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Jan 21, 2011 at 6:46 vote accept Greg Martin
Jan 19, 2011 at 21:08 answer added Mike Spivey timeline score: 5
Jan 19, 2011 at 12:19 comment added Shai Covo And so, the expectation of the maximum distance is $\frac{{H_n }}{n}$, where $H_n$ is the $n$th harmonic number: $H_n = \ln n + \gamma + \frac{1}{{2n}} + O(n^{ - 2} )$.
Jan 19, 2011 at 12:05 comment added Shai Covo In particular, the expectation of the $k$th largest distance is $\frac{1}{n}(\frac{1}{n} + \cdots + \frac{1}{k})$.
Jan 19, 2011 at 11:23 answer added James Martin timeline score: 12
Jan 19, 2011 at 11:01 answer added ansobol timeline score: 0
Jan 19, 2011 at 11:00 comment added Shai Covo For even more general results, see the first two answers at math.stackexchange.com/questions/13959/…
Jan 19, 2011 at 9:31 history edited Greg Martin CC BY-SA 2.5
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Jan 19, 2011 at 9:24 history asked Greg Martin CC BY-SA 2.5