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!
My first thought is that if each crow is independent of the others, it has expected value 5. Thus, each crow on average lands exactly at the 5 meter part, and no wire is ever painted. However, I'm pretty sure that's not what the question is asking.
So, we instead look at the expected value of the difference. I haven't yet formulated a continuous proof, but if we look at the wire as having 11 possible landing spots - 0 through 10 meters - then with highest probability we can expect the difference between two crows to be 1 meter. This probability is about 1/6 - it occurs 20 times in 121 discrete events. This is the most frequent event given 11 discrete landing spots. I'm working on a continuous proof.
Posted by Eric
on 2004-06-07 13:38:59