A ten-sided die, bust past fifteen

Question

You start with a running total of 0. Each turn you may stop and bank your total, or roll a fair ten-sided die (equally likely 1 through 10) and add the result. If a roll pushes your total above 15, you bust and get nothing. Maximize the expected amount you bank. What is the optimal stopping rule, and what is the expected value of the game?

Accepted Answer

Let V(s) be the best expected payoff from safe total s, averaging over ten faces: V(s) = \max\Big(s,\ 	frac{1}{10}	extstyle\sum{d=1}^{10}\big[V(s+d)	ext{ if } s+d \le 15,	ext{ else } 0\big]\Big). Work down from the top. At s = 15 every roll busts, so V(15) = 15, and you keep stopping just below it. Checking each state, you should stop for every total from 9 up. From s = 8 and below, rolling is better, because the ten-sided die adds a healthy amount on average and only the higher faces bust you. Carrying the recursion down to 0, V(0) \approx \9.95, with the optimal rule: keep rolling while…

A ten-sided die, bust past fifteen

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