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April 11[edit]

Probability of having a subset[edit]

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)[reply]

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=2^2ⁿ 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)[reply]

(edit conflict) 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)[reply]

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)[reply]

Oh, now I see the objection — 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) [reply]

This is exactly what I meant :) 31.154.81.64 (talk) 14:09, 11 April 2017 (UTC)[reply]

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)[reply]

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)[reply]

Thanks! 31.154.81.64 (talk) 14:09, 11 April 2017 (UTC)[reply]

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: [1]. --JBL (talk) 02:11, 12 April 2017 (UTC)[reply]