2026-04-04

Problem — 2026-04-04

Medium

Mediumprobability

You roll two fair six-sided dice. What is the expected value of the larger of the two results?

Express as a fraction or decimal

Show solution

Let

M = \max(X,Y)

. Then

P(M = k) = P(\text{both} \le k) - P(\text{both} \le k-1) = \frac{k^2 - (k-1)^2}{36} = \frac{2k-1}{36}

. So

E[M] = \sum_{k=1}^{6} k \cdot \frac{2k-1}{36} = \frac{1+6+15+28+45+66}{36} = \frac{161}{36}