[10] Out of 228 subjects in one study, only 13% chose to switch. 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. The 2/3 chance of finding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is certain to reveal a goat. Therefore, the chance that the host opens door 3 is 50%. 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. It is straightforward: each prisoner has a 1/3 chance of being pardoned. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as 1) the endowment effect,[28] in which people tend to overvalue the winning probability of the already chosen – already "owned" – door; 2) the status quo bias,[29] in which people prefer to stick with the choice of door they have already made; and 3) the errors of omission vs. errors of commission effect,[30] in which, ceteris paribus, people prefer any errors for which they are responsible to have occurred through 'omission' of taking action, rather than through having taken an explicit action that later becomes known to have been erroneous. These probabilities can be determined referring to the conditional probability table below, or to an equivalent decision tree as shown to the right. "Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door. [19] Numerous examples of letters from readers of Vos Savant's columns are presented and discussed in The Monty Hall Dilemma: A Cognitive Illusion Par Excellence. 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. [58][59] Three cards from an ordinary deck are used to represent the three doors; one 'special' card represents the door with the car and two other cards represent the goat doors. In this case, the warden flips a coin chooses one of B and C to reveal the fate of. The original Three Prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1/6 probability of occurring. The Three Prisoners problem appeared in Martin Gardner's Mathematical Games column in Scientific American in 1959. This leaves us with four cases: With the stipulation that the warden will choose randomly, in the 1/3 of the time that A is to be pardoned, there is a 1/2 chance he will say B and 1/2 chance he will say C. This means that taken overall, 1/6 of the time (1/3 [that A is pardoned] * 1/2 [that warden says B]), the warden will say B because A will be pardoned, and 1/6 of the time (1/3 [that A is pardoned] * 1/2 [that warden says C]) he will say C because A is being pardoned. Il s'agit en 3 minutes de trouver le plus grand nombre de mots possibles de trois lettres et plus dans une grille de 16 lettres. 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. 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. p As the question is not directly about A's fate, the warden obliges — either naming the other prisoner to be executed, if A is the one, or secretly flipping a coin to decide which of the remaining names to give A if A is the one being pardoned. 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. 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. Probability and the Monty Hall problem", https://en.wikipedia.org/w/index.php?title=Monty_Hall_problem&oldid=984826074, Short description is different from Wikidata, Use shortened footnotes from October 2020, Creative Commons Attribution-ShareAlike License. The contestant wins (and her opponent loses) if the car is behind one of the two doors she chose. 1 %PDF-1.6 %���� [21] In her book The Power of Logical Thinking,[22] quotes cognitive psychologist Massimo Piattelli Palmarini [it]: "No other statistical puzzle comes so close to fooling all the people all the time [and] 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."  | Dernières modifications. /R0 gs They impose the following constraints on the probability of such propositions. The Three Prisoners problem appeared in Martin Gardner's Mathematical Games column in Scientific American in 1959. There are three criminals Aaron, Bill, and Clark who have committed a very serious crime against the King for which they were imprisoned for several years. On those occasions when the host opens Door 3. endstream endobj 40 0 obj <>>>/ProcSet[/PDF]/XObject<>>>/Subtype/Form/Type/XObject>>stream The second appears to be the first use of the term "Monty Hall problem". They believed the question asked for the chance of the car behind door 2 given the player's initial pick for door 1 and the opened door 3, and they showed this chance was anything between 1/2 and 1 depending on the host's decision process given the choice. If you think about it, the original problem offers you basically the same choice. 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, the opening the door. H�L�K��0��>�N@�p��J66����H��n��πi����������9\ L�ǟ� 8(JJ`��#��%���'�`��� �K3��yL�0��%t\�S3x\�.���z�&S � ��`1!�|�0hԭ#(A�qL:�dV"13^�S��r�lW��׶�7�Gt��Y�X�,��( 5:d�&V�QT�&鸾(���T�uk��r��m|f���Do�z�R�ٟ��2�PS�jᚨ*�j�Da��R�j/����~��`j�cy���Q�MRw@H�:����m����g�o$�����1^e�_ߛ�?�*����NkU�;O@��N���m N�Z�;Ah��� H�L�9r�0s��/@�>��*;Y'N�}C�RP�.Π Ң����@��Z��Rr���}�'�RFpD^J��y]܂��aJ �1��� Kk2��{���y�`װ�SD�G�. Similarly, in case 6, the warden must say B instead of C (the same as case 5). Following Gill,[56] a strategy of contestant involves two actions: the initial choice of a door and the decision to switch (or to stick) which may depend on both the door initially chosen and the door to which the host offers switching. [3] Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance. Les jeux de lettres anagramme, mot-croisé, joker, Lettris et Boggle sont proposés par Memodata. Steve Selvin posed the Monty Hall problem in a pair of letters to the American Statistician in 1975. The warden obliges, (secretly) flipping a coin to decide which name to provide if the prisoner who is asking is the one being pardoned. A quantum version of the paradox illustrates some points about the relation between classical or non-quantum information and quantum information, as encoded in the states of quantum mechanical systems. For instance, one contestant's strategy is "choose door 1, then switch to door 2 when offered, and do not switch to door 3 when offered". ;��v���3��$(5�KBǨ��o�^G���J������NܖS�)�0 ~ʨ The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to … That leaves cases 4 and 5 with 1/3 probability of occurring and leaves us with the same probability as above. The version of the Monty Hall problem published in Parade in 1990 did not specifically state that the host would always open another door, or always offer a choice to switch, or even never open the door revealing the car. Vos Savant asks for a decision, not a chance. 1/3 must be the average probability that the car is behind door 1 given the host picked door 2 and given the host picked door 3 because these are the only two possibilities. 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. Une fenêtre (pop-into) d'information (contenu principal de Sensagent) est invoquée un double-clic sur n'importe quel mot de votre page web. [20], The discussion was replayed in other venues (e.g., in Cecil Adams' "The Straight Dope" newspaper column[14] and reported in major newspapers such as The New York Times.[4]. Another way to understand the solution is to consider the two original unchosen doors together. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. The conditional probability table below shows how 300 cases, in all of which the player initially chooses door 1, would be split up, on average, according to the location of the car and the choice of door to open by the host. endstream endobj startxref 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. ), the player is better off switching in every case. In the article, Hall pointed out that because he had control over the way the game progressed, playing on the psychology of the contestant, the theoretical solution did not apply to the show's actual gameplay. [3] In this case, there are 999,999 doors with goats behind them and one door with a prize. Switching wins the car two-thirds of the time. 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. L'encyclopédie française bénéficie de la licence Wikipedia (GNU). Extending this logic to multiple events, for example A, B and C, we get that we can play with the different subsets of {A, B, C} to calculate the probability of the intersection, as a tool to simplify the calculation of our conditional probability: In our case, since we know that P(H3|C2,X1) = 1, we are in luck: 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. 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. Twelve such deterministic strategies of the contestant exist. The switch in this case clearly gives the player a 2/3 probability of choosing the car. Numerous readers, however, wrote in to claim that Adams had been "right the first time" and that the correct chances were one in two. /Form Do Les lettres doivent être adjacentes et les mots les plus longs sont les meilleurs. Therefore A must judge it safer to try switch his fate with C 's. Indexer des images et définir des méta-données. �E�s���'nߒ�:*h�ԩc��n���qvK���؀�b�Bd��� ���� �_�4 SG�w3#���ș-��~���BsG|�-,��Y-B )�[Āh����Zl� ���S�̖wn�4�@�[0`dH�� ���1�W�Mu�[-���)rf�;��ιâW��)y� C�nA��5l�l> Y�wc�y'��r�>���j�` L pf [26] People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not.