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# Math help - rolling three or more 15+ scores on 4d6 drop lowest arrange by taste

edited February 2012
I'm trying to work out some probabilities connected with class design in TSR-era Dungeons & Dragons, where several classes have stat requirements for entry.

Using 3d6 in order straight down the line, it's trivial to work out the probabilities of getting (say) 15+, 15+, 15+, x x x in exactly that order, because each roll is independent of the next. Using 4d6 straight down the line is equally trivial, just with slightly different rates of getting a 15+.

But when you're looking at "arranged to taste," you're no longer looking at any particular roll in isolation. Instead, you're looking at whether the set of six rolls contains three scores of 15+, doesn't matter where in the set.

Fifteen years ago, when I was in college, I was probably capable of working this out, either from first principles or by writing a program to brute-force it for me. Either feat now is, alas, beyond my feebleminded middle age.

Here's my guess on this:

To calculate the odds of getting at least (multiple) scores of (stat) out of (array size) rolls, it's equal to 100% minus the odds of getting less than (multiple) of that score. So it's 100% minus the chance of getting zero scores of 15+, minus the chance of getting exactly one score of 15+, minus the chance of getting exactly two scores of 15+.

God help me on writing this formula out in plain text:

1 - (Summation from k = 0 to k = (multiple -1)) (combination (array size) choose k) * (odds of getting stat or greater)^k * (odds of getting less than stat)^(array size minus k))

If....
(array size) = 6
(multiple) = 3
(odds of getting 15+ on 4d6 drop low) = 0.2395

Then I end up with 14.13%. But I have no idea if that's correct. Little help?