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
Think in terms of how many combinations get a certain number.
Say your first roll is a 20, how many values of the second die get you a 20 as highest? 1-20, So there are 20 cases.
Say your second roll is a 20, how many values for the first die get you a 20 as highest? 1-20, So there are 20 cases.
So there might seem to be 20 + 20 cases in total, but both sides count the case where both the first and second dice are 20, so you have to take 1 case out.
So in 39/400 cases you roll a 20. (400 = 20 * 20, IE all the possible combinations of 2D20)
So now 19:
if you roll a 19 on the first die how many cases are there where 19 is the highest? 1-19, if they rolled a 20 with the second dice, that wouldn’t count because you would have rolled a 20 not a 19.
19+19 -1 = 37.
So basically it drops by 2 cases every time.
So
the odds of getting a 20: 39/400
19: 37/400
18: 35/400
17: 33/400
16: 31/400
15: 29/400
…..
3: 5/400
2: 3/400
1: 1/400
So just add up the number of the relevant cases, and you get your answer.
EG the odds of you getting a 17 or higher are (39+37+35+33)/400 = 144/400 = 36%.
Also just as a check to be sure, you’ll find that if you add up all the cases, you get 400/400 = 1 .
You’ll also find that if you roll with disadvantage the table is reversed. with 39/400 times you get a 1 , 37/400 you get a 2, etc.
Most of discrete probability (the kind of math that deals with dice and a lot of other things) is just a matter of counting.
EDIT: this is a general purpose approach for dealing with a wide variety of discrete math subjects.
A much more efficient and simple solution for this problem can be found here:
https://old.reddit.com/r/explainlikeimfive/comments/lftx6h/eli5_rolling_with_advantage/gmnxxtm/
For further context I was thinking in these terms because I was just figuring out the expected value for the number of attacks a Warhammer 40,000 Earthshaker cannon makes. (it 2D6 drop the lower number attacks, expected value is 4.472222)
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