# Optimal stopping with dice rolls

Suppose you rolled a 6 sided fair dice and earn how much you roll. If you are unsatisfied, you may roll again (only once though). What is the expected pay-off?

So for the original game with no re-rolls, the payoff is 3.5, so it makes sense to re-roll when you get less than 3.5. Then the expected payoff $E$ satisfies $$E + E + E + 4 + 5 + 6 = 6E,$$ so $3E + 15 = 6E$, or $-3E = -15$, or $E = 5$.

Does this mean my stopping rule is not optimal? Since $E > 4$, we should roll again at 4 right?

• Not sure I am following. using your stopping rule (which is indeed optimal) you expect to get $3.5$ half the time (as you just take what you get on roll $2$) and you get $4,5,6$ each with probability $\frac 16$. Hence $E=\frac 12(3.5)+\frac 16(4+5+6)=4.25$. Swapping out the $4$ in exchange for an expected $3.5$ is suboptimal. – lulu Oct 9 '15 at 18:42
• No. Your expected payoff for the second roll (should you take it) is always $3.5$. Your expected payoff if you don't take the second roll is always what you first rolled. So you roll again on $1,2,3$. – mjqxxxx Oct 9 '15 at 18:43
• If you rolled again at 4, you'd replace a term of average value 4, with one of average value 3.5. Just because it's over 4 doesn't mean you should reroll at 4. – Jason Carr Oct 10 '15 at 5:15

Your equation for $E$ assumes that after rolling again, you're in the initial situation again, i.e. that you're still allowed to re-roll. But you're not, so where you have $E+E+E$ you should have $3.5+3.5+3.5$ for the standard single rolls. Thus the equation is
$$3.5+3.5+3.5+4+5+6=6E\;,$$
so $E=4.25$, as lulu wrote. The optimal strategy is indeed to re-roll if the standard single roll is expected to yield more than you rolled, which is the case if you rolled $1$, $2$ or $3$.