What is the probability that Calvin wins the match of a series of games with a "win by two" rule, in terms of p?

Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 50, p 94.

44. 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 probability that Calvin wins the match (in terms of p) , in two different ways:

(a) by conditioning, using the law of total probability.

Why's my attempt below wrong? What to correct?

• Let P(C) be the probability that Calvin wins the match.
• Let W stands for Winning a game and L stands for losing a game.

So, possible ways of Games for winning the match = WW, WLW, LWW.

(i) WW $\implies$ Calvin wins first two games.

(ii) WLW $\implies$ wins first game, loses second game and wins third game.

(iii) LWW $\implies$ loses first game and wins next two games.

$\begin{align} \implies P(C) & = P(WW) + P(WLW) + P(LWW) \\ & = p*p + p*(1-p)*p + (1-p)*p*p \\ & = p^2(3-2p). \end{align} $

But this is wrong. The right answer is: $P(C) = \frac{p^2}{p^2+q^2}$.

Solution 1:

You are underestimating the number of possible games that can happen. They could swap wins all afternoon and neither of them would ever be ahead by two.

Here's how I would set up the problem:

Let $T$ be the probability that Calvin wins the match from a tie game. Let $A$ be the probability that Calvin wins the match when he is ahead by one point, and let $B$ be the probability that Calvin wins the match when he is behind by one point. Then show that: $$T=pA+(1-p)B\\A=p+(1-p)T\\B=pT$$ and solve that for $T$.