You win the square of two dice summed

Question

A game pays you S^2 dollars, where S = X + Y is the total of two fair six-sided dice. 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 E[S^2]. Rather than squaring all 36 sums, use the identity E[S^2] = \mathrm{Var}(S) + (E[S])^2. For two independent dice, E[S] = 7 and, since variances of independent variables add, \mathrm{Var}(S) = 2 	imes \mathrm{Var}(X) = 2 	imes 	frac{35}{12} = 	frac{70}{12} \approx 5.83. So E[S^2] = \frac{70}{12} + 7^2 = 5.83 + 49 = 54.83, \qquad 	ext{fair price} \approx \54.83. The classic trap is (E[S])^2 = 49. That is wrong by exactly the variance term, \approx 5.83: squaring is convex, so E[S^2] (E[S])^2, and the gap is \mathrm{Var}(S). Notice how variance pooling helped you: …

You win the square of two dice summed

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