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# 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.

## 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 $tge l$.

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

$frac\left\{2\right\}\left\{t\right\},dx.$

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

$frac\left\{2\right\}\left\{pi\right\},dtheta.$

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

$frac\left\{4\right\}\left\{tpi\right\},dx,dtheta.$

The needle crosses a line if

$x le frac\left\{l\right\}\left\{2\right\}sintheta.$

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

$int_\left\{theta=0\right\}^\left\{frac\left\{pi\right\}\left\{2\right\}\right\} int_\left\{x=0\right\}^\left\{\left(l/2\right)sintheta\right\} frac\left\{4\right\}\left\{tpi\right\},dx,dtheta = frac\left\{2 l\right\}\left\{tpi\right\}.$

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

$frac\left\{h\right\}\left\{n\right\} = frac\left\{2 l\right\}\left\{tpi\right\},$

which can be solved for π to get

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

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

$int_\left\{theta=0\right\}^\left\{frac\left\{pi\right\}\left\{2\right\}\right\} int_\left\{x=0\right\}^\left\{m\left(theta\right)\right\} frac\left\{4\right\}\left\{tpi\right\},dx,dtheta ,$
where $m\left(theta\right)$ is the minimum between $\left(l/2\right)sintheta$ and $t/2$.

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

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

## 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 3×10−7. This is an impressive result, but is something of a cheat, as follows.

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".