For a business application, I currently have to provide the probability we are going to have an issue in one application.

  • The combination is composed of N unique elements.
  • Each element is randomly choosed(and randomly choosed until not already contained in the current combination)
  • Each element can be choosed amongst P possibilities
  • I have C combinations

With this given combination, I've to compute the odd to have one element of one combination contained in any other combination.

My stats lessons are a little bit old so I'm pretty sure I'm not taking this the right way.

Currently I was thinking that I've N/P chance to have a specific element. So would I be correct to think that I've (C*N)/P chance to have a common element?


1 Answer 1


The number of ways to choose $C$ sets of $N$ elements from $P$ elements such that all elements are different across combinations is

$$ \binom P{\underbrace{N,\ldots,N}_{C\text{ times}},P-CN}=\frac{P!}{N!^C(P-CN)!}\;. $$

The total number of ways to choose $C$ sets of $N$ elements from $P$ elements is

$$ \binom PN^C=\frac{P!^C}{N!^C(P-N)!^C}\;. $$

Thus the probability of not having a common element is

$$ \frac{(P-N)!^C}{P!^{C-1}(P-CN)!}\;. $$

The probability of having a common element is one minus that.

  • $\begingroup$ Thank you for your answer, I will try that. For now I'm quite stuck with my calculator who doesn't like such high numbers($$P = 2*10^2$$) $\endgroup$
    – J4N
    Aug 17, 2015 at 6:59
  • $\begingroup$ @J4N: Use Wolfram|Alpha. $\endgroup$
    – joriki
    Aug 17, 2015 at 7:01
  • $\begingroup$ Thank you very much, I admit you lost me on the first line, but the solution is containing some things I was expecting. :) $\endgroup$
    – J4N
    Aug 17, 2015 at 7:52
  • $\begingroup$ @J4N: Yes, I wasn't sure how much to explain; if you want to understand the derivation, feel free to ask :-) $\endgroup$
    – joriki
    Aug 17, 2015 at 7:54

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