Buffon's needle

In mathematics, Buffon's needle problem is a question first posed in the 18th century by Georges-Louis Leclerc, Comte de Buffon: suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips?

Using integral geometry, the problem can be solved to get a Monte Carlo method to approximate π.

Mario Lazzarini, an Italian mathematician, performed the Buffon's needle experiment in 1901. Tossing a needle 3408 times, he attained the well-known estimate 355/113 for π, which is a very accurate value, differing from π by no more than 3×10−7. This may be merely a remarkable coincidence; but it has been suggested that Lazzarini fudged or falsified some of the details of the experiment to get this value to come out (proponents of this latter view point out that 3408 is a rather arbitrary number).

Solution

The problem in more mathematical terms is: Given a needle of length [itex]\ell[itex] dropped on a plane ruled with parallel lines t units apart, what is the probability that the needle will cross a line?

Let [itex]t>\ell[itex], let x be the distance from the center of the needle to the closest line, and let θ be the acute angle between the needle and the lines.

The probability density function of x between 0 and t/2 is

[itex] \frac{2}{t}\,dx. [itex]

The probability density function of θ between 0 and π/2 is

[itex] \frac{2}{\pi}\,d\theta. [itex]

The two random variables, x and θ, are independent, so the joint probability density function is the product

[itex] \frac{4}{t\pi}\,dx\,d\theta. [itex]

The needle crosses a line if

[itex]x \le \frac{\ell}{2}\sin\theta.[itex]

Integrating the joint probability density function gives the probability that the needle will cross a line:

[itex]\int_0^{\frac{\pi}{2}} \int_0^{(\ell/2)\sin\theta} \frac{4}{t\pi}\,dx\,d\theta = \frac{2\ell}{t\pi}.[itex]

For n needles dropped with h of the needles crossing lines, the probability is

[itex]\frac{h}{n} = \frac{2\ell}{t\pi},[itex]

which can be solved for π to get

[itex]\pi = \frac{2\ell n}{th}.[itex]

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