Solution
Gold Coin in the Box
Show the problem again
There are three boxes. One holds two gold coins, one holds two silver coins, and one holds one gold coin and one silver coin. You choose a box uniformly at random and then draw one of its two coins uniformly at random, without looking at the other. The coin you drew is gold. What is the probability that the other coin in the same box is also gold? Give your answer as a reduced fraction.
Worked solution
The answer is 2/3. The tempting argument says "the box is either gold-gold or gold-silver, so it is 1/2," but the two boxes are not equally likely to produce a gold draw. Label the six coins. Each is equally likely to be the one drawn, and three of them are gold: two in the gold-gold box and one in the mixed box. Conditioning on a gold draw leaves those three equally likely coins. In two of the three cases the coin came from the gold-gold box, so its companion is gold with probability 2/3. The gold-gold box is twice as likely to show gold, and Bayes' theorem weights it accordingly.
Source: Bertrand's box paradox, from Joseph Bertrand, Calcul des probabilités (1889). Statement written for AxiomIQ.