The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on American television game show Let's Make a Deal and named from its original host, Monty Hall. The problem was originally proposed (and solved) in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question of reader letters quoted in Marilyn vos Savant's "Ask Marilyn" column in the 1990 Parade magazine (vos Savant 1990a):
Suppose you are at a game show, and you are given a choice of three doors: Behind one door is a car; behind the other, the goat. You choose the door, say no. 1, and the host, who knows what's behind the door, opens another door, says No. 3, which has a goat. He then says to you, "Do you want to choose door No. 2?" Does it benefit you to change your choice?
Vos Savant's response was that participants had to turn to another door (vos Savant 1990a). Under standard assumptions, switching contestants have 2 / 3 the chance of winning a car, while the contestant remains on the choice they just have a chance 1 / 3 . A more intuitive explanation of why changing your choices is a winning strategy, is to realize that the most likely event at the beginning of the game is you choose a goat, and then the host reveals to you the other, so, the most likely scenario is the car is in the remaining door. Consider that the door opened by the host has a 100% chance of having a goat, and the door you choose, 66%, so the most likely event is that you have identified two goats.
The probability given depends on the specific assumptions about how the host and the contestant choose their door. A key insight is that, under the conditions of this standard, there is more information about doors 2 and 3 that are not available at the beginning of the game, when door 1 is selected by the player: the deliberate action of the host adds a value to the door he did not choose to omit, but not with those chosen by the original contestant. Another insight is that switching doors is a different act than choosing between two remaining doors at random, because the first action uses the previous and last information not. Other possible behaviors other than those described may disclose additional information differently, or not at all, and result in different probabilities.
Many readers of the vos Savant column refuse to believe that the transition is useful despite its explanation. After problems surfaced in Parade , about 10,000 readers, including nearly 1,000 with PhDs, wrote to magazines, most of them claiming vos Savant wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erd? S, one of the most productive mathematicians in history, remained unsure until he showed computer simulations showing predictable results (Vazsonyi 1999).
The problem is a paradox of the veridical type, since the right choice (which should change the door) is very contrary to intuition, it can seem absurd, but it still proves true. Monty Hall's problem is mathematically closely related to the previous Three Prisoners problem and the paradox of the older Bertrand box.
Video Monty Hall problem
Paradoks
Steve Selvin wrote a letter to the American Statistician in 1975 which described the problem loosely based on the Let's Make A Deal game, (Selvin 1975a), calling it "Monty Hall" the next letter (Selvin 1975b).The problem is mathematically the equivalent of the Three Prisoners Problem described in the "Mathematical Games" Martin Gardner column in 1959a in 1959a and the Three Shells Problems described in the book Gardner "Aha Gotcha" (Gardner 1982).
The same issue was restated in the 1990 letter by Craig Whitaker to the Marilyn vos Savant "Ask Marilyn" column at the Parade :
Suppose you are at a game show, and you are given a choice of three doors: Behind one door is a car; behind the other, the goat. You choose the door, say no. 1, and the host, who knows what's behind the door, opens another door, says No. 3, which has a goat. He then says to you, "Do you want to choose door No. 2?" Does it benefit you to change your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)
Standard assumptions
Under the standard assumption, the probability of winning a car after switching is 2 / 3 . The key to this solution is host behavior. Ambiguity in the Parade version does not explicitly define a host protocol. However, the solution of Marilyn vos Savant (vos Savant 1990a) printed together with Whitaker's question implies, and both Selvin (1975a) and vos Savant (1991a) explicitly define, the role of the host as follows:
- The host should always open the door that is not taken by the contestant (Mueser and Granberg 1999).
- The host should always open the door to reveal a goat and never a car.
- The host should always offer the opportunity to switch between the originally selected door and the remaining closed door.
When one of these assumptions varies, it can change the likelihood of winning by switching the door as described in the section below. It is also usually assumed that the car was originally concealed randomly behind the door and that, if the player originally chose the car, then the host's choice of a goat-hiding door to open is random. (Krauss and Wang, 2003: 9) Some authors, independently or inclusively, consider that the player's initial choice is random as well. Selvin (1975a)
Simple solution
The solution presented by vos Savant (1990b) in Parade shows three possible settings for one car and two goats behind three doors and residence or switch after initially selecting door 1 in each case:
A player who stays with an early choice wins only in one of the three possible possibilities, while a player who switches wins in two out of three.
The intuitive explanation is that, if the contestant originally chose a goat (2 of 3 doors), the contestant will win the car by switching because the other goats can no longer be taken, whereas if the contestant originally chose the car (1 of 3 door), contestants will not win the car by switching (Carlton 2005, closing word). The fact that the host later revealed a goat on one of the unchanging doors did not change the initial probability.
Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that is a 50/50 option. This will be true if the host opens the door at random, but that does not happen; the door opens depending on the player's initial choice, so the assumption of independence does not apply. Before the host opens the door there is a possibility of 1 / 3 the car behind each door. If the car behind the door 1 the host can open door 2 or door 3, so the possibility of the car behind door 1 AND the host opening door 3 is 1 / < span> 3 1 / 2 = 1 / 6 . If the car is behind door 2 (and the player has chosen door 1) the host should open door 3, so the possibility of the car behind door 2 AND the host opening door 3 is 1 / 3 * 1 = 1 / 3 . This is the only case where the host opens the door 3, so if the player has opted for door 1 and the host opens the door 3 cars twice more likely to be behind the 2nd door. The key is that if the car behind the door 2 hosts should open door 3, but if car behind door 1 host can open both doors.
Another way to understand the solution is to consider two original doors that are not selected together (Adams 1990; Devlin 2003, 2005; Williams 2004; Stibel et al., 2008). As Cecil Adams (Adams 1990) says, "Monty says basically: You can keep one door or you can have two other doors." The opportunity to find a car has not been changed by opening one of these doors because Monty, knowing the location of the car, will definitely reveal a goat. So the choice of player after the host opens the door is no different than if the host offers the player an option to switch from the original selected door to set both remaining gates. The switch in this case clearly gives the player the probability of 2 / 3 to select the car.
As Keith Devlin (Devlin 2003) says, "By opening the door, Monty tells the contestants" There are two doors you do not choose, and the possibility that the gift is behind one is 2 < span>/ 3 . I will 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 gift. Now you can take advantage of this additional information. Your door choice A has 1 in 3 chances as the winner. I have not changed it. But by eliminating door C, I have shown you that the probability that door B hides the gift is 2 in 3. ' "
Vos Savant suggested that the solution would be more intuitive with 1,000,000 doors than 3. (vos Savant 1990a) In this case, there are 999,999 doors with goats behind them and one door with a gift. After the player picks up a door, the host opens 999,998 from the remaining door. On average, in 999,999 times out of 1,000,000, the remaining gates will contain prizes. Intuitively, the player must ask how likely he is, given a million doors, he can choose the right one at first. Stibel et al. (2008) proposed that work memory requests were taxed during the Monty Hall issue and that this forced people to "break down" their options into two equally likely options. They report that when the number of options is increased to more than 7 options (7 doors), people tend to switch more frequently; However, most contestants still misjudge the likelihood of success at 50:50.
Maps Monty Hall problem
Vos Savant and media furore
Vos Savant writes in his first column on the Monty Hall issue that players must switch (vos Savant 1990a). He received thousands of letters from his readers - most, including many of the readers with PhD, disagree with his answer. During 1990-1991, three more columns in the Parade were devoted to paradox (vos Savant 1990-1991). Many examples of letters from readers of the Vos Savant column are presented and discussed in The Monty Hall Dilemma: A Cognitive Illusion Par Excellence (Granberg 2014).
The discussion was played elsewhere (for example, in the newspaper column "The Straight Dope" Cecil Adams, (Adams 1990)), and was reported in major newspapers such as The New York Times (Tierney 1991).
In an attempt to clarify his answer, he proposed a game of shells (Gardner 1982) to illustrate: "You look away, and I put the beans under one of three shells, then I ask you to put your finger on the shell that your choice contains beans 3 / 3 , agree? Then I just lift the empty shell from the rest of the other two.As I can ( and will) do this regardless of what you have chosen, we have not learned anything to allow us to revise the odds of the shells under your finger. "He also proposed a simulation similar to three playing cards.
Vos Savant commented that, although some confusion is caused by some readers unaware that they should assume that the host must always reveal a goat, almost all of his correspondents have understood the problem assumptions correctly, and at first still believe that answer vos Savant ("switch") wrong.
Confusion and criticism
Source of confusion
When first presented with the Monty Hall problem, a large majority of people assume that every door has the same probability and concludes that switching is not a problem (Mueser and Granberg, 1999). Of the 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995: 713). In his book The Power of Logical Thinking vos Savant (1996, p.Ã, 15) quotes Massimo Piattelli-Palmarini's cognitive psychologist who says that "no other statistical riddle comes so close to fooling all all people. time, "and" even Nobel physicists systematically give wrong answers, and that they insist, and they are ready to berate those who propose the right answer. "Pigeons repeatedly exposed to problems show that they quickly learn to always switch, unlike humans (Herbranson and Schroeder, 2010).
Most of the problem statements, especially those in Parade Magazine, are inconsistent with the rules of actual gaming performances (Krauss and Wang 2003: 9) and do not fully determine host behavior or that the location of the car is chosen random. (Granberg and Brown, 1995: 712; Tierney 1991; VerBruggen 2015) Krauss and Wang (2003: 10) think that people make standard assumptions even if they are not explicitly stated.
Although this problem is mathematically significant, even when controlling for these factors, almost everyone still thinks each of the two unopened doors has the same probability and concludes that switching is not a problem (Mueser and Granberg, 1999). This "same probability" assumption is deeply rooted intuition (Falk 1992: 202). People are very inclined to think the probabilities are evenly distributed in as many unknown as they exist, whether or not they are (Fox and Levav, 2004: 637). The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, that is, no switching, can be explained by phenomena known in psychological literature as 1) the endowment effect (Kahneman et al., 1991), where people tend to overstate the odds of winning from the chosen ones - already "owned" - doors; 2) status quo bias (Samuelson and Zeckhauser, 1988), where people prefer to stick with the choice of doors they have made; and 3) false negatives vs. commission effect errors (Gilovich et al., 1995), where, ceteris paribus, people prefer errors that are their responsibility to occur through 'negligence' taking action, rather than for having taken explicit action later known to be mistaken. The experimental evidence confirms that this is a plausible explanation that does not depend on the probability of intuition (Kaivanto et al., 2014; Morone and Fiore, 2007). Another possibility is that people's intuitions simply do not deal with textbook versions of the problem, but with real game show settings (EnÃÆ'Ÿlin and Westerkamp, ​​2018). There, it is possible that the performing master played a crime by opening another door only if a door with a goat was originally chosen. Master performances that play half of the evil time change the odds of winning in case one is offered to move on to "the same probability".
Criticisms of simple solutions
As already noted, most sources in the field of probability, including many introductory probability texts, solve the problem by showing the conditional probability that the car is behind door 1 and door 2 is 1 / 3 and 2 / 3 (no 1 / 2 and 1 / 2 ) considering that the contestant initially chose door 1 and the host opened door 3; various ways to derive and understand these results are given in the previous subsections. Among these sources there are some that explicitly criticize the "simple" solution presented in general, saying the solution is "true but... shaky" (Rosenthal 2005a), or not "solving the problem posed" (Gillman 1992), or "incomplete" (Lucas et al., 2009), or "unconvincing and misleading" (Eisenhauer 2001) or (most bluntly) "false" (Morgan et al., 1991).
Sasha Volokh (2015) writes that "any explanation that says something like 'the possibility of door 1 is 1/3, and no one can change it...' is automatically suspicious: the probability is the expression of our ignorance of the world, and new information can change our level of ignorance. "
Some say that this solution answers a slightly different question - one sentence is "You have to announce before a door has opened whether you plan to switch" (Gillman 1992, emphasis on the original).
The simple solution shows in various ways that a contestant who is determined to switch will win the car with the probability of 2 / 3 , and hence the switch is a winning strategy, if the player must choose between "always switch" and "always stay". However, the likelihood of winning with always switching is a logically different concept of the likelihood of winning by switching considering that the player has chosen door 1 and the host has opened the door 3 . As one source said, "the difference between these [questions] seems to confuse many people" (Morgan et al., 1991). The fact that this is different can be demonstrated by varying the problem so that these two probabilities have different numerical values. For example, suppose the contestant knows that Monty did not pick a random second door between all the legal alternatives but instead, when given the opportunity to choose between two defeated doors, Monty will open the one on the right. In this situation, the following two questions have different answers:
- What is the probability of winning a car with always switching?
- What is the probability of winning a car given the player has chosen door 1 and the host has opened the door 3 ?
The answer to the first question is 2 / 3 , as correctly indicated by the "simple" solution. 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 the rightmost door means that he opens door 3 if the car is behind door 1 (initially with the probability of 3 ) or if the car is behind door 2 (also initially with probability 1 / 3 ). For this variation, two questions produce different answers. However, as long as the initial probability that the car is behind each door is 1 / 3 , never the contestant's loss to switch , since the conditional probability of winning with the switch is always at least 1 / 2 . (Morgan et al., 1991)
Four university professors published an article (Morgan et al., 1991) in The American Statistician claiming that vos Savant provided correct advice but false arguments. They believe the question posed for the possibility of the car behind the door 2 given the player's initial choice for door 1 and open door 3, and they indicate this opportunity is anything between 1 / 2 and 1 depending on the host decision process given option. Only when the decision is completely random is the chance of 2 / 3 .
In the comments invited (Seymann 1991) and in the next letter to the editor, (vos Savant 1991c; Rao 1992; Bell, 1992; Hogbin and Nijdam, 2010) Morgan et al. supported by several writers, criticized by others; in each case responses by Morgan et al. published with letters or comments on The American Statistician. In particular, vos Savant defensively vigorously. Morgan et al. complained in their response to vos Savant (1991c) that vos Savant still has not really responded to their own main points. Later in their response to Hogbin and Nijdam (2011), they agree that it is reasonable to assume that the host selects a door to open completely randomly, when he does have a choice, and hence that the probability of being conditional to win by switching (ie, given the player's situation when he must make his choice) has the same value, 2 / 3 , as unconditional probability to win by switching (that is, the average over all possible situations). This equality has been emphasized by Bell (1992), which suggests that mathematically involved solutions Morgan et al. It will only appeal to statisticians, whereas the equality of conditional and unconditional solutions in case of symmetry is intuitively clear.
There is disagreement in the literature as to whether vos Savant's formulation of the problem, as presented in Parade's magazine, asks the first or second question, and whether this difference is significant (Rosenhouse 2009). Behrends (2008) concludes that "One must consider the matter carefully to see that both analyzes are true"; which does not say that they are the same. One analysis for one question, another for another. Some discussion papers by (Morgan et al., 1991), whose contributions were published together with the original paper, strongly criticize the authors for altering the words of vos Savant and misinterpreting his intent (Rosenhouse 2009). One of the discussors (William Bell) considers it a matter of taste whether one explicitly mentions that (under standard conditions), which door opened by the host is independent of whether or not the person wants to switch.
Among the simplest solutions, "joint-door solutions" are closest to conditional solutions, as we see in the discussion of approaches using the concept of opportunity and Bayes's theorem. It is based on deep-rooted intuition that revealing known information does not affect probability . But, knowing that the host can open one of the two unregulated doors to indicate a goat, does not mean that opening a particular door will not affect the possibility of the car being behind the originally chosen door. The point is, although we know in advance that the host will open the door and reveal a goat, we do not know which door he will open. If the host selects a random uniform between the doors that hide a goat (as in the standard interpretation), this probability does remain unchanged, but if the host can choose randomly between the doors, then the special door host open reveals additional information. The host can always open the door to reveal a goat and (in the standard interpretation of the problem) the possibility that the car behind the originally chosen door does not change, but it is not not because of the former the latter is true. The solution is based on the assertion that the action of the host can not affect the probability that the car is behind the originally selected appears persuasive, but the statement is incorrect unless each of the two host choices has the same possibility, if he has a choice (Falk 1992 : 207,213). The statement needs to be justified; without the justification given, the solution is at least complete. The answer can be true but the reason used to justify it is broken.
The solution uses conditional probabilities and other solutions
The simple solution above shows that the player with the transfer strategy wins the car with the overall probability of 2 / 3 , that is, without taking into account which door was opened by the host (Grinstead and Snell 2006: 137-138 Carlton 2005). Instead most sources in the field of probability calculate the conditional probability that the car is behind door 1 and door 2 is 1 / 3 and 3 because the contestant initially chose door 1 and the host opened door 3 (Selvin (1975b ), Morgan et al., 1991, Chun 1991, Gillman 1992, Carlton 2005, Grinstead and Snell 2006: 137-138, Lucas et al., 2009). The solution in this section only considers the cases where the player selects door 1 and the host opens door 3.
Refine the simple solution
If we assume that the host opens the door at random, when given a choice, then which door is opened by the host does not provide any information at all about whether the car is behind the door 1. In a simple solution, we have observed that the possibility of the car being behind door 1, the door originally chosen by the player, initially 1 / 3 . In addition, the host will certainly open the door a (different), so opening the door a (unspecified door ) does not change this. 3 3 should be the average probability that the car is behind door 1 given the opening door of host 2 and given the host chose door 3 because these are two possibilities. However, these two probabilities are the same. Therefore, both are equal to 1 / 3 (Morgan et al., 1991). This indicates that the possibility of the car being behind door 1, given that the player initially chose this door and considering that the host opened door 3, is 1 / 3 , and follow that the possibility of the car is behind door 2, given that the player originally chose door 1 and the host opened door 3, is 2 / 3 . The analysis also shows that the overall success rate of 2 / 3 , achieved by always switching , can not be fixed, and underscores what is already possible intuitively clear: the choice that players face is that between the door that was originally selected, and the other door left by the host, the specific numbers on these doors are irrelevant. Conditional probability with direct count
By definition, the conditional probability of winning by switch given contestant initially takes door 1 and host opens door 3 is the probability for event "car behind door 2 and master opens door 3" divided by probability to "host open door" 3 ". This probability can be determined referring to the conditional probability table below, or to the same decision tree as shown to the right (Chun 1991; Carlton 2005; Grinstead and Snell 2006: 137-138). The conditional probability of winning by switching is 1/3 / 1/3 1/6 , the 2 < span>/ 3 (Selvin 1975b).
The conditional probability table below shows how 300 cases, in which the players originally chose door 1, will be divided on average, according to the location of the car and the door option to be opened by the host.
Bayes.27_theorem "> Bayes Theorem
Many possible textbooks and articles in the field of probability theory obtain conditional probability solutions through formal Bayes theorem applications; between them Gill, 2002 and Henze, 1997. The use of a possible form of Bayes theorem, often called the Bayes rule, makes such derivations more transparent (Rosenthal, 2005a), (Rosenthal, 2005b).
Initially, this car is most likely to be behind one of three doors: the chances at door 1, door 2, and door 3 are 1Ã,: 1Ã,: 1 . This still happens after the player selects door 1, with independence. According to Bayes' rules, the posterior odds on the location of the car, given that the host opens door 3, equal to the previous probability multiplied by the Bayes factor or possibilities, which, by definition, the possibility of a new piece of information (host opens door 3) each hypothesis is considered (location of the car). Now, since the player originally chose door 1, the likelihood that the host opens door 3 is 50% if the car is behind the door 1, 100% if the car behind the door is 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratio of 1 / 2 Ã,: 1Ã,: 0 or the equivalent 1Ã,: 2Ã,: 0 , while the previous opportunity is 1Ã,: 1Ã,: 1 . Thus, the posterior odds become the same as the Bayes factor 1Ã,: 2Ã,: 0 . Given that the host opens door 3, the probability that the car behind door 3 is zero, and that is twice as likely to be behind door 2 rather than door 1.
Richard Gill (2011) analyzes the possibilities for the host to open door 3 as follows. Given that the car is not behind door 1, the odds are behind door 2 or 3. Therefore, the possibility that the host opens door 3 is 50%. Given that the car is behind door 1, the likelihood that the host opens door 3 is also 50%, because, when the host has a choice, the same options are probably the same. Therefore, whether the car is behind door 1, the likelihood that the host opens door 3 is 50%. Information "host opening door 3" contributes to the Bayes factor or the likelihood ratio of 1st: 1 on whether the car is behind the door 1. Initially, the odds against door 1 hide the car 2Ã, : 1 . Therefore, the posterior odds against door 1 hiding the car remain the same as the previous odds, 2Ã,: 1 .
In words, information that is the door is opened by the host (door 2 or door 3?) Reveals no information at all about whether or not the car is behind door 1, and this is precisely what one might expect becomes intuitively clear by the supporters of simple solutions, or using idioms of mathematical proof, "clearly true, by symmetry" (Bell 1992).
Live count
P (H3 | X1) = 1/2 because this expression depends only on X1 , not on Ci anywhere. Thus, in this particular expression, the host selection does not depend on where the car is located, and there are only two doors remaining after X1 is selected (eg, P (H1 | X1) = 0 ); and P (Ci, Xi) = P (Ci) P (Xi) because Ci and Xi are independent events (players do not know where car to make a choice).
Then, if the player initially chose the door 1, and the host opened the door 3, we proved that the probable probability of winning by switching was:
Returning to Nalebuff (1987), Monty Hall's problem is also widely studied in the literature on game theory and decision theory, as well as some popular solutions according to this point of view. Vos Savant asks for a decision, not an opportunity. And the accidental aspects of how the car was hidden and how the unopened door opened is unknown. From this point of view, we must remember that the player has two chances to make a choice: first, which door should be chosen initially; and secondly, whether or not to switch. Because he does not know how the car is hidden or how the host makes a choice, he may be able to take advantage of his first choice opportunity, therefore to neutralize the actions of the team that runs the quiz show, including the host.
Following Gill, 2011, a contestant's strategy involves two actions: the initial choice of a door and the decision to switch (or to a stick) that may depend on both the originally selected door and the door to which the host offers the transfer. For example, one contestant's strategy is "select door 1, then switch to door 2 when offered, and do not switch to door 3 when offered." Twelve deterministic strategies of contestants exist.
The basic comparison of the contestant's strategy shows that, for each strategy A, there is another strategy B "select the door then switch no matter what happens" that dominates it (Gnedin, 2011). It does not matter how the car is hidden and it does not matter who commands the host to use when he has a choice between two goats, if A wins the car then B too. For example, strategy A "select door 1 then always follow with it" dominated by strategy B "select door 1 then always switch after host opens the door": Victory when door 1 hides car, while B wins when one door 2 and 3 conceals car. Similarly, strategy A "select door 1 then switch to door 2 (if offered), but do not switch to door 3 (if offered)" dominated by strategy B "select door 3 then always switch".
Domination is a compelling reason to find solutions among ever-changing strategies, under general assumptions about the environment in which contestants make decisions. In particular, if a car is hidden with a scrambler - such as a symmetrical throw or three asymmetric sides - dominance implies that a strategy that maximizes the likelihood of winning a car will be among three ever-shifting strategies, that would be a strategy that initially selects the most unlikely door and then switches no matter which door to exchange offered by the host.
The strategic dominance attributes Monty Hall's problem to game theory. In the Gill zero-sum game setting, 2011, throwing a non-switching strategy reduces the game to the following simple variant: the host (or TV team) decides on the door to hide the car, and the contestant selects two doors (ie, two doors remaining after the first option , nominal, player). The contestant wins (and his opponent loses) if the car is behind one of the two doors he chooses.
Solution with simulation
A simple way to show that the switching strategy actually wins two out of three times with the standard assumption is to simulate the game with playing cards (Gardner 1959b; vos Savant 1996, p.Ã, 8). Three cards from the deck used to represent three doors; one 'special' card represents the door with the car and two other cards representing the goat door.
The simulation can be repeated several times to simulate several game rounds. The player takes one of three cards, then, sees the two remaining cards 'host' throw away the goat card. If the card remaining in the host's hand is a car card, this is recorded as a switch victory; if the host holds a goat card, the round is recorded as a steady victory. Because this experiment was repeated for several rounds, the observed win rate for each strategy is likely to be close to the probability of its theoretical victory.
Repetitive drama also makes it clear why switching is a better strategy. Once the player picks up his card, it's already already determined whether the switching will win the round for the player. If this is not convincing, the simulation can be done with the whole deck. (Gardner 1959b; Adams 1990). In this variant, the car card goes to host 51 times out of 52, and stays with the host no matter how much non -card car is discarded.
Variant
A common variation of the problem, assumed by some academic writers as a canonical problem, does not make the simplification assumption that the host should uniformly choose the door to open, but instead he uses some other strategy. The confusion about which formalization is authoritative has caused considerable jealousy, especially since this variant makes the evidence more involved without changing the optimality of the always-switched strategy for the player. In this variant, players can have different winning probabilities depending on the choice observed from the host, but in any case the probability wins by switching at least span < 2 (and can be as high as 1), while the overall winning probability with the transition is still exactly 2 / 3 . Variants are sometimes presented in succession in textbooks and articles intended to teach the basics of probability theory and game theory. A large number of other generalizations have also been studied.
Other host behaviors
The version of the Monty Hall issue published at Parade in 1990 did not specifically state that the host would always open another door, or always offer the option to switch, or never even open the door to reveal the car. However, vos Savant makes it clear in the second follow-up column that the hosted host behavior can only be what causes 2 3 he might give as the original answer. "Anything else is a different question". (vos Savant 1991a) "Almost all of my critics understand the intended scenario, I personally read nearly three thousand letters (out of thousands of extra arrivals) and found almost everyone insisting that due to two fixed options (or equivalent mistakes), even possibilities. Very few questions are asked about ambiguity, and the letters actually published in the columns are not the least of them. "(Vos Savant 1996) The answer will arise if a car is placed randomly behind any door, the host must open the door for reveals a goat regardless of the player's initial choice and, if two doors are available, select which ones will be opened randomly (Mueser and Granberg, 1999). The table below shows the various possibilities of other host behaviors and their impact on transition success.
Determining the player's best strategy in a series of other rules that must be followed by the host is the type of problem learned in game theory. For example, if the host is not required to make an offer to divert the player can suspect the host is evil and make the offer more often if the player initially chose the car. In general, the answer to this kind of question depends on the specific assumptions made about the host's behavior, and may range from "ignoring the host completely" to "tossing coins and switching if heads appear"; see the last line of the table below.
Morgan et al. (1991) and Gillman (1992) both show a more general solution in which cars (uniforms) are placed randomly but the host is not limited to randomly choosing uniformly if the player initially chooses the car, in which both interpret the problem statement in > Parade even if there is a writer's rejection. Both of them changed the words of the Parade version to emphasize the moment when they repeated the problem. They consider the scenario in which 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 takes randomly q it will be < span> 1 / 2 and switch win with probability 2 / 3 regardless of which door the host is open. If the player selects door 1 and the host preference for door 3 is q , then the host possibly opens door 3 and the car behind door 2 is 1 / 3 while the host probability opens door 3 and the car behind door 1 is q / 3 . This is the only case where the host opens door 3, so the conditional probability of winning by switching giving the master to open the door 3 is 1/3 / 1/3 q /3 that simplifies 1 / 1 q . Since q can vary between 0 and 1, this conditional probability may vary between 1 / 2 and 1. This means that without even restricting the host to randomly choosing if the player originally chose the car, the player has never been worse than switching. But no source suggests players know how much q means so players can not relate probabilities other than 2 / 3 which Savant considers to be implicit. Do
N
DL Ferguson (1975 in a letter to Selvin quoted in Selvin 1975b) suggests a generalization-door-to-left (N of the original problem where the host opens the p losing the door and then offers the player's opportunity to switch in this switching variant wins with the probability N - 1 / N - p - 1) If the host opens even one door, players are better off, but, if the host is only opening a door, gaining near zero when growing up (Granberg 1996: 188).At the other extreme, if the host opens all the lost doors but one ( p = Ã, N Ã,-2) increased profits such as N growing (possibly winning by switching is 1 / N , which approaches 1 when N growing very big).
Quantum version
The quantum version of the paradox illustrates several points about the relationship between classical information and non-quantum information and quantum information, as coded in the state of quantum mechanical systems. The formula is loosely based on quantum game theory. The three doors are replaced by a quantum system that enables three alternatives; opening the door and looking back translated as making certain measurements. Rules can be expressed in this language, and again the option for the player is fixed by the initial option, or change to other "orthogonal" options. The last strategy turned out to double the odds, as in the classic case. However, if the emcee does not scramble the prize position in the fullest quantum mechanics, players can do better, and can sometimes win prizes with certainty (Flitney and Abbott 2002, D'Ariano et al., 2002).
History
The earliest of several possible puzzles related to the Monty Hall problem is the paradox of the Bertrand box, which was put forward by Joseph Bertrand in 1889 in his book Calcul des probabilità © s (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 each one. After choosing the box randomly and pulling a coin randomly which happens to be a gold coin, the question is what is the probability that the other coins are gold. As in Monty Hall's problem, the intuitive answer is 1 / 2 , but the actual probability is 2 / 3 .
The Three Prisoners problem, published in Martin Gardner's
Steve Selvin filed the Monty Hall problem in a pair of letters to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). The first letter presents the problem in a close version of his presentation at Parade 15 years later. The second seems to be the first use of the term "Monty Hall problem". The real problem is the extrapolation of the game show. Monty Hall did open the wrong door to build excitement, but offered a lesser-known gift - like $ 100 cash - not an option to move the door. As Monty Hall wrote to Selvin:
And if you've ever appeared on my show, the rules apply quickly to you - there's no trading box after the election.
A version of a problem very similar to what appeared three years later at the Parade was published in 1987 in the Puzzles section of The Journal of Economic Perspectives (Nalebuff 1987). Nalebuff, as a later writer in the field of mathematical economics, saw the problem as a simple and funny exercise in game theory.
The Phillip Martin article in the 1989 issue of Bridge Today magazine titled "The Monty Hall Trap" (Martin 1989) presents the Selvin problem as an example of what Martin calls an opportunity trap to treat non-random information as if it were random , and connect this with concepts in the game of the bridge.
The Selvin version problem resurfaced in the Marilyn vos Savant Ask Marilyn Question and answer Parade field in September 1990. (vos Savant 1990a) Although vos Savant gave the correct answer. replied that the transition would win two-thirds of the time, he estimated the magazine received 10,000 letters including nearly 1,000 signed by the PhD, many on letterhead from the mathematics and science department, claiming the solution was wrong. (Tierney 1991) Due to the overwhelming response, Source of the article : Wikipedia
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