Chess match where first to win game wins match. Probability of winning is p and q, of draw is 1-p-q. Find mean, PMF, and variance of match duration. 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 $1 — p — q$.

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*What is the probability that Fischer wins the match?


    Fischer wins = draw the first $(n — 1)^{th}$ games and win the last game.  $$P\text{(Fischer wins)} = \sum_{i=1}^n(1-p-q)^{n-1}\cdot p = \frac{p}{p+q}$$


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*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 $\frac{1-p-q}{(p+q)^2}$
