Ten-sided die, pay two dollars to re-roll

Question

You roll a fair ten-sided die (faces 1 through 10) and will be paid its face value in dollars. After any roll you may keep the current value and stop, or pay \2 to re-roll, replacing the current value. You may re-roll as often as you like, and the \2 costs are subtracted from your winnings. What is your stopping rule, and what is the game worth?

Accepted Answer

Let G be the game's value when you hold a roll and play optimally. Re-rolling is worth G - 2 (a fresh roll worth G, minus the \2 cost). So the value of holding a face x is \max(x,\ G - 2), and G = E\big[\max(X,\ G - 2)\big], \qquad X \sim 	ext{Uniform}\{1,\dots,10\}. Let R = G - 2 be the re-roll threshold: keep x iff x \ge R. Guess R \in (4, 5], so you keep \{5,6,7,8,9,10\} and re-roll \{1,2,3,4\}. Each re-rolled outcome is worth R, and 5+6+7+8+9+10 = 45, so G = \frac{1}{10}\big[4R + 45\big] = \frac{4(G-2) + 45}{10}. Solving: 10G = 4G - 8 + 45 = 4G + 37, so 6G = 37 and G = 	frac{37}{6} \app…

Ten-sided die, pay two dollars to re-roll

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