## A Roll of The Dice

Posted by Karl Sharman on August 14, 2014 – 5:12 am

Suppose n fair 6-sided dice are rolled simultaneously. What is the expected value of the score on the highest valued die?

Posted by Karl Sharman on August 14, 2014 – 5:12 am

August 14th, 2014 at 8:16 am

Hi Karl. If there is a tie for the highest die, do we ignore that trial?

August 14th, 2014 at 3:25 pm

Hi Chris, not going to matter so much….after all, who knows what numbers are on the die?

August 14th, 2014 at 3:58 pm

Hi Karl, ouch I’d be quite surprised if it didn’t affect the solution. Maybe I’ll try both cases (not that I’ve worked out how to do it yet).

August 15th, 2014 at 9:22 am

My son sent me the following:

The expected value of the highest roll of n dice

(6^(n+1)-5^n-4^n-3^n-2^n-1)/6^n

I’m not sure if he’s answered the right question. I haven’t started doing it myself.

August 15th, 2014 at 9:29 am

First we find pn(k), the probability that the highest score is k. Thus, there are kn ways in which n dice can each show k or less. For the highest score to equal k, we must subtract those cases for which each die shows less than k => these number (k − 1)^n.

So, k is the highest score in k^n − (k − 1)^n cases out of 6^n.

Now, I can no longer read what is written on my board, so I shall post this, and carry on tomorrow if they haven’t cleaned my board, or if they have, I have this…

August 15th, 2014 at 6:33 pm

Hi Karl, don’t panic, help is at hand. Thanks for this problem, it’s a quite sweet self-contained exercise.

The following is substantially a copy/paste from my son’s Skype call.

The number of ways you can roll n dice with rolls from 1 to x is x^n. However, that includes ways that don’t have the highest roll as x. So you subtract (x-1)^n to find how many rolls have the highest as x. So that means there are (1^n-0^n) ways to have the highest as 1, (2^n-1^n) ways to have the highest as 2, …, (6^n-5^n) ways to have the highest as 6.

To find the expected value, take the weighted average of the set of rolls =>

(1 + 2(2^n-1) + 3(3^n-2^n) + … + 6(6^n-5^n))/6^n

= (6^(n+1) – 5^n – 4^n – … – 1)/6^n