+0  
 
0
204
2
avatar

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
avatar
-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
avatar+183 
+3

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

25 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.