## Introduction

In the book One Two Three... Infinity: Facts and Speculations of Science by George Gamow, page 229 describes a method to estimate $\pi$ based on a chance experiment. The experiment involves dropping a needle or match stick on a paper containing equidistant horizontal parallel lines. Assuming the length of needle is smaller or equal to the distance between horizontal lines, the probability that the needle or match stick crosses the horizontal line turns out to be an expression involving $\pi$. In the article that follows, I will define the problem and provide two different proofs of it.

## The problem

Let me state the problem more formally as:

A needle or match stick of length $l$ is dropped on a paper that has drawn on it horizontal equidistant parallel lines separated by a distance $d$. The length of the needle or match stick $l$ is less than or equal to $d$, ${l}\le{d}$. The probability that needle crosses turns out to be: $${p=}\frac{2l}{\pi{d}}$$

The above expression involving $\pi$ is a cool thing in that one can do an experiment that involves randomly dropping a needle or match stick several times on a ruled paper with horizontal equidistant parallel lines. By counting the number of times the needle or match crosses the horizontal line to the total number of times the needle or match stick is dropped, one should be able to estimate the value of $\pi$

For simplicity let's assume the needle or the match stick dropped has two unit length and the distance between the line is also of two unit length. Due to symmetry of the figure, we consider a portion of the figure in the proof below.

## A Geometric Proof

Let's consider the cases under which the needle may cross the horizontal line or not. We need to consider two metrics: the orientation of the needle relative to the horizontal lines ($\theta$) and the vertical distance from the middle of the needle to the nearest horizontal line.

If the center of the needle is close to the horizontal line and angle along the direction of horizontal is large, needle crosses the horizontal line. In contrast, the needle doesn't cross the horizontal if the angle it makes the horizontal is small or the distance of the needle from horizontal is large.We can state more precisely that the needle shall intersect if projection of needle along vertical is larger than the distance from the center of needle and no intersection for the opposite case.

Let's draw a graph with X axis as the angle of fallen needle ($\theta$) relative to horizontal and the Y axis as the projected length of half of needle along the vertical direction ($d$). This length is nothing but ${\sin \theta}$ . When the angle is $\frac {\pi}{2}$, the sin value is 1, i.e, the needle is vertical. For other values, this is given by the sin value and this varies between 0 and 1. Thus $ 0 \le y \le $ 1 and $0 \le \theta \le \frac{\pi}{2}$. The area for which $ \sin \theta \gt x $ corresponds to the intersection of needle with the horizontal line (blue shaded region in the figure). The remaining area corresponds to the needle not crossing the horizontal line (shaded red in the figure).

Thus the probability of needle crossing the horizontal line is:$$\frac {\text{Area of the shaded region}} {\text{Area of the rectangle}} = \frac {\int_0^{\frac {\theta} {2}} \sin \theta \,dx} {\frac {\pi} {2} . 1} = \frac {[-cos \theta]_0^{\frac {\theta} {2}}} {\frac {\pi} {2}} = \frac {2}{\pi} $$

## Yet Another Proof

There is yet another interesting proof of above problem.

First let's relate the probability which the needle crosses the horizontal line with the expected value of the number of intersections.

First let's relate the probability which the needle crosses the horizontal line with the expected value of the number of intersections.

The expected value of the number of intersections $E(N)$ of a needle l: $$E(N)= p_1 + 2 * p_2 + 3 * p_3 + .....$$ where $p_i$ represents the probability of making i intersections with the horizontal.

If length of needle l is less than the distance d between the horizontal lines $ l \lt d$, then $$E(N)=p_1$$. Thus we have converted the probability the needle crosses the horizontal as equivalent to the expected value of number of intersections.

Next step is to make use of the linearity of expectation. Consider two needles of length $l_1$ and $l_2$ and their number of intersections $N_1$ and $N_2$ (which are random variables). Since $N_1$ and $N_2$ are independent, the expected value of total number of intersections is

$$E(N_1 + N_2) = E(N_1) + E(N_2)$$.

Now if we join the needles of length $l_1$ and $l_2$ at their ends, still the expectation is additive in nature. Thus a needle of unit length $1$ can be approximated by $m$ equal small needles each of length $\frac{1}{m}$. Thus

$$E(N_1 + N_2 + N_3 + ... + N_m = E(N_1) + E(N_2) + E(N_3) + ... + E(N_m) = m . E(N_1)$$ $$ E(N_i) = \text {length of segment i} . E(N)$$

Thus $E(N)$is length times a constant ($l. c$) and the constant of proportionality $c$ is the probability of getting one crossing from the needle.We make use of this result in the subsequent steps.

Third step is consider the needle as circle C of unit diameter. This circular needle is a needle of length $1.\pi$ (the circumference) and produces exactly two intersections with the horizontal parallel lines.

In the last step, a polygon of m sides can both be inscribed and circumscribed. Let's call the inscribed polygon $P_n$ and the circumscribed polygon $P^n$.

The expected number of intersections satisfy:

$$ E(P_n) \le E(C) \le E(P^n)$$

$$ c . l(P_n) \le 2 \le c. l(P^n) $$

As $n \to \infty$, $P_n$ and $P^n$ approximate C. Thus $$ \lim_{n \to \infty} = 1 . \pi = \lim_{n \to \infty}$$ $$ \text {Therefore,} c\pi \le 2 \le c\pi$$ $$c = \frac {2}{\pi} $$.

This completes the proof.

An consequence of the above proof, if you drop a curve across a system of parallel lines, the expected number of intersections doesn't depend on the shape of the curve instead on it's length. Thus curve is thread of length $l$ which can be made into any shape that is to be dropped. Expected number of Intersections with the parallel lines only depend on it's length.

In the above proof, we have seen that circle crosses the parallel lines the same number of times irrespective where it falls on ruled horizontal parallel lines. If a curve enclosed by two parallel lines has the same width $d$ as the distance between the parallel lines, then it intersects the parallel lines exactly twice. This is a general theorem stated by Emile Barbier. Any closed (convex) curve of constant width $d$ has perimeter $L=\pi \times d$. Examples of shape other than the circle is Reuleaux curve.

## Conclusion

In this article, we have studied a problem known as the Buffon Needle problem, named after it's inventor Georges-Louis_Leclerc,_Comte_de_Buffon in the year 1777. Here are some exercises for the reader to try out.

1. Consider a needle of dimension $l$ being dropped on a rectangular grid of dimensions $a \times b$. If $l \le min(a,b) $. Now find the probability that the needle intersects a boundary.

2. Find it for a regualar hexagon of side $a$ and assume length of needle $l \le a$.

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