Suppose two teams play a series of games, each producing a winner and a loser, until one time has won two more games than the other. Let G be the number of games played until this happens. Assuming your favorite team wins each game with probability p, independently of the results of all previous games, find:
- P(G=n) for n=2,3,…
- E(G)
- Var(G)
I was able to do 1 and 2 relatively easily.
Using linearity of expectation for 2-game sets, I calculated P(G=n) = 0 for odd n and P(g=n) = $(2pq)^\frac{n-2}{2}(p^2+q^2)$ for even n.
Using a recursive method I found E[G] = $\frac{2}{p^2+q^2}$.
Not quite sure how to do this. I tried using $Var(x) = E[X^2] - E[X]^2$ but I didnt get the correct answer (which is $\frac{8pq}{p^2+q^2}$).
If at any moment the series is tied, then we must play two more games. After these two more games, the probability of there being a winner is $p^2+q^2$. If there isn't a winner in two games, then we are back to a tie. Thus if we consider pairs of games, we are in a geometric random variable situation with probability of success being $p^2+q^2$ and probability of failure being $1-p^2-q^2=2pq$ (here $q=1-p$).
Thus $X=\frac{G}{2}$ is geometric with parameter $p^2+q^2$.
The variance of a geometric distribution with parameter $\lambda$ is $\frac{1-\lambda}{\lambda^2}$. You can use this to find $\text{Var}(X)$.
Finally $\text{Var}(G)=\text{Var}(2X)=2^2\text{Var}(X)$