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.)

Guest Jul 27, 2018

#1**-1 **

Here is my attempt at this:

The expected value of the TWO dice is as follows:

1 + 2 + 3 + 4 + 5 + 6 =21/6 =3.5 x 2 =7 - This is the expected value of the 2 dice.

This could happen in 3 different ways: 6+1, 5+ 2, and 4+3. Now, if we average the 3 largest numbers, we get: 4+ 5 + 6 =15/3 =5.

So, I would say that the expected value of the larger number would be = 5.

And as a fraction: 5/7.

**Note: Somebody should check this.**

Guest Jul 27, 2018

#2**+2 **

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

The first die is exactly n, the second die is smaller -- So second die can be 1 ... n-1.

The second die is exactly n, the first die is smaller -- So the first die can be 1 ... n-1.

Both die are the same. This happens in only one way.

So total number of ways in which the larger number is n = 2n-1. Hence the 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 by using formulas for sequences. The answer is 161/36.

-MathCuber

MathCuber
Jul 29, 2018