Fair game paradox

FreeProbabilityEasyJane StreetSIGFive Rings

A fair coin is flipped repeatedly until the first heads appears. The payout is 2n2^n dollars if the first heads occurs on the nnth toss. Determine the fair value of this game. If the expected value is infinite, output −1-1.

Solution

Let NN be the number of coin flips until the first heads appears. Since the coin is fair, NN follows a geometric distribution with success probability 1/21/2: P(N=n)=(1/2)nP(N = n) = (1/2)^n for n=1,2,3,…n = 1, 2, 3, \dots (the first n−1n-1 flips are tails and the nnth flip is heads). The payout when N=nN = n is 2n2^n dollars. The expected value of the payout is

E[payout]=∑n=1∞2n⋅P(N=n)=∑n=1∞2n⋅(12)n=∑n=1∞1=∞.E[\text{payout}] = \sum_{n=1}^{\infty} 2^n \cdot P(N = n) = \sum_{n=1}^{\infty} 2^n \cdot \left(\frac{1}{2}\right)^n = \sum_{n=1}^{\infty} 1 = \infty.

Since the expected value is infinite, the game does not have a finite fair value. Per the problem instructions, output −1-1.

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