If a rod of unit length is broken into $n$ pieces, what is the expected length of the longest piece? The positions at which the rod is broken are chosen randomly uniformly.

This is a generalization of a puzzle I recently encountered, the answer to which I do not know myself:

  • $\begingroup$ Interesting question. I'm tempted to say $1/n$, but I have a feeling it's much more complicated than that. $\endgroup$ – Doorknob Jun 5 '14 at 15:37
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    $\begingroup$ @Doorknob: it has to be greater than that: the only way the greatest is $1/n$ is if they are exactly equal. $\endgroup$ – Ross Millikan Jun 5 '14 at 15:47
  • $\begingroup$ What does "randomly uniformly" mean? How can it be both? $\endgroup$ – DisgruntledGoat Jun 5 '14 at 20:27
  • $\begingroup$ @DisgruntledGoat I mean chosen randomly from a uniform distribution. $\endgroup$ – arshajii Jun 5 '14 at 20:29

This is known as the broken stick rule. The Wolfram Demonstrations Project claims that

If the unit interval is divided into $n $ spaces by $ n-1$ random points, the expected size of the $i^{\text{th}}$ largest space is $$\frac 1n \sum_{j=0}^{n-i} \frac 1{n-j}$$ Setting $i=1$ gives the expected size of the largest space as $$\frac 1n H_{n-1}\approx \frac 1n(\ln(n-1)+\gamma)$$ where $H_{n-1}$ is the $n-1^{\text{st}}$ harmonic number and $\gamma \approx 0.5772$ is the Euler-Mascheroni constant.


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