I am not much of a statistician and I have racked my brains around this problem without success: There are w white balls and b black balls in an urn. I want to get the distribution of the random variable S, the number of white balls obtained after n draws. The tricky part is, when a black ball is drawn, it is replaced in the urn, but when a white ball is drawn it is not replaced.

I have tried to start from the demonstrations of the binomial and hypergeometric distributions, but I get a huge expression that I can't simplify and that is increasingly complex as you increase the number of draws n.

It seems to be a rather simple problem but I couldn't find a solution on the internet. Is there a solution to this problem?

Thank you very much!

  • $\begingroup$ Hmmm interesting. I don't know if there is a nice answer. $\endgroup$ – Lost1 Jan 5 '14 at 18:18
  • $\begingroup$ Did you try applying coupon collector's problem here? $\endgroup$ – Alex Jan 5 '14 at 18:40
  • $\begingroup$ Yes, but as I understand it the coupons are always replaced in the coupon collector problem, whether you already drew them or not, right? $\endgroup$ – jblugagne Jan 5 '14 at 18:44
  • $\begingroup$ You are probably right. Finding the $\textit{mean time}$ until $r$ white balls are sampled is much easier. $\endgroup$ – Alex Jan 5 '14 at 18:51
  • $\begingroup$ Sorry, I am not sure I understand what you mean. Do you suggest using the geometric distribution to get the expected number of black balls before I sample r white balls? Would it give me the distribution? $\endgroup$ – jblugagne Jan 5 '14 at 18:56

The number of samples until each success follows Geometric distribution with parameter $p_k$, e.g. $p_1 = \frac{w}{w+b}, \ p_2 = \frac{w-1}{w+b-1}$, etc. The mean time until the success is $\frac{1}{p_k}$, i.e. $\mathbf{E}X_k=\frac{1}{p_k}=\frac{w+b-k}{w-k}$. Mean time until $r^{\text{th}}$ success is $$ \mathbf{E}X=\sum_{k=1}^{r}\mathbf{E}X_k $$

which you can find using Harmonic sum approximation with logarithms, i.e. $\sum_{k=1}^{n}\frac{1}{k}=O(\log n)$.

  • $\begingroup$ Indeed, working out expectation is not hard, can do variance similarly. This does not really answer the question which distribution it is, but i don't think there is a way working out pmf of the distribution. The only way to characterise it seems to be mgf or pgf. $\endgroup$ – Lost1 Jan 5 '14 at 22:18
  • $\begingroup$ Yes, the way I see it you have to account for the possibility of sampling a white ball in any time between $1$ and $n$ for $P(S_n=r)$ $\endgroup$ – Alex Jan 6 '14 at 1:00

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