4.1.4 problem 4

Key idea is that the expected value of a die roll is $3.5$. If the value of a roll is less than $4$, one should roll again.

For the winnings to be $1$, $2$ or $3$ under the above strategy, the first two dice need to land on a value less than $4$, and the last die needs to land on $1$, $2$, and $3$ respectively.

Thus, $P(W=i)={\left(\frac{3}{6}\right)}^2\ast \frac{1}{6}=\frac{1}{24}$ for $i\in 1,2,3$.

For the winnings to be $i\ge 4$, either the first die lands on $i$, or the first die lands on a value less than $4$, and the second die lands on $i$, or the first two dice land on values less than $4$, and the third die lands on $i$.

Thus, $P(W=i)=\frac{1}{6}+\frac{3}{6}\ast \frac{1}{6}+{\left(\frac{3}{6}\right)}^2\ast \frac{1}{6}=\frac{7}{24}$ for $i\in 4,5,6$.

Thus,