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The Monty Hall problem is a probability puzzle loosely based on the American television game show Let's Make a Deal and named after the show's original host, Monty Hall. The problem, also called the Monty Hall paradox, is a veridical paradox because the result appears impossible but is demonstrably true. The Monty Hall problem, in its usual interpretation, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older Bertrand's box paradox.

The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a) (Selvin 1975b). One published statement of the problem was published 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. If the car is initially equally likely to be behind each door, a player who picks door 1 and doesn't switch has a 1 in 3 chance of winning the car while a player who picks door 1 and does switch has a 2 in 3 chance, because the host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car.

Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that 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). Decision scientist Andrew Vazsonyi described how Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until Vazsonyi showed him a computer simulation confirming the predicted result (Vazsonyi 1999).

The Monty Hall problem has attracted academic interest because the result is surprising and the problem is simple to formulate. Furthermore, variations of the Monty Hall problem are made by changing the implied assumptions, and the variations can have drastically different consequences. For example, if Monty only offered the contestant a chance to switch when the contestant had initially chosen the car, then the contestant should never switch. Variations of the Monty Hall problem are given below.

The Problem

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). In a subsequent letter he dubbed it the "Monty Hall problem" (Selvin 1975b). The problem is mathematically equivalent (Morgan et al., 1991) to the Three Prisoners Problem described in Martin Gardner's Mathematical Games column in Scientific American in 1959 (Gardner 1959a).

In 1990 the same problem was restated in its well-known form in a letter to Marilyn vos Savant's Ask Marilyn column in Parade:

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)

There are certain ambiguities in this formulation of the problem: it is unclear whether or not the host would always open another door, always offer a choice to switch, or even whether he would ever open the door revealing the car (Mueser and Granberg 1999). The standard understanding of the problem (resolving these ambiguities) is that the host is indeed constrained always to open an unchosen door revealing a goat and always to make the offer to switch. Very often it is also assumed that the car is initially hidden completely at random, and that if the host has a choice of door to open (which happens if the player initially picked the car) then this choice is completely random, too. (Krauss and Wang, 2003:9). Some authors, either instead of or together with these probability assumptions, assume that the player's initial choice is completely random (for instance, Selvin 1975a).

Vos Savant and the media furore

"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 correctly responded in her column that the player should switch and that the first door has a 1/3 chance of winning, but the second door offered to switch on has a 2/3 chance as the host always opens a losing door on purpose. She went on to explain her answer by asking the reader to visualise the case where there are a million doors and the player picks #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens all remaining doors except door #777,777.

In response to her correct answer, vos Savant received thousands of letters from her readers, many with PhDs telling her that she was wrong. Of the letters from the general public, 92% were against her answer, and of the letters from universities, 65% were against her answer. Overall, nine out of ten readers completely disagreed with her answer.

Vos Savant replied with further explanation, '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', and she proposed a shell game to illustrate her thinking: '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 of the game with three playing cards.

Even after her further explanations readers continued to write in telling her that she was wrong. Nevertheless people did begin to change their minds about the solution. Of those that actually carried out vos Savant's shell simulation nearly 100% wrote to say that they agreed with her. Of the general public, about 56% came to believe that switching was best and of academic institutions 71%. Vos Savant commented that some confusion was caused by people not realising that the host must always reveal a goat but that most people did understand the question this way.

The little green woman

To help explain why people thought that the probability of winning both sticking and switching was equal, vos Savant asked readers to consider the case where a a little green woman emerges on stage from a UFO at the point that the player has to decide which door to choose, and the host asks her to point to one of the two unopened doors. In that case, vos Savant points out, the chances that she’ll randomly choose the door with the prize are 1/2, but that is because she lacks the information that a player would have from seeing how the two doors were chosen.

Controversy continues

In November 1991 a paper was published in "The American Statistician" by four university professors (Morgan et al) saying that, although, given a certain assumption, vos Savant's answer was numerically correct her methods of proof were not. The authors stated vos Savant actually solved a slightly different problem from that which she herself had formulated. She also had failed to consider the possibility that the host might not choose evenly between the two possible doors that he might open under the rules of the game when the player has initially chosen the car and that a conditional probability solution was required to solve the problem properly.

Vos Savant vigorously defended her solution, stating that she had taken the host to be acting purely as the agent of chance and would therefore choose between the doors available to him at random, thus making her solution correct. Morgan et al responded by saying that, even in the case that the host chose randomly between doors allowed under the rules, a conditional solution was still strictly necessary.

In response to a letter pointing out an error in their paper, Morgan et al insisted that a conditional solution to the problem was still necessary but did acknowledge that, '...had we adopted conditions implicit in the problem, the answer is 2/3, period'.

Simple Solutions

Vos Savant

The solution presented by vos Savant in Parade (vos Savant 1990b) 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

An intuitive explanation is to reason that a player whose strategy is to switch loses 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).

Simply put, if the contestant picks a goat (to which two of the three doors lead) they 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) they will not win the car by switching. So, if you switch, you win the car if you originally picked a goat and you don't win if you originally picked the car. As you have a 2 in 3 chance of originally picking a goat, you have a 2 in 3 chance of winning by switching.

Adams and Devlin

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, as the 2/3 chance of hiding the car has not been changed by the opening of one of these doors.

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 isn't each door 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 1990). 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 hasn't 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):

1. Host reveals either goat 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.

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 (1996:15) 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.

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. Experimental evidence confirms that these are plausible explanations which do not depend on probability intuition (Morone and Fiore, 2007).

Solutions using conditional probability

Most sources in the field of probability, including many introductory probability textbooks, solve the Monty Hall 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 (Selvin 1975b, Morgan et al. 1991, Chun 1991, Gillman 1992, 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.

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; Grinstead and Snell 2006:137-138). Other approaches to getting the same answer, both formal and informal, will be given later. 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 case versus 50)

Extension of vos Savant's table

An equivalent way to present this approach (Rosenhouse, 2009; Krauss and Wang, 2003) is by extending the table we saw at the beginning of the article due to vos Savant. Recall, it's already given that the player has chosen door 1. First of all we reproduce the vos Savant table, with probabilities included:

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

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:

Now we have altogether 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' formula; among them Gill, 2002 and Henze, 1997. This requires introducing quite a lot of mathematical notation, which may distract from the essential simplicity of the calculation.

Consider the discrete random variables, all taking values in the set of door numbers ${\displaystyle \{1,2,3\}}$:

C: the number of the door hiding the Car,

S: the number of the door Selected by the player, and

H: the number of the door opened by the Host.

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

${\displaystyle {\begin{aligned}P(C=2|H=3,S=1)&={\frac {P(H=3,C=2|S=1)}{P(H=3|S=1)}}={\frac {P(H=3|C=2,S=1)P(C=2|S=1)}{\sum _{i=1}^{3}P(H=3|C=i,S=1)P(C=i|S=1)}}\\&={\frac {1\times {\frac {1}{3}}}{{\frac {1}{2}}\times {\frac {1}{3}}+1\times {\frac {1}{3}}+0\times {\frac {1}{3}}}}={\frac {2}{3}},\end{aligned}}}$

as due to the random placement of the car and the independence of the initial choice of the player and the placement of the car:

${\displaystyle P(C=1|S=1)=P(C=2|S=1)=P(C=3|S=1)={\tfrac {1}{3}}.}$

and due to the host's behaviour:

${\displaystyle P(H=3|C=1,S=1)={\tfrac {1}{2}},}$

${\displaystyle P(H=3|C=2,S=1)=1,}$

${\displaystyle P(H=3|C=3,S=1)=0.}$

Note what happens in the big displayed formula for ${\displaystyle P(C=2|H=3,S=1)}$ if we replace C=2 throughout with C=1 and with C=3, in turn. The summation in the denominator of the right hand side doesn't change. The second term in the numerator, ${\displaystyle P(C=2|S=1)}$ gets changed in the obvious way, but in all three cases it remains equal to 1/3. Thus the three probabilities ${\displaystyle P(C=c|H=3,S=1)}$, for c=1,2 and 3, are proportional to the three probabilities ${\displaystyle P(H=3|C=c,S=1)}$, c=1,2 and 3, which as we just saw are equal to 1/2, 1 and 0. Since the three probabilities ${\displaystyle P(C=c|H=3,S=1)}$ have to add to one, they must be equal to 1/3, 2/3 and 0 (same proportions as 1/2, 1, 0, but correct total 1).

This corresponds to the use of Bayes' formula in its formulation for odds, rather than probabilities, to which we now turn.

Odds

Use of the odds form of Bayes' rule can make the above formulas superfluous and the result 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 than door 1.

A similar argument can be used to patch a missing step in Devlin's combined doors solution, (Devlin 2003), described above ("Adams and Devlin" in the section on simple solutions.) 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. We are not just given the opportunity to choose between door 1 and doors 2 plus 3 -- we are also told something specific about those two doors ("no car behind door 3"). Correspondents pointed out this missing step to him 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.

He could however have salvaged his argument as follows. Consider the odds on door 1 hiding the car. Initially, these are 2:1 against. Now, given the car is behind door 1 (the door chosen by the player) the chance the host opens door 3 is 50%, by symmetry (when the host has a choice, the two choices are equally likely). On the other hand, given the car is not behind door 1, the host opens door 3 (and is obliged to do so) when the car is behind door 2, which again has 50% chance, by symmetry (the car is equally likely behind door 2 and door 3, if it is not behind door 1). Thus whether or not the car is behind door 1, the chance is 50% that the host opens door 3 (rather than door 2). The Bayes factor is therefore 50 : 50 or equivalently 1 : 1. Therefore the posterior odds on door 1 hiding the car remain at 2 : 1 against (the prior odds).

In words, the information which door is opened by the host (door 2 or door 3?) tells us nothing about whether or not the car is behind door 1, which is intuitively very obvious from the start (a mathematician would say: is obviously true, by symmetry). On the other hand, which door is opened (door 2 or door 3?) does have a big impact on the chances for doors 2 and 3 themselves! And the fact that the intuitive answer to the Monty Hall problem is "don't switch, because it doesn't make a difference" tells us that we have to be very careful with intuition when solving probability puzzles.

From simple to conditional by symmetry

Formal mathematical derivations can also be given which avoid explicit computations or formula manipulation, and which illustrate various insights into the Monty Hall problem:

From simple solution and partial symmetry to conditional solution

Here we use the symmetry between doors 2 and 3 in the problem.

The chance that the door chosen by the contestant, door 1, hides the car, is 1/3. The conditional probabilities that door 1 hides the car given that the host opens door 2, and that he opens door 3, must be equal to one another, by symmetry. This means that whether or not the host opens door 3 is (statistically) independent of whether or not the car is behind door 1 given the player initially chose door 1. The conditional probability is equal to the unconditional probability, 2/3. (Gill, 2011a), (Bell (1992): "I will leave it to readers as to whether this equivalence of the conditional and unconditional problems is intuitively obvious.").

From simple solution and full symmetry to conditional solution

Imagining for the moment that the initial door chosen by the player is completely random too, we obtain a more full symmetry, namely, symmetry in all three door numbers. From this starting point we can prove that the actual door numbers in any particular case are completely irrelevant to whether the player should stay or switch. Therefore the overall (unconditional) probability of winning by switching (2/3, by the simple solutions) equals the conditional probability of winning by switching in any particular case (door chosen by player, door opened by host).

As before, let ${\displaystyle C}$, ${\displaystyle S}$, ${\displaystyle H}$ stand for the numbers of the doors hiding the Car, Selected by the contestant, and opened by the Host, respectively. Let ${\displaystyle R}$ stand for the Remaining closed door. The triple of door numbers ${\displaystyle (S,H,R)}$ are the specific numbers observed by the player written on the doors selected by the player, opened by the host, and remaining closed (not opened by the host). They form a permutation of the numbers 1, 2, 3. Let ${\displaystyle I}$ stand for the indicator random variable which takes the value 1 if switching would give the car, and 0 if not: thus ${\displaystyle I=1}$ if ${\displaystyle C=R}$, while ${\displaystyle I=0}$ if ${\displaystyle C=S}$. Let's pretend for the moment that the door chosen by the player was actually chosen at random - it just happened to be door 1. With this assumption, an arbitrary renumbering of the doors changes nothing in the probabilistic description of the problem. In mathematical terms, the problem is completely symmetric under permutations of the door numbers. Now, however we renumber the doors, whether or not switching gives the car does not change, ${\displaystyle I}$ is invariant. On the other hand, renumbering the doors in all six possible ways makes ${\displaystyle (S,H,R)}$ take on all of its six possible values, the six different orders in which one can write down the numbers 1, 2, 3. By invariance, the six different orderings all have the same probability, 1/6, and ${\displaystyle I}$ must be (statistically) independent of ${\displaystyle (S,H,R)}$: the probabilities that ${\displaystyle I=0}$ and ${\displaystyle I=1}$ can't depend on which permutation has been realised. The door numbers in a specific case are irrelevant to deciding whether to switch or stay (and it was indeed totally harmless to pretend that the player's initial choice was random). This ties in with Marilyn Vos Savant's almost parenthetical wording "say, door 1", and "say, door 3", and her insistence (vos Savant, 1991b) in response to Morgan et al. (1991) that conditional probability is absolutely not needed to solve the problem she had posed. In Tierney (1991), the mathemagician and Stanford professor Persi Diaconis stands up for vos Savant, see Diaconis (1988) for his work on symmetry in statistics.

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 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 orginal paper, strongly criticized the authors for altering vos Savant's wording and misinterpreting her intention (Rosenhouse 2009). One discussant (William S. 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 doesn't 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 won't 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 don't 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.

Strategic solution by dominance

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 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 doesn't 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.

Simulation

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 1996:8). 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.

This experiment is likely to approximate the probability of winning, and running the experiment over enough rounds should not only verify that the player does win by switching two times out of three, but show why. 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.

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 followup 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.

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)/. 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).

Bapeswara Rao and Rao (1992) suggest a different N door version where the host opens a losing door different from the player's current pick and gives the player an opportunity to switch after each door is opened until only two doors remain. With four doors the optimal strategy is to pick once and switch only when two doors remain. With N doors this strategy wins with probability (N−1)/N and is asserted to be optimal.

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 cohort 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.

— Hall 1975

A version of the problem very similar to the one that appeared three years later in Parade was published in 1987 in the Puzzles section of The Journal of Economic Perspectives (Nalebuff 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:xv). As a result of the publicity the problem earned the alternative name Marilyn and the Goats.

In November 1990, an equally contentious discussion of vos Savant's article took place in Cecil Adams's column The Straight Dope (Adams 1990). Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. After a reader wrote in to correct the mathematics of Adams' analysis, Adams agreed that mathematically, he had been wrong, but said that the Parade version left critical constraints unstated, and without those constraints, the chances of winning by switching were not necessarily 2/3. Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two.

The Parade column and its response received considerable attention in the press, including a front page story in the New York Times in which Monty Hall himself was interviewed. (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.

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. Nowadays, Misplaced Pages is frequently cited as a source for Monty Hall problem (it is cited, for instance, by Ruma Falk, Rosenhouse 2009, Gill 2012, 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 (Huffington Post TV Review). 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 's 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.

See also

• Boy or Girl paradox
• Principle of restricted choice (bridge)
• Sleeping Beauty problem
• Two envelopes problem

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• vos Savant, Marilyn (1996). The Power of Logical Thinking. St. Martin's Press. ISBN 0-312-15627-8.
• vos Savant, Marilyn (2006). "Ask Marilyn" column, Parade Magazine p. 6 (26 November 2006).
• Schwager, Jack D. (1994). The New Market Wizards. Harper Collins. p. 397. ISBN 978-0-88730-667-9.
• Williams, Richard (2004). "Appendix D: The Monty Hall Controversy" (PDF). Course notes for Sociology Graduate Statistics I. Retrieved 2008-04-25.
• Wheeler, Ward C. (1991). "Congruence Among Data Sets: A Bayesian Approach". In Michael M. Miyamoto and Joel Cracraft (ed.). Phylogenetic analysis of DNA sequences. Oxford University Press US. p. 335. ISBN 978-0-19-506698-2.
• Whitaker, Craig F. (1990). . "Ask Marilyn" column, Parade Magazine p. 16 (9 September 1990).
• TV REVIEW: Derren Brown: Svengali - How Does He Do It? No, Really - I'm Asking... ().

External links

• The Game Show Problem–the original question and responses on Marilyn vos Savant's web site
• Template:Dmoz
• "Monty Hall Paradox" by Matthew R. McDougal, The Wolfram Demonstrations Project (simulation)
• The Monty Hall Problem at The New York Times (simulation)
• Three Doors Probability Problem (Simulator to go through 100 simulations in a couple of minutes)

• Probability theory paradoxes
• Let's Make a Deal
• Mathematical problems
• Game theory
• Microeconomics
• Decision theory paradoxes
• Named probability problems