Let $P\_n$ be a "random convex polyhedron" in $\mathbb{R}^3$ of $n$ vertices, where "random" could follow any one
of a number of models:
(1) the convex hull of $n$ points randomly and uniformly distributed *on* a sphere;
(2) the convex hull of $N>n$ points randomly and uniformly distributed *in* a sphere;
(3) analogous definitions but using different distributions, or replacing "sphere" by "a given convex body."
I think my question is largely independent of the precise model:



> 
> Does the expected measure of the minimum face angle $\theta\_{\min}$ 
>  over all faces of $P\_n$ go to zero
>  as $n \rightarrow \infty$?
> 
> 
> 


I am hoping there is a succinct argument that avoids computing the precise expectation
of $\theta\_{\min}$, which might be difficult, and would certainly depend on the model.
I have seen many papers on properties of random convex hulls, but none that I've found
address my specific question. Thanks for ideas/pointers, under any model!



Expected minimum face angle of random convex polyhedron in $\mathbb{R}^3$
The answer is YES. (I am assuming you mean the angle between two adjacent edges on a common face. (The dihedral angles all go to $\pi$.)) The easy and brief reason is that, in a large random point set, everything (that depends on *local* conditions) happens almost surely.


Here is a sketch of a proof for your model (1).


1. Fix $\varepsilon>0$ arbitrarily, and take a (small) triangle $abc$ on the sphere with smallest angle $\varepsilon$.
2. Construct its circumcircle $K\_0$.
3. Let $K$ be a concentric circle twice as large as $K\_0$.
4. Construct some small neighborhoods $A,B,C$ around $a,b,c$ such that any triangle with vertices taken from these neighborhoods
	1. has smallest angle $<2\varepsilon$.
	2. has its circumcircle within $K$.
5. Now we let the number $n$ of points go to infinity. For each $n$:


	1. Construct a scaled-down copy of the configuration $A',B',C',K'$ of $A,B,C,K$ (but still *on* the sphere) such that the expected number of points that falls into $K'$ is 3. (The area of $K'$ is a $3/n$ fraction of the whole sphere.)
	2. Now, the probability that
	exactly 3 points fall into $K'$ is at least some positive probability $p\_0$, independend of $n$. ($p\_0$ is not so small, the number of points is essentially Poisson-distributed with mean 3.)
	3. The probability that




> 
> one point each falls into $A'$, $B'$, and $C'$ but no other point falls into $K'$ 
> 
>  is at least some (small) constant $p\_1>0$ (independent of $n$). The reason is that $A'$, $B'$, $C'$ have some (almost) constant fraction of the area of $K'$.
> 
> 
> 1. If this event happens, there will be a face angle smaller than $2\varepsilon$.
> 2. Now, place const$\cdot n$ disjoint copies $A',B',C',K'$ on the sphere. Then these copies behave essentially like independent Bernoulli experiments with success probability $p\_1$. As $n\to\infty$, the probability of having at least one "success" approaches 1.
> 
> 
> 



