Expected Value of the Higher of Two Dice

Roll two identical dice and keep the larger. What is the average? The direct way to get there is through the distribution of the maximum, and it is short enough to follow. The largest of two dice is at most $k$ exactly when both dice are at most $k$, which happens with probability $(k/d)^{2}$. Differencing that gives the chance the maximum is exactly $k$, summing $k$ times those chances gives the expectation, and the sums of $k$ and $k^{2}$ collapse the whole thing to $(d+1)(4d-1)/(6d)$.

For six-sided dice that is $161/36$, about 4.47, against 3.5 for one die — a gain of just under one pip. On a twenty-sided die it is 13.825 against 10.5, a gain of about 3.3. As the die grows the advantage approaches one sixth of the range, and that ceiling is worth internalising because it is a good deal less than "rolling twice" feels like it should be worth. Most of the time the two dice land close together and the second one adds little or nothing; the large gains are rare, and it is the rare large gains that people remember.

The best check on this formula is not to recompute it but to add it to its partner. The average of the lower of the same two dice is $(d+1)(2d+1)/(6d)$, and the two sum to exactly $d+1$ — 13.825 and 7.175 add to 21 on a twenty-sided die, with nothing left over. That identity has to hold, and the reason is a small piece of reasoning worth having: labelling two dice as "higher" and "lower" is just relabelling the same two dice. Their total is unchanged, so their averages must add to twice the average of one die. Any pair of formulas for the maximum and minimum that fails this test has an arithmetic error in it, and no error can break one of them while leaving the sum intact.

Two extensions and a caution. The same argument generalises: with $n$ dice the maximum is at most $k$ with probability $(k/d)^{n}$, and the expectation approaches $d$ as $n$ grows, with sharply diminishing returns — the third die adds much less than the second. And the identity above generalises too, in a weaker form: the maximum and the minimum of $n$ dice still average to $d+1$ between them, though the dice in between no longer split evenly.

The caution is the usual one, and it applies with force here because a pair of dice looks like a batch rather than two separate events. The two dice are independent. One coming up low tells you nothing about the other, and a pair that has disappointed several times running has exactly this expectation on the next throw.