Question

Two players A and B want respectively m and n points of winning a set of games; their chances of winning a single game are p and q respectively, where the sum of p and q is unity; the stake is to belong to the player who first makes up his set: determine the probabilities in favour of each player.

Answer 1

Suppose that A wins in exactly $m+r$ games; to do this he must win the last game and $m-1$ out of the preceding $m+r-1$ games. The chance of this is ${}^{m+r-1}{C}_{m-1}{p}^{m-1}{q}^{r}p$, or ${}^{m+r-1}{C}_{m-1}{p}^m{q}^r$.
Now the set will necessarily be decided in $m+n-1$ games, and A may win in exactly $m$ games, or $m+1$ games, ....., or $m+n-1$ games; Therefore we shall obtain the chance that A wins the set by giving to $r$ the values $0,1,2,....n-1$ in the expression ${m+r-1}_{m-1}^C{p}^m{q}^r$. Thus A's chance is
$p^m\{1+mq+\frac{m(m+1)}{1.2}{q}^2+....+\frac{\lfloor m+n-2}{\lfloor m-1\lfloor n-1}{q}^{n-1}\};$
similarly B's chance is
$q^n\{1+np+\frac{n(n+1)}{1.2}{p}^2+....+\frac{\lfloor m+n-2}{\lfloor m-1\lfloor n-1}{p}^{m-1}\}$.