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Keep or reroll

ProbabilityHard

You roll a fair $6$ -sided die and are paid the number shown. After seeing any roll, you may reject it and roll again, but you may roll at most $3$ times in total, and the last roll must be kept if you reach it. Playing optimally to maximise the expected payoff, what is the expected payoff? Give an exact fraction in lowest terms.

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Hint

Work backwards: the value of continuing is the value of the game with one fewer roll left.

Solution

With $1$ roll left the expected payoff is $V 1 = \frac{7}{2}$ .
With $2$ rolls left, keep a result only if it beats $\frac{7}{2}$ , which means $4$ , $5$ or $6$ . So $V 2 = \frac{4 + 5 + 6 + 3 \times \frac{7}{2}}{6} = \frac{17}{4}$ .
With $3$ rolls left, keep a result only if it beats $\frac{17}{4}$ , which means $5$ or $6$ . Results $1$ to $4$ are rerolled.
So $V 3 = \frac{5 + 6 + 4 \times \frac{17}{4}}{6} = \frac{28}{6}$ .
This reduces to $\frac{14}{3}$ .

Answer: $\frac{14}{3}$

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