Updated 12 September 2026

Buffon’s needle

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

The answer connects a purely geometric experiment to π:

Rearranging, you get a Monte Carlo estimator for π:

Here needle centers are uniform relative to the parallel lines, orientations are uniform on [0,π), and drops are independent. N counts all drops and C counts crossings. The formula assumes ℓ ≤ d. With C=0 the ratio is undefined; the demo reports this explicitly.

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 $1/\sqrt{N}$ — you need 100× more needles to gain one decimal place.
• Solid crosses, dashed misses. A needle crosses a line when its centre is within $(\ell/2)|\sin\theta|$ 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 $\sin\theta$ over $[0, \pi]$ evaluates to 2, and dividing by the normalising constant $\pi$ is what injects π into the formula. The geometry forces trigonometry, and trigonometry carries π.

From geometry to probability

Let U be the distance from the center to the nearest line. Uniform placement makes U uniform on [0,d/2]. At angle θ, crossing requires U≤(ℓ/2)|sin θ|. Since ℓ≤d, this gives conditional probability (ℓ/d)|sin θ|. Averaging over uniform θ in [0,π) gives:

Longer needles can cross multiple lines, and the conditional probability above can exceed one before being capped. That is why this experiment restricts ℓ/d to at most one; substituting longer lengths into the short-needle formula is invalid.

An interval, not extra decimal places

Each independent drop contributes a Bernoulli crossing indicator, so C has a binomial distribution. The demo constructs a nominal 95% Wilson interval [p_low,p_high] for its success probability and transforms it to [2ℓ/(d·p_high), 2ℓ/(d·p_low)]. The endpoint order reverses because the reciprocal decreases.

When C=0, the point estimate is undefined and the upper interval endpoint is unbounded. Those runs remain in the repeated-experiment display and coverage denominator. The finite plot clips large interval endpoints only for drawing; the table preserves their values.

Wilson coverage is approximate and varies with sample size and crossing probability. The fraction of 60 displayed intervals containing π is itself noisy; it need not be 95%. Increasing N reduces typical sampling error at an asymptotic 1/√N rate, but cannot guarantee that the next estimate improves. The reciprocal estimator also has finite-sample bias.

Reference

NIST: confidence intervals for a binomial proportion describes Wilson score inversion and distinguishes approximate intervals from exact binomial procedures. The reciprocal transformation above applies it to this experiment.