Geometric Expectation problem? Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 25, p 198.



*Calvin and Hobbes play a match consisting of a series of games, where Calvin has
probability $p$ of winning each game (independently). They play with a “win by two”
rule: the first player to win two games more than his opponent wins the match. Find
the expected number of games played.

Hint: Consider the first two games as a pair, then the next two as a pair, etc.

My approach: This is a geometric distribution related problem because you are trying to find the number of games played, until "success" occurs which in this case is winning by two.
So using the hint, I thought about the probability of winning by two. For player 1, the probability of winning two games in a row is $p^2$. For player two, the probability of winning two games in a row is $(1-p)^2$ or $q^2$. So the total probability of winning two games in a row for either player is $p^2+q^2$.
After this, I tried plugging this into the geometric expectation formula which is $E[X]=\frac{1-p}{p}$. But I wasn't getting the right answer. Can anyone explain what I did wrong? Thanks!
The author's solution: $\frac{2}{p^2+q^2}$

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.
