This probability puzzle was published by the mathematician J. Bertrand (no relation to Bertrand Russell) in his 1889 text Calcul des Probabitités. The reader is asked to imagine three desks, with two drawers each. She knows that one desk contains a gold medal in each drawer, one contains a silver medal in each drawer, and one contains one of each. She does not know which desk is which. The question posed is this:

If the reader opens a drawer and discovers a gold medal, what are the chances that the other drawer on that desk also contains gold?

This comes down, then, to figuring out the probability that she picked the gold-gold desk instead of the gold-silver desk. Many people are immediately inclined to say that there are two possibilities, and since the selection was random, it must be 50-50. This is not the case.

Think of the initial selection as picking from among six drawers:
```  before              after:
S   S   G    --->           G
S   G   G               G   G

1   2   3           1   2   3```
So, we have it narrowed down to 3 drawers, with an equal probability of each one being the one that was picked. One of the drawers is in box 2, so there's a 1/3 chance that box #2 was picked. Two of the drawers are in box #3, so there are two 1/3 chanes (a 2/3 chance) that box #3 was picked.

rp is correct that this resembles the Monty Hall problem. It's also isomorphic with a number of familiar puzzles relating to sets of twins and the like. In all these cases, the key is to remember that different choices have different probabilities of providing the piece of evidence you were given. Some people assume that if there are two possible answers, either one has a fifty-fifty chance of being right, but if you know anything about either one, that's no longer necessarily the case. You're not looking for the number of possible answers, but for the probability that each individual answer was the correct one.

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