GIVEN: Positive integers $n,m,L$ and probabilities $p_1, p_2, \ldots, p_n$.

GOAL: Choose $L$ size-$m$ subsets $S_1, S_2, \ldots, S_L$ of $\{1,2,\ldots,n\}$ to maximize $\displaystyle \mathbb{E}[ \max_{1\leq \ell \leq L} |S \cap S_\ell| ]$, where the expectation is taken over (all $2^n$) $S \subseteq \{1,2,\ldots,n\}$ using the pdf $f$ defined as \[ f(S) = \left( \prod_{i \in S} p_i \right) \left( \prod_{i \not\in S} (1-p_i) \right)\ . \]

(Equivalently, $S$ is chosen by examining each $i \in \{1,2,\ldots,n\}$ independently, choosing $i$ to be in $S$ with probability $p_i$.)

Note: the problem is trivial unless $m < n$ and $L < \binom{n}{m}$.

A nice, closed-form expression for $\displaystyle \mathbb{E}[ \max_{1\leq \ell \leq L} |S \cap S_\ell| ]$ (in terms of $n,m,L$ and the $p_i$ and the $S_\ell$) would be a good start.


WLOG, $p_1 \geq p_2 \geq \cdots \geq p_n$.

Special cases:

  • $L=1$: $S_1 = \{ 1, 2, \ldots, m \}$ maximizes $\displaystyle \mathbb{E}[ \max_{1\leq \ell \leq L} |S \cap S_\ell| ]$.

  • $m=1$: Denote by $x_i$ the single element of $S_i$ (i.e., $S_i = \{ x_i \}$) and $X = \{ x_1, x_2, \ldots, x_L \} = \bigcup S_i$. $\displaystyle \mathbb{E}[ \max_{1\leq \ell \leq L} |S \cap S_\ell| ] = \sum_{S \cap X \neq \emptyset} 1 \cdot f(S) = 1 - \sum_{S \subseteq \{ 1, 2, \ldots, n \} \setminus X} f(S) = 1 - \prod_{i \in X} (1-p_i)$, which is maximized by the greedy solution of $S_i = \{ i \}$ (for $1 \leq i \leq L$).

Here's how I examined a few more special cases:

This example ($(n,m,L)=(4,1,2)$) falls under the already discussed $m=1$ case, but I present it to hopefully make the presentation of subsequent examples clearer:

$\binom{\binom{n}{m}}{L} = \binom{\binom{4}{1}}{2} = 6$ possibilities for $(S_1,S_2)$:

1100 1ooo,o1oo,  p1+p2        -  p1p2
1010 1ooo,oo1o,  p1+   p3     -       p1p3
1001 1ooo,ooo1,  p1+      p4  -            p1p4
0110 o1oo,oo1o,     p2+p3     -                 p2p3
0101 o1oo,ooo1,     p2+   p4  -                      p2p4
0011 oo1o,ooo1,        p3+p4  -                           p3p4
  • Column 1: the sum of the subsequent $L=2$ columns (I used "o" for "0" in those for readability)
  • Column 2: represents $S_1$ (the 1st bit is 1 iff $1 \in S$ (0 otherwise), the 2nd bit is 1 iff $2 \in S$, etc.)
  • Column 3: represents $S_2$
  • Column 4: $\displaystyle \mathbb{E}[ \max_{1\leq \ell \leq L} |S \cap S_\ell| ]$ (formatted to put the degree-1 terms all together first, then the degree-2 terms, etc., with spacing so you can look up and down and see which other rows share those terms).

$\binom{\binom{n}{m}}{L} = \binom{\binom{4}{2}}{2} = 15$ possibilities for $(S_1,S_2)$:

2110 11oo,1o1o, p1+p2+p3     -                 p2p3
2101 11oo,1oo1, p1+p2+   p4  -                      p2p4
1210 11oo,o11o, p1+p2+p3     -       p1p3
1201 11oo,o1o1, p1+p2+   p4  -            p1p4
2011 1o1o,1oo1, p1+   p3+p4  -                           p3p4
1120 1o1o,o11o, p1+p2+p3     -  p1p2
1021 1o1o,oo11, p1+   p3+p4  -            p1p4
1102 1oo1,o1o1, p1+p2+   p4  -  p1p2
1012 1oo1,oo11, p1+   p3+p4  -       p1p3
0211 011o,o1o1,    p2+p3+p4  -                           p3p4
0121 011o,oo11,    p2+p3+p4  -                      p2p4
0112 01o1,oo11,    p2+p3+p4  -                 p2p3

1111 11oo,oo11, p1+p2+p3+p4  -       p1p3-p1p4-p2p3-p2p4       +  p1p2p3+p1p2p4+p1p3p4+p2p3p4  -  2p1p2p3p4
1111 1o1o,o1o1, p1+p2+p3+p4  -  p1p2-     p1p4-p2p3-     p3p4  +  p1p2p3+p1p2p4+p1p3p4+p2p3p4  -  2p1p2p3p4
1111 1oo1,o11o, p1+p2+p3+p4  -  p1p2-p1p3-          p2p4-p3p4  +  p1p2p3+p1p2p4+p1p3p4+p2p3p4  -  2p1p2p3p4

I separated the rows into two groups; two rows are in the same group if their first-column entries are permutations of each other.

Notes: the coefficient of 2 in the degree-4 terms of the last three rows seems interesting to me. The sign of a term is the negation of $(-1)$ to the degree of the term.

$\binom{\binom{n}{m}}{L} = \binom{\binom{4}{2}}{3} = 20$ possibilities for $(S_1,S_2,S_3)$:

3111  11oo,1o1o,1oo1,  p1+p2+p3+p4   -                  p2p3-p2p4-p3p4   +                        p2p3p4
1311  11oo,o11o,o1o1,  p1+p2+p3+p4   -        p1p3-p1p4-          p3p4   +                 p1p3p4
1131  1o1o,o11o,oo11,  p1+p2+p3+p4   -   p1p2-     p1p4-     p2p4        +          p1p2p4
1113  1oo1,o1o1,oo11,  p1+p2+p3+p4   -   p1p2-p1p3-     p2p3             +   p1p2p3

2220  11oo,1o1o,o11o,  p1+p2+p3                                          -   p1p2p3
2202  11oo,1oo1,o1o1,  p1+p2+   p4                                       -          p1p2p4
2022  1o1o,1oo1,oo11,  p1+   p3+p4                                       -                 p1p3p4
0222  o11o,o1o1,oo11,     p2+p3+p4                                       -                        p2p3p4

2211  11oo,1o1o,o1o1,  p1+p2+p3+p4   -             p1p4-p2p3-     p3p4   +                 p1p3p4+p2p3p4   -   p1p2p3p4
2121  11oo,1o1o,oo11,  p1+p2+p3+p4   -             p1p4-p2p3-p2p4        +          p1p2p4+       p2p3p4   -   p1p2p3p4
2211  11oo,1oo1,o11o,  p1+p2+p3+p4   -        p1p3-          p2p4-p3p4   +                 p1p3p4+p2p3p4   -   p1p2p3p4
2112  11oo,1oo1,oo11,  p1+p2+p3+p4   -        p1p3-     p2p3–p2p4        +   p1p2p3+              p2p3p4   -   p1p2p3p4
1221  11oo,o11o,oo11,  p1+p2+p3+p4   -        p1p3-p1p4-     p2p4        +          p1p2p4+p1p3p4          -   p1p2p3p4
1212  11oo,o1o1,oo11,  p1+p2+p3+p4   -        p1p3–p1p4-p2p3             +   p1p2p3+       p1p3p4          -   p1p2p3p4
2121  1o1o,1oo1,o11o,  p1+p2+p3+p4   -   p1p2-               p2p4–p3p4   +          p1p2p4+       p2p3p4   -   p1p2p3p4
2112  1o1o,1oo1,o1o1,  p1+p2+p3+p4   -   p1p2-          p2p3-     p3p4   +   p1p2p3+              p2p3p4   -   p1p2p3p4
1221  1o1o,o11o,o1o1,  p1+p2+p3+p4   -   p1p2-     p1p4-          p3p4   +          p1p2p4+p1p3p4          -   p1p2p3p4
1122  1o1o,o1o1,oo11,  p1+p2+p3+p4   -   p1p2-     p1p4-p2p3             +   p1p2p3+p1p2p4                 -   p1p2p3p4
1212  1oo1,o11o,o1o1,  p1+p2+p3+p4   -   p1p2-p1p3-               p3p4   +   p1p2p3+       p1p3p4          -   p1p2p3p4
1122  1oo1,o11o,oo11,  p1+p2+p3+p4   -   p1p2-p1p3-          p2p4        +   p1p2p3+p1p2p4                 -   p1p2p3p4

Note: the sign of a term is the negation of $(-1)$ to the parity of the degree of the term only in the first and third block of rows — the degree-3 terms in the 4 rows in the middle block are all negative.

  • $\begingroup$ Presumably you have tried some special cases? or are you just hoping some here will just know the answer? $\endgroup$
    – Yemon Choi
    Jan 26, 2010 at 7:09
  • $\begingroup$ While it would be great it someone knew the answer, realistically I was hoping at least for some insight/suggestions in how to proceed and/or related results/concepts. I have played with some special cases; I could add the details to the original post, but, as I'm new, I wasn't sure if that was appropriate. (I assume they don't belong in comments, if only because of the character limit.) $\endgroup$
    – Jay Howard
    Jan 26, 2010 at 8:08

1 Answer 1


If you instead ask for a random subset of a fixed size, rather than a random subset of a random size, then being able to maximize the expected maximum size of the intersection would let you solve covering design problems.

"A $(v,k,t)$-covering design is a collection of $k$-element subsets, called blocks, of ${1,2,...,v}$, such that any $t$-element subset is contained in at least one block."

That means the complements of the blocks, of size $v-k$, have maximum intersection of size $v-k$ with each set of size $v-t$.

Let's return to subsets of random size. If you had an oracle which gave you the exact expected value, then setting the $p_i$ equal and close to 1, you can recover the minimum number of $t$-element subsets the complements don't cover as long as the complements cover each set of size at most $t-1$, hence whether a covering design is possible, which appears hard to determine.

So, it doesn't look like a closed form expression is feasible, and you should look at covering designs.


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