. Shame!"

- Scott Smith, Ph.D.

University of Florida

A restated version of Selvin's problem appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990. (vos Savant 1990a) 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. (Tierney 1991) Due to the overwhelming response, Parade published an unprecedented four columns on the problem. (vos Savant 1996, p. xv) As a result of the publicity the problem earned the alternative name Marilyn and the Goats.

In an attempt to clarify her answer she proposed a shell game (Gardner 1982) to illustrate: "You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? 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.

Despite further elaboration, many readers continued to disagree with her, but some changed their minds and agreed. Nearly 100% of those who carried out vos Savant's shell simulation changed their minds. About 56% of the general public and 71% of academics accepted the answer.

To help explain the thought process leading to the equal probability conjecture, vos Savant asked readers to consider the case where a little green woman emerges from a UFO at the point when the player has to decide whether or not to switch. The host asks the little green woman to point to one of the two unopened doors. Vos Savant noted the chance of randomly choosing the door with the prize is 1/2, because she does not know which door the player had initially chosen.

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. (Tierney 1991) 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.

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 being behind door 2 given the player's initial pick for door 1 and the host having opened door 3, and they showed this chance was anything between 1/2 and 1 depending on the host's decision process. Only when the decision is completely randomized is the chance 2/3.

In an invited comment (Seymann, 1991) and in subsequent letters to the editor, (vos Savant, 1991c; Rao, 1992; Bell, 1992; Hogbin and Nijdam, 2010) 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.

## 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 (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996, p. 15) 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."

Even though most statements of the problem, notably the one in Parade Magazine, do not fully specify the host's behavior or that the car's location is randomly selected (Granberg and Brown, 1995:712). Krauss and Wang (2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated. Although those assumptions 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 (Mueser and Granberg, 1999). This "equal probability" assumption is a deeply rooted intuition (Falk 1992:202). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not (Fox and Levav, 2004:637). Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans (Herbranson and Schroeder, 2010).

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 (Kahneman et al., 1991); people tend to overvalue the winning probability of the already chosen – already "owned" – door; 2) the status quo bias (Samuelson and Zeckhauser, 1988); people prefer to stick with the choice of door they have already made; 3) the errors of omission vs. errors of commission effect (Gilovich et al., 1995); all else considered equal, people prefer that any errors that 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 (Kaivanto et al., 2014; Morone and Fiore, 2007).

## Variants

### Host behaviors

The table below shows a variety of possible host behaviors, other than those specified in the standard assumptions, and the resulting probability of success by switching.

Possible host behaviors in unspecified problem
Host behavior Result
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. (Morgan et al. 1991) (Rosenthal, 2005a) (Rosenthal, 2005b). 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. (Tierney 1991) Switching always yields a goat.
"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly (Granberg 1996:185). 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 (Granberg and Brown, 1995:712) (Rosenthal, 2005a) (Rosenthal, 2005b). 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. (Mueser and Granberg 1999) Switching wins 1/2 the time at the Nash equilibrium.
Four-stage two-player game-theoretic (Gill, 2010, Gill, 2011). 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 (Mueser and Granberg 1999). 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.

Morgan et al. (1991) and Gillman (1992) both show a 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. 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 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.

### N-doors

D. L. Ferguson (1975 in a letter to Selvin cited in (Selvin 1975b)) 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(Np−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 (Granberg 1996:188). 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).

### Quantum version

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 (Flitney and Abbott 2002, D'Ariano et al. 2002).

### Recent discussion

Over 75 papers have been published about this problem in academic journals and the popular press. Barbeau 2000 contains a survey of the academic literature pertaining to the Monty Hall problem and other closely related problems as of the year 2000, and contains citations to 40 publications on the problem. At present the book Rosenhouse 2009 has the most recent comprehensive academic survey, and refers to at least 25 publications on the topic which appeared subsequently to Barbeau's book. Since then another 10 or so publications have come out. A number of sources refer to the Wikipedia article on the Monty Hall problem and the editorial discussion accompanying it, e. g. Rosenhouse 2009, Gill 2010 and several 2011 publications, Gnedin 2012.

The problem continues to appear in many venues:

• Derren Brown explains the Monty Hall problem in his stage show Svengali. (Frost 2012) After asking a member of the audience to choose the location of his shoe from three boxes, he reveals an empty box from one of the ones not chosen. He then asks if they would like to change their mind and recommends that they do so, as it will increase their chances of winning. He explains this further by demonstrating on a large screen the same puzzle but with one hundred boxes. The member of the audience decides to stick with their decision and loses. The problem is also addressed in his 2006 book Tricks Of The Mind.
• Penn Jillette explained the Monty Hall Problem on the "Luck" episode of Bob Dylan's Theme Time Radio Hour radio series.
• Economist M. Keith Chen identified a potential flaw in hundreds of experiments related to cognitive dissonance that use an analysis with issues similar to those involved in the Monty Hall problem. (Tierney 2008)
• In 2009 a book-length discussion of the problem, its history, methods of solution, and variations, was published by Oxford University Press (Rosenhouse 2009).
• The problem is presented, discussed, and tested in the television show MythBusters on 23 November 2011. This paradox was not only tested to see if there was an advantage to switching vs. sticking (which, in a repeated sample of 49 "tests", showed a significant advantage to switching), but they also tested the behavior of "contestants" presented with the same situation. All 20 of the common "contestants" tested chose to stay with their original choice.
• The problem was also discussed and tested on the television show James May's Man Lab on 11 April 2013. In this presentation, each test was done by presenting three identical beer cans, two of which had been shaken (with the result that opening it would douse the person in beer foam). James May performed this test 100 times, each time switching his choice from his original choice after one of the shaken cans was removed. In the end, he was doused 40 times, while his colleague Sim, who had to pick the remaining beer can, was doused 60 times. The resulting percentage was roughly what they expected.[1] However, the explanation James May gave was not correct.[citation needed]
• Craig Whitaker's actual letter to vos Savant has been found, and his original question reported in Morgan et al. Response to Hogbin and Nijdam (2011): "I've worked out two different situations (based on Monty's prior behavior i.e. weather [sic.] or not he knows what's behind the doors) in one situation it is to your advantage to switch, in the other there is no advantage to switch." "What do you think?"