2017 AMC 10A Problem 18

The game could go on for many rounds, but each round starts the same way: Amelia is about to toss. Use that.

1. Name the answer. Let $P$ be the chance that Amelia wins, starting from a moment when she is about to toss.
2. Look at one round.
- Amelia gets heads right away: chance $\tfrac13$. She wins.
- Amelia gets tails (chance $\tfrac23$), then Blaine gets heads (chance $\tfrac25$). Amelia loses.
- Amelia gets tails (chance $\tfrac23$), then Blaine gets tails (chance $\tfrac35$). Nobody has won, and Amelia is about to toss again, exactly as at the start. From here her chance is $P$ again.
3. The chance that both miss is $\tfrac23 \times \tfrac35 = \tfrac{6}{15} = \tfrac25$.
4. Write one equation. Amelia wins either now, or after a double miss:
$$P = \tfrac13 + \tfrac25 P.$$
5. Solve. Subtract $\tfrac25 P$ from both sides: $\tfrac35 P = \tfrac13$. Multiply by $\tfrac53$: $P = \tfrac59$.
6. $5$ and $9$ share no common factor, so $p = 5$, $q = 9$, and $q - p = 4$.

The answer is $\boxed{\textbf{(D)}\ 4}$.

In "keep going until someone succeeds" games, a round where nobody succeeds puts you right back at the start. That means the unknown chance appears on both sides of one equation, and you can solve it without adding up endless rounds. As a check, Blaine's chance is $1 - \tfrac59 = \tfrac49$.

Add up the rounds. Amelia wins in round $1$ with chance $\tfrac13$. She wins in round $2$ if both missed once first: $\tfrac25 \times \tfrac13$. In round $3$: $\left(\tfrac25\right)^2 \times \tfrac13$, and so on. Each term is $\tfrac25$ of the one before, so the total is
$$\frac{\tfrac13}{1 - \tfrac25} = \frac{\tfrac13}{\tfrac35} = \frac59.$$