Buffon's Needle README

If you randomly drop needles onto a plane with parallel lines spaced a distance of 2 needles apart, the total number of needles over those that intersect a line approaches $\pi$??? Mysterious, huh? Let's see what's going on!

This is a small project to learn and practice fascinating (geometric) probability theory, specifically Monte Carlo simulations.

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Here is the Math Explained

Let’s say the needle has length $\ell$.

• For intersection, we care about the position $x$ ($y$ in image) of the needle’s center and the angle $\theta$.
• The probability density of the center of the match is
  $P_x = \frac{1}{\ell}$
• The probability density of the angle is
  $P_\theta = \frac{2}{\pi}$
• The probability for the match to cross a line is given by the double integral
  $P = \iint \bigl(P_x ,P_\theta\bigr),dx,d\theta$
• The condition for the match to cross the line is
  $x < \frac{\ell}{2},\sin(\theta)$
• Therefore, the $x$ integration bounds are from $0$ to $\frac{\ell}{2},\sin(\theta)$, and the $\theta$ integration goes from $0$ to $\frac{\pi}{2}$.

Following through with these integrals, we get:

$P =\ \int_{\theta=0}^{\frac{\pi}{2}} \int_{x=0}^{\frac{\ell}{2}\sin(\theta)} \left( \frac{1}{\ell} \times \frac{2}{\pi} \right),dx),d\theta =\ \frac{1}{\pi}$

Going back to the problem, if you drop $N_{\text{tot}}$ total needles, we expect approximately $N_{\text{cross}} =\ N_{\text{tot}} \times \frac{1}{\pi}$ needles to cross a line. Rearranging this, we get $\frac{N_{\text{tot}}}{N_{\text{cross}}} \approx\ \pi$ As $N_{\text{tot}}$ becomes very large, this ratio converges to $\pi$ (by the Law of Large Numbers), making it a neat way to estimate $\pi$ empirically.