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## 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?

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This post is under “Tom” and has 6 respond so far.

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1. 1. Chris Said：

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

2. 2. Karl Sharman Said：

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

3. 3. Chris Said：

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).

4. 4. Chris Said：

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.

5. 5. Karl Sharman Said：

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…

6. 6. Chris Said：

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

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