Expected Value of the Second-Highest of Several Dice Rolls

Worked Solution

How to Think About It: Use the tail-sum formula for expectations of nonnegative integer variables: $E[Y] = \sum_{k \ge 1} P(Y \ge k)$. Let $Y$ be the second-highest of $n$ rolls. The event $\{Y \ge k\}$ means at least two of the rolls are $\ge k$. Compute that via the complement (zero or one roll is $\ge k$).

Quick Estimate: With 4 rolls the second-highest typically sits a bit below the maximum. The max of 4 dice averages about 5.2, so the second-highest should be around 4.5.

Formal Solution: Let $p_k = P(\text{one die} \ge k) = (7 - k)/6$ for $k = 1, \ldots, 6$. With $n = 4$ independent rolls, the number that are $\ge k$ is Binomial$(4, p_k)$. The second-highest is $\ge k$ iff at least 2 dice are $\ge k$:

$P(Y \ge k) = 1 - (1 - p_k)^4 - 4 p_k (1 - p_k)^3.$

Then $E[Y] = \sum_{k=1}^{6} P(Y \ge k)$. Evaluating $p_k = 6/6, 5/6, 4/6, 3/6, 2/6, 1/6$ for $k=1,\ldots,6$ and summing the tail terms gives

$E[Y] \approx 1 + 0.984 + 0.890 + 0.688 + 0.407 + 0.132 \approx 4.10.$

Answer: For $n = 4$ rolls, $E[\text{second-highest}] \approx 4.10$.