Defeated, I come to you for help.
I'm trying to figure out the probability to roll over a target number with a certain number of dice. The dice are not added together, but the number of times you roll over the target number is of interest as the number of successes.
Take a dice pool of 3 dice as an example. I know that the total number of possibilites is 6*6*6=216.
The chance of getting 3 successes of target number 2+ is 5/6 * 5/6 * 5/6 (57%). of getting 3 successes of target number 3+ is 4/6 * 4/6 * 4/6 (29%). etc.
The change of getting 1 success of target number 2+ is 1-(1/6 * 1/6 * 1/6) = 99%. of getting 1 success of target number 3+ is 1-(2/6 * 2/6 * 2/6) = 96%. etc.
It is the in between number of successes that has me baffled. How do you calculate the probability of rolling 2 successes of a target number on 3 dice?
So you can check your math, I've done it long hand:
Target Number 2+ = 92%
Target Number 3+ = 74%
Target Number 4+ = 50%
Target Number 5+ = 26%
Target Number 6+ = 7%
Let's say you need two or more sixes on 3d6. That should be (5/216 x 3 = 15/216, + 1/216 =) 16/216, or 7.4% I'll try to explain how I did that.
Say you've got a red die, a blue die, and a yellow die. If they all come up six, you've got your two successes. The chance of the red die and the blue die both scoring successes is 1/36; the chance the yellow die will not also be a six is 5/6. (That part's important, because we don't want to count all-sixes more than once.) Five in six times one in thirty-six equals 5/216. You do the same thing for if the red and yellow dice are six and the blue is not and if the yellow and blue dice are six and the red is not. Thus, (3x5)/216 + 1/216.
It's important to remember that if only two results are important out of three dice, those results can appear in three arrangements. Three results on four dice can come up four ways, four on five in five ways, and so on. Two successes on four dice can crop up in six different ways.
For example, your dice are red, blue, yellow, and green. Your successes could be RB, RY, RG, BY, BG, or YG. Thus, if you needed two fives-or-better on four dice, you'd have (4/36 x 16/36 = 64/1296, x 6 =) 384/1296, plus (8/216 x 4/6 = 32/1296, x 3 =) 96/1296, plus 16/1296, which comes to 594/1296 = 46.0%.
I wrote a formula on a napkin one time that yields the number of color combinations given the number of successes that matter and the number of dice, but I forget what it is and where I put the napkin.
Edit: In the original post, the last calculation had 25/36 where 16/36 is now; this was obviously wrong (yielding a percent result exactly 1/6 higher than it should have been), and has been corrected.
Aha! Fantastic! Got it.