Probability that a vertex in the spanning tree of an $N$ x $N$ grid graph is a leaf Suppose we have an $N$ x $N$ grid graph $G(V,E)$ and we construct a spanning tree of this graph in the following way.
Start with a set $S$ which contains only the vertex at the top left corner of the grid graph (i.e location $(0,0)$), and an empty set $T$ which will contain our spanning tree.
Now pick an edge at random (with equal probability) from the set of  edges $(p,q)\in E(G)$ such that $p \in S$ and $q \not \in S$. Add this edge to our set $T$ and add $q$ to set $S$, repeat this process until we obtain a spanning tree.
Now, the problem is this: 
What is the probability that a given vertex of the grid graph at location $(x,y)$ is a leaf in our spanning tree? In general, what is the expected number of leafs in a spanning tree constructed in this way? If this problem is too difficult, what can we say about a smaller grid? For example if we consider an $N$ x $2$, $N$ x $3$, ... e.t.c grid graph  
Note:
This was inspired by a more general question asked by Nick Wu on Quora
 A: You asked some questions related to the Uniform Spanning Tree model in Statistical Mechanics and Combinatorics.  Let's try to address your first question:

What's the probability a given vertex of the grid $(x,y) \in N \times N$ is a leaf in our spanning tree?

First, we need to count all the spanning trees of the $N \times N$ grid.  This is in the Online Encyclopedia of Integer Sequences A007341
1, 4, 192, 100352, 557568000, ...

This number grows large very quickly.  And we have an exact formula:
$$ a_n = \frac{2^{n^2-1} }{n^2} \prod_{m_1, m_2 =1}^{n-1} \left(2 - \cos \frac{\pi m_1}{n} - \cos \frac{\pi m_2}{n}\right) \in \mathbb{Z}$$
One way to check this is an integer is to verify this is invariant under the Galois group of $\mathbb{Q}[\omega]$ with $\omega^n = 1$ is a root of unity.
We can also tell that from the $2^{n^2}$ factor that we need $n^2$ bits in order to uniquely describe our spanning tree.
This formula is disccussed on MathOverflow: https://mathoverflow.net/questions/8497/number-of-spanning-trees-in-a-grid
Deriving the Product Formula for the Number of Spanning Trees
We can use the Matrix Tree Theorem which states that number of trees of any graph - in our case the $N \times N$ grid - is equal to the determinant of the Laplacian of the grid.  
$$ \# \{ \text{spanning trees}\} = \det \Delta_G  = \frac{1}{n}\lambda_1\dots \lambda_{n-1}$$
In the case of the $\{ 1, 2, \dots n \} \times \{ 1, 2, \dots n \}$ we can write down explicit eigenfunctions so that
$$ \lambda \; f(x,y) = \frac{1}{4} \big[ f(x+1,y) + f(x-1,y) + f(x,y+1) + f(x,y-1)\big] $$
The functions we get are $(x,y) \mapsto \exp\left(\frac{\pi m_1 x}{n}\right)\exp\left(\frac{\pi m_2 x}{n}\right) $ with eigenvalues $\lambda = 2 - \cos \frac{\pi m_1}{n} - \cos \frac{\pi m_2}{n}$.
So, the number of spanning trees on the grid has something to do with Harmonic analysis on $\mathbb{Z}_N^2$, random walks and the roots of unity.
A: Not sure if it helps and it's a bit late but for the case of the $N\times N$ grid in the limit of $N\rightarrow \infty$ the following exact result for the leaf-probability has been derived here based on an equivalence to the Abelian sandpile model:
$$\frac{8}{\pi^2}\left( 1 - \frac{2}{\pi}\right) \approx 0.29454$$
