Wednesday, October 7, 2009

Bertrand's Box Paradox

The Setup: There are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each.

The Problem: After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, what is the probability the remaining coin is gold?

The Solution: The correct answer is two-thirds. It may seem that the probability that the remaining coin is gold has a probability of 1⁄2; in fact, the probability is actually 2⁄3. This is because of the six coins on the table, there are only three that could have been drawn. We know all three are gold and of those three, two are in the box with another gold coin. The coin seen is equally likely to be any of the three gold coins, only one of which is opposite a silver coin

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