Bank before busting past twelve

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 12, 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: V(s) = \max\Big(s,\ 	frac16	extstyle\sum{d=1}^{6}\big[V(s+d)	ext{ if } s+d \le 12,	ext{ else } 0\big]\Big). Work down from the top. At s = 12 every roll busts, so V(12) = 12, and you keep stopping just below it. Checking each state, you should stop for every total from 8 up, since rolling from 8 or higher exposes a large total to too much bust risk. From s = 7 and below, rolling is better. At s = 7, a roll gives 	frac16[V(8)+V(9)+V(10)+V(11)+V(12)+0] = 	frac16[8+9+10+11+12] = 	frac{50}{6} \approx 8.33 7, so you roll. Optimal rule:…

Bank before busting past twelve

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