## Birds on a Wire

### By Moshe Eliner

Let's pass through all birds, starting, say, from the left, stopping at every fourth bird. I'll call this segments 4b. Notice a delicate point:

Denote the shortest connecting segments $sc$ (i.e. yellow lines). Each four birds has three inner connecting segments (average $2).$

Since:

the states of $sc=1$ and $sc=3\,$ force inner point positions on the neighboring the fourth bird.

on a uniform wire birds tend to have uniform distribution, hence inner point positions cannot be predefined.

so there cannot be $sc=1$ or $3$ more than average. This implies that each four birds have on average $sc=2.$

Now it's enough to calculate the density when $sc=2:$

There are three possible $sc=2's,$ but, since one can rearrange segments inside the four birds group, all have the same density, let's choose one of them, e.g., the one with two shortest at first:

Integrate $ydxdy:$

$y$ from $\displaystyle y=\frac{1}{2}$ to $\displaystyle \frac{2}{3},$ $x$ from $2y-1$ to $1-y$

$y$ from $0$ to $\displaystyle \frac{1}{2},$ $x$ from $0$ to $y.$

This yields $\displaystyle \frac{7}{108},$ but, since the total integration area is $\displaystyle \frac{1}{6},$ the resulting density is: $\displaystyle \frac{7}{18}.$

|Contact| |Front page| |Contents| |Probability| |Up|

Copyright © 1996-2018 Alexander Bogomolny