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<l} 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.

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