HMMT-Nov 2010 guts, Problem 3

Jacob flipped a fair coin five times. In the first three flips, the coin came up heads exactly twice. In the last three flips, the coin also came up heads exactly twice. What is the probability that the third flip was heads?

Solution: How many sequences of five flips satisfy the conditions, and have the third flip be heads? We have \_\textit{- H}-- , so exactly one of the first two flips is heads, and exactly one of the last two flips is heads. This gives $2 \times 2=4$ possibilities. How many sequences of five flips satisfy the conditions, and have the third flip be tails? Now we have \_- $T_{--}$, so the first two and the last two flips must all be heads. This gives only 1 possibility. So the probability that the third flip was heads is $\frac{4}{(4+1)}=\frac{4}{5}$ $\fbox{\frac{4}{5}}$.

Answer: \frac{4}{5}