The likelihood X of rolling precisely okay successes on n cube, the place every die can succeed with likelihood p is given by the binomial distribution:
X(okay,n,p) = nCk * p^okay * (1-p)^(n-k)
the place nCk is “n select okay” is n! / (okay! * (n-k)!)
Since there are three sword sides (melee successes) on the die, p for melee hits is 1/2. Two bows (ranged successes) means p for ranged hits is 1/3. And two shields means p for protection is 1/3.
So then it is only a matter of figuring the product of every attacking X(okay,n,p) worth in opposition to the decrease protection successes.
With one melee assault die and nil protection cube, the one technique to succeed is with successful (success):
X(1,1,1/2) = 0.5
With 1 melee die vs 1 protection die, you verify for 1 vs. 0:
X(1,1,1/2) * X(0,1,1/3) = 0.5 * 0.67 = 0.33
With 2 melee cube vs 1 protection die, you verify for all of the methods to attain a number of profitable hits: 1 vs. 0, 2 vs. 0, and a pair of vs.1, and add them up.
X(1,2,1/2)*X(0,1,1/3) + X(2,2,1/2)*X(0,1,1/3) + X(2,2,1/2)*X(1,1,1/3)
And so forth.
To wrap all of it up in on summation equation, use a operate to symbolize the variety of successes anticipated on n cube:
X ~ B(n,p)
so E[X] = n*p
After which work out the likelihood of the attacker’s rating X being larger than the defender’s rating Y
P[X>Y]
which is identical as
P[X-Y > 0]
which is the topic answered over on this query on math change.
