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Let:

$\phantom{2}N = \{ 1 .. n \}$

$\phantom{2}J \subset N$, where $\vert J \vert = j$

$\phantom{2}K \subset N$, where $\vert K \vert = k$ and $k \le j$

$\phantom{2}L$ any $\subset\ K$, where $\vert L \vert = l$

For example: $n = 45$, $j = 7$, $k = 6$, $l = 5$

What is the probability that any $L \subset J$?

N.B. the 'any' is important, implying that J K are specific instances, but L is all instances (I'm not sure how to notate this, advice welcome).

So the case $k = j$ is well known:

$\phantom{2}$ $P = \binom{j}{l}.\binom{n-j}{j-l}/\binom{n}{j}$

But I've had no luck finding any results or discussion on the more general case where $k \le j$.

For context this is basically a lottery problem. j is the number of numbers picked by an entrant, $l=k$ is the case of winning the main prize, and $l<k$ are the cases for other lesser prizes. Many lotteries limit the number of picks to the number drawn, but there are those which will allow a greater number of picks.

Asaf Karagila
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CmdrVim
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  • Intersecting or being a subset? – Henry Feb 11 '21 at 11:10
  • yes subset, subject is wrong sorry – CmdrVim Feb 11 '21 at 11:16
  • So, there are $n$ possible numbers, you put $j$ on your ticket, the lottery selects $k$ numbers, and you want to find the probability that there are at least $\ell$ numbers in common between your ticket and the lottery selection? So, if $n=100,j=10, k = 8, l = 5$, then want the probability of getting $5,6,7$, or $8$ numbers in common? Or do you want the probability of getting exactly $\ell=5$ numbers only? – Mike Earnest Feb 11 '21 at 15:26
  • I agree with Henry's answer. His answer is for exactly $l$ numbers. For an explanation of it see this. – Fabius Wiesner Feb 11 '21 at 17:58

1 Answers1

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Doing this as a lottery problem you get a hypergeometric probability that with $n$ possibilities you get an overlap size exactly $l$ between a random set size $j$ and a random set size $k$ $($in fact one of the random sets can be fixed and it does not change the probability, but both cannot as this would make the answer $0$ or $1)$. This is $$\frac{{k\choose l}{n-k\choose j-l}}{{n\choose j}}$$ and with $n = 45$, $j = 7$, $k = 6$, $l = 5$ you get $\frac{6\times 741}{45379620}$ which is just under $10^{-4}$

Henry
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