Playing chess until one party wins 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.
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.
Probability of A winning(before B) is =$0.3\sum_{i=0}^{\infty}(0.45)^i=0.3\times \frac{1}{1-0.45}=\frac{0.3}{0.55}=30/55$

Reason: If A wins after the first game it must be with prob $0.3$, if he wins after the 2nd game then the first game must be a draw and A must win the 2nd game, this happens with prob. $0.45\times 0.3$in this way it goes on. Ultimately all these prob. must be added to get the required prob.

Yes it is always possible to calculate the prob. of A winning given that somebody wins.

In the same way you can calculate the prob. of B winning before A.Adding both these prob. you will get the prob. of somebody winning.Then using the formula for conditional prob. you can find the above cond. prob.