hypergeometric distribution - expected number of draws until k successes are drawn

Question

Hypergeometric distribution describes the probability of k successes in n draws from population of N with K successes. But what if I want expected number of draws until k successes are drawn from the same population?

Answer 1

As per my comments $$P(n \text{ draws until }k\text{ successes})=P(\text{last draw is success})\cdot P(n-1 \text{ draws have }k-1\text{ successes})$$

$P(n-1 \text{ draws have }k-1\text{ successes})$ is the standard hyper geometric probability $$P(n-1 \text{ draws have }k-1\text{ successes})=\frac{\binom{K}{k-1}\binom{N-K}{n-1-k+1}}{\binom{N}{n-1}}=\frac{\binom{K}{k-1}\binom{N-K}{n-k}}{\binom{N}{n-1}}$$ and (everything is without replacement) $$P(\text{last draw is success})=\frac{K-k+1}{N-n+1}$$ $E[X]$ is a function of $k$, also considering $\binom{n}{k}=\binom{n}{n-k}$ $$\color{red}{E[X]=}\sum\limits_{n\geq k}nP(n \text{ draws until }k\text{ successes})=\sum\limits_{n\geq k}n\frac{\binom{K}{k-1}\binom{N-K}{n-k}}{\binom{N}{n-1}}\frac{K-k+1}{N-n+1}=\\ \sum\limits_{n\geq k}n\frac{\binom{K}{K-k+1}\binom{N-K}{n-k}}{\binom{N}{N-n+1}}\frac{K-k+1}{N-n+1}=\\ \binom{K}{K-k+1}(K-k+1)\sum\limits_{n\geq k}n\frac{\binom{N-K}{n-k}}{\binom{N}{N-n+1}(N-n+1)}=\\ \frac{K!}{(K-k+1)!(k-1)!}(K-k+1)\sum\limits_{n\geq k}n\frac{\binom{N-K}{n-k}}{\frac{N!}{(N-n+1)!(n-1)!}(N-n+1)}=\\ \frac{K!}{(K-k)!(k-1)!}\sum\limits_{n\geq k}n\frac{\binom{N-K}{n-k}}{\frac{N!}{(N-n)!(n-1)!}}=\\ k\frac{K!}{(K-k)!k!}\sum\limits_{n\geq k}\frac{\binom{N-K}{n-k}}{\frac{N!}{(N-n)!n!}}= \color{red}{k\binom{K}{k}\sum\limits_{n\geq k}\frac{\binom{N-K}{n-k}}{\binom{N}{n}}}$$

I have tried the following Python $3$ code

from scipy.special import comb
from decimal import *

N = 10000
K = 1000

k = 1000

res = Decimal(k * comb(K, k, exact=True))
sum = Decimal('0.0')
one = Decimal('1.0')

def combinations(b, a):
    if (b >= a):
        return comb(b, a, exact=True)
    else:
        return 0

for n in range(k, N - K + k + 1):
    sum = sum + (Decimal(combinations(N-K, n-k)) * one) / Decimal(combinations(N, n))

res = res * sum
print(res)

producing

k=1     E[X]=9.991008991008991008991008979
k=10    E[X]=99.91008991008991008991008991
k=100   E[X]=999.1008991008991008991008979
k=1000  E[X]=9991.008991008991008991008992

all within 8 minutes (altogether, so $\approx 2$ minutes per run) on my MacBook Pro. And

N = 10000 K=5000 k=3000 E[X]=5999.400119976004799040191959

took less than $2$ minutes. Larger values of $N$ take longer to compute (E.g. $N=100000$ I had to stop the program after 1 hour).

Although, it looks like empirically $$E[X]\approx k\cdot \frac{N}{K}$$