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. What is the expected number of games played?
My attempt: I use Markov chains. Let $S$ denote that starting state, $W$ and $L$ denote the states where Calvin and Hobbes win respectively. Then, I have $$\mu_S = 1+ p\mu_W+q\mu_L $$ $$\mu_W = 1+ q\mu_S $$ $$\mu_L = 1+ p\mu_S $$ where $\mu_j$ denote the expected number of games from state $j$. Solving the system, I get the solution $\mu_S = \frac{2}{1-2pq}$ which I think is correct. I however would like to know how to approach it with geometric distribution instead of Markov chains. I read one solution where the number of games is given as $1+\text{Geom}(p^2+q^2)$ but I do not understand it. My understanding is that for the game to stop, the required probability is $p^2 + q^2$ since this is the probability that either Calvin or Hobbes win two games in a row. How to use this information in a Geometric random variable?
Thanks.
Consider the first two games. There are three possible outcomes: $C$ wins, $H$ wins, the game restarts. Phrased differently, there are two outcomes each of which takes two turns: either someone wins (probability $p^2+q^2$) or the game restarts. Thus you can think of the game as a binomial process where each move consists of two games. It follows that the desired process (waiting for a winner to emerge) is: $$2\times \text {Geom}(p^2+q^2)$$ which is to say: $$P(X=2n)=P(\text {Geom}(p^2+q^2)=n)$$ which implies that the expectation is $$2\times \text E[\text {Geom}(p^2+q^2)]$$ and the desired expectation is therefore $$\frac 2{p^2+q^2}$$ which is easily seen to agree with your Markov result.