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.

Solution

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

Let $t>\backslash ell$, 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

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

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

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

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

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

The needle crosses a line if

$x\; \backslash le\; \backslash frac\{\backslash ell\}\{2\}\backslash sin\backslash theta.$

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

$\backslash int\_0^\{\backslash frac\{\backslash pi\}\{2\}\}\; \backslash int\_0^\{(\backslash ell/2)\backslash sin\backslash theta\}\; \backslash frac\{4\}\{t\backslash pi\}\backslash ,dx\backslash ,d\backslash theta\; =\; \backslash frac\{2\backslash ell\}\{t\backslash pi\}.$

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

$\backslash frac\{h\}\{n\}\; =\; \backslash frac\{2\backslash ell\}\{t\backslash pi\},$

which can be solved for π to get

$\backslash pi\; =\; \backslash frac\{2\{\backslash ell\}n\}\{th\}.$

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 3×10⁻⁷. 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

π ≈ 5/3 · 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 · 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

• 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)

Integral geometry | Pi algorithms

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