However, I'm wondering
> if there are some general guidelines for *which* gaps to fill if
> the resulting distributions are the same. For example, suppose
> you have three checkers left, and they're on the 6, 4, and 2 points.
> You have a 1 to play. What should you do?
I don't think there are any real simple guidelines here, but one thing you
definitely want to consider is whether you expect to get one or two
further rolls to get off. If you only get one more shot at it, obviously
you play to maximize the good doubles that will bear you off -- so play
the ace from 4 to 3 in order to activate 3-3 as a winner.
If you need to win in the next two rolls, it's trickier.
1)I would tend to shift to the ace point in the example you give so that
assuming I got only one checker off in the next shake i wouldn't be in the
position of losing with "any ace".
2) I think you have to look at each possible distribution and consider the
bad numbers that would get *no* checkers off, probably ranking it last.
3) Another consideration is looking at where the non-bear-off numbers
would play to -- try to avoid leaving a checker where such a number would
wind up stacking it atop another one --especially on that deuce point.
Maybe not too helpful -- I'd like to hear others sharpen up these ideas or
offer further ones.
Albert
The two-checker case is easy to work out: if there's a man on the
6-point, move it unless the two checkers would end up on the same
point. If the 6-point is already clear (or you'd stack up), move
the lower checker instead. Are there similar rules -- or at least
heuristics -- for guiding play in the general case?
Carl Tait
: Carl Tait
I worked all this out several years ago, and if I remember correctly the
following rules handled almost all situations correctly:
1) Don't stack
2) Aim for the four point
3) Aim for the ace point
Of course this assumes that you have two rolls to get off. If you need
to get off in the next roll you will have to play for specific doubles,
which may require an otherwise unnatural play.
Kit
Interesting; thanks. What's the priority on rules (2) and (3)?
In the two-checker case, it's variable:
Checkers on 6 and 3 with a 2 to play => 4 3 is best
Checkers on 5 and 2 with a 1 to play => 5 1 is best
Does the 2-checker rule of preferring to clear the 6-point hold up
in the many-checker case? For example, with 6 5 3 and a 2 to play,
is 5 4 3 better than 6 5 1? (This particular example can obviously
be worked out directly, but I'm hoping for a general rule.)
Carl Tait
The chances of bearing off with a checker on 5 and 1 are equal to the
chances of bearing off with a checker on 4 and 2.
--
Kevin Cline
Right; sorry. The simple two-checker rule of "Clear the 6 point if
you don't stack; in all other cases, move the lower checker" will
always give you an optimal move, but it won't tell you if the two
moves are equally good.
But I'm still curious about the priority of Kit's rules 2 and 3.
Another example from the two-checker case: with checkers on 5 3
and a 1 to play, 5 2 is better than 4 3 (19 winning rolls vs. 17).
Here, *aiming* for 1 is better than *landing* on 4.
Carl Tait
--------------------------------------------
Subject: When you can't bear off a checker
From: ta...@news.cs.columbia.edu (Carl Tait)
Date: 24 Mar 1996 15:15:02 -0500
Message-ID: <4j4ag6$4...@ground.cs.columbia.edu>