4.1.6 problem 6

Let $N$ be the number of games played. Then the probability than $N=i$ is the probability of exactly $3$ wins in the first $i-1$ games, and the last game being a win. $P(N=i)=2\left(\frac{i-1}{3}\right){\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{i-1-3}\frac{1}{2}=2\left(\frac{i-1}{3}\right){\left(\frac{1}{2}\right)}^i.$ Note that the factor of $2$ in $P(N=i)$ is to account for either of the two players winning after $i$ games.

Then,