Homepage › Solution manuals › Joseph Blitzstein › Introduction to Probability › Exercise 4.6

Exercise 4.6

Answers

Let $N$ be the number of games played. Then the probability than $N = i$ is the probability of exactly $3$ wins in the first $i - 1$ games, and the last game being a win. $P (N = i) = 2 (\binom{i - 1}{3}) {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{i - 1 - 3} \frac{1}{2} = 2 (\binom{i - 1}{3}) {(\frac{1}{2})}^{i} .$ Note that the factor of $2$ in $P (N = i)$ is to account for either of the two players winning after $i$ games.

Then,

E (N) = 2 \sum_{i = 4}^{7} i (\binom{i - 1}{3}) {(\frac{1}{2})}^{i} \approx 5.81

Var (N) = E (N^{2}) - {(E (N))}^{2} \approx 1.06

fifthist

2021-12-05 00:00

Exercise 4.6

Answers

Comments

Add answer