Comment by programjames

There's a rather simple proof for this "add one to the top, n to the bottom" posterior. Take a unit-perimeter circle, and divide it into n regions for each of the possible outcomes. Then lay out your k outcomes into their corresponding regions. You have n dividers and k outcomes for a total of n + k points. By rotational symmetry, the distance between any two points is equal in expectation. Thus, the expected size of any region is (1 + # outcomes in the region) / (n + k). So, if you were to take one more sample

E[sample in region i] = (1 + # outcomes in the region) / (n + k)

But, the indicator variable "sample in region i" is always either zero or one, so this must equal the probability!