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: $\frac{2}{p^2+q^2}$
 A: 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.
A: 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,
$$
\begin{align}
N
&=p(1+p+(1-p)(N+1))+(1-p)(1+1-p+p(N+1))\\[6pt]
&=2+2p(1-p)N\\
&=\frac2{1-2p+2p^2}
\end{align}
$$
A: Denote by $E_k$ the expected number of additional games when Calvin is $k$ wins ahead. Then  $E_{-2}=E_2=0$ and
$$E_k=1+p E_{k+1}+(1-p)E_{k-1}\qquad(-1\leq k\leq1)\ .$$
Solving for $E_0$ gives
$$E_0={2\over p^2+(1-p)^2}\ .$$
A: 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 $$2/(p^2+q^2).$$
