Birds on a Wire

By Nathan Bowler

The following argument, although non-rigorous, gives the right answer and shows why it is true (and I guess with a bit of effort you could make it rigorous). Consider a wire with lots of birds on it, and scale it up so that there is one bird per unit length. Consider a bird at b. I shall assume that b is far enough away from the ends that they make no difference. For any interval , then number of birds in that interval has about a poisson distribution with parameter x. So in particular the interval is empty with probability 1 - e^-x. So the distance to the nearest bird to the right has an exponential distribution with parameter 1. The same may be said for the nearest bird to the left. So the distance to the nearest bird overall has an exponential distribution with parameter 1+1 = 2. W.l.o.g. the neasrest bird, at b' satisfies b' - b = x > 0. Then b is the nearest bird to b' with probability e^-x.

Give bird b a paintbrush, and let it paint half of the interval (b, b+x) if it is the nearest bird to b', and the whole interval otherwise. A simple combinatorial argument shows that if each bird on the wire does this, then the wire will be painted as in the question. So we are interested in the expected length painted by bird b since this is (after scaling) the required value. However, the comments in the first paragraph allow us to evaluate it as a simple integral:

p	= Int<0 to infinity, x·e^-2x·(1/2(e^-x + (1 - e^-x))dx>
	=Int<0 to infinity, x·(2*e^-2x - e^-3x)dx>
	= 2/4 - 1/9
	= 7/18,

using Int = 1/k².

This value is 0.38888888..., which seems to fit with the results produced by the applet. I must admit, at first I thought the answer was going to be something nice like 1/e, but 7/18 is what the algebra gives.

Geometric Probability

49551885