Question

The problem of points. Two player $A$ and $B$ want respectively $m$ and $n$ points of winning a single game are $p$ and $q$ respectively where $p+q=1$. The stake is to belong to the player who first makes up his set; find the probabilities in favour of each player.

• A. $P^m[1+mq+\frac{m(m+1)}{1.2}]q^2+...\frac{(m+n-2)!}{(m-1!)(n-1)!}{q}^{n-1}$,$q^n[1+np+\frac{n(n+1)}{1.2}{p}^2+......+\frac{(m+n-2)}{(m-1)!(n-1)!}{p}^{m-1}]$
• B. $P^m[1+mq+\frac{m(m+1)}{1.2}]q^2+...\frac{(m+n-1)!}{(m-1!)(n-1)!}{q}^{n-1}$,$q^n[1+np+\frac{n(n+1)}{1.2}{p}^2+......+\frac{(m+n-2)}{(m-1)!(n-1)!}{p}^{m-1}]$
• C. $P^m[1+mq+\frac{m(m+1)}{1.2}]q^2+...\frac{(m+n-2)!}{(m-1!)(n-1)!}{q}^{n-1}$,$q^n[1+np+\frac{n(n+1)}{1.2}{p}^2+......+\frac{(m+n-1)}{(m-1)!(n-1)!}{p}^{m-1}]$
• D. $P^m[1+mq+\frac{m(m+1)}{1.2}]q^2+...\frac{(m+n-1)!}{(m-1!)(n-1)!}{q}^{n-1}$,$q^n[1+np+\frac{n(n+1)}{1.2}{p}^2+......+\frac{(m+n-1)}{(m-1)!(n-1)!}{p}^{m-1}]$

Answer 1

Answer: (A)

$P^m[1+mq+\frac{m(m+1)}{1.2}]q^2+...\frac{(m+n-2)!}{(m-1!)(n-1)!}{q}^{n-1}$,$q^n[1+np+\frac{n(n+1)}{1.2}{p}^2+......+\frac{(m+n-2)}{(m-1)!(n-1)!}{p}^{m-1}]$

Suppose $A$ wins in exactly $m+r$ games.
To do so he must win the game and $m-1$ out of the preceding ${}^{m+r-1}{C}_{m-1}{P}^{m-1}{P}^{m-1}{q}^{r}p,i.e{.}^{m+r-1}{C}_{m-1}{P}^m{q}^r.$
Now the set will definitely be decided in $m+n-1$ games; and $A$ may win these $m$ games in exactly $m$ games, or $m+1$ games,...., or $m+n-1$ games.
Hence we shall obtain the chance of $A$'s winning the set by giving to $r$ values $0,1,2,.....,n-1$ in succession in the expression ${}^{m+r-1}{C}_{m-1}{P}^m{q}^r$.
Hence $A$'s probability to win the set is $P^m[1+mq+\frac{m(m+1)}{1.2}]q^2+...\frac{(m+n-2)!}{(m-1!)(n-1)!}{q}^{n-1}$;
similarly $B$'s probability to win the set is $q^n[1+np+\frac{n(n+1)}{1.2}{p}^2+......+\frac{(m+1-2)}{(m-1)!(n-1)!}{p}^{m-1}]$