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
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.
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