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Bertrand's box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his ''Calcul des probabilités'', published in 1889. There are three boxes: # a box containing two gold coins, # a box containing two silver coins, # a box containing one gold coin and a silver coin. After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, it may seem that the probability that the remaining coin is gold is ; in fact, the probability is actually . Two problems that are very similar are the Monty Hall problem and the Three Prisoners problem. These simple but counterintuitive puzzles are used as a standard example in teaching probability theory. Their solution illustrates some basic principles, including the Kolmogorov axioms. ==Box version== There are three boxes, each with one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (GG), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold? The following reasoning appears to give a probability of : : *Originally, all three boxes were equally likely to be chosen. : *The chosen box cannot be box SS. : *So it must be box GG or GS. : *The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is . The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically: : *The probability that GG would produce a gold coin is 1. : *The probability that SS would produce a gold coin is 0. : *The probability that GS would produce a gold coin is . Initially GG, SS and GS are equally likely. Therefore, by Bayes rule the conditional probability that the chosen box is GG, given we have observed a gold coin, is: :::: The correct answer of can also be obtained as follows: : *Originally, all six coins were equally likely to be chosen. : *The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS. : *So it must come from the G drawer of box GS, or either drawer of box GG. : *The three remaining possibilities are equally likely, so the probability that the drawer is from box GG is . Alternatively, one can simply note that the chosen box has two coins of the same type of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?". Bertrand's point in constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This condition is correctly applied in the second solution method, but not in the first. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Bertrand's box paradox」の詳細全文を読む スポンサード リンク
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