On Mar 16, 8:39*pm, " wrote:

First, what Schaeffer did is to show that freestyle (unrestricted)
checkers is an absolute draw.

He has not analyzed all of the 3-move restriction openings so he has
not proven that tournament checkers is a draw. *He has proven that
some of the 3-movers are a draw, and will likely eventually show that
all 156 of the accepted tournament choices are a draw (the other few
are almost certain losses and are not used). *Of course, there could
be a deeply-buried surprise in one or more of the 156, but with the
amount of other computer analysis done to date, it's not very likely
--- but it hasn't been categorically proven yet.

I wouldn't be all that surprised if instead he finds that some of the
156 accepted three-move tournament choices are not draws. After all,
the ones that were rejected were obvious losses, so openings that
provided one player sufficient advantage to narrowly force a win with
perfect play might not have been noted.

I realize that this is a fairly large advantage, though, and that
might mean it would have been suspected, but then the fact that
freestyle checkers was a draw wasn't known for certain until it was
proved.

John Savard

On Mar 25, 10:18 am, (Torben Ægidius Mogensen)
wrote:
samsloan writes:
One factor to be considered is that the number of possible moves in a
backgammon games is infinite. The players could easily just keeping
hitting each other to infinity.

That doesn't matter, as long as the number of possible board positions
is finite (which it is).

The main difference between Backgammon and, say, Checkers is not the
possibility of infinite play but the fact that Backgammon involves
random elements, so few positions are definitely winning or definitely
losing -- all you can say is the probability of winning with perfect
play (i.e., always picking the move that gives you the best winning
probability after moving).

You can solve Backgammon by for each possible position have edges to
every other position that it is possible to get to in one move, and
label each edge with the dice outcome that allow this move).

This can be translated into a set of equations that you can solve to
find the probability of each possible position being winning or
losing. The set of equations is huge, but finite.

Torben

Even that is not obvious. There are 21 possible rolls of the dice (6!
= 21) and three possible positions of the doubling cube plus 24
possible slots for each checker.

The average chess position has 27 moves and most chess games are over
in 50 moves.

I have written a chess playing computer program and a shogi playing
computer program. However, I once tried to write a backgammon playing
computer program and I quickly gave it up as hopeless. It is much
harder than it looks.

Although backgammon seems to be an easier game than chess, I am not
sure that this is really true.

Sam Sloan

samsloan wrote:

Even that is not obvious. There are 21 possible rolls of the dice (6!
= 21) and three possible positions of the doubling cube plus 24
possible slots for each checker.


BZZZT!

Are the checkers in your backgammon set marked in some way?

--
Kenneth Sloan
Computer and Information Sciences +1-205-932-2213
University of Alabama at Birmingham FAX +1-205-934-5473
Birmingham, AL 35294-1170 http://KennethRSloan.com/

On Mar 25, 11:18*am, (Torben Ægidius Mogensen)
wrote:
samsloan writes:
One factor to be considered is that the number of possible moves in a
backgammon games is infinite. The players could easily just keeping
hitting each other to infinity.

That doesn't matter, as long as the number of possible board positions
is finite (which it is).

The main difference between Backgammon and, say, Checkers is not the
possibility of infinite play but the fact that Backgammon involves
random elements, so few positions are definitely winning or definitely
losing -- all you can say is the probability of winning with perfect
play (i.e., always picking the move that gives you the best winning
probability after moving).

You can solve Backgammon by for each possible position have edges to
every other position that it is possible to get to in one move, and
label each edge with the dice outcome that allow this move).

This can be translated into a set of equations that you can solve to
find the probability of each possible position being winning or
losing. *The set of equations is huge, but finite.

* * * * Torben

Agree for backgammon played without a cube, for backgammon played
with a cap on the cube, or for match play. If you are talking about
money backgammon though then the cube position and value makes the
number of positions infinite. Now one might say that its position is
all that matters since if you know the correct theoretical play
holding a 2 cube then you also know the correct theoretical play
holding a 4 or any higher value cube. There is a problem though in
that the equations might not have a solution. As a simple example of
how this might come about suppose we are betting on the flip of a
coin. The first person to toss a head wins. Before each flip a
doubling cube may be used as in backgammon. Solving the equations
gives that every turn is a double and take but this leads to undefined
equities. How to prove there is no situation like that possible in
backgammon?

Bob Koca

On Mar 26, 1:45*am, bob wrote:
On Mar 25, 11:18*am, (Torben Ægidius Mogensen)
wrote:





samsloan writes:
One factor to be considered is that the number of possible moves in a
backgammon games is infinite. The players could easily just keeping
hitting each other to infinity.

That doesn't matter, as long as the number of possible board positions
is finite (which it is).

The main difference between Backgammon and, say, Checkers is not the
possibility of infinite play but the fact that Backgammon involves
random elements, so few positions are definitely winning or definitely
losing -- all you can say is the probability of winning with perfect
play (i.e., always picking the move that gives you the best winning
probability after moving).

You can solve Backgammon by for each possible position have edges to
every other position that it is possible to get to in one move, and
label each edge with the dice outcome that allow this move).

This can be translated into a set of equations that you can solve to
find the probability of each possible position being winning or
losing. *The set of equations is huge, but finite.

* * * * Torben

* Agree for backgammon played without a cube, for backgammon played
with a cap on the cube, or for match play. If you are talking about
money backgammon though then the cube position and value makes the
number of positions infinite. Now one might say that its position is
all that matters since if you know the correct theoretical play
holding a 2 cube then you also know the correct theoretical play
holding a 4 or any higher value cube. There is a problem though in
that the equations might not have a solution. As a simple example of
how this might come about suppose we are betting on the flip of a
coin. The first person to toss a head wins. Before each flip a
doubling cube may be used as in backgammon. Solving the equations
gives that every turn is a double and take but this leads to undefined
equities. How to prove there is no situation like that possible in
backgammon?

Bob Koca- Hide quoted text -

- Show quoted text -

Interesting example but I'm surprised by "How to prove there is no
situation like that possible in
backgammon?"

Almost surely, there _are_ situations like that in backgammon. Can't
you mimic the above scenario in backgammon by postulating 18 monster
rolls for each side?
But, suppose there are such situations in backgammon, why does it then
follow that the value of a cube, as well as its position, can affect
theoretical money play? This seems to be a hole in your argument. So
let us assume that your scenario is exactly replicated in backgammon
where there's a stalemate but each side has 18 monster winning rolls,
and no gammons are possible. Please explain why this scenario leads
to the conclusion that the scenario with a 2 cube is essentially
different than the scenario with a 1024 cube.

Paul Epstein

samsloan wrote:

Even that is not obvious. There are 21 possible rolls of the dice (6!
= 21) and three possible positions of the doubling cube plus 24
possible slots for each checker.

I haven't been following this thread so maybe I've missed something...
BUT
from what I read above: What do you mean by "possible rolls of the
dice"?
Combinations or Permutations? There are 21 of the former but 36 of
the
latter. Also 6! (6 factorial) is 720 not 21. Now, where did I go
wrong???

LRB

samsloan writes:

On Mar 25, 10:18 am, (Torben Ægidius Mogensen)
wrote:
samsloan writes:
One factor to be considered is that the number of possible moves in a
backgammon games is infinite. The players could easily just keeping
hitting each other to infinity.

That doesn't matter, as long as the number of possible board positions
is finite (which it is).
[...]
This can be translated into a set of equations that you can solve to
find the probability of each possible position being winning or
losing. The set of equations is huge, but finite.

Torben

Even that is not obvious. There are 21 possible rolls of the dice (6!
= 21)

6! is 720, actually. But you are right that the number of different
rolls is 21 = 6*7/2. This is still finite, though.

and three possible positions of the doubling cube

I can't see how this would affect the winning probability.

plus 24 possible slots for each checker.

You forgot the bar and home, so there are 26 possible positions. But
since the checkers are not distinct, and since black and white pieces
can't coexist (except on the bar and in the home), you get a lot fewer
than the 30^26 different positions you imply.

In any case, my point was that the number of positions is finite (but
huge), so arguing that there are many possible rolls and positions of
doubling cubes and pieces doesn't change that, unless you can show
something is infinute.

The doubling cube is normally limited to 7 possible positions (absent
or 2, 4, ..., 64), but even if you allow unbounded doubling, this
doesn't change the probability of winning.

Torben

On Mar 26, 4:49*pm, (Torben Ægidius Mogensen)
wrote:
...

The doubling cube is normally limited to 7 possible positions (absent
or 2, 4, ..., 64), ....

This is completely false. There is nothing wrong or problematic with
a double to 128. Admittedly, this can't be recorded on a normal
doubling cube but players can just write it down. Also, doubling
"cubes" are available which reach a max value of 256. Doubles beyond
64 are rare among good players, but if you're talking about what is
rare rather than what is legal, you shouldn't include 64 since good
players don't often let the cube get even that high.

Paul Epstein

On Mar 26, 4:49*am, (Torben Ægidius Mogensen)
wrote:

and three possible positions of the doubling cube

I can't see how this would affect the winning probability.

plus 24 possible slots for each checker.

You forgot the bar and home, so there are 26 possible positions. *But
since the checkers are not distinct, and since black and white pieces
can't coexist (except on the bar and in the home), you get a lot fewer
than the 30^26 different positions you imply.

In any case, my point was that the number of positions is finite (but
huge), so arguing that there are many possible rolls and positions of
doubling cubes and pieces doesn't change that, unless you can show
something is infinute.

The doubling cube is normally limited to 7 possible positions (absent
or 2, 4, ..., 64), but even if you allow unbounded doubling, this
doesn't change the probability of winning.

* * * * Torben- Hide quoted text -

- Show quoted text -

If we agree that owning a 2-cube has the same theoretical meaning at
owning a 4-cube or 16-cube (or whatever level), how do you get 7
possible cube positions? I count only three: centered, owned by me,
or owned by the opponent.

--
Gregg C.

I wouldn't be all that surprised if instead he finds that some of the
156 accepted three-move tournament choices are not draws. After all,
the ones that were rejected were obvious losses, so openings that
provided one player sufficient advantage to narrowly force a win with
perfect play might not have been noted.

Obviously I can't say for sure until the computer analysis is
complete. You could very well be correct. But the trend has been in
the other direction. It used to be that there were 144 accepted 3-
move ballots. 12 more were added a few years back based to a large
extent on computer analysis that shows that despite their seeming
unbalance (which was why they originally were left out), they are
likely to be draws. One of them, "The Black Hole," a notoriously
difficult ballot, under computer analysis is showing more and more
drawing lines.

As checkers is analyzed more deeply, more drawing resources seem to be
found all the time. But it would be a rather exciting find if one of
the accepted ballots could be shown to be a win!