Geometric Expectation problem? 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.

Solution: $\frac{2}{p^2+q^2}$

Can anyone explain what I did wrong? Thanks!
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.
First of all number of games played will be even.
now consider games $1-$2 as box1 , $2-$4 as box two ..........
for probability that any of one of the player wins the both the games in box will be $p^2+q^2$
so the game will stop if any such box comes in the sequence 
$X$~first_sucess$(p^2+q^2)$ 
Expected number of such boxes $$E(X)=1/(p^2+q^2).$$
therfore number of games played is $$2/(p^2+q^2).$$