# How to intuit p = Calvin's probability of winning each game independently = $1/2 \implies$ P(Calvin wins the match) = 1/2?

Please see the sentence beside my red line. The notion of a "sanity check" suggests that these resultant integers should be obvious, without calculation or contemplation. But why's it plain and intuitive that $p = 1/2 \implies P(C) = 1/2$?

Indubitably, a game differs from a match. Just because $p = 1/2$ doesn't automatically entail $P(C) = 1/2$.

- 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.

(b) by interpreting the problem as a gambler's ruin problem.

## Solution:

(a) Let C be the event that Calvin wins the match, $X \thicksim Bin(2, p)$ be how many of the first 2 games he wins, and $q = 1 - p$. Then

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

p 18 in the publicly downloadable PDF of curbed solutions.

## 1 answer

The intuition at $p = \frac{1}{2}$ is based on symmetry.

If Calvin wins a game with probability $p = \frac{1}{2}$, then Hobbes also wins a game with *the same probability* $q = 1 - p = \frac{1}{2}$. Then the respective probabilities $P(C)$ and $P(H)$ of each player winning the whole match *must also be same*: $P(C) = P(H)$.

Then the particular value $P(C) = P(H) = \frac{1}{2}$ follows from the assumption that either Calvin wins and Hobbes does not, or else Hobbes wins and Calvin does not (the events are *mutually exclusive* and *exhaustive*), so that $P(C) + P(H) = 1$.

## 0 comment threads