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→Comparing estimators of {{pi}}: avoid doubly definite phrase "the original Buffon's needle experiment" |
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: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?
Buffon's needle was the earliest problem in [[geometric probability]] to be solved;<ref>{{Cite journal|author1-link=Eugene Seneta |last1=Seneta |first1=Eugene |last2=Parshall |first2=Karen Hunger |last3=Jongmans |first3=François |date=2001 |title=Nineteenth-Century Developments in Geometric Probability: J. J. Sylvester, M. W. Crofton, J.-É. Barbier, and J. Bertrand |url=https://www.jstor.org/stable/41134124 |journal=Archive for History of Exact Sciences |volume=55 |issue=6 |pages=501–524 |doi=10.1007/s004070100038 |jstor=41134124 |s2cid=124429237 |issn=0003-9519}}</ref> it can be solved using [[integral geometry]]. The solution for the sought probability {{mvar|p}}, in the case where the needle length {{mvar|l}} is not greater than the width {{mvar|t}} of the strips, is
:<math>p=\frac{2}{\pi} \cdot \frac{l}{t}.</math>
This can be used to design a [[Monte Carlo method]] for approximating the number [[pi|{{pi}}]], although that was not the original motivation for de Buffon's question.<ref>{{cite web|last1=Behrends|first1=Ehrhard|title=Buffon: Hat er Stöckchen geworfen oder hat er nicht?|url=https://www.mathematik.de/ger/presse/ausdenmitteilungen/artikel/dmvm-2014-0022.pdf|accessdate=14 March 2015}}</ref> The seemingly unusual appearance of [[pi|{{pi}}]] in this expression occurs because the underlying probability distribution function for the needle orientation is rotationally symmetric.
== Solution ==
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Let {{mvar|x}} be the distance from the center of the needle to the closest parallel line, and let {{mvar|θ}} be the acute angle between the needle and one of the parallel lines.
The uniform [[probability density function]] (PDF) of {{
:<math>f_X(x)=
\begin{cases}
\dfrac{2}{t} &:\ 0 \le x \le \dfrac{t}{2}\\[4px]
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The uniform probability density function of {{mvar|θ}} between 0 and {{math|{{sfrac|''π''|2}}}} is
:<math>f_\Theta(\theta)=
\begin{cases}
\dfrac{2}{\pi} &:\ 0 \le \theta \le \dfrac{\pi}{2}\\[4px]
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Here, {{math|''θ'' {{=}} 0}} represents a needle that is parallel to the marked lines, and {{math|''θ'' {{=}} {{sfrac|''π''|2}}}} [[radian]]s represents a needle that is perpendicular to the marked lines. Any angle within this range is assumed an equally likely outcome.
The two [[random variables]], {{mvar|x}} and {{mvar|θ}}, are independent,<ref>The problem formulation here avoids having to work with [[Regular conditional probability|Regular conditional probability densities]].</ref> so the [[Joint probability distribution|joint probability density function]] is the product
:<math>f_{X,\Theta}(x,\theta)=
\begin{cases}
\dfrac{4}{t\pi} &:\ 0 \le x \le \dfrac{t}{2}, \ 0 \le \theta \le \dfrac{\pi}{2}\\[4px]
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:<math>\begin{align}
P &= \left(\int_{\theta=0}^{\arcsin \frac{t}{l}} \int_{x=0}^{\frac{l}{2}\sin\theta} \frac{4}{t\pi}dxd{\theta}\right) + \left(\int_{\arcsin\frac{t}{l}}^\frac{\pi}{2} \frac{2}{\pi}d{\theta}\right) \\[6px]
&= \frac{2l}{t\pi} - \frac{2}{t\pi}\left(\sqrt{l^2 - t^2} + t\arcsin \frac{t}{l}\right) + 1
\end{align}</math>
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* [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.metablake.com/pi.swf Estimating PI Visualization (Flash)]
* [http://www.slideshare.net/cypztm/buffons-needle-fun-and-fundamentals Buffon's needle: fun and fundamentals (presentation)] at [[slideshare]]
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