Question

The Die is Cast: Imagine a fair die with six faces numbered 1,2,3,4,5,6. You will get to spin the die 3 times. After the first spin, you can choose to be paid dollars equal to the number shown on the die. If you choose to accept the money, the game ends. If you choose to continue, the die is spun again, and again you can accept dollar payment equal to the number showing on the die, at which point the game ends. If you refuse the money on both the first two spins, then you can spin the die a third time, and receive dollars equal in number to the number shown on the die. Then the game ends. What is your optimal strategy to get the highest expected payoff? How much do you expect to be paid?

Answer 1

The expected payoff for any roll of the die is

$\frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$

Thus, a sensible strategy to begin with would be to quit the game as soon as the die rolls $\geq 4$ . If we got to spin the die only twice, this strategy would win us an expected amount of

$\frac{1}{6} \cdot \left (4 \right ) + \frac{1}{6} \cdot \left (5 \right ) + \frac{1}{6} \cdot \left (6 \right ) + \frac{3}{6} \cdot (3.5) = 4.25$

In the LHS of the equation above, the first three values are the pay-offs if we roll $\geq 4$ , while the fourth value is the expected pay-off should we choose to continue to the second roll.

In a game where we get to spin the die thrice, if we choose to continue playing after the first roll, it essentially reduces to a game of two rolls like above. This, as calculated above, has an expected pay-off of .

Thus, the optimal strategy would be to stop playing if we get $\geq 5$ in the first roll or $\geq 4$ in the second roll.

The expected amount we would be paid with this strategy is

$\frac{1}{6} \cdot \left (5 \right ) + \frac{1}{6} \cdot \left (6 \right ) + \frac{4}{6} \cdot (4.25) \approx 4.66$