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The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive.

A well-known statement of the problem was published in Parade magazine:

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)

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.

When the problem and the solution appeared in Parade, approximately 10,000 readers, including several hundred mathematics professors, wrote to the magazine claiming the published solution was wrong. Some of the controversy was because the Parade statement of the problem was technically ambiguous. However, even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.

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 1959).

Selvin's Monty Hall 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 analysis of the problem assumes that the host is indeed constrained always to open a door revealing a goat, always to make the offer to switch, and to open one of the remaining two doors randomly if the player initially picked the car (Barbeau 2000:87). Hence a more exact statement of the problem is as follows:

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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? (Krauss and Wang 2003:10)

Note that the player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors. It is assumed that the player is trying to win the car.

Solution

The overall probability of winning by switching is determined by the location of the car. Assuming the problem statement above and that the player initially picks Door 1:

• The player originally picked the door hiding the car. The game host must open one of the two remaining doors randomly.
• The car is behind Door 2 and the host must open Door 3.
• The car is behind Door 3 and the host must open Door 2.

Player picks Door 1
Car behind Door 1		Car behind Door 2	Car behind Door 3
Host opens either goat door		Host must open Door 3	Host must open Door 2
Switching loses with probability 1/6	Switching loses with probability 1/6	Switching wins with probability 1/3	Switching wins with probability 1/3

Players who choose to switch win if the car is behind either of the two unchosen doors. In two cases with 1/3 probability switching wins, so the overall probability of winning by switching is 2/3.

The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).

Referring to the figure above or to an equivalent decision tree as shown to the right (Grinstead and Snell 2006:137-138) and considering only the cases where the host opens Door 2, switching loses in a 1/6 case where the player initially picked the car and otherwise wins in a 1/3 case. Similarly if the host opens Door 3 switching wins twice as often as staying, so the conditional probability of winning by switching given either door the host opens is 2/3 — the same as the overall probability. A formal proof of this fact using Bayes' theorem is presented below (see Bayesian analysis).

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."

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. 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).

A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207). This intuition is the basis of solutions to the problem that assert the host's action of opening a door does not change the player's initial 1/3 chance of selecting the car. For the fully explicit problem this intuition leads to the correct numerical answer, 2/3 chance of winning the car by switching, but leads to the same solution for other variants where this answer is not correct (Falk 1992:207).

Another source of confusion is that the usual wording of the problem statement asks about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open if the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3. In its usual form the problem statement does not specify this detail of the host's behavior, making the answer that switching wins the car with probability 2/3 mathematically unjustified. Many commonly presented solutions address the unconditional probability, ignoring which door the host opens; Morgan et al. call these "false solutions" (1991).

Aids to understanding

Why the probability is not 1/2

The most commonly voiced objection to the solution is that the past can be ignored when assessing the probability—that it is irrelevant which doors the player initially picks and the host opens. However, in the problem as originally presented, the player's initial choice does influence the host's available choices subsequently.

This difference can be demonstrated by contrasting the original problem with a variation that appeared in vos Savant's column in November 2006. In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed. Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006).

In this version of the puzzle, the player has an equal chance of winning whether switching or not. Assuming the player picks Door 1 there are six possible outcomes that can occur, each with probability 1/6:

In two cases above, the host reveals the car. What might happen in these cases is unknown—perhaps the contestant immediately wins or immediately loses. However, in the problem as stated, the host has revealed a goat, so only four of the six cases remain possible, and they are equally likely. In two of these four cases, switching results in a win, and in the other two, switching results in a goat. Staying with the original pick gives the same odds: a loss in two cases and a win in two others.

The player's probability of winning by switching increases to 2/3 in the original problem because in the two cases above where the host would reveal the car, he is forced to reveal the remaining goat instead. In the table below, these two cases are highlighted:

This change in the host's behavior causes the car to be twice as likely to be behind the "third door", and is what makes switching to be twice as likely to win in the "host knows" variation of the problem.

Increasing the number of doors

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. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, 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. A rational player should switch.

Stibel et al. (2008) propose 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.

Combining doors

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, and will not, choose the opened door (Adams 1990; Devlin 2003; Williams 2004; Stibel et al., 2008). The player therefore has the choice of either sticking with the original choice of door with a 1/3 chance of winning the car, or choosing the sum of the contents of the two other doors with a 2/3 chance as shown.

The game assumptions play a role here—switching is equivalent to taking the combined contents if and only if the game host knows what is behind the doors, must open a door with a goat, and chooses between two losing doors randomly with equal probabilities.

The only difference between trading for both doors and the trade that is actually offered is whether the host opens one of the two doors. Opening one shows which of these doors the car must be behind if it is behind either. At least one of the two unpicked doors contains a goat, and the host is equally likely to open either of these doors so opening one gives the player no additional information; opening one does not change the 2/3 probability that the car is behind one of them (Devlin 2003).

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 2001:243). 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 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.

By the law of large numbers, 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. Two times out of three, after one card has been dealt to the player, the Ace of Spades is in the host's hand. At that point, it is already determined whether staying or switching will win the round for the player.

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 (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.

Variants

Other host behaviors

In some versions of the Monty Hall problem, the host's behavior is not fully specified. For example, the version 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. Without specifying these rules, the player does not have enough information to conclude that switching will be successful two-thirds of the time (Mueser and Granberg, 1999). The table shows possible host behaviors and the impact on the success of switching.

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(N-p-1). If the host opens even a single door the player is better off switching, but 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 probability of winning by switching approaches 1.

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.

This problem appears similar to the television show Deal or No Deal, however with each selection the Deal or No Deal player is just as likely to open the winning box as a losing one. Monty on the other hand, knows the contents and is forbidden from revealing the winner. Assuming the grand prize is still left with two boxes remaining, the Deal or No Deal player has a 50/50 chance that the initially selected box contains the grand prize.

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 of the problem

An essentially identical problem appeared as the Three Prisoners Problem in Martin Gardner's Mathematical Games column in Scientific American in 1959 (Gardner 1959). Gardner's version makes the selection procedure explicit, avoiding the unstated assumptions in the Parade version. This puzzle in probability theory involves three prisoners, a random one of whom has been secretly chosen to be executed in the morning. The first prisoner begs the guard to tell him which of the other two will go free, arguing that this reveals no information about whether the prisoner will be the victim; the guard responds by claiming that if the prisoner knows that a specific one of the other two prisoners will go free it will raise the first prisoner's subjective chance of being executed from 1/3 to 1/2. The question is whether the analysis of the prisoner or the guard is correct. In the version given by Martin Gardner, the guard then performs a particular randomizing procedure for selecting which name to give the prisoner; this gives the equivalent of the Monty Hall problem without the usual ambiguities in its presentation.

In 1975, Steve Selvin wrote a pair of letters to the American Statistician (Selvin 1975a, Selvin 1975b) regarding the Monty Hall problem. The first presented the problem in a version close to its most popular form; the version presented in Parade 15 years later is a restatement of Selvin's version. The second letter 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)

Phillip Martin's article in a 1989 issue of Bridge Today magazine titled "The Monty Hall Trap" (Martin 1989) presented Selvin's problem, with the correct solution, as an example of how one can fall into the trap of treating non-random information as if it were random. Martin then gives examples in the game of bridge where players commonly miscalculate the odds by falling into the same trap, such as the Principle of Restricted Choice. Given the controversy that would arise over this problem a year later, Martin showed a lack of prescience when he stated, "Here [in the Monty Hall problem] the trap is easy to spot. But the trap can crop up more subtly in a bridge setting."

A restated version of Selvin's problem statement appeared in Marilyn vos Savant's Ask Marilyn question-and-answer column of Parade in September 1990 (vos Savant 1990). Though vos Savant gave the correct answer that switching would win two-thirds of the time, vos Savant estimates 10,000 readers including several hundred mathematics professors wrote in to declare that her solution was wrong. 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 (Tierney 1991) in which Monty Hall himself was interviewed. He appeared to understand the problem quite well, giving the reporter a demo with car keys and explaining how actual game play on Let's Make a Deal differed from the rules of the puzzle.

Over 40 papers have been published about this problem in academic journals and the popular press (Mueser and Granberg 1999).

The problem continues to resurface outside of academia. 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). An account of mathematician Paul Erdos's first encounter of the problem can be found in The Man Who Loved Only Numbers—like many others, he initially got it wrong. 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) and in Derren Brown's 2006 book Tricks Of The Mind. 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).

Bayesian analysis

An analysis of the problem using the formalism of Bayesian probability theory (Gill 2002) makes explicit the role of the assumptions underlying the problem. In Bayesian terms, probabilities are associated to propositions, and express a degree of belief in their truth, subject to whatever background information happens to be known. For this problem the background is the set of game rules, and the propositions of interest are:

${\displaystyle C_{i}\,}$ : The car is behind Door i, for i equal to 1, 2 or 3.

${\displaystyle H_{ij}\,}$ : The host opens Door j after the player has picked Door i, for i and j equal to 1, 2 or 3.

For example, ${\displaystyle C_{1}\,}$ denotes the proposition the car is behind Door 1, and ${\displaystyle H_{12}\,}$ denotes the proposition the host opens Door 2 after the player has picked Door 1. Indicating the background information with ${\displaystyle I\,}$, the assumptions are formally stated as follows.

First, the car can be behind any door, and all doors are a priori equally likely to hide the car. In this context a priori means before the game is played, or before seeing the goat. Hence, the prior probability of a proposition ${\displaystyle C_{i}\,}$ is:

${\displaystyle P(C_{i}|I)\,={\frac {1}{3}}.}$

Second, the host will always open a door that has no car behind it, chosen from among the two not picked by the player. If two such doors are available, each one is equally likely to be opened. This rule determines the conditional probability of a proposition ${\displaystyle H_{ij}\,}$ subject to where the car is — i.e., conditioned on a proposition ${\displaystyle C_{k}\,.}$ Specifically, it is:

${\displaystyle P(H_{ij}|C_{k},\,I)\,\,=\,{\begin{cases}\,\\\,\\\,\\\,\end{cases}}}$	${\displaystyle \,0\,}$		if i = j, (the host cannot open the door picked by the player)
	${\displaystyle \,0\,}$		if j = k, (the host cannot open a door with a car behind it)
	${\displaystyle \,1/2\,}$		if i = k, (the two doors with no car are equally likely to be opened)
	${\displaystyle \,1\,}$		if i ${\displaystyle \neq }$k and j ${\displaystyle \neq }$ k, (there is only one door available to open)

The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors. Without loss of generality, assume, by re-numbering the doors if necessary, that the player picks Door 1, and that the host then opens Door 3, revealing a goat. In other words, the host makes proposition ${\displaystyle H_{13}\,}$ true.

The posterior probability of winning by not switching doors, subject to the game rules and ${\displaystyle H_{13}\,}$, is then ${\displaystyle P(C_{1}|H_{13},\,I)}$. Using Bayes' theorem this is expressed as:

${\displaystyle P(C_{1}|H_{13},\,I)={\frac {P(H_{13}|C_{1},\,I)\,P(C_{1}|I)}{P(H_{13}|I)}}.}$

By the assumptions stated above, the numerator of the right-hand side is:

${\displaystyle P(H_{13}|C_{1},\,I)\,P(C_{1}|I)={\frac {1}{2}}\times {\frac {1}{3}}={\frac {1}{6}}.}$

The normalizing constant at the denominator can be evaluated by expanding it using the definitions of marginal probability and conditional probability:

${\displaystyle {\begin{array}{lcl}P(H_{13}|I)&{}=&P(H_{13},\,C_{1}|I)+P(H_{13},\,C_{2}|I)+P(H_{13},\,C_{3}|I)\\&{}=&P(H_{13}|C_{1},\,I)\,P(C_{1}|I)\,+\\&&P(H_{13}|C_{2},\,I)\,P(C_{2}|I)\,+\\&&P(H_{13}|C_{3},\,I)\,P(C_{3}|I)\\&{}=&{\displaystyle {\frac {1}{2}}\times {\frac {1}{3}}+1\times {\frac {1}{3}}+0\times {\frac {1}{3}}}\ =\ {\displaystyle {\frac {1}{2}}\ .}\end{array}}}$

Dividing the numerator by the normalizing constant yields:

${\displaystyle P(C_{1}|H_{13},\,I)={\frac {1}{6}}\,/\,{\frac {1}{2}}={\frac {1}{3}}.}$

Note that this is equal to the prior probability of the car's being behind the initially chosen door, meaning that the host's action has not contributed any novel information with regard to this eventuality. In fact, the following argument shows that the effect of the host's action consists entirely of redistributing the probabilities for the car's being behind either of the other two doors.

The probability of winning by switching the selection to Door 2, ${\displaystyle P(C_{2}|H_{13},\,I)}$, can be evaluated by requiring that the posterior probabilities of all the ${\displaystyle C_{i}\,}$ propositions add to 1. That is:

${\displaystyle 1=P(C_{1}|H_{13},\,I)+P(C_{2}|H_{13},\,I)+P(C_{3}|H_{13},\,I).}$

There is no car behind Door 3, since the host opened it, so the last term must be zero. This can be proven using Bayes' theorem and the previous results:

${\displaystyle {\begin{aligned}P(C_{3}|H_{13},\,I)&={\frac {P(H_{13}|C_{3},\,I)\,P(C_{3}|I)}{P(H_{13}|I)}}\\&=\left(0\times {\frac {1}{3}}\right)/\,{\frac {1}{2}}=0\ .\end{aligned}}}$

Hence:

${\displaystyle P(C_{2}|H_{13},\,I)=1-{\frac {1}{3}}-0={\frac {2}{3}}.}$

This shows that the winning strategy is to switch the selection to Door 2. It also makes clear that the host's showing of the goat behind Door 3 has the effect of transferring the 1/3 of winning probability a-priori associated with that door to the remaining unselected and unopened one, thus making it the most likely winning choice.

See also

Similar problems

• Bertrand's box paradox (also known as the Three cards problem)
• Boy or Girl
• Three Prisoners Problem
• Two envelopes problem

References

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• Stibel, Jeffrey, Dror, Itiel, & Ben-Zeev, Talia (2008). "The Collapsing Choice Theory: Dissociating Choice and Judgment in Decision Making," Theory and Decision. Published online at http://www.springerlink.com/content/v65v2841q3820622/.
• Tierney, John (1991). "Behind Monty Hall's Doors: Puzzle, Debate and Answer?", The New York Times, 1991-07-21. Retrieved on 2008-01-18.
• Tierney, John (2008). "And Behind Door No. 1, a Fatal Flaw", The New York Times, 2008-04-08. Retrieved on 2008-04-08.
• vos Savant, Marilyn (1990). "Ask Marilyn" column, Parade Magazine p. 16 (9 September 1990).
• vos Savant, Marilyn (1996). The Power of Logical Thinking. St. Martin's Press. ISBN 0-612-30463-3.
• vos Savant, Marilyn (2006). "Ask Marilyn" column, Parade Magazine p. 6 (26 November 2006).
• Williams, Richard (2004). "Appendix D: The Monty Hall Controversy" (PDF). Course notes for Sociology Graduate Statistics I. Retrieved 2008-04-25.
• Whitaker, Craig F. (1990). . "Ask Marilyn" column, Parade Magazine p. 16 (9 September 1990).

External links

2. Part 2

• The Monty Hall Problem at letsmakeadeal.com
• The Game Show Problem–the original question and responses on Marilyn vos Savant's web site
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• "Monty Hall Paradox" by Matthew R. McDougal, The Wolfram Demonstrations Project (simulation)
• The Monty Hall Problem at The New York Times (simulation)

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• Probability theory paradoxes
• Let's Make a Deal