# Monty Hall problem

(Redirected from Monty hall problem)
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)

The argument relies on assumptions, explicit in extended solution descriptions given by Selvin (1975b) and by vos Savant (1991a), that the host always opens a different door from the door chosen by the player and always reveals a goat by this action—because he knows where the car is hidden. Leonard Mlodinow stated: "The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated." (Mlodinow 2008)

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

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 Monty Hall problem has attracted academic interest from the surprising result and simple formulation. Variations of the Monty Hall problem are easily made by changing the implied assumptions, creating drastically different consequences. If Monty only offers the contestant a chance to switch when having initially chosen the car, then the contestant should never switch. If Monty opens another door randomly and happens to reveal a goat, then it makes no difference (Rosenthal, 2005a), (Rosenthal, 2005b).

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.

The most well known statement of the problem is:

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? (Whitaker, 1990, as quoted by vos Savant 1990a)

Vos Savant's response was:

Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here's a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what's behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You'd switch to that door pretty fast, wouldn't you?

Players initially have a 2/3 chance of picking a goat. Those who swap always get the opposite of their original choice, so those who swap have 2/3 chance of winning the car (Carlton 2005).

The correct answer, that players who swap have a 2/3 chance of winning the car and players who stick have a 1/3 chance of winning the car, is based on the premise that the host knows which door hides the car and intentionally reveals a goat. If the player selected the door hiding the car (1/3), then both remaining doors hide goats and the host may choose either door at random, and switching doors loses in 1/3. On the other hand, if the player initially selected a door that hides a goat (a 2-in-3 chance), then the host's choice is no longer at random, as he is forced to show the second goat only, and switching doors wins for sure in 2/3.

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

## Origin and full assumptions

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, (Selvin 1975a), dubbing it the "Monty Hall problem" in a subsequent letter (Selvin 1975b). The problem is mathematically equivalent to the Three Prisoners Problem described in Martin Gardner's "Mathematical Games" column in Scientific American in 1959 (Gardner 1959a) and the Three Shells Problem described in Gardner's book "Aha Gotcha" (Gardner 1982). (Morgan et al., 1991) The same problem was famously restated in a 1990 letter by Craig Whitaker to Marilyn vos Savant's "Ask Marilyn" column in Parade also.

Ambiguities in Craig Whitaker's quoted description of game play does not explicitly confine the requirement of the host to opening one of the two other doors containing a goat with the opportunity for the player to switch doors. However Marilyn vos Savant's (1990a) solution printed alongside Whitaker's question implies this interpretation (Mueser and Granberg 1999) likewise for the Monty Hall problem explicitly defined by both Selvin (1975a) and vos Savant (1991a).

It is also typically presumed that the car is initially hidden behind a random door. If the player initially picked the car, then the host's choice of 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)

## Vos Savant and the media furor

"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

Vos Savant wrote in her first column on the Monty Hall problem that the player should switch since the first door has a 1/3 chance of winning, so the second door has a 2/3 chance as the host always opens a losing door on purpose. (vos Savant 1990a) She went on to explain her answer by asking the reader to visualize the player selecting #1 amongst a million doors. Following the constraints, the host opens all remaining doors except door #777,777. Her conclusion: "You'd switch to that door pretty fast, wouldn't you?"

She received thousands of letters from her readers; 92% of the general public, 65% of universities, and many with PhDs, were against her answer. During 1990-1991 three more of her columns in Parade were devoted to debates (vos Savant 1990–1991), replayed in other venues (e.g., in Cecil Adams popular "The Straight Dope" newspaper column, Adams, 1990), and reported in major newspapers such as the New York Times (Tierney, 1991).

In an attempt to clarify by stating "The winning odds of 1/3 on the first choice can't go up to 1/2 just because the host opens a losing door," 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.

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, still, almost all of her numerous correspondents had correctly understood the problem assumptions, and despite this were initially convinced that vos Savant's answer ("switch") was wrong.

### The little green woman

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 and which one was offered by the host to switch to.

### A second controversy

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 (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.

## Solutions

### Vos Savant

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:

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

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. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

### Carlton

Player's pick has a 1/3 chance on the car and 2/3 chance on a goat. As each one has 1/3 chance on the car, the combined chance of door 2 and door 3 hiding the car is 2/3.
With the usual assumptions, after the host deliberately opened a door to intentionally show a goat, but without any further information, the player's pick still retains its 1/3 chance, likewise the other two doors still retain their combined 2/3 chance:
Null for the door opened, but 2/3 for the host's second still unopened door.

An intuitive explanation is to reason that players whose strategy is to switch lose if and only if they initially pick the car; that happens with probability 1/3, so switching must win with probability 2/3 (Carlton 2005, concluding remarks). The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial 1/3 probability.

Simply put, if the contestant picks a goat (to which two of the three doors lead) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (to which one door leads) the contestant will not win the car by switching. So, if contestants switch, they will win the car if they originally picked a goat and they will not win if they originally picked the car. As they have a 2 in 3 chance of originally picking a goat, they have a 2 in 3 chance of winning by switching.

Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door (Adams 1990; Devlin 2003; 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 player therefore has the choice of either sticking with the original choice of door, or choosing the sum of the contents of the two other doors. The 2/3 chance of hiding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is equally likely to reveal a goat whether the player originally picked the car or the goat (in fact: in both cases, he is certain to do so). If the host did not know where the car is hidden, but opened a door revealing a goat by chance, the revelation would indeed change the probability of the player's original choice being correct to 1/2. The critical difference is that an ignorant host is more likely to successfully reveal a goat when the player's original choice is the car than he is when the original choice is a goat, so the revealing of a goat by chance does provide evidence in favor of the player's original choice winning the car.

In other words, offering the player to switch doors after revealing a goat is no different than if the host offered the player to switch from their original chosen door to both remaining doors, and only after the player chooses, the host shows that one of the two doors contains a goat (which at least one always will). The switch in this case clearly gives the player a 2/3 probability of chosing the car, and the revelation that a goat is behind one of the player's two doors, which is always the case, does not change this probability.

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 – Increasing the number of doors

That switching has a probability of 2/3 of winning the car runs counter to many people's intuition. If there are two doors left, then why is each door not 1/2? It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three. (vos Savant 1990a) In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat – the chance that the player's door is correct has not changed. A rational player should switch.

Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. It's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens 999,998 for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.

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.

### 'The Economist'

In this analysis the words, 'say No. 1', and 'say No. 3' in the question are not taken to mean that we are only to consider the case where the player has chosen door 1 and the host has revealed a goat behind door 3. The problem is solved by considering the equally likely events that the player has initially chosen the car, goat A, or goat B (Economist 1999). In the first case, switching doors loses, but in the other two cases, switching doors wins, see illustration:

1.
Player picks car
(probability 1/3)
Switching loses.
2.
Host must
reveal Goat B

Player picks Goat A
(probability 1/3)
Switching wins.
3.
Host must
reveal Goat A

Player picks Goat B
(probability 1/3)
Switching wins.
The player has an equal chance of initially selecting the car, Goat A, or Goat B. Switching results in a win 2/3 of the time.

### Simulations

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 on the average 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 such as the Ace of Spades should represent the door with the car, and ordinary cards, such as the two red twos, represent the goat doors.

The simulation, using the following procedure, can be repeated several times to simulate multiple rounds of the game. One card is dealt face-down at random to the 'player', to represent the door the player picks initially. Then, looking at the remaining two cards, at least one of which must be a red two, the 'host' discards a red two. If the card remaining in the host's hand is the Ace of Spades, this is recorded as a round where the player would have won by switching; if the host is holding a red two, the round is recorded as one where staying would have won.

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 one card has been dealt to the player, it is already determined whether switching will win the round for the player; and two times out of three the Ace of Spades is in the host's hand.

If this is not convincing, the simulation can be done with the entire deck, dealing one card to the player and keeping the other 51 (Gardner 1959b; Adams 1990). In this variant the Ace of Spades goes to the host 51 times out of 52, and stays with the host no matter how many non-Ace cards are discarded.

Another simulation, suggested by vos Savant, employs the "host" hiding a penny, representing the car, under one of three cups, representing the doors; or hiding a pea under one of three shells.

### Solutions using conditional probability

The player is only asked whether or not he would like to switch after the host has opened a particular door, different from the player's initial choice, and revealed a goat behind it. In order to be sure of having the best possible chance of getting the car, the player should clearly make his decision of whether or not to switch on the basis of the probability that the car is behind the other closed door, given all the information which he has at this moment. That information consists, in the example given by Marilyn vos Savant, of the fact that it was door 1 which was initially chosen by himself, and that it was door 3 which was opened by the host to reveal a goat.

#### Refining the simple solution

Given the player's initial choice, door 1, the host might also have opened door 2 to reveal a goat. Suppose we assume that the host is equally likely to open either door 2 or 3 if the car is behind door 1. We already assumed that the car is equally likely behind any of the three doors, and this remains so after the player has made his initial choice. It follows that the probability that the car is behind door 1 given the host opens door 3 and the player initially chose door 1 must equal the probability that the car is behind door 1 given the host opens door 2 and the player initially chose door 1, since all elementary probabilities in the problem (probabilities of locations of car given the player initially chose door 1, and probabilities of which door the host will open if the car is behind door 1, the door chosen by the player) are unchanged by interchanging the door numbers 2 and 3.

In other words, given that the player initially chose door 1, whether the host opens door 2 or door 3 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 to reveal a goat, so opening a door (which door, door 2 or door 3, unspecified) does not change this probability. This probability, 1/3, must be the average of the probability that the car is behind door 1, given the host opens door 2 and the player chose door 1, and the probability that the car is behind door 1, given the host opens door 3 and the player chose door 1, because these are the only two possibilities. But these two probabilities are the same. Therefore they are both equal to 1/3.

This more refined analysis (which can be found in the published discussion following the paper of Morgan et al. 1991) 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

Most sources in the field of probability, including many introductory probability textbooks, solve the Monty Hall problem by showing from first principles that 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 (Selvin (1975b), Morgan et al. 1991, Chun 1991, Gillman 1992, Carlton 2005, Grinstead and Snell 2006:137–138, Lucas et al. 2009). In contrast, 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 (Grinstead and Snell 2006:137–138 Carlton 2005). For example, if we imagine 300 repetitions of the show, in all of which the player initially chooses door 1, a strategy of "always switching" will win about 200 times – namely every time the car is not behind door 1. The solutions in this section consider just those cases in which not only the player picked door 1, but moreover the host went on to open door 3. After all, it is only after a door was opened to reveal a goat that the player was asked if he wants to switch or not. Of our total of 300 repetitions, door 3 will be opened by the host about 150 times: the other 150 times the host will open door 2. The following solutions show that when we restrict attention to the first mentioned 150 cases (player initially picked door 1, host went on to reveal a goat behind door 3), the car will be behind door 2 about 100 times, but behind door 1 only about 50 times.

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).

Assuming the player picks door 1, the car is behind door 2 and the host opens door 3 with probability 1/3. The car is behind door 1 and the host opens door 3 with probability 1/6. These are the only possibilities given the player picks door 1 and the host opens door 3. Therefore, 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)

#### Extension of vos Savant's table

An equivalent way to present this approach (Rosenhouse, 2009; Krauss and Wang, 2003) is by extending vos Savant's table above, where it is given that the player has chosen door 1. Here it is with probabilities included:

Case Prob. behind door 1 behind door 2 behind door 3 result if staying at door #1 result if switching to the door offered
1 1/3 Car Goat Goat Car Goat
2 1/3 Goat Car Goat Goat Car
3 1/3 Goat Goat Car Goat Car

Next we expand row 1 according to the host's behaviour - if he has a choice, into two equally likely subcases 1a and 1b:

Case Prob. behind door 1 behind door 2 behind door 3 opened door result if staying at door #1 result if switching to the door offered
1a 1/6 Car Goat Goat 2 Car Goat
1b 1/6 Car Goat Goat 3 Car Goat
2 1/3 Goat Car Goat 3 Goat Car
3 1/3 Goat Goat Car 2 Goat Car

Finally we expand the remaining rows, too; imagine a fair coin being tossed to separate each outcome of probability 1/3 into two outcomes of probability 1/6. This splits case 2 into two equally likely subcases 2a and 2b, and case 3 into two equally likely subcases 3a and 3b:

Case Prob. behind door 1 behind door 2 behind door 3 opened door result if staying at door #1 result if switching to the door offered
1a 1/6 Car Goat Goat 2 Car Goat
1b 1/6 Car Goat Goat 3 Car Goat
2a 1/6 Goat Car Goat 3 Goat Car
2b 1/6 Goat Car Goat 3 Goat Car
3a 1/6 Goat Goat Car 2 Goat Car
3b 1/6 Goat Goat Car 2 Goat Car

Now we have six equally likely cases, and can compute probabilities and conditional probabilities simply by counting. In three of the six equally likely cases, the host opens door 3. In two of those three cases, the car is behind door 2. Therefore the conditional probability of winning by switching given the host opened door 3 (and the player initially chose door 1) is 2/3.

#### Bayes' theorem

Many probability text books and articles in the field of probability theory (and the teaching 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 more likely to be behind door 2 than door 1.

Richard Gill (2011) uses a similar argument to patch a missing step in Devlin's combined doors solution (Devlin 2003), described above ("Adams and Devlin"). Devlin does not explain why the information of which door had been opened by the host did not change the odds on door 1 hiding the car. The contestant is not just given the opportunity to choose between door 1 and doors 2 plus 3 – he is also told something specific about those two doors ("no car behind door 3"). Correspondents pointed out this missing step to Devlin and he later (Devlin 2005) retracted his "combined doors" solution, giving a direct calculation using Bayes theorem instead, saying that though unintuitive, it did at least give one the guarantee of obtaining the right answer in an automatic way.

The justification provided by Gill is as follows. Given the car is not behind door 1 (the door initially chosen by the contestant), it is equally likely that it is behind door 2 or door 3. In those cases, the host is forced to open the other door. Therefore, the chance that the host opens door 3 given that the car is not behind door 1 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 50 : 50, or 1 : 1, concerning the question 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).

### 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 (for instance, that were published in The Economist, see above, among the simple 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 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 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 (and her opponent loses) if the car is behind one of the two doors she chose.

## 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 (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." Interestingly, pigeons make mistakes and learn from mistakes, and experiments show that they rapidly learn to always switch, unlike humans (Herbranson and Schroeder, 2010).

Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show (Krauss and Wang, 2003:9), and 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. (From the point of view of subjective probability, the standard assumptions can be derived from the problem statement: they follow from our total lack of information about how the car is hidden, how the player initially chooses a door, and how the host chooses a door to open if there's a choice.)

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 (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). 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 (Marc C. Steinbach, 2000).

The problem continues to attract the attention of cognitive psychologists. The typical behaviour 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).

### 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" (Rosenthal 2005a), or do not "address the problem posed" (Gillman 1992), or are "incomplete" (Lucas et al. 2009), or are "unconvincing and misleading" (Eisenhauer 2001) or are (most bluntly) "false" (Morgan et al. 1991). 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" (Gillman 1992, 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" (Morgan et al. 1991). 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:

1. What is the probability of winning the car by always switching?
2. 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. (Morgan et al. 1991)

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 (Rosenhouse 2009). Behrends (2008) 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 (Morgan et al. 1991), whose contributions were published alongside the original paper, strongly criticized the authors for altering vos Savant's wording and misinterpreting her intention (Rosenhouse 2009). 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 (Falk 1992:207,213). 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.

## Variants

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". (vos Savant 1991a) "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." (vos Savant 1996) 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 (Mueser and Granberg, 1999). 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 behaviour, 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. (1991) and Gillman (1992) 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 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
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 organisers (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.

### 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).

## 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 who will 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.

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 1987). 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" (Martin 1989) 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. (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.

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.

### 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:

• The syndicated NPR program Car Talk featured it as one of their weekly "Puzzlers", and the answer they featured was quite clearly explained as the correct one (Magliozzi and Magliozzi, 1998).
• Accounts of the Hungarian mathematician Paul Erdős's first encounter of the problem can be found in The Man Who Loved Only Numbers and Vazsonyi 1999; like so many others, Erdős initially got it wrong.
• The problem is presented in fictional form in the first chapter of the novel Mr Mee (2000) by Andrew Crumey.
• The problem is discussed, from the perspective of a boy with Asperger syndrome, in The Curious Incident of the Dog in the Night-Time, a 2003 novel by Mark Haddon.
• The problem is also addressed in a lecture by the character Charlie Eppes in an episode of the CBS drama NUMB3RS (Episode 1.13).
• 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.
• The problem is featured in Ian McEwan's 2012 novel Sweet Tooth.
• Penn Jillette explained the Monty Hall Problem on the "Luck" episode of Bob Dylan's Theme Time Radio Hour radio series.
• The Monty Hall problem appears in the film 21 (Bloch 2008).
• 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, but 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?"