Roll and bank before you bust past ten

Question

You start with a running total of 0. Each turn you may stop and bank your total, or roll a fair die and add the result to your total. But if a roll pushes your total above 10, you bust and get nothing. You play to maximize the expected amount you bank. What is the optimal stopping rule, and what is the expected value of the game?

Accepted Answer

Work backwards from the states nearest the danger zone. Let V(s) be the best expected payoff when your safe total is s. From s, rolling gives 	frac16 on each face, and any face that lands you above 10 pays 0: V(s) = \max\Big(\underbrace{s}{	ext{stop}},\ \underbrace{	frac16	extstyle\sum{d=1}^{6}\big[V(s+d)	ext{ if } s+d \le 10,	ext{ else } 0\big]}{	ext{roll}}\Big). Fill in from the top. At s = 10 any roll busts, so you stop: V(10) = 10. Coming down, at s = 9 only a 1 survives, at s = 8 only 1 or 2, and so on, so you keep stopping. Checking each, you should stop for every total from 6 up…

Roll and bank before you bust past ten

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