I would like to sample a uniformly random point in a polygon...
If sample a large number they'd be equally likely to fall into two regions if they have the same area.
This would be quite simple if it were a square since I would take two random numbers in [0,1] as my coordinates.
The shape I have is a regular polygon, but I'd like it to work for any polygon.
https://stackoverflow.com/questions/3058150/how-to-find-a-random-point-in-a-quadrangle
One easy way is to find the bounding box for your polygon and use rejection sampling: sample from the bounding box and accept if it falls within the polygon, which will happen with probability $1/2$ at least (I think).
Another possibility is to triangulate your polygon. First sample a triangle in a proportionate way, then sample a random point in the triangle. The latter is simple: up to affine transformations, all triangles are of the form $\{(x,y) : x,y \geq 0, x+y \leq 1\}$. To sample uniformly a point from that distribution, first sample $x \in [0,1]$ according to the density $2(1-x)$ (i.e. sample a uniform $r \in [0,1]$ and compute $x = 1-\sqrt{1-r}$) and then sample $y \in [0,1-x]$ uniformly (i.e. sample a uniform $s \in [0,1]$ and compute $y = (1-x)s$). An even simpler method is to sample $x,y \in [0,1]$, and if $x+y > 1$ replace $(x,y)$ with $(1-x,1-y)$.
This is a little crazy, but should work well even if your polygon is very weird.
Use the Reimann mapping theorem to find a conformal map from the unit disc to your polygon, viewing it as a subset of $\mathbb{C}$. See, for example the references in:
http://siam.org/pdf/news/1297.pdf
Then use the pushforward of a uniform density on the disc as the proposal density in Metropolis-Hastings MCMC sampling.
