In 1773, Georges Louis Leclerc, le Comte de Buffon devised a
mathematical experiment...

To perform this experiment at home, you'll need:

• Needles. (Preferably substitute something less sharp.)

• A grid of equally spaced parallel lines (unit distance apart).

How often does a randomly tossed needle cross a line?
Call this probability $E$ (depends on length $\ell$ of a needle).

\[\frac{2}{\text{average number of crossings}} \approx \frac{2}{E}\]

\[\frac{2}{\text{average number of crossings}} \approx \frac{2}{E} = \pi?\]

By breaking a needle in two, one checks that \[E=C \ell,\] where $\ell$ is the length of the needle and $C$ is some constant.



But how can we determine this constant?

$E=C\ell$ only depends on $\ell$, and doesn't change for a "noodle".

$E=C\ell$ only depends on $\ell$, and doesn't change for a "noodle".
Since we can approximate a noodle by straight needles.

$E=C\ell$ only depends on $\ell$, and doesn't change for a "noodle".

The length $\ell$ of such a noodle is our definition of $\pi$.

Such a noodle will always have exactly two intersections:

The length $\ell$ of such a noodle is our definition of $\pi$.

Such a noodle will always have exactly two intersections:
Since $E = C \ell$, we can now determine our constant $C$: \[2 = C \pi, \text{ or } C = \frac{2}{\pi}.\]

\[\frac{2}{\text{average number of crossings}} \approx \frac{2}{E} = \frac{2}{C\ell} = \frac{2}{\pi/2} = \pi.\]

Approximating $\pi$ using Buffon's needles is slow. There are more efficient ways....

• Inscribe/circumscribe regular polygons in a circle.





• Infinite series.

\[ \frac{\pi}{4} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\] \[ \pi = \sum_{k=0}^\infty \frac{1}{16^k} \left( \frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right)\] \[ \frac{\pi}{2} = \frac{ 2\cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdots}{ 1\cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdots}\] \[ \frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \cdots\]