## Thursday, July 29, 2010

### 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

${n \choose p} {o - 1 \choose o - p} {m + n - p - 1 \choose m}$

#### 17 comments:

1. I did the math and calculated the total number of positions in a backgammon game. It is a stagering:
3458085312432494932095

or
three sextillion ,
four hundred fifty eight quintillion ,
eighty five quadrillion ,
three hundred twelve trillion ,
four hundred thirty two billion ,
four hundred ninety four million ,
nine hundred thirty two thousand ,
ninety five

2. Hi Daan.

I would like to reprint this article in my backgammon magazine, Bibafax. Would you kindly give your permission?

You can checkout my references at www.backgammon-biba.co.uk.

Michael Crane
info@backgammon-biba.co.uk

3. Daan, since the above comment I have formatted your content into my magazine and I would like to send you a pdf so that you can preview it before (hopefully) giving your permission.

Please email me and I will send it by return.

Regards.

Michael

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