Wikipedia:Reference desk/Archives/Mathematics/2017 April 11

= April 11 =

Probability of having a subset
Let A be a set with $$|A|=n$$. Let B be a set such that $$B\subseteq 2^A, |B|=k$$. What's the probability of the event $$\exists b_1, b_2 \in B : b_1\subseteq b_2$$? 31.154.81.65 (talk) 08:23, 11 April 2017 (UTC)
 * So you have a finite set A of n elements, and an arbitrary subset B of the powerset of A. Then for any non-empty B you can always pick b1 = b2. Assuming that all subsets of the powerset are equally likely, the powerset has 2n elements, so s=22 n  subsets, and your chance of getting a non-empty set is s-1 in s. With your extra constraint on k, the chance is 0 if k=0, 1 otherwise. I'm not sure if that is what you want. --Stephan Schulz (talk) 08:50, 11 April 2017 (UTC)


 * Do you require $$b_1$$ and $$b_2$$ to be distinct ? If not, can we select $$b_1=b_2$$ ? And how is B chosen - is it selected from all subsets of $$2^A$$, and just happens to have size k - in which case you are expecting an answer that is independent of k ? Or is it selected from only those subsets of $$2^A$$ that have size k - in which case you are expecting an answer that is a function of k ? Gandalf61 (talk) 08:54, 11 April 2017 (UTC)
 * I think the question is clear as written, with the usual understanding of what is meant in such questions. The OP wants to select k elements from the powerset of A, without replacement.  What is the probability that one of these elements is a subset of another?  That's a perfectly well-defined question, with an answer that depends on k and n.  Off the top of my head I don't know what the answer is, but the question seems clear. --Trovatore (talk) 09:51, 11 April 2017 (UTC)
 * Oh, now I see the objection &mdash; right, I'm sure the question meant the probability that there are two distinct elements b1 and b2. It's not literally what was written, but it's pretty obviously what was meant. --Trovatore (talk) 09:54, 11 April 2017 (UTC)
 * This is exactly what I meant :) 31.154.81.64 (talk) 14:09, 11 April 2017 (UTC)
 * If you do add the requirement b1≠b2, then for k=2 the problem reduces to finding the number of pairs of subsets of A which are not comparable. This is oeis sequence A038721 for which the formula is 4n+1 - 2*3n+1 + 2n+1. Perhaps some sort of inclusion-exclusion counting method could be used to generalize to larger k. --RDBury (talk) 10:03, 11 April 2017 (UTC)
 * Correction, the formula should be 4n - 2*3n + 2n; the oeis was using n+1 for what we're calling n. Also, I looked at k=3 and the results are not encouraging. An inclusion-exclusion formula would run over the quasi-orders on A and just enumerating these seems difficult, see oeis A000798. It does seem clear though that if you fix k and let n->infinity then the probability of there being two distinct but comparable sets approaches 0. --RDBury (talk) 11:12, 11 April 2017 (UTC)
 * Thanks! 31.154.81.64 (talk) 14:09, 11 April 2017 (UTC)
 * Essentially this boils down to counting antichains in the Boolean lattice. There is literature on this, and it is nt a trivial problem.  Here's one paper whose references are probably of nterest: . --JBL (talk) 02:11, 12 April 2017 (UTC)