Buffon's needle

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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.

1 Solution
2 Lazzarini's estimate
3 See also
4 External links and references

[ Solution

The a needle lies across a line, while the b needle does not.

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 $m (θ)$ is the minimum between $(l / 2)sinθ$ 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 3×10⁻⁷. 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".