A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.
Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.
On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.
Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?
And if a thousand crows landed?
A computergenerated solution could be found, but bonus points will be awarded for a formal proof!
(In reply to
Abstract Answer by Richard)
I find that one way to write the integral as an iterated integral is like this:
n!*integral from 0 to 1 over dx1 (integral from x1 to 1 over dx2 (integral from x2 to 1 over dx3 (... (integral from x(n1) to 1 of (xn  x1) over dxn)...).
For n=3 this gives 6*integral from 0 to 1 over dx1 (integral from x1 to 1 over dx2 (integral from x2 to 1 of (x3x1) over dx3)) and evaluates (rather messily) to 1/2.
Changing the limits to [0,1], [0,x1],...,[0,x(n1)] and changing sign gives an equivalent expression which should be simpler to evaluate.
Edited on June 8, 2004, 8:00 pm

Posted by Richard
on 20040608 05:22:06 