Buffon's needle 667479 220783959 2008-06-21T15:43:03Z DOI bot 6652755 Citation maintenance. You can [[WP:DOI|use this bot]] yourself! Please [[User:DOI_bot/bugs|report any bugs]]. {{Unreferenced|date=October 2007}} 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 (geometry)|parallel]] strips of [[wood]], each the same width, and we drop a [[Sewing needle|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 [[pi|&pi;]]. == Solution == [[Image:Buffon needle.gif|right|frame|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 <math>l</math> 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 <math>t\ge l</math>. The [[probability density function]] of ''x'' between 0 and ''t'' /2 is :<math> \frac{2}{t}\,dx. </math> The probability density function of θ between 0 and π/2 is :<math> \frac{2}{\pi}\,d\theta. </math> The two [[random variables]], ''x'' and ''θ'', are independent, so the joint probability density function is the product :<math> \frac{4}{t\pi}\,dx\,d\theta. </math> The needle crosses a line if :<math>x \le \frac{l}{2}\sin\theta.</math> Integrating the joint probability density function gives the probability that the needle will cross a line: :<math>\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}.</math> For ''n'' needles dropped with ''h'' of the needles crossing lines, the probability is :<math>\frac{h}{n} = \frac{2 l}{t\pi},</math> which can be solved for ''π'' to get :<math>\pi = \frac{2{l}n}{th}.</math> Now suppose <math>t < l</math>. In this case, integrating the joint probability density function, we obtain: :<math>\int_{\theta=0}^{\frac{\pi}{2}} \int_{x=0}^{m(\theta)} \frac{4}{t\pi}\,dx\,d\theta ,</math> where <math>m(\theta) </math> is the minimum between <math>(l/2)\sin\theta</math> and <math>t/2 </math>. Thus, performing the above integration, we see that, when <math>t < l</math>, the probability that the needle will cross a line is :<math>\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.</math> ==Lazzarini's estimate== [[Mario Lazzarini]], an [[Italy|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<sup>&minus;7</sup>. This is an impressive result, but is something of a cheat. 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; &asymp; 5/3 &middot; ''n''/''x'' π is very nearly 355/113; in fact, there is no better rational [[Pi#Numerical approximations|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)] * {{cite journal|last = Ramaley|first = J. F.|title = Buffon's Noodle Problem|journal = The American Mathematical Monthly|volume = 76|issue = 8|date = Oct 1969|pages = 916–918|url = http://links.jstor.org/sici?sici=0002-9890%28196910%2976%3A8%3C916%3ABNP%3E2.0.CO%3B2-9&size=LARGE|doi = 10.2307/2317945}} * {{cite book|last = Mathai|first = A. M.|title = An Introduction to Geometrical Probability|date = 1999|publisher = Gordon & Breach|address = Newark|url = http://books.google.com.au/books?id=FV6XncZgfcwC}} p. 5 [[Category:Integral geometry]] [[Category:Pi algorithms]] [[cs:Buffonova jehla]] [[es:Aguja de Buffon]] [[fr:Aiguille de Buffon]] [[it:Ago di Buffon]] [[pl:Igła Buffona]] [[pt:Agulha de Buffon]] [[sv:Buffons nål]] [[zh:布豐投針問題]]