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You play the following game.

  1. You get $4n$ gold coins and have to arrange them in the unit square in general position (no two coins have the same x or the same y coordinate). Call this set of coins $S_1$.
  2. There is a lottery in which each coin in $S_1$ is selected with probability 1/2, independently of the others. Call the set of selected coins $S_2$ (it is a subset of $S_1$).
  3. You select an axis-parallel rectangle $C$, with the constraint that it must contain exactly $2n$ original coins ($|C\cap S_1| = 2n$).
  4. Your prize is the coins in $C\cap S_2$.

What is the largest number of coins that you can guarantee to yourself in probability at least $1-1/n$, by a sophisticated arrangement of the coins and selection of $C$?

NOTES:

  • A. If you have to select $C$ before the lottery (steps 2 and 3 are swapped), then your prize is a binomial variable distributed like $Binom[2n,1/2]$. Its expected value is $n$, and by known tail-bounds, with probability at least $1-1/n$ the prize is at least $n-O(\sqrt{n\log n})$. Here you can select $C$ after seeing the lottery outcomes, so you can potentially do better.
  • B. In contrast, if $C$ does not have to be convex, then of course you can just select a set which contains $2n$ coins of $S_2$ and your prize is $\min(|S_2|,2n)$. So the answer should be between $n-O(\sqrt{n\log n})$ and $2n$.
  • C. I do not know the answer even in the simpler case in which $C$ must be a square or a 1-dimensional interval (about the latter case I asked in Math.Se).
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    $\begingroup$ If the points are in convex position, say on a circle, you can select an arbitrary subset to be the intersection with a convex set, including $\max(2n,|S_2|)$ elements of $S_2$. If $C$ is restricted to be an axis-parallel rectangle or square, the question is more interesting. $\endgroup$ Jan 16, 2016 at 22:15
  • $\begingroup$ @DouglasZare You are right. I edited the question accordingly. $\endgroup$ Jan 17, 2016 at 7:57

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One quick upper bound for the axis-parallel rectangle case for large $n$: Fix an arbitrary arrangement of $4n$ points. Then there are at most $cn^4$ distinct subsets of $2n$ points that lie inside axis-parallel rectangles containing exactly $2n$ points (the points in a rectangle are determined by an uppermost, lowermost, leftmost, and rightmost point).

For any fixed rectangle containing $2n$ points, the proportion of lotteries for which that rectangle beats $n+C\sqrt{n \log n}$ is $o(n^{-4})$ for sufficiently large $C$, by the same tail bounds you linked to. By the union bound, it follows that with probability approaching $1$ none of the axis-parallel rectangles beat $n+C \sqrt{n \log n}$.

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  • $\begingroup$ By "there are at most $cn^4$ possible axis-parallel rectangles containing exactly $2n$ of them" you mean "there are at most $cn^4$ different subsets of $2n$ points that are defined by containment in an axis-parallel rectangle"? (since there are infinitely many different rectangles). $\endgroup$ Jan 17, 2016 at 10:54
  • $\begingroup$ Yes. I've edited to try and clarify. $\endgroup$ Jan 17, 2016 at 20:14

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