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: Concrete Answer is (n-1)/(n+1) by Charlie)

Googling about, I came across

 http://mathworld.wolfram.com/StatisticalRange.html

which gives the distribution function of the range statistic R for N samples of a uniform variate on [0,1] as N*R^(N-1)-(N-1)*R^N. The density is then the derivative N*(N-1)*R^(N-2)-(N-1)*N*R^(N-1). The mean of R is then the integral of R times that density over [0,1]which comes out to N*(N-1)*[(1/N)-(1/(N+1))]=(N-1)/(N+1). The birds do tend to occupy the whole wire on average as their number increases without bound. Your simulation must be unduly constraining them in some way, Charlie.

Posted by Richard on 2004-06-08 21:23:16