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 &pi;.

Solution


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

Let x be the distance from the center of the needle to the closest line, let θ be the acute angle between the needle and the lines, and let $$t\ge l$$.

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


 * $$ \frac{2}{t}\,dx. $$

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


 * $$ \frac{2}{\pi}\,d\theta. $$

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


 * $$ \frac{4}{t\pi}\,dx\,d\theta. $$

The needle crosses a line if


 * $$x \le \frac{l}{2}\sin\theta.$$

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


 * $$\int_{\theta=0}^{\frac{\pi}{2}} \int_{x=0}^{(l/2)\sin\theta} \frac{4}{t\pi}\,dx\,d\theta = \frac{2 l}{t\pi}.$$

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


 * $$\frac{h}{n} = \frac{2 l}{t\pi},$$

which can be solved for π to get


 * $$\pi = \frac{2{l}n}{th}.$$

Now suppose $$t < l$$. In this case, integrating the joint probability density function, we obtain:


 * $$\int_{\theta=0}^{\frac{\pi}{2}} \int_{x=0}^{m(\theta)} \frac{4}{t\pi}\,dx\,d\theta ,$$

where $$m(\theta) $$ is the minimum between $$(l/2)\sin\theta$$ and $$t/2 $$.

Thus, performing the above integration, we see that, when $$t < l$$, the probability that the needle will cross a line is


 * $$\frac{h}{n} = \frac{2 l}{t\pi} - \frac{2}{t\pi}\left\{\sqrt{l^2 - t^2} + t\sin^{-1}\left(\frac{t}{l}\right)\right\}+1.$$

Lazzarini's estimate
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&times;10&minus;7. This is an impressive result, but is something of a cheat.

Lazzarini chose needles whose length was 5/6 of the width of the strips of wood. In this case, the probability that the needles will cross the lines is 5/3π. Thus if one were to drop n needles and get x crossings, one would estimate π as


 * &pi; &asymp; 5/3 &middot; n/x

π is very nearly 355/113; in fact, there is no better rational approximation with fewer than 5 digits in the numerator and denominator. So if one had n and x such that:


 * 355/113 = 5/3 &middot; n/x

or equivalently,


 * x = 113n/213

one would derive an unexpectedly accurate approximation to π, simply because the fraction 355/113 happens to be so close to the correct value. But this is easily arranged. To do this, one should pick n as a multiple of 213, because then 113n/213 is an integer; one then drops n needles, and hopes for exactly x = 113n/213 successes.

If one drops 213 needles and happens to get 113 successes, then one can triumphantly report an estimate of π accurate to six decimal places. If not, one can just do  213 more trials and hope  for  a total of 226 successes; if not, just repeat as necessary. Lazzarini performed 3408 = 213 · 16 trials, making it seem likely that this is the strategy he used to obtain his "estimate".

External links and references

 * Buffon's Needle at cut-the-knot
 * Math Surprises: Buffon's Noodle at cut-the-knot
 * MSTE: Buffon's Needle
 * Buffon's Needle Java Applet
 * Estimating PI Visualization (Flash)
 * p. 5
 * p. 5

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