Here's something you can try if the attacks between characters are independent.

Character A's attack has the binomial function f(kA; 3, 0.21), character B's is f(kB; 7, 0.07), and character C's is f(kC; 2, 0.35). The probability that all three characters do 'K' wounds to the target is f(kA; 3, 0.21)*f(kB; 7, 0.07)*f(kC; 2, 0.35), where kA + kB + kC = K. So I believe that if you calculate that probability for all permutations of kA/kB/kC that sum to 'K' and add them together, you'll get the probability you're looking for.

Disclaimer, I am a little rusty [grin]

You can use generating functions to compute this. The idea is that the number of injuries as a result of an attack is a random variable over the natural numbers. You can represent that using a power series. Since you have a finite number of possible injuries, you just have to deal with polynomials.

This is how the computation would go:

An attack by A is represented by the polynomial
P_A = .79+.21*x
Similarly,
P_B = .94+.06*x
P_C = .65+.35*x

The beauty of representing things in this bizarre way is that in order to combine attacks, you just have to multiply their corresponding polynomials. So the combined attack in your example can be described by:
P = P_A^3 * P_B^7 * P_C^2 = (.79+.21*x)^3 * (.94+.06*x)^7 * (.65+.35*x)^2

If you write the polynomial as P = P_0 + P_1*x + P_2*x^2 + P_3*x^3 + ..., then P_i is the probability that the number of injuries is exactly i.

[Edited by - alvaro on July 28, 2008 12:21:51 PM]

Quote:Original post by Calexus
And Alvaro, what can I say. That looks complicated. Generating functions, don't think I've ever heard of does. Maybe I have but I don't remember. Power series I do remember but not much more then the name...
Anyway it seems to be a pretty elegant way of doing it but polynomes in code, that cant be funny.

Maybe I should give it a try. Just so that I don't misunderstand anything now x in your example would represent what?

You can read about this on Wikipedia. It's not the easiest thing to digest, but I gave you enough details to actually implement it. Multiplying low-degree polynomials is not that hard.

I'll try to explain the method in simple terms. For any given attack or combination of attacks, we want to know the probability that the number of resulting injuries is 0, 1, 2, ... If we just list those probabilities as a sequence, we'll write:
Attack from A = (.79, .21, 0, 0, 0, 0, 0, ...)
Attack from B = (.94, .06, 0, 0, 0, 0, 0, ...)
Attack from C = (.65, .35, 0, 0, 0, 0, 0, ...)

Now the question is how we combine attacks. It turns out that if the attacks are independent (which you are probably assuming they are), you can combine two attacks by a formula that is identical to series multiplication.

As an example, let's combine one attack from A and one attack from B. What's the probability that you get 0 injuries as a result? Well, for that to happen, both attacks must have resulted in 0 injuries, so the probability of this happening is .79*.94 = .7426 . What's the probability that you get exactly 1 injury? There are two ways that can happen: Either A hit and B missed or A missed and B hit. The probability is then .21*.94+.79*.06 = .2448 . What's the probability that you get exactly 2 injuries? Both A and B have to hit, so it's .21*.06 = .0126 . Writing that as a sequence, it's (.7426, .2448, .0126, 0, 0, 0, ...).

Now we'll introduce the magic of polynomials. If we associate a sequence like above with a polynomial, by mapping the sequence (a0, a1, a2, a3, ...) to the polynomial a0 + a1*x + a2*x^2 + a3*x^3 + ..., we can just combine attacks by multiplying the polynomials. In the example of combining one attack from A and one from B, we have
(.79 + .21*x) * (.94 + .06*x) = .7426 + .2448*x + .0126*x^2

So `x' doesn't really mean anything. Well, you can read the term .0126*x^2 as "the probability of getting 2 injuries is .0126".

I hope my previous post makes more sense now.