I've been working on probabilities in nWoD on and off for several months, and so far came up with the following formulae:
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P(s=0 | d=1) = 0.7 (1) the probability of getting no successes on a single die (not a luck die)
P(s=0 | d=n) = 0.7^n n>0 (2) the probability of getting no successes on n dice - generalization of eq. (1)
P(s=n | d=1) = ((0.1^n)*(0.7))+((0.1^(n-1))*(0.2)) n>0 (3) the probability of getting exactly n successes on a single die (not a luck die)
equivalent = (2.7)*(10^-n)
Es(d=n) = n*SUM [i*P(s=i | d=1)] over i=1 to inf n>=0 (4) the expected (average) value of the number of successes obtained from n dice
Another way of approaching the expected value is this:
Let x be the number of successes obtained from a single die. Let r be the reroll chance (r= .1 for '10-again').
Then x = .7*0+(.3-r)*1+(1+x)*r
Solving for x gives gives x= .3/(1-r)
Since this is the expected value for a single die, we can simply multiply by n to get:
Es(d=n) = n*(.3/(1-r)) (4a) the expected (average) value of the number of successes obtained from n dice
P(s<=n | d=1) = SUM [P(s=i | d=1)] over i=0 to n n>=0 (5) the probability of getting at most n successes on a single die (not a luck die)
P(s>=n | d=1) =1- SUM [P(s=i | d=1)] over i=0 to n-1 n>0 (6) the probability of getting at least n successes on a single die (not a luck die)
equivalent = 0.1^(n-1)*0.3 = 3*10^-n
P(s=1 | d=n) = n*P(s=1 | d=1)*[P(s=0 | d=1)]^n-1 n>=0 (7) the probability of getting exactly 1 success on n dice
P(s<=n | d=m) = SUM [P(s=i | d=m)] over i=0 to n (8) the probability of getting at most n successes on m dice
P(s>=n | d=m) = 1- SUM [P(s=i | d=m)] over i=0 to n-1 (9) the probability of getting at least n successes on m dice
Besides that, I've been using Excel to run simulations and plot nifty graphs based on them, but as far as the OP is concerned we need only look at equation (9) above. For small dice pools and successes needed, Ceasar's formula works quite well, but it becomes quite a monster when you start thinking about realistic dice pools and target numbers.
For example, a character with 4 dex and 5 firearms (plus specialization in assault rifles) firing a long burst from an AK will roll 4+5+1+3+4 = 17 dice. If she wants to down a character in one go, she'll likely need 7 or 8 successes (let's say 8 ).
The problem is that even P(s=7 | d=17) - the first term of the sum in eq. (9) - is quite difficult, because those 7 successes can come from any combination of the 17 dice. In theory, you could roll 7 successes on 1 die and zero on the other 16, or any combination of successes from a single die that sums to 7, as illustrated here:
(7,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(17 permutations)
(6,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(272 permutations)
(5,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(272 permutations)
(5,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(2040 permutations)
(4,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(272 permutations)
(4,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(4080 permutations)
(4,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(9520 permutations)
(3,3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(2040 permutations)
(3,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(2040 permutations)
(3,2,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(28560 permutations)
(3,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0) =>(30940 permutations)
(2,2,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0) =>(9520 permutations)
(2,2,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0) =>(61880 permutations)
(2,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0) =>(74256 permutations)
(1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0) =>(19448 permutations)
The probabilities for each of the above sequences can be calculated using eq. (2) and (3), and then multiplied by the number of permutations and added up.. but that's only the first term for use with eq. (9)! A similar exercise will have to be done for s=6 through 0 to find the other terms, which, while getting easier as s gets lower, is still quite a feat.
I have an Excel sheet which does the calculations up to n=5 and m=20 (including interaction with the 9-again and 8-again rules) but that may be too large to post. If you want to take a look at it, drop me a pm with your email.