Problem 90

We first want to find out which sequences of coin flips satisfy the given condition. For Debra to see the second tail before the second head, her first flip can't be heads, as that would mean she would either end with double tails before seeing the second head, or would see two heads before she sees two tails. Therefore, her first flip must be tails. The shortest sequence of flips by which she can get two heads in a row and see the second tail before she sees the second head is $THTHH$, which has a probability of $\frac{1}{2^5} = \frac{1}{32}$. Furthermore, she can prolong her coin flipping by adding an extra $TH$, which itself has a probability of $\frac{1}{2^2} = \frac{1}{4}$. Since she can do this indefinitely, this gives an infinite geometric series with a first term of $\frac{1}{32}$ and a common ratio of $\frac{1}{4}$, which means the answer (by the infinite geometric series sum formula) is $\frac{\frac{1}{32}}{1-\frac{1}{4}} = \boxed{\textbf{(B) }\frac{1}{24}}$.