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 computer-generated solution could be found, but bonus points will be awarded for a formal proof!

It looks to me like the answer (in units of 10 meters) is n! times the integral of xn - x1 over the volume described by 0<x1<...<xn<1, whatever that evaluates to. This is based on the n! orders that x1,...,xn can have such as x1<x2<...<xn and then using symmetry. How to convert this abstract volume integral into a concrete iterated integral with the right limits so far has eluded me, I'm sorry to say. Help, anyone?

Posted by Richard on 2004-06-07 20:15:36