I was trying to figure out how to quantify the benefits of rolling with Advantage in D&D and I’m coming up short.
For anyone unfamiliar with the idea, it’s simply rolling a 20-sided die twice and taking the more favorable roll.
Correct me if I’m wrong, but with a single roll my probability of success would be:
Rolls resulting in success ÷ total possible rolls.
My question, and I apologize for the rambling, is how would I figure out the aggregate chance of success with a second roll?
In: Mathematics
I find that making a table helps with problems like these. I’m not sure if there’s a way to think of it abstractly, and 20×20 possible results isn’t too many to calculate.
I just whipped it up in Excel; you can probably do it too. Put 1 through 20 on both axes, these represent the two dice rolls. Then in every cell on the table, just pick which value is higher; that will give you all 400 possible results of 2 D20s. Using my table, I calculate that the average value of the advantage roll is 13.825, while the average value of only one roll is 10.5.
Think of it this way: when you roll with advantage, how many ways are there to roll 1? There’s only one way: 1-1. But how many ways can you roll 20? There are 39: 19 ways you can roll 20 with the first die (20-1, 20-2, 20-3, etc) and 19 ways you can roll 20 second, and one way you can roll 20 on both. This doesn’t get you all the probabilities but it illustrates how much easier it is to roll higher numbers.
Just for shits and giggles, I applied this same logic on a disadvantage roll (i.e. roll two D20s, pick the lower value) and the average roll is just 7.175. Same logic.
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