Expected minimum face angle of random convex polyhedron in $\mathbb{R}^3$ 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!
 A: Section 8.2.4  of 

Rolf Schneider, Wolfgang Weil:
  Stochastic and Integral Geometry,
  Springer  Verlag 2008

may be a good place to start.    Roughly,  there they  select   $n$ random points in a  given convex body (say the unit ball)    and they  describe  the large $n$ behavior of  support function of the expected   convex hull.   There are lots of references  and  historical remarks following this subsection  and maybe you get lucky.
A: For i.i.d. points chosen in a bounded subset of $\mathbb{R}^3$ (or $\mathbb{R}^d$) it seems to me that $\theta_\min(n)\to 0$ is ensured when the support of the distribution has a smooth boundary. This covers the case of the uniform distribution on an Euclidean ball, and a uniform spherical distribution as well. (I'm not quite sure about how to state a converse). 
A: 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).

*

*Fix $\varepsilon>0$ arbitrarily, and take a (small) triangle $abc$ on the sphere with smallest angle $\varepsilon$.

*Construct its circumcircle $K_0$.

*Let $K$ be a concentric circle twice as large as $K_0$.

*Construct some small neighborhoods $A,B,C$ around $a,b,c$ such that any triangle with vertices taken from these neighborhoods

*

*has smallest angle $<2\varepsilon$.

*has its circumcircle within $K$.


*Now we let the number $n$ of points go to infinity. For each $n$:


*

*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.)

*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.)

*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 $K'$.
  
  
*
  
*If this event happens, there will be a face angle smaller than $2\varepsilon$.
  
*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.
  

