If you roll two dice, then what is the expected value of the larger of the two dice?

Source:

This is from my curiosity. I was looking for a one or two step argument but couldn’t really think of any, so here’s a more lengthy solution.

Solution:

We break up the problem into different values for the larger die:

If the maximum is 1, then you had to roll two 1’s. This occurs with probability.

If the maximum is 2, then you rolled either (1,2) or (2,1) or (2,2), which gives three possibilities, each with a probability. Thus the probability that the maximum is 2 is .

For a maximum of 3, 4, 5, and 6, you can count (*see note below) that the probabilities are .

Therefore, the expected value of the maximum is:

Or .

*a fast way to count would be to realize that for a maximum of , you had to roll , which is possibilities, or , which is possibilities. The total is possible rolls.

Extensions:

The maximum of two -sided dice is quite easy to find, given what we have above. In the little note, we saw that for a maximum of , there are possible rolls. Therefore, the probability that the maximum is is . The expected value is

.

Now we have to get some intimidating algebra done. Using the summation formulas

the expected value summation turns into

where obtaining the last step involved factoring . We can confirm that this formula indeed gives our original problem, .

What if more than two dice are rolled? Most generally, what is the expected value of the maximum of -sided dice? This is a headache because we now need a way to count the number of dice roll tuples such that the maximum is is needed. There are at least two ways; the first way is bit lengthy and is given here (but it’s a nice exercise in counting and combinatorics, if you’re up for the extra reading).

The second way gives the answer immediately:

Each die shows a number that is or less; with dice, that’s possible combinations.

or less; with dice, that’s possible combinations. However, we’ve also counted the number of combinations in which none of the dice gives . The highest number in this case is . We have choices for each die, so the total combinations over dice is .

. The highest number in this case is . We have choices for each die, so the total combinations over dice is . Subtracting these two should give us the number of combinations with maximum : .

The total number of tuples for rolling -sided dice is . So the probability that the maximum of -sided dice is is . The expected value can then be written as:

and I’m quite sure there’s no general way to perform this summation, nor would I want to fathom how complex such an expression would be, so I’ll just leave the answer like this. See this article for more info on power sums.