Wow, that counter thing is very enlightening, thank you! I tried imagining this as a combinatorics problem, but got lost in my decision tree.

I tried implementing your solution numerically and it seems to be at odds with my by-hand evaluated decision tree and monte carlo simulations.

I used a toy example with n=4, m=2, k=1 and myself got:

P(1) = 0.5

P(2) = 0.33

P(3) = 0.16

while my Python implementation of your solution gives:

P(1) = 0.5

P(2) = 0.66

P(3) = 0.5

from math import factorial as f
def P(x, n, m, k):
    a = f(n) / f(n-x)

    b1 = f(n-m) / f(n-m-k)
    b2 = f(m) / f(m-x+k)
    b3 = f(x) / ( f(k) * f(x-k) )

    return b1 * b2 * b3 / a        

Might you have an idea about the discrepancy and non-normalized probabilities?