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$3$ die rolls

Expected valueMedium

You may roll a fair $6$ -sided die up to $3$ times. After each roll you either stop and receive the face value in dollars, or roll again; the last roll must be kept. With optimal play, what is the expected payout?

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Hint

Work backwards from the last roll.

Solution

With one roll left the value is $\frac{7}{2}$ .
With $2$ rolls left, keep $4$ , $5$ or $6$ and otherwise roll again, so the value is $\frac{15}{6} + \frac{1}{2} \times \frac{7}{2} = \frac{17}{4}$ .
With $3$ rolls left, keep only $5$ or $6$ (above $\frac{17}{4}$ ), so the value is $\frac{11}{6} + \frac{4}{6} \times \frac{17}{4} = \frac{11}{6} + \frac{17}{6}$ .
That equals $\frac{14}{3}$ .

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

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