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Pick $m$ numbers in $[0,1]$, independently and uniformly at random. It is known that the expected value of the smallest number is $1/(m+1)$ and of the largest number is $m/(m+1)$.

Now, partition the numbers into two sets, such that sum of numbers in the set with the smaller sum is as large as possible (in other words: solve the partition problem on the numbers).

What is the expected value of the smaller sum?

Erel Segal-Halevi
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  • Is it helpful to prove $\lim\limits_{m\to \infty} \frac{4a_m}{m} = 1$ ? where $a_m$ is the number we want. – Asinomás May 18 '21 at 19:51
  • @Onir let $a_m$ be the expected value of the smaller sum and $b_m$ the expected value of the larger sum. Is your claim equivalent to $\lim_{m\to\infty}a_m/b_m = 1$? – Erel Segal-Halevi May 19 '21 at 05:18
  • I think I can prove both in a similar way. – Asinomás May 19 '21 at 12:10
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    Aren't they both immediate consequences of LLN? You can just put m/2 elements into two sets, and label them small and large appropriately. – E-A May 19 '21 at 20:09
  • @E-A yes ${}{}{}{}$ – Asinomás May 22 '21 at 21:31
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    Have you seen this paper https://arxiv.org/pdf/cond-mat/0009230.pdf ? In the intro to sec. 2. they point to some references that describe the asymptotic behaviour of the parition difference (exponentially vanishing) - I haven't read the rest of the paper fully but it seems to go into some details about deriving the actual distribution for the partition difference (difference between $a_m$ and $b_m$) – QCD_IS_GOOD May 25 '21 at 16:55
  • @JoshuaLin this is very useful, thanks – Erel Segal-Halevi May 25 '21 at 18:58

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