gll buffon

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.


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


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


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


The needle crosses a line if

x le frac{l}{2}sintheta.

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)sintheta} frac{4}{tpi},dx,dtheta = frac{2 l}{tpi}.

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

frac{h}{n} = frac{2 l}{tpi},

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}{tpi},dx,dtheta ,
where m(theta) is the minimum between (l/2)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{h}{n} = frac{2 l}{tpi} - frac{2}{tpi}left{sqrt{l^2 - t^2} + tsin^{-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 3×10−7. 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".

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