Shortly I posted a similar question to this (here: Probability for "drawing balls from urn"). Actually, the following question was included in this topic as well::

In an urn there are $N$ balls, of which $N-2$ are red and the remaining are blue. Person $A$ draws $k$ balls, so that the first $k-1$ are red and the $k$ ball is blue. Then Person $B$ draws $m$ balls $\dots$:

What's the probability for Person $A$ to draw the first blue ball after drawing $k$ red ones? Meant is, that the drawing stops as soon as the blue ball has been drawn. So the first blue ball must be the last one for Person $A$

My ideas - following the answer from Jimmy R. (= direct approach):

  1. $P(X=1) = \dfrac{2}{N}$
  2. $P(X=2) = \dfrac{N-2}{N} \cdot \dfrac{2}{N-1}$

Following this pattern I finally get

$$P(X=k) = \frac{\dbinom{N-2}{k}}{\dbinom{N}{k}} \cdot \frac{2}{N-k}$$

In fact, this solution differs only slightly from doing it via "Hypergeometric Distribution", which I did first. However, if I got it right then the Hypergeometric Distribution should be the wrong thing in this case. What remains is an explanation for Hypergeometric Distribution being not what I want, but here I stuck:

I'm citing wiki ("Hypergeometric distribution"): In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the probability of k successes in n draws, without replacement, from a finite population of size N that contains exactly K successes, wherein each draw is either a success or a failure.

Let's see what I have here:

  • $k$ successes in $n$ draws? Yes, if I see a success as "drawing blue ball" so I have $k=1$ success
  • without replacement? Yes, since I won't put any ball back.
  • a finite population of Size $N$? Yes, I have $N-2$ red ones and $2$ blue ones.
  • exactly $K$ successes? Yes, in my case $K = 2$. I have two blue ones, so two successes.

So why won't this help me?

EDIT: Possible answer: It won't help, since I'm not interested in "$1$ success among $k$ draws" but in "$1$ success at the END of $k$ draws"?


1 Answer 1


The hypergeometric distribution can actually help, but in an indirect (and tricky and therefore preferably to be avoided) way if you think of this as follows: For any $k$ in $\{1,\dots, N-1\}$ you want to calculate the probability $P(X=k)$ where $X$ is the number of trials until the first success. Now fix $k$ and consider the following random variable

  • $Y=$ number of successes (i.e. blue balls) in the first $k-1$ trials.

Now $Y$ indeed follows a hypergeometric distribution with parameters $N, K=2, n=k$ and therefore you can write $$P(X=k)=P(Y=0)\cdot\frac{2}{N-k}$$ In words: For a fixed $k$ you can express the probability of needing $k$ draws until you first success as the probability of $0$ successes in the first $k-1$ draws and success in the $k-$th draw. But be careful (the tricky part) that for each $k$ the random variable $Y$ will be different (since the parameters depend on $k$ by $n=k$ in the distribution of $Y$) so actually it is a $Y_k$.

The pattern that you found from the direct approach is correct but note that it holds for $k\ge 2$. For $k=1$ the closed formula does not work, but that is a detail, you have it correct.

Your possible answer in your edit seems valid. The point where the hypergeometric fails to solve this problem is in the first of your bullets. You actually do not have $1$ success in $n$ draws, but $1$ success at the end of $k$ draws. Moreover you are not interested in the number of successes but in the number of trials until the first success. Different things.

  • $\begingroup$ Once again, I'd like to thank you very much for your detailed help. It really gets clear by reading your explanations! Thanks! $\endgroup$
    – Vazrael
    Mar 31, 2016 at 15:47
  • $\begingroup$ @Vazrael Oh, thanks, I am glad to hear this! $\endgroup$
    – Jimmy R.
    Mar 31, 2016 at 16:51

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