Recently I came up with an equation for determining the chance of success when rolling for tech in A&A Revised. It is as follows:
(1 – (5/6)^n) * 100
Where n is the number of dice purchased.ex. 3 dice = (1 - (5/6)3) * 100 = (1 - 125/216) * 100 = 42% (rounded down)
I have written a proof for this, but the scanned image is too large to attach per forum standards (I think the max is around 256Kb, and the file is 1.16Mb). If anyone is interested I can type the proof and post it later, but if not thats ok b/c I am lazy by nature.
even some experts around here like <cmdrjennifer>have incorrect statistical proofs in some of their posts. here’s the proof for this one (nice job <syntaxerror111>). P() is the probability function:
P( success ) = 1 - P( all fail )
and
P( all fail ) = P( trial#1 fails AND trial#2 fails AND … )
assuming rolls of a die are independent and identically distributed, then:
P( all fail ) = [ P( single fail ) ]number_of_trials
assuming a fair die:
P( single fail ) = 5/6
let n=number_of_trials, and
P( success ) = 1 - (5/6)n
if you prefer probabilities in %, multiply by 100%. i personally do not.
here’s a table:
trials P( all fail ) P( success )
0 1.00 0.00
1 0.83 0.17
2 0.69 0.31
3 0.58 0.42
4 0.48 0.52
5 0.40 0.60
6 0.33 0.67
7 0.28 0.72
8 0.23 0.77
9 0.19 0.81
10 0.16 0.84</syntaxerror111></cmdrjennifer>
