Probability of Winning in Sports In sports such as volleyball (prior to 1999), you can score a point only if it is your turn, with turns alternating until a point is scored. Suppose your probability of scoring a point when it is your turn is $p$, and your opponent's probability of scoring a point when it is her turn is $q$. Find a formula for the probability $S$ that you are the first to score the next point, assuming it is currently your turn.
I have little experience with probability problems, so any help would be well appreciated!

Let's denote your answer by $\psi$.
To do it with series, let's compute the probability that you win on your $n^{th}$ turn.  In order for that to happen, both you and your opponent must lose $n-1$ times.  The probability of that happening is $(1-p)^{n-1}(1-q)^{n-1}$.  After that, you must actually score, which happens with probability $p$.  Thus the probability that you win on the $n^{th}$ trial is $$\psi_n=(1-p)^{n-1}(1-q)^{n-1}p$$  It follows that the probability that you win eventually is the sum of these, hence $$\sum_{n=1}^{\infty}\psi_n=\sum_{n=1}^{\infty}(1-p)^{n-1}(1-q)^{n-1}p=p \sum_{n=1}^{\infty}(1-p)^{n-1}(1-q)^{n-1}=p\sum_{n=0}^{\infty}(1-p)^{n}(1-q)^{n}$$  But this is just a geometric series, which we can sum to get $$\psi=\frac p{1-(1-p)(1-q)}=\frac p{p+q-pq}$$
Sanity checks:  As the formula isn't exactly intuitive, it's worth thinking about some special cases.  If, say, $p=1$ then this gives $\psi=1$ as it should.  Also, if $q=0$, this correctly gives $\psi=1$ (ignoring the hopeless case in which both $p$ and $q$ are $0$).  If $p=\frac 12 = q$ we get $\psi=\frac {\frac 12}{\frac 12+\frac 12-\frac 14}=\frac 23$ which may be recognized as the answer to the question "if two people alternate tosses of a fair coin, what is the probability that the first player gets the first $H$", so again the formula holds up.
