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
Real simple way to calculate this.
Don’t think of it as two chances for success, think of it as two chances for failure.
Say you have to roll an 11 or higher with one of the dice
What are the odds that you roll **UNDER** a 11?
10/20
So what are the odds **BOTH** rolls are numbers under a 11?
(10/20)*(10/20) = 100/400
So what are the odds of that not happening?
1 – (100/400) = 300/400 = 3/4
That not happening is having one or both dice succeding.
Each throw of the dice is independent so the probability of the first dice been a success is the same as for the second dice.
The correct way to look at the problem is that at least one die should be a success. That is the same as all dice being a failure.
The probability of a failure is 1- the probability of a success
Let’s call the probability of success for a single die a = (Rolls resulting in success /total possible rolls)
The probability of a failure is 1-a
The probability of two failures is (1-a)*(1-a) =(1-a)^2
The probability of at least one success will be 1 -(probability of two failures) so you get
1-(1-a)^2
If you calculate the probability to roll equal to or below for 1-20 the probability will be
1 dice Advantage
1 0,05 0,0975
2 0,10 0,1900
3 0,15 0,2775
4 0,20 0,3600
5 0,25 0,4375
6 0,30 0,5100
7 0,35 0,5775
8 0,40 0,6400
9 0,45 0,6975
10 0,50 0,7500
11 0,55 0,7975
12 0,60 0,8400
13 0,65 0,8775
14 0,70 0,9100
15 0,75 0,9375
16 0,80 0,9600
17 0,85 0,9775
18 0,90 0,9900
19 0,95 0,9975
20 1,00 1,0000
You can do the calculation for successes too if you like but it is harder to remember.
(probability first dice success)+(probability first dice failure)*)probability second dice success)
a +(1-a)*a
If you would go to 3 dice the result would be
1-(1-a)^3
vs
a +(1-a)*a +(1-a)*(1-a)*a
Then it is clear why looking at the chase of 3 failures is simpler.
Check for failure. If you need to roll a 6 to beat the check, then you fail 5/20 times, or 25% of the time.
If you roll with advantage, you square that chance of failure. (5/20)*(5/20)=fail 6.25% of the time.
If it’s a straight coin flip, then a normal roll is 1/2 and an advantaged roll would be 1/2 * 1/2, or 1/4 chance of failure.
The benefit is very different depending on what you need to roll.
Edit: math is hard
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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Wouldn’t the probability of successes would be exactly the same, since the first time does not influence the result of the second.
As long as you can’t “double succeed,” then there’s no effective difference between succeeding once and succeeding twice. Which means really, you just need to figure out the chance of failing twice in a row, which is simpler, mathematically.
For example, if a d20 roll needs a 17 or higher to succeed (or an 80% of failure, 20% chance of success), the chance of failing twice is (.8)^2=.64, or 64% chance of failure, 36% chance of success.
If you want to map out all combinations, it would be
(success)(success) = (4/20)*(4/20)
(success)(fail) = (4/20)*(16/20)
(fail)(success) = (16/20)*(4/20)
(fail)(fail) = (16/20)*(16/20)
For a 256/400 chance of failure, and a (16/400) + (64/400) + (64/400) = 144/400 chance of success, since single and double successes are identical for our purposes.
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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