Question Fischer and Spassky play a sudden-death chess match whereby the first player to win a game wins the match. Each game is won by Fischer with probability $p$, by Spassky with probability $q$, and is a draw with probability $1 — p — q$.
- What is the probability that Fischer wins the match?
Fischer wins = draw the first $(n — 1)^{th}$ games and win the last game.
$$P\text{(Fischer wins)} = \sum_{i=1}^n(1-p-q)^{n-1}\cdot p = \frac{p}{p+q}$$
- What is the $PMF$, the mean, and the variance of the duration of the match?
Doubt : I know that $Var[X] = E[X^2] -E[X]^2$, however, can't follow the arithmetic calculation provided in solution. Help me understand how $E[X]$ end up as $(P+1)^{-1}$ ? I supposed $E[X]^2$ is $(p+q)^{-2}$ in this case.
Can someone please explain in more steps?
The expressions are wrong. P(Fisher wins)=$p\sum_{n=0}^\infty (1-p-q)^n=p\frac{1}{1-(1-p-q)}=\frac{p}{p+q}$
Let $X$ be the duration of the match . $E(X)=(p+q)\sum _{k=1}^\infty k(1-p-q)^{k-1}=\frac{1}{p+q}$, $E(X^2)=(p+q)\sum_{k=1}^\infty k^2(1-p-q)^{k-1}=\frac{2-p-q}{(p+q)^2}$ The variance is $\frac{1-p-q}{(p+q)^2}$