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!
 A: Assume the probability of you scoring the next point if it is your turn now is $x$. 
The probability is p that you score immediately. The probability is $(1-p)(1-q)$ that you don't score, then your opponent doesn't score, and it's your turn again to win with probability $x$. 
So $x = p + (1-p)(1-q) x$, and you calculate $x$ from that getting $x  = \frac p{ p + q - pq}$
A: 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.
A: If you manage to score on your first turn, then the probability is just $p$. However if you don't, then you have to hope your opponent does not score. However if you fail to score again, then once again you have to hope your opponent doesn't score.
If I am interpreting this right, the equation would be $$P(S) = (1-p)^{n-1}*(1-q)^{n-1}*p$$
where you score on your $n^{th}$ turn.
