Suppose Bob starts with 1 credit. Everyday Bob can choose to go to a casino. On Day \(i\) (with \(i\) starting from 1), he has a \(\frac{101-i}{100}\) chance to win his gambling. If he wins, he wins a prize of \(\frac{1}{2^i}\) credits. If he loses, then he loses all of the credits accumulated so far, and he has no more money to gamble anymore. Optimally, how many days of gambling should Bob do?
Solution
Answer: \(4\) days
Consider the marginal profit from each day. Bob should continue only while the expected marginal profit is positive; otherwise, the expected amount of money he has decreases.
On day \(i\), the expected revenue is \(\frac{101-i}{100}\cdot\frac1{2^i}\). His accumulated money before that day is \(1+\frac12+\cdots+\frac1{2^{i-1}}=2-\frac1{2^{i-1}}\), so the expected loss is \(\frac{i-1}{100}\left(2-\frac1{2^{i-1}}\right)\).
\[\text{Marginal profit}=\frac{101-i}{100\cdot2^i}-\frac{i-1}{100}\left(2-\frac1{2^{i-1}}\right)=\frac{99+i-(i-1)2^{i+1}}{100\cdot2^i}.\]
This is positive only for \(i\leq4\). Thus, optimally, Bob should gamble for \(4\) days.