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 4b has 3 inner connecting segments (average 2).

Since:

1. the states of sc=1 and sc=3 force inner point positions on the neighboring 4b's.

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

3. so there cannot be sc=1 or 3 more than average. This implies that each 4b has on average sc=2.

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

There are 3 possible sc=2's , but since one can rearrange segments inside the 4b, all have the same density, let's choose one of them, e.g., the one with two shortest at first:

Integrate ydxdy:

1. y=1/2..2/3, x=2y-1..1-y

2. y=0..1/2, x=0..y.

This yields 7/108, but since the total integration area is 1/6, the result density is: 7/18.

Geometric Probability

• Geometric Probabilities
• Are Most Triangles Obtuse?
  • Eight Selections in Six Sectors
• Barycentric Coordinates and Geometric Probability
  • Stick Broken Into Three Pieces (Trilinear Coordinates)
  • Stick Broken Into Three Pieces. Solution in Cartesian Coordinatess
• 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
  • Averaging Raindrops, Part 2
• Rectangle on a Chessboard: an Introduction

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Copyright © 1996-2012 Alexander Bogomolny