First the Coin, Then the Dice

Question

Flip a fair coin. Heads: you roll two fair dice. Tails: you roll four fair dice. Either way you are paid the total of your rolls, in dollars. What is the fair price of this game?

Accepted Answer

Condition on the coin. Each die is worth 3.5 in expectation, so: mathbbE = tfrac12 (2 times 3.5) + tfrac12 (4 times 3.5) = tfrac12(7) + tfrac12(14) = 10.5. Faster: the number of dice N has mathbbE[N] = frac2+42 = 3, and for a sum of a random number of iid terms, mathbbEleft[sum_i=1^N X_iright] = mathbbE[N] cdot mathbbE[X] = 3 times 3.5 = 10.5, provided N is decided independently of the rolls (here the coin doesn't peek at the dice). This is the simplest instance of Wald's identity, and it is the engine behind pricing anything of the form "random count times random size": number of fills times edge per fill, number of claims times claim size, number of trades times P&L per trade. The trap to avoid in harder versions: the multiply-the-means shortcut needs the count to not depend on the outcomes themselves. A rule like "keep rolling until the dice run hot" breaks independence, and then the identity needs the full stopping-time treatment. Here the coin moves first, so the two-line answer is airtight.

First the Coin, Then the Dice

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