Drop a Needle Enough Times and You Get Pi

Rule a sheet with parallel lines a fixed distance d apart. Drop a needle of length L (where L ≤ d) onto the sheet at random. What is the probability the needle crosses one of the lines?

Georges-Louis Leclerc, Comte de Buffon — naturalist, director of the Jardin du Roi, author of the 36-volume Histoire Naturelle — posed this question in 1733 and worked on it intermittently for forty-four years. His solution appeared in the fourth supplement to the Histoire in 1777, in a section titled Essai d'Arithmétique Morale. The answer is 2L / (πd). When L = d, that simplifies to 2/π, about 0.6366.

The derivation is one of those calculations that looks like a card trick the first time. Let θ be the angle between the needle and the lines, distributed uniformly on [0, π]. Let x be the distance from the needle's center to the nearest line, uniform on [0, d/2]. The needle crosses a line iff x ≤ (L/2) sin θ. Integrate the indicator over the joint uniform density and the answer is 2L/(πd). π enters through the integral of sin θ, which is rooted in the fact that orientation is rotationally symmetric — there is no preferred direction in which to drop a needle.

That last point is what makes the result more than a curio. Buffon had founded a sub-field. Geometric probability — probability over continuous geometric configurations — became its own area, with Laplace extending the needle to a grid of perpendicular lines in 1812 and Crofton building it into integral geometry in the 1860s.

You can use the formula in reverse: drop n needles, count k crossings, estimate π ≈ 2nL / (kd). It works, slowly. Convergence is roughly as 1/√n, which is why nobody computes π this way for real. The interesting part is that you can.