Fix a $V\in\mathbb{N}$ with $V\ge 4$. Randomly pick $V$ points on a sphere (independently and uniformly with respect to the surface area measure). You may think of the convex hull of these $V$ points. With probability 1, this constitutes a nondegenerate convex polyhedron. In fact, all faces would be triangles (probability 1)(*), so this should be a $(2V-4)$-hedron (from Euler's $F-E+V=2$ when $3F=2E$).
But what is the probability for each topological type?
I think(**) the first "interesting" case is for $V=6$ random points on the sphere, so random octahedra, so let me ask about that.
What is the probability that 6 randomly chosen points on a sphere span an octahedron which
is topologically like a regular octahedron (so all 6 vertices have same vertex order).
Both an exact answer and an approximate value from simulating the random process would be interesting.
Notes:
(*) I learn that a polyhedron all of whose faces are triangles can be called a simplicial polyhedron.
(**) A bit more research leads to the beautiful table Counting Polyhedra which confirms that out of the 257 distinct octahedra, there are 2 distinct ones with $V=6$. One is the regular octahedron. The other one seems to be called the biaugmented tetrahedron or bicapped tetrahedron although I am not sure if it has other names.
Best Answer
The two types of triangular octahedra can be characterized by the degree sequences of their vertices. The regular octahedron has six vertices with $4$ incident edges, whereas the other one has degree sequence $(3,3,4,4,5,5)$. A vertex $v$ having degree $d$ means that $v$ can “see” $d$ other vertices; that is, if you project from $v$, the convex hull of the other $5$ vertices is a $d$-gon, with the other $5-d$ vertices inside the hull and thus occluded from view.
In the following, one vertex will always be considered as the north pole, and the remaining vertices will be considered in stereographic projection, with the edges between them mapped to straight lines in the plane; in particular, all triangles referred to are triangles in the plane, not spherical triangles (nor projections of spherical triangles, since the projections of the edges are not projections of great circles).
Denote the probability of the convex hull of $5$ points in stereographic projection to consist of $k$ points by $x_k$ and the probability of $k$ points in stereographic projection to form a convex $k$-gon by $p_k$. In How is the number of points in the convex hull of five random points distributed? I derived the $x_k$ in terms of the $p_k$:
\begin{align} x_3&=\frac32-\frac52p_4+p_5\;,\\ x_4&=-\frac12+\frac52p_4-2p_5\;,\\ x_5&=\vphantom{\frac12}p_5\;. \end{align}
(This holds irrespective of the specific way in which the points are chosen; at the time I wasn’t considering choosing them on a sphere and projecting them.)
Interestingly, in the present case, since any selection of $6$ points yields the same number of vertices of degree $3$ and $5$ (either zero or two), we have $x_3=x_5$. Substituting this above yields $p_4=\frac35$. So without writing a single integral, we can deduce that the probability for four points uniformly randomly chosen on the sphere to form a convex quadrilateral in stereographic projection is $\frac35$.
(Update: I just realized that there’s a simpler derivation for this: The only possible degree sequence for a polyhedron of $5$ vertices uniformly randomly chosen on the sphere, which with probability $1$ has $6$ triangular faces and $9$ edges, is $(3,3,4,4,4)$, so for three out of five points the quadrilateral formed by the other four points is convex.)
The expected number of vertices of degree $3$ is $6x_3$, and in terms of the probability $q$ for the resulting octahedron to be topologically irregular it is $2q$, so $q=3x_3$. Unfortunately I don’t see a way to derive $x_3$ from first principles, so we’ll have to resort to integration after all; but knowing the exact value of $p_4$ will yet prove valuable.
The $\binom53=10$ events that the triangle formed by a given triple of points contains the other two points are disjoint, so the probability of each of them is $\frac1{10}x_3=\frac1{30}q$. Also, if we consider $4$ points at a time, the $\binom43=4$ events that the triangle formed by a given triple of them contains the fourth are disjoint, so the probability of each of them is $\frac14(1-p_4)=\frac14\left(1-\frac35\right)=\frac1{10}$.
We can view this latter probability as the mean, taken over all positions of the three points, of the fraction $X$ of the surface area of the sphere in which the fourth point would need to be chosen in order to lie in the triangle the three points form in stereographic projection. And the former probability is a similar expected value, namely of $X^2$ (since two points need to lie in the fraction $X$ of the surface area). So we know $\langle X\rangle=\frac1{10}$, and we want to find $\langle X^2\rangle$.
The spherical area element in terms of the coordinates $x$ and $y$ of the stereographic projection is
$$ \mathrm dA=\frac4{\left(1+x^2+y^2\right)^2}\,\mathrm dx\mathrm dy $$
(see Wikipedia). Dividing by $4\pi$ to transform to probabilities and transforming to polar coordinates $r,\phi$ in the plane yields
$$ \mathrm dp=\frac1\pi\frac1{\left(1+r^2\right)^2}r\,\mathrm dr\,\mathrm d\phi\;, $$
and integrating over $r$ from $r=0$ yields
\begin{eqnarray*} \mathrm dp &=& \int_0^r\frac1\pi\frac1{\left(1+r'^2\right)^2}r'\mathrm dr'\mathrm d\phi \\ &=& \frac1{2\pi}\left[-\frac1{1+r'^2}\right]_0^r\mathrm d\phi \\ &=& \frac1{2\pi}\cdot\frac1{1+r^{-2}}\,\mathrm d\phi\;. \end{eqnarray*}
We want to integrate over triangles, which we can regard as composed of three triangles with one vertex at the origin. If the triangle contains the origin, this decomposition is straightforward; if it doesn’t, then one or two of the component triangles contribute negatively.
So we need the contribution of a triangle with one vertex at the origin. The equation of a line in polar coordinates is $r=r_0/\cos(\phi-\phi_0)$, where the closest approach to the origin at distance $r_0$ occurs at $\phi_0$. Given two vertices $(r_1,\phi_1)$ and $(r_2,\phi_1)$, we can determine $r_0$ and $\phi_0$ for the line that joins them and then integrate:
\begin{eqnarray*} p &=& \int_{\phi_1}^{\phi_2}\mathrm dp \\ &=& \int_{\phi_1}^{\phi_2}\frac1{2\pi}\cdot\frac1{1+r^{-2}}\mathrm d\phi \\ &=& \int_{\phi_1}^{\phi_2}\frac1{2\pi}\cdot\frac1{1+\frac{\cos^2(\phi-\phi_0)}{r_0^2}}\mathrm d\phi \\ &=& \frac1{2\pi}\left[\sqrt{\frac1{1+r_0^{-2}}}\arctan\,\left(\sqrt{\frac1{1+r_0^{-2}}}\tan\phi\right)\right]_{\,\phi_1-\phi_0}^{\,\phi_2-\phi_0} \end{eqnarray*}
(Wolfram|Alpha computation).
I didn’t find a way to make further progress analytically, so I performed the remaining integration over the three vertices of the triangle numerically, using the largest published Lebedev quadrature rule of order $131$ with $5810$ points on the sphere.
If the same quadrature grid is used for all three points of the triangle, a lot of singular triangles with coinciding or colinear points result, so I transformed the grid for each point by a random rotation. This leads to a sort of hybrid between quadrature and Monte Carlo integration, where each quadrature with random rotations of the grids yields a sample.
This is where we get back to the fact that we’re trying to determine the mean of the square of a quantity whose mean we know exactly. In a purely random setting where we draw independent random samples from a random variable $X$, we could use this information by subtracting out the mean. Equivalently, we could estimate the variance $\sigma^2$ using the differences from the known value of $\langle X\rangle$ and then estimate $\langle X^2\rangle$ using $\langle X^2\rangle=\langle X\rangle^2+\sigma^2$:
\begin{eqnarray*} \hat{\langle X^2\rangle} &=& \langle X\rangle^2+\hat\sigma^2 \\ &=& \langle X\rangle^2+\overline{(X-\langle X\rangle)^2} \\ &=& \langle X\rangle^2+\overline{X^2}-2\langle X\rangle\overline X+\langle X\rangle^2 \\ &=& \overline{X^2}+2\langle X\rangle(\langle X\rangle-\overline X)\;. \end{eqnarray*}
That is, we know by how much the sample mean $\overline X$ deviates from the correct mean $\langle X\rangle$, and we can apply a correction proportional to that deviation to $\overline{X^2}$ (which would otherwise have been our best estimate of $\langle X^2\rangle$). In this case the factor of proportionality for the correction is $2\langle X\rangle$.
In the present setup, we’re not taking independent samples of $X$, but the situation is similar in that if we rotate the quadrature grids in a way that makes $X$ come out too high, $X^2$ will probably also come out too high. Here are plots of the quadrature result for $X$ versus $X^2$. The first plot is with lots of results on a smaller grid (of order $53$ with $974$ points), and the second one with fewer results on the full $5810$-point grid. (Each of the latter runs has to evaluate $5810^3\approx2\cdot10^{11}$ triangles, so I only did $108$ of those.)
The slope of the regression line is about $0.7$, so the effect is considerably stronger than in the purely random case (where it would have been $2\langle X\rangle=0.2$). Substracting out the fitted linear function reduces the sample standard deviation from $1.2\cdot10^{-6}$ to $3.7\cdot10^{-7}$ for the smaller grid and from $3.6\cdot10^{-8}$ to $6.4\cdot10^{-9}$ for the larger grid, so we gain almost a full significant digit by making use of the known exact value of $\langle X\rangle$. The result for the larger grid is
$$\langle X^2\rangle=0.02066818545(62)\;,$$
so the desired probability for the octahedron to be topologically regular is
\begin{eqnarray*} 1-q &=& 1-30\langle X^2\rangle \\ &=& \fbox{0.379954436(18)}\;. \end{eqnarray*}
To check the result, I also did a direct simulation by generating quintuples of points and checking whether the convex hull of any three of them contains the other two. It would take quite some time to get the same accuracy with this approach, so I only took $9\cdot10^{10}$ samples. The result is $\langle X^2\rangle=0.2066820(13)$, in good agreement.
Here’s the Java code for the quadrature, and here’s the Java code for the simulation.
Update: You also asked more generally about the probability for each topological type. Here’s the distribution of degree sequences from a sample of $10^{10}$ decahedra with $7$ vertices (and thus $15$ edges):
\begin{eqnarray*} (4,4,4,4,4,5,5) : 0.387264(4)\\ (3,4,4,4,5,5,5) : 0.364633(4)\\ (3,3,4,4,5,5,6) : 0.156604(3)\\ (3,3,4,4,4,6,6) : 0.061653(2)\\ (3,3,3,5,5,5,6) : 0.029843(2) \end{eqnarray*}
The Counting Polyhedra site you linked to lists $5$ types, so these can all be distinguished by degree sequence. (This is no longer the case for dodecahedra with $8$ vertices, where the site lists $14$ types but only $12$ different degree sequences are generated, so apparently in that case there are two pairs of non-isomorphic graphs with identical degree sequences.)
Note that the probability decreases with increasing variance of the degrees (the last two degree sequences have the same variance). This is in contrast to the octahedra, where the topologically regular octahedron with zero degree variance is the less probable type.
Unfortunately there aren’t enough linear constraints to obtain these five probabilities by the sort of integration used above, but one linear relationship between them can be obtained: A vertex of degree $3$ occurs only when one triangle contains the three remaining points, and the probability for this is $\langle X^3\rangle$. Thus we must have
$$ p_{(3,4,4,4,5,5,5)}+2p_{(3,3,4,4,5,5,6)}+2p_{(3,3,4,4,4,6,6)}+3p_{(3,3,3,5,5,5,6)}=140\langle X^3\rangle $$
(where $140=7\cdot\binom63$ is the number of vertices times the number of triples formed by the other $6$ vertices). Indeed, the simulation yields $0.890680(8)$ for the left-hand side and the above quadrature scheme yields $0.890690(1)$ for the right-hand side, in good agreement.
I also created 3D visualizations of the different types of simplicial octahedra and decahedra. Unfortunately they can’t be embedded here, but you can view them here (octahedra) and here (decahedra). The structure is easier to see if you select “Show Edges” in the preferences. You can click on the cube icon in the left sidebar to get a list titled “Meshes”, where you can see the degree sequences and switch individual polyhedra on or off (eye icon) or zoom in on one of them (four-arrow icon). Here’s a snapshot of the decahedra: