Pi approximation with Montecarlo: Why not just evenly distributed points?
Let us compare the efficiency of both methods. The first method generates random points on the unit square $[-1,1]^2$, and checks how many of them lie inside the unit circle – the fraction should be $\pi/4$. If we sample $N$ points, our estimate has roughly normal distribution with mean $\pi/4$ and variance $\pi/4(1-\pi/4)/N$, and so we expect an error of order $1/\sqrt{N}$.

The second method generates an $M \times M$ grid inside $[-1,1]^2$ (we can take $M = \sqrt{N}$ for later comparison), and checks how many of the $M^2$ points lie inside the unit circle. If we think of each point as standing in for a grid square, then we can estimate the number of problematic squares (those intersecting the circle) as $\Theta(M)$ out of the $M^2$, leading to an overall error of order $1/M = 1/\sqrt{N}$, just as before.

We can speed up the second method, however: for a given value of $x$, we can use binary search to locate the $y$ coordinates at which the circle crosses the grid. This gives an $O(M\log M)$ algorithm, an almost quadratic speedup over the two algorithms mentioned above. This is very similar to evaluating the integral $2\int_{-1}^1 \sqrt{1-x^2} \, dx$, implementing the square root using a binary search.