Bank before busting past eight

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 it. If a roll pushes your total above 8, 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, with V(s) = \max\Big(s,\ 	frac16	extstyle\sum{d=1}^{6}\big[V(s+d)	ext{ if } s+d \le 8,	ext{ else } 0\big]\Big). Work down from the top. At s = 8 every roll busts, so V(8) = 8, and similarly you keep stopping just below it. Checking each state, you should stop for every total from 5 up: rolling from 5 or higher risks a bust that outweighs the modest expected gain. From s = 4 and below, rolling wins. At s = 4, for instance, a roll yields 	frac16[V(5)+V(6)+V(7)+V(8)+0+0] = 	frac16[5+6+7+8] = 	frac{26}{6} \approx 4.33 4, so you roll. …

Bank before busting past eight

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