The probability that Calvin wins the match of a series of games with a “win by two” rule, in terms of p

conditional probabilityprobability

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

  1. 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: $P(C) = \frac{p^2}{p^2+q^2}$.

Best Answer

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 $T$.

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