What is the probability that the sum of outcomes of three dice thrown together is 9?

There are 6^3 = 216 possible outcomes from all three dice.

For those whose sum is 9, there are five groups of possibilities in numeric order: [1 2 6], [1 3 5], [2 3 4], [1 4 4], and [3 3 3]. The first three groups have distinct values and each have 3! = 6 permutations. [1 4 4] has a repeating 4 and therfore has only 3 permutations ([1 4 4], [4 1 4], and [4 4 1]). And [3 3 3] has the same value and therefore has only 1 permutation. So the number of outcomes that sum to 9 = 3x6 + 3 + 1 = 22, and the probability for this = 22/216 = **11/108**.