Buffon’s needle

In 1777, Georges-Louis Leclerc, Comte de Buffon asked: if you drop a needle of length $ℓ$ onto a floor with parallel lines spaced $d$ apart ( $ℓ \leq d$ ), what is the probability the needle crosses a line?

The answer connects a purely geometric experiment to π:

P (cross) = \frac{2 ℓ}{π d}

Rearranging, you get a Monte Carlo estimator for π:

π \approx \frac{2 ℓ N}{d \cdot C}

where N is the total number of needles dropped and C is the number that crossed a line.

π ≈ 3.07692 error: 0.06467

325 crossings / 500 needles

needles 500

Red needles cross a line; blue do not. P(cross) = 2/π, so π ≈ 2N/crossings. Showing last 250 of 500.

What to notice

The estimate converges slowly. After 500 needles the estimate might still be off by 0.02–0.1. Monte Carlo estimators for π converge at rate $1/ N$ — you need 100× more needles to gain one decimal place.
Red crosses, blue misses. A needle crosses a line when its centre is within $(ℓ /2) ∣ sin θ ∣$ of the nearest line, where θ is the needle’s angle.
Resample to see a new realisation and how much the estimate varies from run to run.

Why π appears

The crossing probability comes from integrating over all possible needle positions and angles. The integral of $sin θ$ over $[0, π]$ evaluates to 2, and dividing by the normalising constant $π$ is what injects π into the formula. The geometry forces trigonometry, and trigonometry carries π.