Ben rolls two fair six-sided dice. What is the expected value of the larger of the two numbers rolled

Sorry, Full question: Ben rolls two fair six-sided dice. What is the expected value of the larger of the two numbers rolled? Express your answer as a fraction. (If the two numbers are the same, we take that number to be the "larger" number.)

Can anyone help with this problem? Thanks so much!! :D

First calculate the probability that the larger number is n (1 <= n <= 6). This is exactly (2n-1)/36. Why? There are three possibilities:

First die is exactly n, second die is smaller -- So second die can be 1 ... n-1.
Second die is exactly n, first die is smaller -- So the first die can be 1 ... n-1.
Both die are n. This happens in only one way.

So total number of ways in which the larger number is n = 2n-1. Hence probability that the larger number is n is (2n - 1)/36.

Consequently, the expected value of the larger of the two numbers is

Sum [n = 1 to 6] (2n -1)/36 * n
= 1/36 (Sum [n = 1 to 6] 2n^2 - Sum [n = 1 to 6] n)

You can compute this either directly or using formulas for sequences. The answer is 161/36.

I calculated all 36 values and took the average, getting 4.47222, which is equivalent to the previous answer.