Geometric Expectation problem?

Solution 1:

Expectation of a geometric random variable with parameter $p$ is given by $\frac{1}{p}$. So in this case, we would have an expectation of $\frac{1}{p^2+q^2}$ trials. But each trial corresponds to 2 games, so we get an expected value of $\frac{2}{p^2+q^2}$ games played.

Solution 2:

Let $N$ be the expected duration.

The probability that Calvin wins is $p$, after which the expected duration is $$ \overbrace{\ \ \ \ \ \ 1\ \ \ \ \ \ }^{\substack{\text{step taken}\\\text{to get here}}}+\overbrace{\ \ \ \ \ \ p\vphantom{1}\ \ \ \ \ \ }^{\substack{\text{probability}\\\text{of one more}}}+\overbrace{\ (1-p)\ }^{\substack{\text{probability of}\\\text{$N+1$ more}}}(N+1) $$ The probability that Calvin loses is $1-p$, after which the expected duration is $$ \overbrace{\ \ \ \ \ \ 1\ \ \ \ \ \ }^{\substack{\text{step taken}\\\text{to get here}}}+\overbrace{\ \ \ 1-p\ \ \ }^{\substack{\text{probability}\\\text{of one more}}}+\overbrace{\ \ \ \ \ \ p\vphantom{1}\ \ \ \ \ \ }^{\substack{\text{probability of}\\\text{$N+1$ more}}}(N+1) $$ Thus, $$ \begin{align} N &=p(1+p+(1-p)(N+1))+(1-p)(1+1-p+p(N+1))\\[6pt] &=2+2p(1-p)N\\ &=\frac2{1-2p+2p^2} \end{align} $$