>>43333233>>43332224>>43331892here's how probability works (prepare yourself)
the way dice probabilities work (i.e. each roll is independent, there are a finite number, each die has same prob of getting a hit) indicates that die roll probabilities will follow a binomial distribution
the probability of getting a hit is 1/3 (hit =5 or 6 (2 options out of 6 possibilities))
the probability of getting EXACTLY a certain number of hits is calculated by the equation:
binomial(n,k,p) = ( n! / (k! * (n-k)! ) ) * p^k * (1-p) ^ (n-k)
where:
p = prob. of a hit (1/3)
n = number of dice being rolled
k = number of hits to be calculated
! = factorial
here's an example: prob. of getting exactly 2 hits with 3 dice
3! / (2! * (3 - 1)! ) * (1/3)^2 * (1 - 1/3)^(3-2) = 3 * (1/3)^2*(2/3) = 0.222222 = 22.22222%
now remember, this is for calculating the prob. of EXACTLY k hits occur with n dice. For the probability that AT LEAST k hits occur with n dice, you need to sum all the binomial probabilities from k to n (i.e if you have 3 dice and want the prob of AT LEAST 2 hits, you sum the equations with k=2 and k=3)
sum of binomial(3,x,p) for (x=2,3) = 0.22222222 + ( 1 ) * (1/3)^3 * (2/3)^0 = 0.2222222 + 0.037037
= .259259 = 25.93%
which can be confirmed on the table here
>>43331892 for 3 dice and at least 2 hits
but this can get tedious for even small n's and k's recalculating all of the time, so often times people just calculate an entire table of k values for binomial(n,k,p) and using the logic that the (prob. of x>=k = 1 - prob. of x<k )
so instead they sum binomial(n,x,p) from x=0 to x=(k-1) and subtract it from 1 (the total probability function can't exceed 1 or rather 100%)
so for the case of 3 dice and at least 2 hits:
1 - binomial(3,x,(1/3)) for x = 0,1 = 1 - (3! / (0! * (3-0)!) * (1/3)^0 * (2/3)^3 - (3! / (1! * (3-1)!)) * (1/3)^1 * (2/3)^2 = 1 - (2/3)^3 - 3*(1/3)*(2/3)^2 = .259259 = 25.93%
i'll post python code for these equations for anyone here to use
