2005 AMC 10A Problem 18

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

Conditioning restricts the sample space to the outcomes satisfying the given information; the answer is the probability of the target outcomes divided by the probability of the restricted space. The subtlety is that series of different lengths are not equally likely, so you must weight each listed sequence by $2^{-\text{length}}$ rather than count sequences. Equivalently, imagine all five games always being played: $ABAA$ then stands for two five-game strings.

• Counting the four qualifying series as equally likely and answering 1/4; the four-game series ABAA is twice as probable as any five-game series.
• Answering $\frac{1}{2}$ because "each game is a coin flip," ignoring the conditioning on the series result.