Dungeons & Dragons

This week’s Riddler Classic is a probability problem about the game Dungeons & Dragons. Here it goes:

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?

Here is a detailed derivation of the relevant probabilities:
[Show Solution]

And here are the results:
[Show Solution]

One thought on “Dungeons & Dragons”

  1. I really like that using the continuous version, you were able to find the crossing point. It’s cute but curious that it’s 1/phi. Any intuition on why it should be?

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