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!

(In reply to re: Abstract Answer by Richard)

n!*integral from 0 to 1 over dx1 (integral from 0 to x1 over dx2 (integral from 0 to x2 over dx3 (... (integral from 0 to x(n-1) of (x1 - xn) over dxn)...) evaluates to simply (n-1)/(n+1).  The integral itself without the n! factor evaluates to [1/(n+1)*(n-1)!] - 1/(n+1)! = (n-1)/(n+1)!

Therefore I claim that the concrete answer is (n-1)/(n+1) in units of 10 meters.  This does not agree with Charlie's simulation for n>3. As n gets large, the birds tend to occupy the whole wire, which makes sense to me. Thus I have concretized the "Abstract Answer" I gave below.

Posted by Richard on 2004-06-08 18:11:35