A considerable number of other generalizations have also been studied. [67] 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. [58][14] In this variant, the car card goes to the host 51 times out of 52, and stays with the host no matter how many non-car cards are discarded. 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: From the Bayes' rule, we know that P(A,B) = P(A|B)P(B) = P(B|A)P(A). An analysis using Bayesian probability theory begins by expressing the problem in terms of statements, or propositions, that may be true or false. 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. This probability is always greater than As Keith Devlin says,[15] "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. He then says to you, "Do you want to pick door No. /MWFOForm Do The same problem was restated in a 1990 letter by Craig Whitaker 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. Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. As already remarked, most sources in the field of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities that the car is behind door 1 and door 2 are 1/3 and 2/3 (not 1/2 and 1/2) given that 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. After the player picks his card, it is already determined whether switching will win the round for the player. Before the host opens a door there is a 1/3 probability the car is behind each door. The three prisoners’ problem has always been a highly debated topic. 0 Lettris est un jeu de lettres gravitationnelles proche de Tetris. Le dictionnaire des synonymes est surtout dérivé du dictionnaire intégral (TID). There is enough mathematical illiteracy in this country, and we don't need the world's highest IQ propagating more. − %PDF-1.6 %���� [3] In this case, there are 999,999 doors with goats behind them and one door with a prize. 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. [1][2] The first letter presented the problem in a version close to its presentation in Parade 15 years later. Whether you change your selection or not, the odds are the same. [21][4][24] Krauss and Wang conjecture that people make the standard assumptions even if they are not explicitly stated.[25]. 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 + q/3 which simplifies to 1/1 + q. %%EOF A key insight is that, under these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. 1, and the host, who knows what's behind the doors, opens another door, say No. Assuming the warden's truthfulness, there are now only two possibilities for who will be pardoned: A, and whichever of B or C the warden did not name. Tous droits réservés. This is because Monty's preference for rightmost doors means that 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). 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". After the host reveals a goat, you now have a one-in-two chance of being correct. Monty is saying in effect: you can keep your one door or you can have the other two doors, one of which (a non-prize door) I'll open for you." This means his chances of being pardoned, now knowing B isn't, again are 1/3, but whoever between B and C is not being executed has a 2/3 chance of being pardoned. 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. She also proposed a similar simulation with three playing cards. [1][2] It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990:[3]. The warden obliges, (secretly) flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. 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. However, the warden in that case may not reveal the fate of a pardonned prisoner. It is important to note that this problem is intended to denote a one-time situation. Several critics of the paper by Morgan et al,[38] whose contributions were published alongside the original paper, criticized the authors for altering vos Savant's wording and misinterpreting her intention. [48][49] In contrast most sources in the field of probability calculate the conditional probabilities that the car is behind door 1 and door 2 are 1/3 and 2/3 given the contestant initially picks door 1 and the host opens door 3. Moreover, the host is certainly going to open a (different) door, so opening a door (which door unspecified) does not change this. ", 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".[39]. [64] "That's the same assumption contestants would make on the show after I showed them there was nothing behind one door," he said. Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. You pick a door, say No. In Morgan et al,[38] four university professors published an article in The American Statistician claiming that vos Savant gave the correct advice but the wrong argument. In particular, vos Savant defended herself vigorously. Now you're offered this choice: open door #1, or open door #2 and door #3. Similarly, in case 6, the warden must say B instead of C (the same as case 5). Very few raised questions about ambiguity, and the letters actually published in the column were not among those few. 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 host opens door 3 and the car is behind door 1 is q/3. 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. To make the analogy to the Monty Hall problem more explicit: if the warden says "B will be executed" and A could switch fates with C, should he? 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. Another way to understand the solution is to consider the two original unchosen doors together. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. h�b```�V��B cc`a�x��Ю����ճ&0����N叫43��Z| �����=���E�ew8: ���B { )����@����CH��Xd� � C�5�.�\Fn0i&;�3n1�a����y �r�|������3��zo�)�W��!�Q����2Ĕ@����YԀ4#]�b���P�@� J;= After the player picks a door, the host opens 999,998 of the remaining doors. Strategic dominance links the Monty Hall problem to the game theory. Therefore, they are both equal to 1/3. 3, which has a goat. The Three Prisoners problem appeared in Martin Gardner's Mathematical Games column in Scientific American in 1959. [33] There, the possibility exists that the show master plays deceitfully by opening other doors only if a door with the car was initially chosen. 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. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. Only when the decision is completely randomized is the chance 2/3. The problem is a paradox of the veridical type, because the correct choice (that one should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. [46] One analysis for one question, another analysis for the other question. N In November 1990, an equally contentious discussion of vos Savant's article took place in Cecil Adams's column "The Straight Dope". 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. endstream endobj 35 0 obj <> endobj 36 0 obj <>/Border[0 0 0]/F 4/Rect[241 747 316 760]/Subtype/Link/Type/Annot>><>/Border[0 0 0]/F 4/Rect[29 730 536 760]/Subtype/Link/Type/Annot>><>/Border[0 0 0]/F 4/Rect[323 730 422 742]/Subtype/Link/Type/Annot>><>/Border[0 0 0]/F 4/Rect[29 701 538 728]/Subtype/Link/Type/Annot>><><><><><><><><><><><><>67 0 R 68 0 R 69 0 R 70 0 R 71 0 R 72 0 R 73 0 R 74 0 R]/Contents[54 0 R 55 0 R 56 0 R 57 0 R 58 0 R 59 0 R 60 0 R 61 0 R]/CropBox[0 0 595 842]/MediaBox[0 0 595 842]/Parent 32 0 R/Resources<>/Font<>/ProcSets[/PDF/Text/ImageB/ImageC/ImageI]>>/Rotate 0/Type/Page>> endobj 37 0 obj <>>>/ProcSet[/PDF]/XObject<>>>/Subtype/Form/Type/XObject>>stream Between brackets is the name of the prisoner revealed as being executed by the warden when A asks.

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