I recently watched Matt Parker’s video “The Unexpected Logic Behind Rolling Multiple Dice and Picking the Highest” (which—for two dice—players of D&D 5e might know as “rolling with advantage”). In this video he calculates the expectation value for rolling 2, 3, and 4 dice and picking the highest value. True to the title, I found the answer quite unexpected!
Once I’d seen the result, though, it seemed to me that there was a geometric way of arriving at it that’s more straightforward than the one Matt used. It also points toward a proof of the general case, which is something Matt leaves as an exercise to the viewer.
UPDATE: My friend Will has pointed out an error in this work. He’s writing up a post with the correct derivation, which I’ll link to here when he posts it. For now, let me explain my mistake.
My basic argument relied on the fact that, for any n-sided die, the expectation value is the midpoint of the number line interval from 1 to n. This is true.
I then said that, when rolling two n-sided dice, you could fix the higher value v_{high} in place and treat the lower value as the roll of a v_{high}-sided die. Then the expectation value for v_{low} is halfway between 1 and v_{high}. This is also true.
I further argued that you could reverse this, holding v_{low} fixed, and determining that the expectation value for v_{high} is halfway between v_{low} and n. This, too, is true.
I then said that you could use these facts together to conclude that the expectation values must partition the interval from 1 to n exactly into thirds. This step is false, or at least doesn’t follow from the previous two facts. The expectation values are not the same as individual rolls of the dice, and I can’t necessarily treat them as rolled values. This last step swaps out the specific values I was holding constant for the expectation values of those dice. There’s no reason to believe I can legitimately do that, so the whole of the rest of the argument falls apart.
(There was also a trivial condition with 2-sided dice not fitting one of the equations from Matt’s video that I was working off of, which I’m sure Will will include in his writeup.)
UPDATE 2: Here’s Will’s post, wherein he derives the general case formula for the expected value of rolling m dice with n sides each with advantage to be n - \sum\limits_{i=1}^{n-1}(\frac{i}{n})^m. More commentary on this, and how expectation values partition the interval from 1 to n, when I have time to play around with it more.