Random chord longer than the side of a regular n-gon, Geometric Probability

Problem 24.1

For an integer $n \geq 3$ , two points are chosen independently and uniformly on the unit circle, defining a chord. What is the probability the chord is longer than the side of the regular $n$ -gon inscribed in the circle?

Solution. A regular $n$ -gon inscribed in the unit circle has side length $2 \sin(\pi / n)$ (by standard trigonometry: each side subtends a central angle of $2\pi/n$ , so its length is $2 \sin(\pi / n)$ ).

As in Chapter 13, let $\phi \in [0, 2\pi)$ be the angular separation of the two endpoints; $\phi$ is uniform on $[0, 2\pi)$ . The chord length is $2 \sin(\phi / 2)$ , which is monotone on $[0, \pi]$ and symmetric about $\pi$ . The reduced angular separation $\phi^* = \min(\phi, 2\pi - \phi) \in [0, \pi]$ is uniform on $[0, \pi]$ .

The chord exceeds $2 \sin(\pi/n)$ iff $\sin(\phi^*/2) > \sin(\pi/n)$ , i.e., iff $\phi^* > 2\pi/n$ . The set $\{\phi^* : \phi^* > 2\pi/n\}$ has measure $\pi - 2\pi/n$ , so $\boxed{\mathbb{P}(\text{chord} > \text{side of inscribed } n\text{-gon}) = \frac{\pi - 2\pi/n}{\pi} = 1 - \frac{2}{n}.} \qedhere$

Specialisations:

$n = 3$ (equilateral triangle, side $\sqrt 3$ ): $\mathbb{P} = 1/3$ (Bertrand’s first sampling).
$n = 4$ (square, side $\sqrt 2$ ): $\mathbb{P} = 1/2$ .
$n = 6$ (hexagon, side $= 1 =$ radius): $\mathbb{P} = 2/3$ , recovering Chapter 15.
$n \to \infty$ (side $\to 0$ ): $\mathbb{P} \to 1$ (the chord almost always exceeds a vanishing side).

This simple pattern $1 - 2/n$ is one of the cleanest families in classical geometric probability.

Three inscribed regular polygons in the unit circle. A uniform random chord exceeds the polygon’s side with probability 1 - 2/n : \tfrac13 for the triangle, \tfrac12 for the square, \tfrac23 for the hexagon. — Three inscribed regular polygons in the unit circle. A uniform random chord exceeds the polygon’s side with probability $1 - 2/n$ : $\tfrac13$ for the triangle, $\tfrac12$ for the square, $\tfrac23$ for the hexagon.

import numpy as np
for n in [3, 4, 6, 8]:
    theta = np.random.rand(2, 10**6) * 2*np.pi
    chord = 2 * np.abs(np.sin((theta[0]-theta[1])/2))
    thresh = 2*np.sin(np.pi/n)
    print(f"n={n}: sim={(chord>thresh).mean():.5f}  exact={1 - 2/n:.5f}")
# n=3: sim=0.33415  exact=0.33333
# n=4: sim=0.50046  exact=0.50000
# n=6: sim=0.66734  exact=0.66667
# n=8: sim=0.75101  exact=0.75000