# Why doesn't P(Calvin wins the match by winning 2 more games than his opponent) = $P(C|W_1)P(W_1)+P(C|L_1)P(L_1) = (p)(p) + (1-q)(q)$?

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.

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

## My attempt

Let event $$C$$ be "Calvin wins."

$$P(C) = P(C|W_1)P(W_1)+P(C|L_1)P(L_1) = (p)(p) + (1-q)(q)$$

This doesn't match the answer key, $$\frac{p^2}{p^2+q^2}$$.

I think the conceptual misunderstanding is $$P(C|W_1)$$, $$P(C|L_1)$$. For $$P(C|W_1)$$, my thinking is: if we interpret this problem as the gambler ruin's problem, $$i$$ is originally at 2, but given that $$W_1$$, we have that $$i=3$$. Thus, because C occurring corresponds to $$i=4$$, $$P(C|W_1) = \text{ (the probability that we move from } i=3 \text{ to } i = 4) = p$$ The same goes for $$P(C|L_1)$$: $$i$$ is originally at 2, but given that $$L_1$$, we have that $$i=1$$. Thus, because C not occurring corresponds to $$i=0$$, $$P(C|L_1) = 1 - P(C^c|L_1) = 1 -\text{ (the probability that we move from } i=1 \text{ to } i = 0) = 1-q$$

I know this is wrong because the gambler ruin's problem would interpret, for example, $$P(C|W_1) = p_{i+1}$$. When I first learned the gambler ruin's problem, this made sense. However, now that I'm actually doing a problem that's kind of related, I'm left questioning this: why can't we just $$P(C|W_1) = p^{n-i-1}$$, where $$n$$ represents the # of wins to win the series, $$i$$ is the original starting spot, and one represents winning the first game

And yet, I also don't think the gambler's ruin problem totally applies here; the biggest difference is that given $$W_1$$ or $$L_1$$, the important thing is that we are one spot away from the edge, so the probability of winning or losing one spot from the edge is either $$p$$ or $$q$$, whereas if we were at an $$i$$ in the "middle," $$i$$ can shift left, right, and back left, and left some more.

You are missing the chance that Calvin can win the first game, lose the second, and still win the match. Your conversion of $$P(C|W_1)P(W_1)$$ to $$p^2$$ says the only way Calvin can win given that he won the first game is to win the second as well. You should use the fact that the games are independent to say if Calvin wins the first game, either he wins the second and wins the match or he loses the second and has the same chance of winning the match as he started with. Similarly, if he loses the first game and wins the second he is back where he started.
• Ahhh. Thank you. I correctly redid the problem by solving it as a gambler ruin's problem. But, there's a second part to this problem which is to condition using the LOTP. How else can I condition $P(C)$? Commented Aug 9, 2021 at 5:33