Roll two dice, keep the higher, fair price?

Question

You roll two fair dice and are paid the larger of the two faces, in dollars. Solve in your head. What is the fair price of this game? Aim: under 60 seconds.

Accepted Answer

Use the tail-sum formula with P(\max \le k) = (k/6)^2: E[\max] = \sum{k=0}^{5} P(\max k) = \sum{k=0}^{5}\left(1 - \frac{k^2}{36}\right) = 6 - \frac{0+1+4+9+16+25}{36} = 6 - \frac{55}{36} = \frac{161}{36}. Fast arithmetic finish: \frac{161}{36} = 4\frac{17}{36} \approx 4.47. The brute-force route (probability \frac{2k-1}{36} for each max value k, then \sum k(2k-1) = 161) also works but juggles six products; the tail sum needs only the squares 1,4,9,16,25, which you already know. Takeaway. For maxima of independent draws, go through the CDF: P(\max \le k) factorises into a power, and the tail-s…

Roll two dice, keep the higher, fair price?

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