Homepage › Solution manuals › Joseph Blitzstein › Introduction to Probability › Exercise 4.11

Exercise 4.11

Answers

Note that $31 = 2^{4} + 2^{3} + 2^{2} + 2^{1} + 1$ . Thus, Martin can play at most $5$ rounds. For every possible win, Martin makes $1$ dollar. If the game reaches the fifth round, it is also possible that Martin loses and walks away with nothing.

Let $X$ be Martin’s winnings.

Then,

E (X) = \sum_{i = 1}^{5} (\frac{1}{2^{i}} 1) + (\frac{1}{2^{5}} 0) \approx 0.97

fifthist

2021-12-05 00:00

Exercise 4.11

Answers

Comments

Add answer