Paid the square of your die roll

Question

You roll a fair six-sided die and are paid the square of its face value in dollars (so a $4$ pays $16). Before being paid, you may reroll once; if you do, you must accept the second roll.

What is the fair value of this game under optimal play?

Accepted Answer

The continuation value is the expected payout of a fresh roll, and because the payout is X^2 you need the mean of the squares: E[X^2] = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6} \approx 15.17. Keep the first roll only if its square beats 15.17. Since 3^2 = 9 < 15.17 but 4^2 = 16 15.17, keep 4, 5, 6 and reroll on 1, 2, 3. Value the game at the start: - With probability 	frac{3}{6}, you keep 4, 5, 6, with average payout \dfrac{16 + 25 + 36}{3} = \dfrac{77}{3} \approx 25.67. - With probability 	frac{3}{6}, you reroll, worth \dfrac{91}{6} \approx 15.1…

Paid the square of your die roll

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