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.
 A: 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.
A: By independence, the probability the series goes on more than $n$ games is $(0.45)^n$, so the probability the game goes on forever is $0$.
Because of this, draws are irrelevant, so effectively we could assume that we are playing a game in which the probability A wins is $p=\frac{0.3}{0.55}$, and B wins with probability $1-p$.
In the modified game, A wins if and only if she wins the first game. This has probability $p$. 
A: By the law of total Probability
you have $$P(A\mathrm{\,wins})=P(\mathrm{someone\,wins})P(A\mathrm{\,wins}|\mathrm{someone\,wins})+P(\mathrm{nobody\,wins}|)P(A\mathrm{\,wins}|\mathrm{nobody\,wins})$$
$P(\mathrm{nobody\,wins})$ means that all the games are draw so $P(\mathrm{nobody \, wins})\leq P(\mathrm{The \,first \,n\,games\,are\,draw})=(0,45)^n$ since this holds for every $n$ it implies $P(\mathrm{nobody\,wins})=0$. Which makes your assertion correct.
A: 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.
