One reroll on an eight-sided die, played optimally

Question

You roll a fair eight-sided die (faces 1 through 8). You may keep its value, or reroll once and must take the second roll. Playing to maximize your expected score, what score should you expect?

Accepted Answer

Decide the rule first, then condition on the first roll (the tower rule). Step 1, the reroll's worth. A fresh eight-sided die averages E[	ext{reroll}] = \frac{1 + 8}{2} = 4.5. So keep the first roll only when it beats 4.5, that is when it is 5, 6, 7, or 8, and reroll on 1, 2, 3, or 4. Step 2, condition on the first roll. - First roll is 5, 6, 7, or 8 (probability 	frac12): keep it, averaging \frac{5+6+7+8}{4} = 6.5. - First roll is 1, 2, 3, or 4 (probability 	frac12): reroll, averaging 4.5. Step 3, combine: E[	ext{score}] = 	frac12 	imes 6.5 + 	frac12 	imes 4.5 = 3.25 + 2.25 = 5.5. Yo…

One reroll on an eight-sided die, played optimally

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