1. Only in Ruby 2.4, and (at time of writing) Codewars uses Ruby 2.3.0
2. Yep, this method will never be exact. The distribution will always sum to a power of two, which obviously can't ever be evenly divisible by 3. I probably can't really count this as a solution because it will only pass a not-very-stringent randomness test.

Very cool idea! Thanks.

However, I have a couple of comments:

1. reduce(:+) can be replaced by sum.
2. I believe the resulting probability distribution is not precisely uniform - 1, 2 and 3 are produced in proportions 341 : 341 : 340. This can be shown by noticing that a sum of Bernoulli random variables follows a Binomial distribution. With your array size of 10 you therefore produce sums
   10, 11, 12, ... 20
   with (Binomial) proportions
   1 : 10, : 45 : 120 : 210 : 252 : 210 : 120 : 45 : 10 : 1
   therefore (%3 + 1) values 1, 2, 3 occur with proportions
   1+120+210+10 : 10+210+120+1 : 45+252+45
   i.e.
   341 : 341 : 342

The deviation from uniformity is small, and gets smaller with larger array size.

Didn't think about using abs! Nice!