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
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|
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 Probabilities
- Are Most Triangles Obtuse?
- Barycentric Coordinates and Geometric Probability
- Bertrand's Paradox
- Birds On a Wire (Problem and Interactive Simulation)
- Birds on a Wire: Solution by Nathan Bowler
- Birds on a Wire. Solution by Mark Huber
- Birds on a Wire: a probabilistic simulation. Solution by Moshe Eliner
- Birds on a Wire. Solution by Stuart Anderson
- Buffon's Noodle Simulation
- Averaging Raindrops - an exercise in geometric probability
- Rectangle on a Chessboard: an Introduction
- Marking And Breaking Sticks
- Random Points on a Segment
Copyright © 1996-2017 Alexander Bogomolny