I have not changed that. [14][15][16][17][18] As Cecil Adams puts it,[14] "Monty is saying in effect: you can keep your one door or you can have the other two doors." 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. The switch in this case clearly gives the player a 2/3 probability of choosing the car. Publication list for In general, the answer to this sort of question depends on the specific assumptions made about the host's behavior, and might range from "ignore the host completely" to "toss a coin and switch if it comes up heads"; see the last row of the table below. [26] People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not.[27]. 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. This problem involves three condemned prisoners, a random one of whom has been secretly chosen to be pardoned. Maximum weighted matching 299 ... game theory. 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. 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. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. The odds that your choice contains a pea are 1/3, agreed? The goal of game theory is to understand these 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. ", "The 'Monty Hall' Problem: Everybody Is Wrong", "An 'easy' answer to the infamous Monty Hall problem", University of California San Diego, Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game, "Stick or switch? Existence of Nash equilibria and fixed points, 5.3. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. Undergraduate and graduate students and professors N Steve Selvin wrote a letter to the American Statistician in 1975 describing a problem based on the game show Let's Make a Deal,[1] dubbing it the "Monty Hall problem" in a subsequent letter. theorems, and probabilistic arguments. A solution 293 16.3.3. Is it to your advantage to switch your choice?[9]. Algorithms for finding stable matchings, 13.5. 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. Vos Savant wrote in her first column on the Monty Hall problem that the player should switch. Ambiguities in the Parade version do not explicitly define the protocol of the host. Other partisan games played on graphs, 2.4. [31][32] Another possibility is that people's intuition simply does not deal with the textbook version of the problem, but with a real game show setting. [14] Adams initially answered, incorrectly, that the chances for the two remaining doors must each be one in two. So, in this particular expression, the choosing of the host doesn't depend on where the car is, and there's only two remaining doors once X1 is chosen (for instance, P(H1|X1) = 0); and P(Ci,Xi) = P(Ci)P(Xi) because Ci and Xi are independent events (the player doesn't know where is the car in order to make a choice). 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. If the host picks randomly q would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. The given probabilities depend on specific assumptions about how the host and contestant choose their doors. A formula for the expected payment, 14.10.1. p evolutionary stability), and learning theory. [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. Among these sources are several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but ... shaky",[34] or do not "address the problem posed",[35] or are "incomplete",[36] or are "unconvincing and misleading",[37] or are (most bluntly) "false". "Game 3, which has a goat. It offers an appealing and versatile selection of topics [25], 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. agents interacting and myriad opportunities for conflict and [1], 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:[12]. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as: Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition. "Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly. 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. All homeworks should be uploaded to bSpace, unless otherwise stated. [70] As a result of the publicity the problem earned the alternative name Marilyn and the Goats. As this experiment is repeated over several rounds, the observed win rate for each strategy is likely to approximate its theoretical win probability, in line with the law of large numbers. https://www.ams.org/exam-desk-review-request?&eisbn=978-1-4704-3667-4&pisbn=978-1-4704-1982-0&epc=MBK/101.E&ppc=MBK/101&title=Game%20Theory%2C%20Alive&author=Anna%20R.%20Karlin%3B%20Yuval%20Peres&type=DE, https://www.ams.org/exam-desk-review-request?&eisbn=978-1-4704-3667-4&pisbn=978-1-4704-1982-0&epc=MBK/101.E&ppc=MBK/101&title=Game%20Theory%2C%20Alive&author=Anna%20R.%20Karlin%3B%20Yuval%20Peres&type=R, https://www.copyright.com/openurl.do?isbn=9781470419820&WT.mc.id=American%20Mathematical%20Society, Part I: Analyzing games: Strategies and equilibria, 1.2.2.

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