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
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
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