Let's say I have $N$ balls, of which $k$ are red and the rest are blue. The probability that I pick all the red balls without replacement in $n$ attempts is:

$\frac{\binom{k}{k}*\binom{N-k}{n-k}}{\binom{N}{n}}$ right? But what if I peek into the bag and discard one blue ball after each random draw, regardless of what was chosen?

What would the probability be then?

  • $\begingroup$ Discard a blue no matter which ball is selected? $\endgroup$ – David G. Stork Jan 17 at 1:16
  • $\begingroup$ Yes, discard a blue regardless $\endgroup$ – Eliot Jan 17 at 1:19
  • $\begingroup$ You first probability should be $\binom{k}n\binom{N-k}0\big/\binom{N}n$. $\endgroup$ – Mike Earnest Jan 17 at 1:36
  • $\begingroup$ A clearer statement of the procedure proposed in the body of your Question would improve it. While the title includes the phrase "with replacement" (as also the phrase "non-distinct ball"), it invites confusion to have critical information that appears only in the title and not in the body. In particular it is unclear from what collection you intend to "discard one blue ball after each random draw." $\endgroup$ – hardmath Jan 17 at 1:43
  • $\begingroup$ "without replacement" sorry! ugh very unfortunate typo. $\endgroup$ – Eliot Jan 17 at 1:59

The probability the first ball is red is $\frac{k}N$. The probability the second ball is red is $\frac{k-1}{N-2}$. And so on. Continuing for $n$ balls, the probability is $$ \frac{k(k-1)\cdots (k-n+1)}{N(N-2)\cdots(N-2(n-1))}=\frac{k!\cdot (N-2n)!!}{(k-n)!\cdot N!!}. $$ where $N!!=N(N-2)(N-4)\cdots$, with the product stopping at $2$ or $1$ according to whether $N$ is even or odd, respectively.


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