Buffon’s needle: estimating pi

Required math: (very simple) probability & calculus

Required physics: none

This interesting little problem serves well to illustrate the notion of a probability density and its application to an experiment which can be done at home. It is known as Buffon’s needle, since it is believed that Georges-Louis Leclerc, Comte de Buffon, first posed the problem in the 18th century.

Suppose we have a needle of length ${l}$ and we drop this needle onto a sheet of paper on which there are a number of parallel lines spaced a distance ${l}$ apart. What is the probability that the dropped needle will cross a line?

One way of analyzing this problem is to begin by considering the needle as the hypotenuse of a right triangle, with sides parallel and perpendicular to the parallel lines. Then if the needle makes an angle ${\theta}$ with the lines, the sides of this triangle are ${l\cos\theta}$ for the parallel side and ${l\sin\theta}$ for the perpendicular side. The side of the triangle parallel to the lines is of no interest here; what we are interested in the perpendicular component.

To see this, suppose we placed a needle of length ${x on the paper in such a way that it was perpendicular to the lines. Such a needle would cover a fraction ${x/l}$ of the distance between two adjacent lines. Thus the probability that a line would cross this specially dropped needle is just ${x/l}$. (Note that if ${x=l}$ the probability is 1, since such a needle covers the entire distance between the lines.)

Therefore, the probability that a needle that makes an angle ${\theta}$ with the lines crosses a line is ${(l\sin\theta)/l=\sin\theta}$.

How likely is the needle to drop at an angle ${\theta}$? Since ${\theta}$ is a continuous variable, we need a probability density, rather than just a simple probability. Assuming that the needle is equally likely to drop at any angle between 0 and ${\pi\;(180^{\circ})}$, the density must be a constant, and must integrate to 1 over the range of valid ${\theta}$, so it must be ${\rho(\theta)=\frac{1}{\pi}}$.

The probability ${P(\mathrm{crosses\; line})}$ that a randomly dropped needle crosses a line is therefore

 $\displaystyle P(\mathrm{crosses\; line})$ $\displaystyle =$ $\displaystyle \int_{0}^{\pi}P(\mathrm{angle}\;\theta\;\mathrm{crosses\; line})\rho(\theta)d\theta\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi}\int_{0}^{\pi}\sin\theta\; d\theta\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\pi}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 0.6366 \ \ \ \ \ (4)$

Besides being another of those curious situations where ${\pi}$ pops up unexpectedly, this result offers the possibility of an interesting way of whiling away a rainy afternoon. Simply by dropping a needle repeatedly onto lined paper, you can do an experiment that will determine the value of ${\pi}$, since

 $\displaystyle \pi$ $\displaystyle =$ $\displaystyle \frac{2}{P(\mathrm{crosses\; line})}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{(\mathrm{Fraction\; of\; needles\; crossing\; lines})}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2(\mathrm{Number\; of\; needles\; dropped})}{\mathrm{Number\; of\; needles\; crossing\; lines}} \ \ \ \ \ (7)$

OK, you would need to drop a lot of needles to get a decent value of ${\pi}$, but it’s interesting that there is such a simple method for getting even an approximate value of ${\pi}$ without using circles, triangles, angles, measurement or anything more than just counting.