Counting Backgammon End Positions

In this blog I will count the number of end positions in backgammon.

Recently I picked up backgammon again. But I could not put my mathematical mind to rest. I got interested in the question of how many end positions backgammons has.
In order to count the number of positions I will introduce the following terminology. The number of stones you own in the end position will be denoted by \(m\). The number of points used available is \(n\). \(o\) is the number of stones owned by the opponent, which occupy \(p\) points of the available \(n\).
So in the following example, playing as black, the following equalities apply \(m=8,\,n=6,\,o=2,\,p=1\).

Let's look at a special case first. Lets assume we have a contact-less end position. So \(o=0\). I will proof that the number of backgammon end positions with \(m\) stones and \(n\) points is

\[
m + n - 1 \choose m
\]

To see this equality, study the following diagram, which corresponds with the figure above if you ignore the opponent stones:

Every dot corresponds with a stone. Every bar with the division between points. A moments reflection will bring the insight that every diagram of this kind corresponds to a backgammon end position and vice versa, every backgammon end position can be described by such a diagram. This proofs the stated equality.

With the result of this special case we can answer the main question: How many backgammon end positions exist with \(m\) stones distributed over \(n\) points of which \(p\) are occupied by \(o\) opponent stones? This number is exactly

\[
{n \choose p} {o - 1 \choose o - p} {m + n - p - 1 \choose m}
\]

The proof of the above equality comes from the following insight. The opponent stones are distributed over \(p\) points. there are \(n \choose p\) ways of picking the occupied points.
For a point to be occupied it must at least contain one opponent stone. So of the \(o\) opponents stones, we can freely distribute o - p opponent stones over \(p\) points. This is the special case we counted already, so this can be achieved in \({(o-p) + (p-1) \choose o - p} = {o - 1 \choose o - p}\) ways.
This brings us to the final factor. It represent the \(m\) stones which should be distributed over the remaining \(n-p\) points. Again this is given by our preliminary result of \({m + n - p - 1 \choose m}\) ways. This proofs the stated result.

In conclusion: the number of backgammon end point positions with \(m\) stones distributed over \(n\) points of which \(p\) are occupied by \(o\) opponent stones equals

\[
{n \choose p} {o - 1 \choose o - p} {m + n - p - 1 \choose m}
\]