I am self-learning basic undergrad calculus-based probability. I would like someone to verify if my expression for the expected waiting time is correct. I am posting my attempt/solution below:
[BH 4.25] Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability of $p$ 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 consider the next two as a pair.
Solution. (My Attempt)
The result of the match is always decided in an even number of games, $2n$.
We consider the first two games as a pair, the next two games as pair and so forth,.
Let $W$ be the event that Calvin wins the game and let $L$ be the event that Calvin loses the game to Hobbes.
The match continues until the first $(L,L)$ or $(W,W)$ occurs.
Let $N$ be the number of games played.
\begin{align*}
P \{N = 2k \} = \hat{q}^{k-1} \cdot \hat{p}
\end{align*}
where $\hat{q}=2p(1-p)$ and $\hat{p}=p^2 + (1-p)^2$. Note that, $\hat{p} + \hat{q} = 1$. Thus, $N/2$ follows a first success distribution with success probability $\hat{p}$.
Thus,
\begin{align*}
E(N/2) &= \frac{1}{\hat{p}} = \frac{1}{p^2 + (1-p)^2}
\end{align*}
Consequently,
\begin{align*}
E(N) &= \frac{2}{p^2 + (1-p)^2}
\end{align*}
Best Answer
I have used a different method.
Taking win probability of Calvin as $p$
let $x$ = additional games needed from start (= tied) for Calvin to win
$y$ = additional games needed if Calvin leads by $1$ game
$z$ = additional games needed if Calvin trails by $1$ game,
then $x = 1 +py +qz, y = 1+qx, z = 1 + px$
Solving, $x = \frac{2}{2p^2-2p+1}= \frac{2}{1-2pq},$
and by result symmetry, expected # of games needed by Hobbes to win is also $\;\frac{2}{1-2pq}$
Thus for one or the other to win, $\Bbb{E}[N] = \frac{2}{1-2pq}$