Dice Game: 3 Throws

You have the option to throw a fair six-sided die up to three times. You earn the face value of the die from the final throw you decide to keep. After each throw, you may stop and walk away with the current result, or you can choose to roll again (until at most three throws). What is the expected payoff of this game?

Answer

First, recall that the game in which you can throw the die twice (and stop or continue after the first throw) has an expected value of $$4.25$$ (see Dice Game 1). Hence the optimal strategy here, after the first throw is:

\text{Keep the first roll if it is }5 \text{ or }6,\quad \text{otherwise roll again.}

Now we condition on the first outcome (by also incorporating our optimal strategy):

\mathbb{E}[\text{final payoff}] \;=\; \underbrace{\frac{4}{6}}_{\substack{\text{prob of }1,2,3,4}} \times 4.25 \;+\; \underbrace{\frac{2}{6}}_{\substack{\text{prob of }5,6}} \times 5.5.

Thus, the expected payoff of this three-throw game is $\displaystyle 4.67$ . This of course contingent upon optimal play (goes without saying).