Monty Hall Problem Analysis

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Monty Hall Problem Analysis

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The Monty Hall Problem

Steve Selvin wrote a letter to the American Statistician in describing a problem based on the game show Let's Make a Deal , [1] dubbing it the "Monty Hall problem" in a subsequent letter. The key to this solution is the behavior of the host. Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution [3] printed alongside Whitaker's question implies, and both Selven [1] and vos Savant [5] explicitly define, the role of the host as follows:. When any of these assumptions is varied, it can change the probability of winning by switching doors as detailed in the section below. It is also typically presumed that the car is initially hidden randomly behind the doors and that, if the player initially picks the car, then the host's choice of which goat-hiding door to open is random.

The solution presented by vos Savant 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: [11]. A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three.

An intuitive explanation is that, if the contestant initially picks a goat 2 of 3 doors , the contestant will win the car by switching because the other goat can no longer be picked, whereas if the contestant initially picks the car 1 of 3 doors , the contestant will not win the car by switching. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold.

These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3, the car is twice as likely to be behind door 2 as door 1. The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door. Another way to understand the solution is to consider the two original unchosen doors together. So the player's choice after the host opens a door is no different than if the host offered the player the option to switch from the original chosen door to the set of both remaining doors.

I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3. Vos Savant suggests that the solution will be more intuitive with 1,, doors rather than 3. After the player picks a door, the host opens , of the remaining doors.

On average, in , times out of 1,,, the remaining door will contain the prize. Intuitively, the player should ask how likely it is that, given a million doors, he or she managed to pick the right one initially. Stibel et al [17] proposed that 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 the number of options is increased to more than 7 choices 7 doors , people tend to switch more often; however, most contestants still incorrectly judge the probability of success at 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. Scott Smith, Ph. University of Florida [3]. Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. During —, three more of her columns in Parade were devoted to the paradox. The discussion was replayed in other venues e. In an attempt to clarify her answer, she proposed a shell game [8] 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. 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. Vos Savant commented that, though some confusion was caused by some readers' not realizing they were supposed to assume that the host must always reveal a goat, almost all her numerous correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer "switch" was wrong. 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.

Most statements of the problem, notably the one in Parade , do not match the rules of the actual game show [10] and do not fully specify the host's behavior or that the car's location is randomly selected. 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 that switching does not matter. The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, i. Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition.

A show master playing deceitfully half of the times modifies the winning chances in case one is offered to switch to "equal probability". Among these sources are several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but 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". However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given that 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". 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:. For this variation, the two questions yield different answers. In Morgan et al. In an invited comment [40] and in subsequent letters to the editor, [41] [42] [43] [44] 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 [41] that vos Savant still had not actually responded to their own main point. Later in their response to Hogbin and Nijdam, [44] 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.

This equality was already emphasized by Bell , who suggested that Morgan et al' s mathematically involved solution would appeal only to statisticians, whereas the equivalence of the conditional and unconditional solutions in the case of symmetry was intuitively obvious. There is disagreement in the literature regarding whether vos Savant's formulation of the problem, as presented in Parade , is asking the first or second question, and whether this difference is significant. 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 that the host can open one of the two unchosen doors to show a goat does not mean that opening a specific door would not affect the probability that the car is behind the initially chosen door. The point is, though we know in advance that the host will open a door and reveal a goat, we do not know which door he will open. If the host chooses uniformly at random between doors hiding a goat as is the case in the standard interpretation , this probability indeed remains unchanged, but if the host can choose non-randomly between such doors, then the specific door that the host opens reveals additional information.

The host can always open a door revealing a goat and in the standard interpretation of the problem the probability that the car is behind the initially chosen door does not change, but it is not because of the former that the latter is true. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially chosen appear persuasive, but the assertion is simply untrue unless each of the host's two choices are equally likely, if he has a choice. The answer can be correct but the reasoning used to justify it is defective. If we assume that the host opens a door at random, when given a choice, then which door the host opens gives us no information at all as to whether or not the car is behind door 1.

Moreover, the host is certainly going to open a different door, so opening a door which door unspecified does not change this. But, these two probabilities are the same. By definition, the conditional probability of winning by switching given the contestant initially picks door 1 and the host opens door 3 is the probability for the event "car is behind door 2 and host opens door 3" divided by the probability for "host opens door 3". These probabilities can be determined referring to the conditional probability table below, or to an equivalent decision tree as shown to the right.

The conditional probability table below shows how cases, in all of which the player initially chooses door 1, would be split up, on average, according to the location of the car and the choice of door to open by the host. Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem ; among them books by Gill [51] and Henze. Initially, the car is equally likely to be 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 that 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. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.

Richard Gill [54] analyzes the likelihood for the host to open door 3 as follows. Given that the car is not behind door 1, it is equally likely that it is behind door 2 or 3. The information "host opens door 3" contributes a Bayes factor or likelihood ratio of 1 : 1 , on whether or not the car is behind door 1. Initially, the odds against door 1 hiding the car were 2 : 1. Therefore, the posterior odds against door 1 hiding the car remain the same as the prior odds, 2 : 1. In words, the information which door is opened by the host door 2 or door 3? Consider the event Ci , indicating that the car is behind door number i , takes value Xi , for the choosing of the player, and value Hi , for the host opening the door.

Then, if the player initially selects door 1, and the host opens door 3, we prove that the conditional probability of winning by switching is:. Going back to Nalebuff, [55] the Monty Hall problem is also much studied in the literature on game theory and decision theory , and also some popular solutions correspond to this point of view. Vos Savant asks for a decision, not a chance. And the chance aspects of how the car is hidden and how an unchosen door is opened are unknown. From this point of view, one has to remember that the player has two opportunities to make choices: first of all, which door to choose initially; and secondly, whether or not to switch. Since he does not know how the car is hidden nor how the host makes choices, he may be able to make use of his first choice opportunity, as it were to neutralize the actions of the team running the quiz show, including the host.

Following Gill, [56] 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" that dominates it.

For example, strategy A "pick door 1 then always stick with it" is dominated by the strategy B "pick door 1 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 2 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 that 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, [56] discarding the non-switching strategies reduces the game to the following simple variant: the host or the TV-team decides on the door to hide the car, and the contestant chooses two doors i. The contestant wins and her opponent loses if the car is behind one of the two doors she chose. A simple way to demonstrate that a switching strategy really does win two out of three times with the standard assumptions is to simulate the game with playing cards. The simulation can be repeated several times to simulate multiple rounds of the game. The player picks one of the three cards, then, looking at the remaining two cards the 'host' discards a goat card.

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