Post History
#3: Post edited
by
DNB
·
2022-01-17T08:54:59Z (almost 3 years ago)
- 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 $p = 1/2 \implies P(C) = 1/2$?
- >50. 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.
- 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 $P(C) = 1/2$.
- >50. 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.
#2: Post edited
by
DNB
·
2021-12-31T08:33:56Z (almost 3 years ago)
- 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 $p = 1/2 \implies P(C) = 1/2$?
- >50. 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.
- 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 $p = 1/2 \implies P(C) = 1/2$?
- >50. 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: Initial revision
by
DNB
·
2021-12-31T08:31:07Z (almost 3 years ago)
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 $p = 1/2 \implies P(C) = 1/2$? >50. 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.