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 $y$ (for horiziontal lines) of the needle’s center and the angle $\theta$.
• The probability density of the center of the needle is
  $P_y = \frac{1}{\ell}$
• The probability density of the angle is
  $P_\theta = \frac{2}{\pi}$
• The probability for the needle to cross a line is given by the double integral
  $P = \iint \bigl(P_y ,P_\theta\bigr),dy,d\theta$
• The condition for the needle to cross the line is
  $y < \frac{\ell}{2}\sin(\theta)$
• Therefore, the $y$ 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:

$$ P = \int_{0}^{\pi/2} \int_{0}^{(\ell/2)\sin(\theta)} \left(\frac{1}{\ell} \cdot \frac{2}{\pi}\right) ,dy,d\theta $$

First, combine the constants:

$$ P = \int_{0}^{\pi/2} \int_{0}^{(\ell/2)\sin(\theta)} \frac{2}{\pi \ell} ,dy,d\theta $$

Now evaluate the inside integral with respect to $y$:

$$ P = \int_{0}^{\pi/2} \frac{2}{\pi \ell} \left[ y \right]_{0}^{(\ell/2)\sin(\theta)} d\theta $$

Plug in the bounds for $y$:

$$ P = \int_{0}^{\pi/2} \frac{2}{\pi \ell} \left( \frac{\ell}{2}\sin(\theta) - 0 \right) d\theta $$

Simplify:

$$ P = \frac{1}{\pi} \int_{0}^{\pi/2} \sin(\theta) ,d\theta $$

Now integrate:

$$ P = \frac{1}{\pi} \left[ -\cos(\theta) \right]_{0}^{\pi/2} $$

Evaluate the bounds:

$$ P = \frac{1}{\pi} \left( -\cos(\pi/2) + \cos(0) \right) $$

Since $\cos(\pi/2)=0$ and $\cos(0)=1$,

$P =\frac{1}{\pi}$

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.

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How Fast Does It Approach $\pi$?

The estimate approaches $\pi$, but slowly. Each needle drop is a Bernoulli trial: it either crosses a line or it does not. Since the crossing probability is

$$p = \frac{1}{\pi}$$

the number of crossings after $N_{\text{tot}}$ drops follows a binomial distribution:

$$N_{\text{cross}} \sim \text{Binomial}\left(N_{\text{tot}}, p\right)$$

The observed crossing probability is

$$\hat{p} = \frac{N_{\text{cross}}}{N_{\text{tot}}}$$

and since $p = 1/\pi$, our estimate of $\pi$ is

$$\hat{\pi} = \frac{1}{\hat{p}} = \frac{N_{\text{tot}}}{N_{\text{cross}}}$$

For a binomial experiment, the standard error of the observed proportion is

$$\text{SE}(\hat{p}) = \sqrt{\frac{p(1-p)}{N_{\text{tot}}}}$$

So the error in the crossing probability shrinks like

$$\frac{1}{\sqrt{N_{\text{tot}}}}$$

To understand how this affects the estimate of $\pi$, think of $\hat{\pi}$ as a function of $\hat{p}$:

$$f(\hat{p}) = \frac{1}{\hat{p}}$$

The derivative is

$$f'(\hat{p}) = -\frac{1}{\hat{p}^2}$$

This tells us how sensitive the $\pi$ estimate is to small errors in the crossing probability. Near the true value $p = 1/\pi$, this derivative is approximately

$$f'(p) = -\pi^2$$

So a small error in $\hat{p}$ becomes about $\pi^2$ times larger when converted into an error in $\hat{\pi}$. However, multiplying by a constant does not change the overall convergence rate.

Therefore, Buffon's Needle has the usual Monte Carlo convergence rate:

$$\text{error} = O\left(\frac{1}{\sqrt{N_{\text{tot}}}}\right)$$

This means that to make the estimate about 10 times more accurate, we need about 100 times more needle drops.