Buffon’s needle, Geometric Probability

Problem 53.1

A plane is ruled with parallel lines at equal spacing $d$ . A needle of length $\ell \leq d$ is dropped “at random” on the plane. What is the probability that the needle crosses one of the lines?

Solution. By translation symmetry, the needle’s position is determined by two parameters:

$y$ , the distance from the needle’s centre to the nearest line, uniform on $[0, d/2]$ ;
$\theta$ , the angle the needle makes with the lines, uniform on $[0, \pi/2]$ (by symmetry).

The needle crosses the nearest line iff the needle’s centre is within $\tfrac{\ell}{2} \sin\theta$ of that line, i.e., iff $y \leq \tfrac{\ell}{2} \sin\theta$ . Hence $\begin{align*} \mathbb{P}(\text{crossing}) &= \frac{1}{(d/2)(\pi/2)} \int_0^{\pi/2} \!\! \int_0^{\ell \sin\theta / 2} dy \, d\theta \\ &= \frac{4}{\pi d} \int_0^{\pi/2} \frac{\ell \sin\theta}{2} \, d\theta. \end{align*}$ The integral equals $\tfrac{\ell}{2} \cdot [-\cos\theta]_0^{\pi/2} = \tfrac{\ell}{2}$ , so $\boxed{\mathbb{P}(\text{crossing}) = \frac{2 \ell}{\pi d}.} \qedhere$

Buffon raised the franc-carreau coin-on-tiles problem in 1733, which marks the birth of geometric probability; the needle problem itself came later, in the work usually cited from his 1777 Essai d’arithmétique morale. A Monte Carlo consequence: if one tosses a needle many times and computes the empirical crossing frequency $f$ , then $\pi \approx \tfrac{2 \ell}{f d}$ . Mario Lazzarini reported in 1901 an experiment yielding $\pi \approx 355/113$ , an absurdly good result that is today suspected to have been selected from a larger run of trials.

Buffon’s needle. Crossings (rose) occur when the needle spans a ruled line; non-crossings (copper) do not. The probability 2\ell/(\pi d) gives a Monte Carlo estimator for \pi . — Buffon’s needle. Crossings (rose) occur when the needle spans a ruled line; non-crossings (copper) do not. The probability $2\ell/(\pi d)$ gives a Monte Carlo estimator for $\pi$ .

import numpy as np
L, d, N = 0.8, 1.0, 10**7
theta = np.random.rand(N) * np.pi/2
y = np.random.rand(N) * d/2  # centre to nearest line
p = (y <= L/2 * np.sin(theta)).mean()
print(f"sim: {p:.5f}   exact: {2*L/(np.pi*d):.5f}")
# sim: 0.50946   exact: 0.50930