2024 AMC 10A Problem 17

Game 1 is at A's home, so A wins it with chance $\tfrac23$ (and loses with chance $\tfrac13$). Games 2 and 3 are away, so A wins each with chance $p$ (and loses with chance $1 - p$).

1. List every way Team A can win the playoff. A needs $2$ wins before B gets $2$:
- Win game 1, win game 2 (no game 3). Chance: $\tfrac23 \cdot p$.
- Win game 1, lose game 2, win game 3. Chance: $\tfrac23 \cdot (1 - p) \cdot p$.
- Lose game 1, win game 2, win game 3. Chance: $\tfrac13 \cdot p \cdot p$.
2. Add them and set the total to $\tfrac12$.
$$\tfrac23 p + \tfrac23 p - \tfrac23 p^2 + \tfrac13 p^2 = \tfrac12.$$
Combine like terms: $\tfrac43 p - \tfrac13 p^2 = \tfrac12$.
3. Clear the fractions. Multiply every term by $6$: $8p - 2p^2 = 3$, which rearranges to
$$2p^2 - 8p + 3 = 0.$$
4. Solve with the quadratic formula.
$$p = \frac{8 \pm \sqrt{64 - 24}}{4} = \frac{8 \pm \sqrt{40}}{4} = \frac{8 \pm 2\sqrt{10}}{4} = \frac{4 \pm \sqrt{10}}{2}.$$
5. Pick the right root. A probability is at most $1$. With the plus sign, $p$ is about $3.6$, which is impossible. With the minus sign, $p = \tfrac12(4 - \sqrt{10})$, about $0.42$. So $m = 4$ and $n = 10$.

$m + n = 14$.

The answer is $\boxed{\textbf{(E)}\ 14}$.

A short series has only a few winning paths. List them, multiply the chances along each path (the games do not affect each other), and add the paths. The home and away games just mean each game uses its own chance. The answer form $\tfrac12(m - \sqrt n)$ is a hint that a quadratic appears and that you keep the smaller root.