I was wondering, what’s the probability that you’ll land AT LEAST some number of hits in d20. Turns out, it’s a lot easier to ask the question than to answer it. Start with at least 1 hit…easy enough:
P(at least 1 hit) = 1- P(all misses) = 1 – ∏i(1 – Pi(hit))
That was so easy! Plus, it makes it super duper easy to calculate P(at least 2 hits):
P(at least 2 hits) = P(1 hit) × (1 – P(all the rest misses)) = ∑iP(hit)×[1 - ∏j≠i(1-Pj(hit))]
And so on, until you exhaust all of your BAB-(5 per swing), at which point you get
P(all hit) = ∏iPi(hit)
Which is again easy. So, computing that stuff is very time consuming, but straightforward. It’s not as bad as it could be, though, because your max hit chance is P(hit) = .95, and min is P(hit) = .05, and at -0.25 per swing, you’ll only have at most 6 distinct Pi(hit), including several easy swings at P(hit) = .95 and hard swings at P(hit) = .05, at one each of the other 4. So if you get many many attacks per round, many of the terms in those sums will be identical. Maybe I’ll write a program to compute these things, maybe I’ll work out a rough approximation that’ll be easier to compute. /shrug
