Exercise 2.63

This problem is similar to the variations on example 2.2.5 (Two Children) in the textbook.

It is true that conditioned on specific two of the three coins matching, the probability of the third coin matching is $\frac{1}{2},$ but the way the problem statement is phrased, at least two of the coins match. According to the Two Children problem, the result is no longer $\frac{1}{2}.$ In fact, the probability of all the coins matching given at least two match is $\frac{1}{4}.$