Random Unfoldings of the Cube Motivated by unfoldings of the dodecahedron in How To Fold It --
How many (labeled or unlabeled) unfoldings of the 1 x 1 x n stack of unit cubes are there?

JORourke (4Nov16): John's original image is now lost:
http://www.freeimagehosting.net/newuploads/f3dkg.gif
JDM (11/15/16): The original image is certainly lost from 4 years ago.  Here is a quick example with a 1 × 1 × 1 cube


The dissections of the cube are related to each other by "hinge" moves (which I've tried to draw).  The graph of unfolding of the cube seems connected by these hinge moves.

It's equivalent to consider cutting the cube along the edges.  Then we're looking for uniform spanning trees on the surface of the cube. 

JDM (11/15/16) I've added some figures of the shapes I was trying to unfold.

For my  4 (unlabeled) (incomplete list of) shapes, how often do they appear among (uniformly random) unfoldings of the labeled cube?? JDM (11/15/16) This part of the question is also ambiguous.  I had also been concerned in the 1 × 1 × n case the diagrams are not planar.  David Speyer's answer looks possibly correct.  The graph of unfoldings connected by hinge moves looks non-trivial even in the 1 × 1 × 1 case.)
 A: Not an answer; just a remark: 
Your inventory of the unfoldings of the $1 \times 1 \times 1$ cube is missing two unfoldings:
there are 11.
           


      
(Figure from How To Fold It: The Mathematics of Linkages, Origami, and Polyhedra, p.166)
A: Let $a_n$ be defined by the recursion
$$a_{n+2} + 4 a_{n+1} + a_n =0,\ a_1=-4,\ a_2=15.$$
Let $b_n$ be defined by the recursion
$$b_{n+2} + 6 b_{n+1} + b_n =0,\ b_1 =-6,\ b_2=35.$$
The number of unfoldings of the labeled $1 \times 1 \times n$ cube is $-4 a_n^2 b_n$. (I might have a sign error here.)
We count spanning trees using the matrix tree theorem. Let $G$ be the dual graph to the $1 \times 1 \times n$ cube (so $G$ has $4n+2$ vertices). Let $V$ be the vector space of functions on the vertices of $G$. Let $\Lambda: V \to V$ be the Laplacian operator. Then $\Lambda$ has a one dimensional kernel -- the constant function -- and we are supposed to compute the coefficient of $t$ in $\det(\Lambda - t \mathrm{Id})$. 
The cyclic group of order $4$ acts on the $1 \times 1 \times n$ cube by rotation around the long axis. This action commutes with $\Lambda$. So we can compute $\Lambda$ separately on each of the four eigenspaces of this rotation. The corresponding matrices are
$$\begin{pmatrix}
-4 & 1 & 0 & 0 & 0 & \cdots & 0 \\
 1 &-4 & 1 & 0 & 0 & \cdots & 0 \\
 0 & 1 &-4 & 1 & 0 & \cdots & 0 \\
 0 & 0 & 1 &-4 & 1 & \cdots & 0 \\
 0 & 0 & 0 & 1 &-4 & \cdots & 0 \\
   &   &   &   &   & \ddots &   \\
 0 & 0 & 0 & 0 & 0 & \cdots & -4 \end{pmatrix} \ \mathrm{on\ the} \pm i \ \mathrm{eigenspace}$$
$$\begin{pmatrix}
-6 & 1 & 0 & 0 & 0 & \cdots & 0 \\
 1 &-6 & 1 & 0 & 0 & \cdots & 0 \\
 0 & 1 &-6 & 1 & 0 & \cdots & 0 \\
 0 & 0 & 1 &-6 & 1 & \cdots & 0 \\
 0 & 0 & 0 & 1 &-6 & \cdots & 0 \\
   &   &   &   &   & \ddots &   \\
 0 & 0 & 0 & 0 & 0 & \cdots & -6 \end{pmatrix} \ \mathrm{on\ the} -1 \ \mathrm{eigenspace}$$
$$\begin{pmatrix}
-4 & 4 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & -2 & 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0\\
0 & 1 &-2 & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\
0 & 0 & 1 &-2 & 1 & 0 & \cdots & 0 & 0 & 0\\
0 & 0 & 0 & 1 &-2 & 1 & \cdots & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 &-2 & \cdots & 0 & 0 & 0\\
   &   &   &   &   & \ddots &   \\
0 & 0 & 0 & 0 & 0 & 0 & \cdots & 1 & -2 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 4 & -4 \end{pmatrix} \ \mathrm{on\ the}\ 1 \ \mathrm{eigenspace}$$
The first two matrices are invertible and obey the recursions $a_n$ and $b_n$ respectfully. (See the recursive formula displayed in this article.) 
The last matrix has a one dimension kernel. Taking the cofactor by deleting the last row and column, one gets a matrix whose determinant is $-4$, as can be checked from the same recursion as above.
Putting it all together, the number of spanning trees is $-4 a_n^2 b_n$.
I don't understand the statistical question you want to ask about the unfolded shapes.

Remark: Using the cyclic symmetry broght this computation down into the range of reasonable computation. But, even without that, I immediately knew that the answer would obey some linear recursion. Consider building a spanning tree of $G$ one layer at a time as we travel along the long axis of the cube. At the $k$-th partial stage, there are $4$ vertices of $G$ on the exposed boundary; call them $u_k$, $v_w$, $w_v$, $x_k$. The graph so far is a forest, each connected component of which contains at least one of the four exposed vertices. There are $15$ ways to group these $4$ vertices into connected components, one of which is actually impossible for planarity reasons. So there is some $14 \times 14$ matrix which records, for example, if the vertices on one level are grouped as $(\{ u_k, v_k \}, \{ w_k \}, \{ x_k \})$, how many ways are there to extend the forest to one where the vertices on the next level are grouped as $( \{ u_{k+1} \}, \{ v_{k+1},  w_{k+1}, x_{k+1} \})$. Taking powers of this $14 \times 14$ matrix and then pairing with some row and column vectors to handle end effects, you get this count.
This is what I learned to call the "transfer matrix method" (although Wikipedia seems to call something else by that name) and it solves almost all "linear" combinatorics problems.
