## 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: $\displaystyle \frac{7}{18}.$

### Geometric Probability

- 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
- Birds on a Wire. Solution by Bogdan Lataianu

- 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
- Semicircle Coverage
- Hemisphere Coverage
- Overlapping Random Intervals

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