# Picking random points in the volume of sphere with uniform probability

I have a sphere of radius $R_{s}$, and I would like to pick random points in its volume with uniform probability. How can I do so while preventing any sort of clustering around poles or the center of the sphere?

Since I’m unable to answer my own question, here’s another solution:

Using the strategy suggested by Wolfram MathWorld for picking points on the surface of a sphere: Let $\theta$ be randomly distributed real numbers over the interval $[0,2\pi]$, let $\phi=\arccos(2v−1)$ where $v$ is a random real number over the interval $[0,1]$, and let $r=R_s (\mathrm{rand}(0,1))^\frac13$. Converting from spherical coordinates, a random point in $(x,y,z)$ inside the sphere would therefore be: $((r\cos(\theta)\sin(\phi)),(r\sin(\theta)\sin(\phi)),(r\cos(\phi)))$.

A quick test with a few thousand points in the unit sphere appears to show no clustering. However, I’d appreciate any feedback if someone sees a problem with this approach.

## Answer

Let’s say your sphere is centered at the origin $(0,0,0)$.

For the distance $D$ from the origin of your random pointpoint, note that you want $P(D \le r) = \left(\frac{r}{R_s}\right)^3$. Thus if $U$ is uniformly distributed between 0 and 1, taking $D = R_s U^{1/3}$ will do the trick.

For the direction, a useful fact is that if $X_1, X_2, X_3$ are independent normal random variables with mean 0 and variance 1, then

is uniformly distributed on (the surface of) the unit sphere. You can generate normal random variables from uniform ones in various ways; the Box-Muller algorithm is a nice simple approach.

So if you choose $U$ uniformly distributed between 0 and 1, and $X_1, X_2, X_3$ iid standard normal and independent of $U$, then

would produce a uniformly distributed point inside the ball of radius $R_s$.

Attribution
Source : Link , Question Author : MHK , Answer Author : Nate Eldredge