If I have a graph composed of two communities of vertices, say divided into one community with 6 white vertices and another community of 6 black vertices. If I then pair the vertices off into 6 pairs at random, and then form an edge between each pair, the final graph will have 6 edges, with each vertex having a degree of 1.
If I then define a variable which keeps track of the number of edges which have one white vertex and one black vertex (edges with one end in each community), I am assuming it should follow a binomial distribution. However, I am having a hard time being able to demonstrate this, would someone be able to give me an insight?
Edit:
Sorry, in your question I missed the fact that your vertices had degree 1 after you chose your edges. So my answer is not valid in this case.
As paw88789 said in his answer, in this case the distribution is probably not binomial.
Suppose your graph $G = (V, E)$ has $|V| = N$, and $V = B \cup W$ where $B$ represents the set of black vertices, and $W$ represents the set of white vertices. Let $|B| = b$ and $|W| = w$ be the number of black vertices and the number of white vertices respectively (note that $b + w = N$). You want to place $M$ edges at random between your vertices.
Denote by $X$ the random variable counting the number of edges between $B$ and $W$. We will compute the expression $\mathbb{P}[X= k]$ which is the probability that there are $k$ edges (among $M$) between $B$ and $W$.
We start by computing the probability that an edge lies between $B$ and $W$. It is given by the probability that one end is in $B$ and the other end is in $W$. Hence it is given by $$p = \frac{b}{N} \cdot \frac{w}{N} = \frac{b \cdot w}{N^2}.$$ Now, we look at $\mathbb{P}[X=k]$. You have to choose $k$ edges from $M$, there are $M \choose k$ ways to do so. You want these $k$ edges to be between $B$ and $W$, hence the probability to have this is $p^k$. Moreover you want the $M-k$ remaining edges to be between two vertices in $B$ or two vertices in $W$. The probability to have this is $(1-p)^{M-k}$. Hence $$\mathbb{P}[X = k] = \binom{M}{k}p^k(1-p)^{M-k}. $$
So indeed, $X$ the variables that counts the edges between $B$ and $W$ follows a binomial distribution of parameters $M$ and $p$.
So in your example, keeping my notations: \[\mathbb{P}[X = k] = \binom{6}{k} \left( \frac{6 \cdot 6}{12 \cdot 12} \right) ^k \left(1 - \frac{1}{4}\right)^{6-k} = \binom{6}{k} \left( \frac{1}{4} \right)^k \left( \frac{3}{4} \right)^{6-k}.\]