If C = Calvin wins the match, and $X \thicksim Bin(2, p) =$ how many of the first 2 games he wins — then why P(C|X = 1) = P(C)?
The author's solution doesn't expatiate why $\color{red}{P(C|X = 1) = P(C)}$? This similar question on Math Stack Exchange has 0 answers, as at 4 January 2022.
- 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 17 in the publicly downloadable PDF of curbed solutions.
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The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| whybecause | (no comment) | Jan 5, 2022 at 16:15 |
| DNB | (no comment) | Jan 9, 2022 at 23:51 |
The key is
the first player to win two games more than his opponent wins the match
$P(C \mid X = 1)$ describes the situation when Calvin has won 1 of the first two matches: so the opponent won the other match, and they're currently even. Since a 1-1 score is equivalent to a 0-0 score for the purposes of winning, the probability that Calvin wins from a 1-1 situation is the same as the probability that Calvin wins from a 0-0 situation.
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