A real doozy - which situation produces the highest expected roll?

Question

The fifth edition of Dungeons & Dragons introduced a system of “advantage and disadvantage.” When you roll a die “with advantage,” you roll the die twice and keep the higher result. Rolling “with disadvantage” is similar, except you keep the lower result instead. The rules further specify that when a player rolls with both advantage and disadvantage, they cancel out, and the player rolls a single die. Yawn!

There are two other, more mathematically interesting ways that advantage and disadvantage could be combined. First, you could have “advantage of disadvantage,” meaning you roll twice with disadvantage and then keep the higher result. Or, you could have “disadvantage of advantage,” meaning you roll twice with advantage and then keep the lower result. With a fair 20-sided die, which situation produces the highest expected roll: advantage of disadvantage, disadvantage of advantage or rolling a single die?

Extra Credit: Instead of maximizing your expected roll, suppose you need to roll N or better with your 20-sided die. For each value of N, is it better to use advantage of disadvantage, disadvantage of advantage or rolling a single die?

Answer 1

It is much easier to derive this result computationally than to attempt a closed-form solution.

You just have 4 nested loops of the values 1 to 20, then calculate the advantage of disadvantage and disadvantage of advantage: if the variables for looping for a, b, c, and d, advantage of disadvantage would be

max(min(a,b),min(c,d)); disadvantage of advantage would be min(max(a,b),max(c,d))

you set up counters for each of the possible outcomes 1 to 20. This will allow you to answer both parts of the problem.

There are various ways to generate the loops in Excel. I looped the values 0 to 19 instead and just added 1 to my answer. I did the looping by taking the values 0 to 20^4-1 = 159999, and getting the base 20 value; you can get this by taking the value mod 20 (this generates a), subtracting a from value to get new value, dividing by 20, taking the mod 20 of new value(this generates b); repeat last 3 steps two more times to generate c and d (the last time, you don't need to take mod as your value will be between 0 and 19.

Minmax (disadvantage of advantages) has an expected value of 11.1666625. Maxmin(advantage of disadvantages) has an expected value of 9.8333375. Note the pattern of the minmax and maxmin results: P(x) for minmax = P(21-x) for maxmin. Of course, the two expected values add to 21.

The expected value for one roll is 10.5 (Either add values 1 to 20 and divide by 20 or note 1 and 20, 2 and 19, ..., 10 and 11 are equally likely, and each average to 10)

The final question asks which strategy is better for >= N. For N = 1, it doesn't matter; for N > 1, minmax is always better than maxmin. For N 14 or above, pick the single roll; for N 2 to 13, pick the minmax (disadvantage of advantages). (Calculate the cumulative probabilities starting at the bottom; for the single roll, each value has probability .05).

I could not copy the table from Excel. I am giving you the counts for N = 1 to 20 for minmax and maxmin, You can easily calculate everything from this. (divide by 160000 for probability; E(x) is the sum of x * P(x)

Minmax: 1: 799, 2: 2385, 3: 3935, 4: 5425, 5: 6831, 6: 8129, 7: 9295, 8: 10305, 9: 11135, 10: 11761, 11: 12159, 12: 12305, 13: 12175, 14: 11745, 15: 10991, 16: 9889, 17: 8415, 18: 6545, 19: 4255, 20: 1521

Maxmin: 1: 1521, 2: 4255, 3: 6545, 4: 8415, 5: 9889, 6: 10991, 7: 11745, 8: 12175, 9: 12305, 10: 12159, 11: 11761, 12: 11135, 13: 10305, 14: 9295, 15: 8129, 16: 6831, 17: 5425, 18: 3935, 19: 2385, 20: 799