Roll two ten-sided dice, keep the higher

Question

You roll two fair ten-sided dice (faces 1 through 10) and are paid the larger face, in dollars. Solve in your head. What is the fair price of this game? Aim: under 60 seconds.

Accepted Answer

Apply the tail sum with P(\max \le k) = (k/10)^2: E[\max] = \sum{k=0}^{9}\left(1 - \frac{k^2}{100}ight) = 10 - \frac{1}{100}\sum{k=1}^{9} k^2 = 10 - \frac{285}{100}. The squares 1 through 9 sum to 285, so E[\max] = 10 - 2.85 = 7.15. For comparison, a single d10 averages 5.5; taking the higher of two lifts that to 7.15. And the general formula \frac{n}{n+1} scaling appears if you rescale to [0,1]: two uniforms average 	frac23, and 	frac23 	imes 10 \approx 6.7, close but not exact because the die is discrete. Takeaway. The tail-sum method is scale-free: swap the 6 for a 10 and P(\max \le k…

Roll two ten-sided dice, keep the higher

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