Pay to play, you win the square of a die roll

Question

A game pays you X^2 dollars, where X is one roll of a fair six-sided die. What is the fair (zero-EV) price to play? Would you actually pay that much, and how would you make a market on the game?

Accepted Answer

The fair price is the expected payout: 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. A classic trap is to compute (E[X])^2 = 3.5^2 = 12.25. That's wrong by Jensen's inequality: squaring is convex, so E[X^2] (E[X])^2; the gap is exactly the variance, \mathrm{Var}(X) = 	frac{91}{6} - 	frac{49}{4} = 	frac{35}{12} \approx 2.92. Convex payoffs love volatility, the same reason option prices increase with vol. Would you pay 15.17? At the fair price the game is zero EV with real variance (payoffs from 1 to 36, standard deviation \sqr…

Pay to play, you win the square of a die roll

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