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Keep the best $2$

Expected valueHard

$4$ fair $6$ -sided dice are rolled. You are paid the sum of the $2$ highest results, in dollars. What is the expected payout?

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Hint

Write the sum of the top $2$ as a count of thresholds: the sum of min( $2$ , N_m) over m, where N_m counts dice at least m.

Solution

The sum of the $2$ highest dice equals the sum over $m = 1, \dots, 6$ of $min (2, N_{m})$ , where $N_{m}$ is the number of dice showing at least $m$ , and $E [min (2, N)] = P (N \geq 1) + P (N \geq 2)$ .
With success chance $q = \frac{7 - m}{6}$ and $N_{m}$ binomial on $4$ dice, the contributions over $1296$ are $2592$ , $2570$ , $2432$ , $2106$ , $1568$ and $842$ for $m = 1, \dots, 6$ .
Their sum is $2592 + 2570 + 2432 + 2106 + 1568 + 842 = 12110$ .
The expected payout is $\frac{12110}{1296} = \frac{6055}{648}$ .

Answer: $\frac{6055}{648}$