length of a random segment Take three random points, iid in [0,1].
The first two points devide the interval in three.
What is the expected length of the segment where the third point is lying?
Generalise for n.
 A: Join $0$ and $1$ to make a circle with circumference $1$; place a black dot at the join.  Then place the next $n$ points independently at random, marking the first $n-1$ with blue dots and the last with a red dot.  Rotate the circle by a random amount.  You now have $n+1$ dots at independent random positions, dividing the circle into $n+1$ arcs, and you want to know the expected combined length of the arcs flanking the red dot.  The arcs are all equivalent by symmetry, so each has expected length $1/(n+1)$, and the answer is $2/(n+1)$.
A: A full description of the distribution (i.e., not just the expected value) is of interest in classic distributed hash table systems (e.g., BitTorrent).
Cecile Penland "The distribution of Random Segment Lengths" has the answer:
Segmenting a line of length L to N segments yields segment length distribution $P_{L,N}(d)=(N-1)\frac{(L-d)^{N-2}}{L^{N-1}}$
Interestingly, taking N and L to infinite yields:
$P_{L,N}(d) \rightarrow \frac{N}{L} exp ( - \frac{N}{L}d)$. I.e., the exponential distribution with the natural expected value. 
