Forum Archive : Theory

Number of distinct positions

From:   Walter Trice
Date:   17 June 1997
Subject:   Re: Total Possible Possition

In you write:

> I'm new to backgammon, and was wondering if anyone has figured out
> these numbers.
> I calculate only 54,264 possible positions with 15 or fewer checkers
> on the last 6 rows.


> This is a very small number, so is there a table
> showing all the expected probilities for bearing off?

Yes. Lots of people have done this, e.g. Hal Heinrich more than
10 years ago. A one-sided table with 54264 records gives you the
probabilities of bearing off in N rolls, N=1,2,3,..., subject to
some assumption about checker play (usually that you play to
minimize mean rolls to bear off.) From these numbers you can
compute the probability of winning a 2-sided position given
the checker-play assumption.

There are also 2-sided tables for subsets of the 54264 X 54264 positions
that give exact equities taking the cube into account. E. g., the
Sconyers database.

> Also, what is the total number of possible legal positions?  My quick
> guesstimate is on the order of 10^10.

Actually, it's 18,528,584,051,601,162,496. This is still a rather
small number -- smaller by several orders of magnitude, for instance,
than the total number of stars in the universe. If you can get the
54264 number you can calculate this one by an extension of the
same method. In brief...

  C(n,m) = n!/(m!(n-m)!) = number of ways to select m objects from
    a set of n.

  D(n,m) = C(n+m-1,m) = # ways to distribute m checkers of the same
    color over n points.

  E. g., D(7,15) = C(21,15) = 54264.

  Suppose Black occupies m of the 24 regular points. There are
  C(24,m) ways to select the points. This accounts for m checkers.
  The other 15-m can be on any of the m selected points OR on the
  bar or off, so the number of ways to distribute them is D(m+2,15-m).
  The White checkers have 26-m places left to go, so they can be
  distributed in D(26-m,15) ways.

  So putting it all together you sum

    C(24,m) X D(m+2,15-m) X D(26-m,15)

  over all values of m from 0 to 15, and get

  18,528,584,051,601,162,496, assuming the programming language
  you use can handle big integers.

> It seems very small compared to
> chess or go, so has this game already been 'solved' by computers yet?

No, but a reduced version of backgammon with only 3 checkers for
each side was solved by a Hugh Sconyers program.

                         -- Walter Trice
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