TITLE: baricentric distributions
QUESTION [5 upvotes]: Let $\mathcal{P}$ be the convex hull of a point set $p_1, \dotsc, p_n$  (for simplicity, assume that no $p \in P$ lies in the convex hull of the other points.) 
Now, pick a  point uniformly at random from the simplex $x_1 + \dotsc + x_n = 1, x_i \geq  0,$ and take $\phi=\sum x_i p_i.$ The point $\phi$ will be contained in $\mathcal{P}$ and in the simple case where $\mathcal{P}$ is a triangle, $\phi$ is uniformly distributed. In general, however, it is not: The two graphics are for the case $\mathcal{P}$ is a square. The first is a simple plot of a million points sampled from the distribution - the distribution looks like it is trying to be supported on the quadrangle spanned by the midpoints of the sides of the square.

The second is the density histogram:

which seems to indicate that the distribution has a bit of a peak at the center of gravity. 
Any deep thoughts?

REPLY [2 votes]: The problem seems to lie in the way you sample the unit simplex.
Let $(U_i)_{1\leq i\leq n}$ be a sequence of i.i.d. uniform random variables, $S=\sum U_i$ and define $X_i=U_i/S.$ Then non-intuitively, $(X_1,\dots,X_n)$ is not uniformly distributed over the unit simplex.
The following plot shows 300000 points on the unit square with barycentric coordinates generated using this method:
First method
The method can easily be fixed: if the $U_i$s defined above are i.i.d. exponential random variables, then  $(X_1,\dots,X_n)$ is uniformly distributed over the simplex. Again, the following plot shows 300000 points generated according to this second method:
Second method
It should be noted that uniformly distributed $X_i$s doesn't imply uniformly distributed $\phi,$ as said by Mateusz Kwaśnicki in a comment.