# Stat: Probability to have one element of a combination identical to one element of another combination

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?

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.

• 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($$P = 2*10^2$$)
– J4N
Aug 17, 2015 at 6:59
• @J4N: Use Wolfram|Alpha. Aug 17, 2015 at 7:01
• Thank you very much, I admit you lost me on the first line, but the solution is containing some things I was expecting. :)
– J4N
Aug 17, 2015 at 7:52
• @J4N: Yes, I wasn't sure how much to explain; if you want to understand the derivation, feel free to ask :-) Aug 17, 2015 at 7:54