Throws | |

Hits | |

π |

We can approximate *π* using nothing more than random numbers and some
simple geometry: we randomly throw darts at a square board of
side *r*; within the square we inscribe a quadrant of a circle of radius *r*
with its centre at *(0, 0)*. We count all of the 'throws'; if a dart lands
within the quadrant, we also count a 'hit'.

For a large number of throws, we see that:

$\frac{\mathrm{hits}}{\mathrm{throws}}=\frac{\mathrm{area-of-quadrant}}{\mathrm{area-of-square}}$

Some half-remembered geometry tells us that:

$\frac{\mathrm{area-of-quadrant}}{\mathrm{area-of-square}}=\frac{{\mathrm{\pi r}}^{2}/4}{{r}^{2}}=\frac{\pi}{4}$

Or:

$\pi =4\left(\frac{\mathrm{area-of-quadrant}}{\mathrm{area-of-square}}\right)=4\left(\frac{\mathrm{hits}}{\mathrm{throws}}\right)$

I first solved this problem as an undergraduate sometime in 1994 as part of a Computational Physics module. Using FORTRAN 77.