A four-sided die, bust past six

Question

You start with a running total of 0. Each turn you may stop and bank your total, or roll a fair four-sided die (equally likely 1, 2, 3, 4) and add the result. If a roll pushes your total above 6, 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, now averaging over four faces: V(s) = \max\Big(s,\ 	frac14	extstyle\sum{d=1}^{4}\big[V(s+d)	ext{ if } s+d \le 6,	ext{ else } 0\big]\Big). Work down from the top. At s = 6 every roll busts, so V(6) = 6, and you keep stopping just below it. Checking each state, you should stop for every total from 4 up. From s = 3 and below, rolling wins. At s = 3, a roll gives 	frac14[V(4)+V(5)+V(6)+0] = 	frac14[4+5+6] = 	frac{15}{4} = 3.75 3, so you roll. Optimal rule: keep rolling while your total is 3 or below, and stop once it reaches 4. Carryi…

A four-sided die, bust past six

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