2005 AMC 10A Problem 18

We are told two facts: B won game 2, and A won the series. List every series that fits both facts, find how likely each one is, and see what share of that has B winning game 1.

Each game is a fair coin flip, so a series that lasts $4$ games has probability $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$, and one that lasts $5$ games has probability $\frac{1}{32}$.

1. B wins game 1. Then B leads $2$ to $0$. For A to win the series, A must win the next three: $BBAAA$. Probability $\frac{1}{32}$.
2. A wins game 1. After $A$, $B$ it is $1$ to $1$. A needs two more wins before B gets two more:
- $ABAA$: probability $\frac{1}{16} = \frac{2}{32}$.
- $ABABA$: probability $\frac{1}{32}$.
- $ABBAA$: probability $\frac{1}{32}$.
3. All series that fit the facts. $\frac{1}{32} + \frac{2}{32} + \frac{1}{32} + \frac{1}{32} = \frac{5}{32}$.
4. Share where B won game 1. Only $BBAAA$, which is $\frac{1}{32}$:
$$\frac{1/32}{5/32} = \frac{1}{5}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{1}{5}}$.

"Given that" means you keep only the outcomes that match the facts and ask what fraction of them you want. The catch is that series of different lengths are not equally likely: a shorter series needs fewer coin flips to go right. Weight each listed series by its probability instead of just counting the list.

Pretend all five games are always played, even after someone has won. Now every one of the $32$ strings of five results is equally likely, and you can simply count. Keep strings with B in game 2 where A reaches three wins first. With B in game 1: only $BBAAA$, so $1$ string. With A in game 1: $ABAA$ followed by either result ($2$ strings), plus $ABABA$ and $ABBAA$, so $4$ strings. The answer is $\frac{1}{1 + 4} = \frac{1}{5}$.

We want $P(B \text{ wins game 1} \mid B \text{ wins game 2 and } A \text{ wins the series})$. Enumerate every series consistent with the condition, remembering that a series of $k$ games has probability $\left(\tfrac{1}{2}\right)^k$.

Case 1: $B$ wins game 1. Then $B$ leads $2$-$0$ and $A$ must take the next three: $BBAAA$, probability $\frac{1}{32}$.

Case 2: $A$ wins game 1. After $AB$ the series is $1$-$1$ and $A$ needs two more wins before $B$ gets two:
- $ABAA$ (four games): $\frac{1}{16}$,
- $ABABA$: $\frac{1}{32}$,
- $ABBAA$: $\frac{1}{32}$.

Total probability of the condition: $\frac{1}{32} + \frac{2}{32} + \frac{1}{32} + \frac{1}{32} = \frac{5}{32}$.

The conditional probability is
$$\frac{1/32}{5/32} = \frac{1}{5}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{1}{5}}$.