- Buffon's needle
In
mathematics , Buffon's needle problem is a question first posed in the 18th century byGeorges-Louis Leclerc, Comte de Buffon ::Suppose we have afloor made of parallel strips ofwood , each the same width, and we drop a needle onto the floor. What is theprobability that the needle will lie across a line between two strips?Using
integral geometry , the problem can be solved to get aMonte Carlo method to approximate π.Solution
The problem in more mathematical terms is: Given a needle of length 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 .
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
:
Integrating the joint probability density function gives the probability that the needle will cross a line:
:
For "n" needles dropped with "h" of the needles crossing lines, the probability is
:
which can be solved for "π" to get
:
Now suppose . In this case, integrating the joint probability density function, we obtain:
:where is the minimum between and .
Thus, performing the above integration, we see that,when ,the probability that the needle will cross a line is
:
Lazzarini's estimate
Mario Lazzarini , an Italianmathematician , 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" = 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] atcut-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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