I think this is different than the traditional gambler's ruin question answered here: Geometric Expectation problem?
In that question, Calvin and Hobbes play a deuces match, in which Calvin wins an individual point with $p$ probability, independently. To win the match, Calvin must win by two points.
Let's change the formulation just a tiny bit. Now, instead of a win by two rule, the winner is the first person to win two points in a row. I'm getting stuck in the recursions here, because now, if Calvin wins one point and loses the next, he's no longer back at the start. Hobbes has one point under his belt. He only needs one more to win the match, not two. Similarly, if Calvin loses the first point but wins the second, he's not back at zero. He now only needs to win once more to win the match.
I can simulate this fairly easily, but is there a closed-form solution? What is Calvin's overall probability of winning the match?
Best Answer
Let $P$/$Q$ be the probability C wins if he has respectively won/lost the previous game.
Then $P=p+(1-p)Q$ and $Q=pP$.
Then $P=\frac{p}{p^2-p+1},Q=\frac{p^2}{p^2-p+1}$.
Probability P wins is $$pP+(1-p)Q=\frac{2p^2-p^3}{p^2-p+1}.$$