Players $A$ and $B$ decide to play chess until one of them wins. Assume games are independent with $P(A\text{ wins})=0.3$, $P(B\text{ wins})=0.25$, $P(\text{draw})=0.45$ on each game. If the game ends in a draw another game will be played. Find the probability $A$ wins before $B$.
Since the games are independent, I can simply calculate $P(A \text{ wins} \mid \text{somebody wins})$ right? The textbook does not have a solution.
Best Answer
Yes. This is because we have:
$$\Pr(A \text{ wins first}) = \sum_{n \ge 0} 0.3 \cdot 0.45^n$$
which by the geometric series can be evaluated to:
$$\sum_{n \ge 0} 0.3\cdot 0.45^n = 0.3\cdot \sum_{n \ge 0} 0.45^n = 0.3 \cdot \frac1{1-0.45} =\frac{0.3}{1-0.45}$$
and the latter expression equals $\Pr(A \text{ wins}\mid \text{someone wins})$ because "$A$ wins" and "nobody wins" are mutually exclusive.