The box paradox
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Three boxes look identical. Each has two drawers, and each drawer contains two medals. Those in the first box are gold; those in the second are silver; the third box contains one gold medal and one silver medal. A box is selected; what is the probability of finding a gold coin and a silver coin in its drawers?
"Only one case is favorable, so the probability is 1/3," the mathematician observes. He then offers another line of reasoning, which is of course fallacious: once the box has been selected, one drawer is opened. Whatever medal is found, there are two possibilities. The closed drawer contains either a gold or a silver medal. Only one of these two possibilities is favorable, so the probability would be one-half. Bertrand adds: "Yet how can we believe that simply opening a drawer could change the probability from 1/3 to 1/2?"
The trick is obvious, but we know how much ink has been spilled over the Monty Hall problem (see page 28), which is quite similar and cleverly framed. This box problem can be reduced to the Monty Hall problem by noting that, once the drawer has been opened and found to contain, say, a gold medal, the box containing two silver medals can be eliminated; the choice is then between the two remaining boxes.
There is no ambiguity in this problem; only faulty reasoning leads to the answer 1/2.
