average valence of vertex in tetrahedral mesh It is well-known that the average valence of a triangle mesh is 6: this can be derived from Euler's formula
$$V-E+F=\chi;$$
for sufficiently large meshes of sufficiently small genus and boundary, $\chi$ is negligible and $2E\approx 3F$, so that $E/V \approx 3$ and since every edge is shared by two vertices, the average vertex has valence 6.
Is there a similar calculation that can be done for tetrahedral meshes? In this case
$$V-E+F-C\approx 0$$
and again $2F\approx 4C$,  but the issue is that we still need one more relation between the variables in order to estimate $E/V$.
 A: Let $d$ be a positive integer and $M$ a $d$-dimensional triagularizable geometric object.  Let $F_j$ denote the number of $j$-dimensional faces of a triangularization $T$ of $M$.  (For example, $F_0$ is the number of vertices and $F_1$ is the number of edges.)  Each time a new vertex is added into the interior of an $n$-simplex in $T$, we see that the new triangularization $T'$ satisfies
$$F_j'=\begin{cases}F_j+\binom{d+1}{j}\,,&\text{ if }j=0,1,2,\ldots,d-1\,,\\
F_d+d\,,&\text{ if }j=d\,,
\end{cases}$$
where $F'_j$ is the number of $d$-dimensional faces in $T'$ for each $j=0,1,2,\ldots,d$.  Hence, if you refine the mesh on $M$ nicely (i.e., avoid fiddling with boundaries of all $n$-simplices), then the ratio
$$F_0:F_1:\ldots:F_{d-1}:F_d$$
should tend to
$$\binom{d+1}{0}:\binom{d+1}{1}:\ldots:\binom{d+1}{d-1}:d\,,$$
as the number of vertices increases.  In particular, for $d=3$, one would expect $$\frac{F_1}{F_0}\approx \frac{\binom{3+1}{1}}{\binom{3+1}{0}}=4\,.$$  If we are allowed to play with the boundaries of $n$-simplices, then the ratios $\frac{F_{j}}{F_{j-1}}$ for $j=1,2,\ldots,d$ may not have limits, or can tend to arbitrarily large values, provided that $d\geq 3$.
