Buffon's needle problem

The problem in more mathematical terms is: Given a needle of length {{mvar|l}} dropped on a plane ruled with parallel lines {{mvar|t}} units apart, what is the probability that the needle will lie across a line upon landing?

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 {{mvar|x}} between 0 and {{math|{{sfrac|t|2}}}} is

:$f\_X(x)=$

\begin{cases}

\dfrac{2}{t} &:\ 0 \le x \le \dfrac{t}{2}\\[4px]

0 &: \text{elsewhere.}

\end{cases}

Here, {{math|x {{=}} 0}} represents a needle that is centered directly on a line, and {{math|x {{=}} {{sfrac|t|2}}}} represents a needle that is perfectly centered between two lines. The uniform PDF assumes the needle is equally likely to fall anywhere in this range, but could not fall outside of it.

The uniform probability density function of {{mvar|θ}} between 0 and {{math|{{sfrac|π|2}}}} is

:$f\_\backslash Theta(\backslash theta)=$

\begin{cases}

\dfrac{2}{\pi} &:\ 0 \le \theta \le \dfrac{\pi}{2}\\[4px]

0 &: \text{elsewhere.}

\end{cases}

Here, {{math|θ {{=}} 0}} represents a needle that is parallel to the marked lines, and {{math|θ {{=}} {{sfrac|π|2}}}} radians 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,The problem formulation here avoids having to work with Regular conditional probability densities. so the joint probability density function is the product

:$f\_\{X,\backslash Theta\}(x,\backslash theta)=$

\begin{cases}

\dfrac{4}{t\pi} &:\ 0 \le x \le \dfrac{t}{2}, \ 0 \le \theta \le \dfrac{\pi}{2}\\[4px]

0 &: \text{elsewhere.}

\end{cases}

The needle crosses a line if

:$x\; \backslash le\; \backslash frac\; \{l\}\{2\}\; \backslash sin\backslash theta.$

Now there are two cases.

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

:$P\; =\; \backslash int\_\{\backslash theta=0\}^\backslash frac\{\backslash pi\}\{2\}\; \backslash int\_\{x=0\}^\{\backslash frac\{l\}\{2\}\backslash sin\; \backslash theta\}\; \backslash frac\{4\}\{t\backslash pi\}\backslash ,dx\backslash ,d\backslash theta\; =\; \backslash frac\{2\; l\}\{t\backslash pi\}.$

Suppose {{math|l > t}}. In this case, integrating the joint probability density function, we obtain:

:$\backslash int\_\{\backslash theta=0\}^\backslash frac\{\backslash pi\}\{2\}\; \backslash int\_\{x=0\}^\{m(\backslash theta)\}\; \backslash frac\{4\}\{t\backslash pi\}\backslash ,dx\backslash ,d\backslash theta,$

where {{math|m(θ)}} is the minimum between {{math|{{sfrac|l|2}} sinθ}} and {{math|{{sfrac|t|2}}}}.

Thus, performing the above integration, we see that, when {{math|l > t}}, the probability that the needle will cross at least one line is

:$P\; =\; \backslash frac\{2l\}\{t\backslash pi\}\; -\; \backslash frac\{2\}\{t\backslash pi\}\backslash left(\backslash sqrt\{l^2\; -\; t^2\}\; +\; t\backslash arcsin\; \backslash frac\{t\}\{l\}\; \backslash right)+1$

or

:$P\; =\; \backslash frac\{2\}\{\backslash pi\}\; \backslash arccos\; \backslash frac\{t\}\{l\}\; +\; \backslash frac\{2\}\{\backslash pi\}\backslash cdot\backslash frac\{l\}\{t\}\; \backslash left(1\; -\; \backslash sqrt\{1\; -\; \backslash left(\; \backslash frac\; t\; l\; \backslash right)^2\; \}\; \backslash right).$

In the second expression, the first term represents the probability of the angle of the needle being such that it will always cross at least one line. The right term represents the probability that the needle falls at an angle where its position matters, and it crosses the line.

Alternatively, notice that whenever {{mvar|θ}} has a value such that {{math|l sin θ ≤ t}}, that is, in the range {{math|0 ≤ θ ≤ arcsin {{sfrac|t|l}}}}, the probability of crossing is the same as in the short needle case. However if {{math|l sin θ > t}}, that is, {{math|arcsin {{sfrac|t|l}} < θ ≤ {{sfrac|π|2}}}} the probability is constant and is equal to 1.

:$\backslash begin\{align\}$

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}

The following solution for the "short needle" case, while equivalent to the one above, has a more visual flavor, and avoids iterated integrals.

We can calculate the probability {{mvar|P}} as the product of two probabilities: {{math|P {{=}} P₁ · P₂}}, where {{math|P₁}} is the probability that the center of the needle falls close enough to a line for the needle to possibly cross it, and {{math|P₂}} is the probability that the needle actually crosses the line, given that the center is within reach.

Looking at the illustration in the above section, it is apparent that the needle can cross a line if the center of the needle is within {{math|{{sfrac|l|2}}}} units of either side of the strip. Adding {{math|{{sfrac|l|2}} + {{sfrac|l|2}}}} from both sides and dividing by the whole width {{mvar|t}}, we obtain {{math|P₁ {{=}} {{sfrac|l|t}}}}.

File:Buffon's needle corrected.PNG

Now, we assume that the center is within reach of the edge of the strip, and calculate {{math|P₂}}. To simplify the calculation, we can assume that $l\; =\; 2$.

Let {{mvar|x}} and {{mvar|θ}} be as in the illustration in this section. Placing a needle's center at {{mvar|x}}, the needle will cross the vertical axis if it falls within a range of {{math|2θ}} radians, out of {{pi}} radians of possible orientations. This represents the gray area to the left of {{mvar|x}} in the figure. For a fixed {{mvar|x}}, we can express {{mvar|θ}} as a function of {{mvar|x}}: {{math|θ(x) {{=}} arccos(x)}}. Now we can let {{mvar|x}} range from 0 to 1, and integrate:

:$\backslash begin\{align\}$

P_2 &= \int_0^1 \frac{2\theta(x)}{\pi}\,dx \\[6px]

&= \frac{2}{\pi}\int_0^1 \cos^{-1}(x)\,dx \\[6px]

&= \frac{2}{\pi}\cdot 1 = \frac{2}{\pi}.

\end{align}

Multiplying both results, we obtain {{math|P {{=}} P₁ · P₂ {{=}} {{sfrac|l|t}} · {{sfrac|2|π}} {{=}} {{sfrac|2l|tπ}}}} as above.

There is an even more elegant and simple method of calculating the "short needle case". The end of the needle farthest away from any one of the two lines bordering its region must be located within a horizontal (perpendicular to the bordering lines) distance of {{math|l cos θ}} (where {{mvar|θ}} is the angle between the needle and the horizontal) from this line in order for the needle to cross it. The farthest this end of the needle can move away from this line horizontally in its region is {{mvar|t}}. The probability that the farthest end of the needle is located no more than a distance {{math|l cos θ}} away from the line (and thus that the needle crosses the line) out of the total distance {{mvar|t}} it can move in its region for {{math|0 ≤ θ ≤ {{sfrac|π|2}}}} is given by

: $\backslash begin\{align\}$

P &= \frac{\displaystyle\int_0^\frac{\pi}{2} l\cos\theta \, d\theta}{\displaystyle\int_0^\frac{\pi}{2} t \, d\theta} \\[6px]

&= \frac l t\cdot\frac{\displaystyle\int_0^\frac{\pi}{2} \cos\theta \, d\theta}{\displaystyle\int_0^\frac{\pi}{2} d\theta} \\[6px]

&= \frac l t\cdot\frac{1}{\,\frac{\pi}{2}\,} \\[6px]

&=\frac{2l}{t\pi}.

\end{align}

The short-needle problem can also be solved without any integration, in a way that explains the formula for {{mvar|p}} from the geometric fact that a circle of diameter {{mvar|t}} will cross the distance {{mvar|t}} strips always (i.e. with probability 1) in exactly two spots. This solution was given by Joseph-Émile Barbier in 1860{{Cite book|title=Proofs from THE BOOK|last1=Aigner|first1=Martin|last2=Ziegler|first2=Günter M.|edition=2nd|year=2013|pages=189–192|publisher=Springer Science & Business Media}} and is also referred to as "Buffon's noodle".

File:Streicholz-Pi.jpg

File:Buffon needle experiment compressed.gif based simulation using Matplotlib to sketch Buffon's needle experiment with the parameters {{math|1=t = 5.0}}, {{math|1=l = 2.6}}. Observe the calculated value of {{pi}} ({{mvar|y}}-axis) approaching 3.14 as the number of tosses ({{mvar|x}}-axis) approaches infinity.]]

In the first, simpler case above, the formula obtained for the probability {{mvar|P}} can be rearranged to

:$\backslash pi\; =\; \backslash frac\{2l\}\{tP\}.$

Thus, if we conduct an experiment to estimate {{mvar|P}}, we will also have an estimate for {{pi}}.

Suppose we drop {{mvar|n}} needles and find that {{mvar|h}} of those needles are crossing lines, so {{mvar|P}} is approximated by the fraction {{math|{{sfrac|h|n}}}}. This leads to the formula:

:$\backslash pi\; \backslash approx\; \backslash frac\{2l\backslash cdot\; n\}\{t\; h\}.$

In 1901, Italian mathematician Mario Lazzarini performed Buffon's needle experiment. Tossing a needle 3,408 times, he obtained the well-known approximation {{sfrac|355|113}} for {{pi}}, accurate to six decimal places.{{cite journal |first=M. |last=Lazzarini |title=Un'applicazione del calcolo della probabilità alla ricerca sperimentale di un valore approssimato di {{pi}} |language=it |trans-title=An Application of Probability Theory to Experimental Research of an Approximation of {{pi}}|journal=Periodico di Matematica per l'Insegnamento Secondario |date=1901 |volume=4 |pages=140–143 |url=https://books.google.com/books?id=n3FVAAAAYAAJ&pg=RA1-PA140}}

Lazzarini's "experiment" is an example of confirmation bias, as it was set up to replicate the already well-known approximation of {{sfrac|355|113}} (in fact, there is no better rational approximation with fewer than five digits in the numerator and denominator, see also Milü), yielding a more accurate "prediction" of {{pi}} than would be expected from the number of trials, as follows:

Lee Badger, [http://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/1995/Badger.pdf 'Lazzarini's Lucky Approximation of {{pi}}'], Mathematics Magazine 67, 1994, 83–91.

Lazzarini chose needles whose length was {{sfrac|5|6}} of the width of the strips of wood. In this case, the probability that the needles will cross the lines is {{math|{{sfrac|5|3π}}}}. Thus if one were to drop {{mvar|n}} needles and get {{mvar|x}} crossings, one would estimate {{pi}} as

:$\backslash pi\; \backslash approx\; \backslash frac\; 53\; \backslash cdot\; \backslash frac\; nx$

So if Lazzarini was aiming for the result {{sfrac|355|113}}, he needed {{mvar|n}} and {{mvar|x}} such that

:$\backslash frac\{355\}\{113\}\; =\; \backslash frac\; 53\; \backslash cdot\; \backslash frac\; nx,$

or equivalently,

:$x\; =\; \backslash frac\{113\; n\}\{213\}.$

To do this, one should pick {{mvar|n}} as a multiple of 213, because then {{math|{{sfrac|113n|213}}}} is an integer; one then drops {{mvar|n}} needles, and hopes for exactly {{math|x {{=}} {{sfrac|113n|213}}}} successes. If one drops 213 needles and happens to get 113 successes, then one can triumphantly report an estimate of {{pi}} 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 {{nowrap|3,408 {{=}} 213 × 16}} trials, making it seem likely that this is the strategy he used to obtain his "estimate".

The above description of strategy might even be considered charitable to Lazzarini. A statistical analysis of intermediate results he reported for fewer tosses leads to a very low probability of achieving such close agreement to the expected value all through the experiment. This makes it very possible that the "experiment" itself was never physically performed, but based on numbers concocted from imagination to match statistical expectations, but too well, as it turns out.

Dutch science journalist Hans van Maanen argues, however, that Lazzarini's article was never meant to be taken too seriously as it would have been pretty obvious for the readers of the magazine (aimed at school teachers) that the apparatus that Lazzarini said to have built cannot possibly work as described.Hans van Maanen, [https://skepsis.nl/lazzarini/ 'Het stokje van Lazzarini' (Lazzarini's stick)], "Skepter" 31.3, 2018.

Now consider the case where the plane contains two sets of parallel lines orthogonal to one another, creating a standard perpendicular grid. We aim to find the probability that the needle intersects at least one line on the grid. Let {{mvar|a}} and {{mvar|b}} be the sides of the rectangle that contains the midpoint of the needle whose length is {{mvar|l}}. Since this is the short needle case, {{math|l < a}}, {{math|l < b}}. Let {{math|(x,y)}} mark the coordinates of the needle's midpoint and let {{mvar|φ}} mark the angle formed by the needle and the {{mvar|x}}-axis. Similar to the examples described above, we consider {{mvar|x}}, {{mvar|y}}, {{mvar|φ}} to be independent uniform random variables over the ranges {{math|0 ≤ x ≤ a}}, {{math|0 ≤ y ≤ b}}, {{math|−{{sfrac|π|2}} ≤ φ ≤ {{sfrac|π|2}}}}.

To solve such a problem, we first compute the probability that the needle crosses no lines, and then we take its complement. We compute this first probability by determining the volume of the domain where the needle crosses no lines and then divide that by the volume of all possibilities, {{mvar|V}}. We can easily see that {{math|V {{=}} πab}}.

Now let {{math|V*}} be the volume of possibilities where the needle does not intersect any line. Developed by J.V. Uspensky,J.V. Uspensky, [https://ia601407.us.archive.org/17/items/in.ernet.dli.2015.263184/2015.263184.Introduction-To.pdf 'Introduction to Mathematical Probability'], 1937, 255.

:$V^*\; =\; \backslash int\_\{-\backslash frac\{\backslash pi\}\{2\}\}^\{\backslash frac\{\backslash pi\}\{2\}\}\; F(\backslash varphi)\backslash ,d\backslash varphi$

where {{math|F(φ)}} is the region where the needle does not intersect any line given an angle {{mvar|φ}}. To determine {{math|F(φ)}}, let's first look at the case for the horizontal edges of the bounding rectangle. The total side length is {{mvar|a}} and the midpoint must not be within {{math|{{sfrac|l|2}} cos φ}} of either endpoint of the edge. Thus, the total allowable length for no intersection is {{math|a − 2({{sfrac|l|2}} cos φ)}} or simply just {{math|a − l cos φ}}. Equivalently, for the vertical edges with length {{mvar|b}}, we have {{math|b ± l sin φ}}. The ± accounts for the cases where {{mvar|φ}} is positive or negative. Taking the positive case and then adding the absolute value signs in the final answer for generality, we get

:$F(\backslash varphi)\; =(a\; -\; l\backslash cos\backslash varphi)(b\; -\; l\backslash sin\backslash varphi)\; =\; ab\; -\; bl\; \backslash cos\; \backslash varphi\; -\; al|\backslash sin\; \backslash varphi|+\; \backslash tfrac\{1\}\{2\}l^2|\backslash sin\; 2\backslash varphi|.$

Now we can compute the following integral:

:$V^*\; =\; \backslash int\_\{-\backslash frac\{\backslash pi\}\{2\}\}^\{\backslash frac\{\backslash pi\}\{2\}\}\; F(\backslash varphi)\backslash ,d\backslash varphi\; =\; \backslash pi\; ab-2bl-2al+l^2\; .$

Thus, the probability that the needle does not intersect any line is

:$\backslash frac\{V^*\}\{V\}=\; \backslash frac\{\backslash pi\; ab-2bl-2al+l^2\}\{\backslash pi\; a\; b\}\; =\; 1\; -\; \backslash frac\{2l(a+b)-l^2\}\{\backslash pi\; a\; b\}.$

And finally, if we want to calculate the probability, {{mvar|P}}, that the needle does intersect at least one line, we need to subtract the above result from 1 to compute its compliment, yielding

:$P\; =\; \backslash frac\{2l(a+b)-l^2\}\{\backslash pi\; a\; b\}$.

As mentioned above, Buffon's needle experiment can be used to estimate {{pi}}. This fact holds for Laplace's extension too since {{pi}} shows up in that answer as well. The following question then naturally arises and is discussed by E.F. Schuster in 1974.E. F. Schuster, [https://www.jstor.org/stable/2318907 Buffon's Needle Experiment], The American Mathematical Monthly, 1974, 29-29.

Is Buffon's experiment or Laplace's a better estimator of the value of {{pi}}? Since in Laplace's extension there are two sets of parallel lines, we compare {{mvar|N}} drops when there is a grid (Laplace), and {{math|2N}} drops in Buffon's original experiment.

Let {{mvar|A}} be the event that the needle intersects a horizontal line (parallel to the {{mvar|x}}-axis)

:$x\; =$

\begin{cases}

1 &: \text{intersection occurs}\\

0 &: \text{no intersection}

\end{cases}

and let {{mvar|B}} be the event that the needle intersects a vertical line (parallel to the {{mvar|y}}-axis)

:$y\; =$

\begin{cases}

1 &: \text{intersection occurs}\\

0 &: \text{no intersection}

\end{cases}

For simplicity in the algebraic formulation ahead, let {{math|a {{=}} b {{=}} t {{=}} 2l}} such that the original result in Buffon's problem is {{math|P(A) {{=}} P(B) {{=}} {{sfrac|1|π}}}}. Furthermore, let {{math|N {{=}} 100}} drops.

Now let us examine {{math|P(AB)}} for Laplace's result, that is, the probability the needle intersects both a horizontal and a vertical line. We know that

:$P(AB)\; =\; 1\; -\; P(AB\text{'})\; -\; P(A\text{'}B)\; -\; P(A\text{'}B\text{'}).$

From the above section, {{math|P(A′B′)}}, or the probability that the needle intersects no lines is

:$P(A\text{'}B\text{'})\; =\; 1\; -\; \backslash frac\{2l(a+b)-l^2\}\{\backslash pi\; a\; b\}\; =\; 1\; -\; \backslash frac\{2l(4l)-l^2\}\{4\; l^2\; \backslash pi\}\; =\; 1\; -\; \backslash frac\{7\}\{4\backslash pi\}\; .$

We can solve for {{math|P(A′B)}} and {{math|P(AB′)}} using the following method:

:$\backslash begin\{align\}$

P(A) &= \frac{1}{\pi} = P(AB) + P(AB') \\[4px]

P(B) &= \frac{1}{\pi} = P(AB) + P(A'B).

\end{align}

Solving for {{math|P(A′B)}} and {{math|P(AB′)}} and plugging that into the original definition for {{math|P(AB)}} a few lines above, we get

:$P(AB)\; =\; 1\; -\; 2\backslash left(\backslash frac\{1\}\{\backslash pi\}-P(AB)\backslash right)\; -\; \backslash left(1\; -\; \backslash frac\{7\}\{4\backslash pi\}\backslash right)\; =\; \backslash frac\{1\}\{4\backslash pi\}$

Although not necessary to the problem, it is now possible to see that {{math|P(A′B) {{=}} P(AB′) {{=}} {{sfrac|3|4π}}}}. With the values above, we are now able to determine which of these estimators is a better estimator for {{pi}}. For the Laplace variant, let {{mvar|p̂}} be the estimator for the probability that there is a line intersection such that

:$\backslash hat\{p\}\; =\; \backslash frac\{1\}\{100\}\backslash sum\_\{n=1\}^\{100\}\backslash frac\{x\_n+y\_n\}\{2\}$.

We are interested in the variance of such an estimator to understand the usefulness or efficiency of it. To compute the variance of {{mvar|p̂}}, we first compute {{math|Var(x_n + y_n)}} where

:$\backslash operatorname\{Var\}(x\_n\; +\; y\_n)\; =\; \backslash operatorname\{Var\}(x\_n)\; +\backslash operatorname\{Var\}(y\_n)\; +\; 2\backslash operatorname\{Cov\}(x\_n,y\_n).$

Solving for each part individually,

:$\backslash begin\{align\}$

\operatorname{Var}(x_n) = \operatorname{Var}(y_n) &= \sum_{i = 1}^2 p_i\bigl(x_i-\mathbb{E}(x_i)\bigr)^2 \\[6px]

&= P(x_i=1)\left(1-\frac{1}{\pi}\right)^2 + P(x_i=0)\left(0-\frac{1}{\pi}\right)^2 \\[6px]

&= \frac{1}{\pi}\left(1-\frac{1}{\pi}\right)^2 + \left(1-\frac{1}{\pi}\right)\left(-\frac{1}{\pi}\right)^2 = \frac{1}{\pi}\left(1-\frac{1}{\pi}\right). \\[12px]

\operatorname{Cov}(x_n,y_n) &= \mathbb{E}(x_ny_n) - \mathbb{E}(x_n)\mathbb{E}(y_n)

\end{align}

We know from the previous section that

:$\backslash mathbb\{E\}(x\_ny\_n)\; =\; P(AB)\; =\; \backslash frac\{1\}\{4\backslash pi\}$

yielding

:

Thus,

:$\backslash operatorname\{Var\}(x\_n+y\_n)\; =\; \backslash frac\{1\}\{\backslash pi\}\backslash left(1-\backslash frac\{1\}\{\backslash pi\}\backslash right)\; +\; \backslash frac\{1\}\{\backslash pi\}\backslash left(1-\backslash frac\{1\}\{\backslash pi\}\backslash right)\; +\; 2\backslash left(\backslash frac\{\backslash pi-4\}\{4\backslash pi^2\}\backslash right)\; =\; \backslash frac\{5\backslash pi\; -\; 8\}\{2\backslash pi^2\}$

Returning to the original problem of this section, the variance of estimator {{mvar|p̂}} is

:$\backslash operatorname\{Var\}(\backslash hat\{p\})\; =\; \backslash frac\{1\}\{200^2\}(100)\backslash left(\backslash frac\{5\backslash pi\; -\; 8\}\{2\backslash pi^2\}\backslash right)\; \backslash approx\; 0.000\backslash ,976.$

Now let us calculate the number of drops, {{mvar|M}}, needed to achieve the same variance as 100 drops over perpendicular lines. If {{math|M < 200}} then we can conclude that the setup with only parallel lines is more efficient than the case with perpendicular lines. Conversely if {{mvar|M}} is equal to or more than 200, than Buffon's experiment is equally or less efficient, respectively. Let {{mvar|q̂}} be the estimator for Buffon's original experiment. Then,

:$\backslash hat\{q\}\; =\; \backslash frac\{1\}\{M\}\backslash sum\_\{m=1\}^M\; x\_m$

and

:$\backslash operatorname\{Var\}(\backslash hat\{q\})\; =\; \backslash frac\{1\}\{M^2\}(M)\backslash operatorname\{Var\}(x\_m)\; =\; \backslash frac\{1\}\{M\}\backslash cdot\backslash frac\{1\}\{\backslash pi\}\backslash left(1-\backslash frac\{1\}\{\backslash pi\}\backslash right)\; \backslash approx\; \backslash frac\{0.217\}\{M\}$

Solving for {{mvar|M}},

:$\backslash frac\{0.217\}\{M\}\; =\; 0.000\backslash ,976\; \backslash implies\; M\; \backslash approx\; 222.$

Thus, it takes 222 drops with only parallel lines to have the same certainty as 100 drops in Laplace's case. This isn't actually surprising because of the observation that {{math|Cov(x_n,y_n) < 0}}. Because {{mvar|x_n}} and {{mvar|y_n}} are negatively correlated random variables, they act to reduce the total variance in the estimator that is an average of the two of them. This method of variance reduction is known as the antithetic variates method.

• Bertrand paradox (probability)

{{Reflist}}

• {{cite journal|doi = 10.2307/2690682|last = Badger|first = Lee|title=Lazzarini's Lucky Approximation of {{pi}}|journal= Mathematics Magazine|volume=67|issue=2|date=April 1994|pages=83–91|jstor=2690682|publisher = Mathematical Association of America}}
• {{cite journal|last = Ramaley|first = J. F.|title = Buffon's Noodle Problem|journal = The American Mathematical Monthly|volume = 76|issue = 8|date=October 1969|pages = 916–918|doi = 10.2307/2317945|jstor = 2317945|publisher = Mathematical Association of America}}
• {{cite book|last = Mathai|first = A. M.|title = An Introduction to Geometrical Probability|year = 1999|publisher = Gordon & Breach|location = Newark|url = https://books.google.com/books?id=FV6XncZgfcwC | isbn=978-90-5699-681-9 |page=5}}
• {{cite journal| last = Dell|first = Zachary| author2 = Franklin, Scott V.|title = The Buffon-Laplace needle problem in three dimensions|journal = Journal of Statistical Mechanics: Theory and Experiment|date=September 2009| volume = 2009| issue = 9| pages = 010| doi = 10.1088/1742-5468/2009/09/P09010| bibcode = 2009JSMTE..09..010D| s2cid=32470555 }}
• Schroeder, L. (1974). "Buffon's needle problem: An exciting application of many mathematical concepts". Mathematics Teacher, 67 (2), 183–6.
• Uspensky, James Victor. "Introduction to mathematical probability." (1937).

{{commons category|Buffon's needle}}

• [http://www.cut-the-knot.org/fta/Buffon/buffon9.shtml Buffon's Needle Problem] 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]
• [https://www.angelfire.com/wa/hurben/buff.html Buffon's Needle Java Applet]
• [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
• [https://yihui.name/animation/example/buffon-needle/ Animations for the Simulation of Buffon's Needle] by Yihui Xie using the R package [https://cran.r-project.org/package=animation animation]
• [http://www.ventrella.com/Buffon/index.html 3D Physical Animation] by Jeffrey Ventrella
• {{cite web|last=Padilla|first=Tony|title=π Pi and Buffon's Needle|url=http://www.numberphile.com/pi/pi_matches.html|work= Numberphile |publisher=Brady Haran|access-date=2013-04-09|archive-url=https://web.archive.org/web/20130517072808/http://www.numberphile.com/pi/pi_matches.html|archive-date=2013-05-17|url-status=dead}}

Category:Applied probability

Category:Integral geometry

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