Buffon's needle

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{2}{t},dx.

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

: frac{2}{pi},d heta.

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

: frac{4}{tpi},dx,d heta.

The needle crosses a line if

:x le frac{l}{2}sin heta.

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

:int_{ heta=0}^{frac{pi}{2 int_{x=0}^{(l/2)sin heta} frac{4}{tpi},dx,d heta = 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_{ heta=0}^{frac{pi}{2 int_{x=0}^{m( heta)} frac{4}{tpi},dx,d heta ,where m( heta) is the minimum between(l/2)sin heta 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} ight) ight}+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;10−7. 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

:&pi; ≈ 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" = 113"n"/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 113"n"/213 is an integer; one then drops "n" needles, and hopes for exactly "x" = 113"n"/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".

See also

* Buffon's noodle

External links and references

* [http://www.cut-the-knot.org/fta/Buffon/buffon9.shtml Buffon's Needle] at cut-the-knot
* [http://www.cut-the-knot.org/ctk/August2001.shtml Math Surprises: Buffon's Noodle] at cut-the-knot
* [http://www.mste.uiuc.edu/reese/buffon/buffon.html MSTE: Buffon's Needle]
* [http://www.angelfire.com/wa/hurben/buff.html Buffon's Needle Java Applet]
* [http://www.metablake.com/pi.swf Estimating PI Visualization (Flash)]
*
* p. 5


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