2019 AMC 10B Problem 21

Trace what the conditions force. Until she stops, the flips must alternate (a repeat ends the game).

• Flip 1 must be $T$: if it were $H$, the second tail could only come after a second head or via $HTT$, which stops with tails.
• Flip 2 must be $H$ (a $T$ would end the game with $TT$).
• Flip 3 must be $T$: an $H$ gives $THH$, ending with only one tail seen.
• Flip 4 must be $H$: a $T$ gives $THTT$, ending with tails.

After $THTH$, which has probability $\frac{1}{16}$, she has seen two tails and two heads, with the second tail (flip 3) before the second head (flip 4). The condition is met; she just needs to eventually finish with $HH$.

Let $p$ be the probability of finishing with $HH$ when the last flip was $H$ and the game is still running. The next flip is $H$ (done, probability $\frac12$) or $T$; after a $T$, the following flip must be $H$ to survive (probability $\frac12$), returning to the same state. So
$$p = \frac12 + \frac12 \cdot \frac12 \cdot p \quad\Longrightarrow\quad p = \frac23.$$

The answer is $\frac{1}{16} \cdot \frac23 = \frac{1}{24}$, so $\boxed{\textbf{(B)}\ \frac{1}{24}}$.

Stopping-rule problems often have a long forced prefix: each early flip has exactly one survivable value. Once the prefix satisfies the "order of events" condition, only the outcome of the remaining game matters, and that is a one-equation state computation. Separate "getting into the right situation" from "finishing correctly."

List the successful sequences: $THTHH$, $THTHTHH$, $THTHTHTHH, \ldots$ with probabilities $\frac{1}{32}, \frac{1}{128}, \frac{1}{512}, \ldots$. The geometric sum is $\frac{1/32}{1 - 1/4} = \frac{1}{24}$.

• Counting only the single sequence THTHH (probability 1/32) and missing the longer alternating continuations THTHTHH, THTHTHTHH, and so on.
• Allowing sequences starting with $H$ such as $HTHTHH$, in which the second head (flip 3) comes before the second tail (flip 4).