On Mar 17, 7:08*am, Sanny wrote:
* Am I missing something in my intuitive argument or is one of the
calculations incorrect? The Trice argument looks solid to me.

Are you the one who used to play with username "Bob" at GetClub? I was
searching for Bob for long time. You left playing long before. Play a
few games at GetClub and see how well it plays now.

Play Chess at:http://www.GetClub.com/Chess.html

Help Bot used to say that you use Computer's help while playing
against GetClub is that True?

Your games are remarkable. Only Zebediah matches your game style.

Bye
Sanny

Play Chess at:http://www.GetClub.com/Chess.html

A different Bob than me.

Bob Koca

On Mar 17, 8:46*pm, bob wrote:
On Mar 17, 7:08*am, Sanny wrote:





* Am I missing something in my intuitive argument or is one of the
calculations incorrect? The Trice argument looks solid to me.

Are you the one who used to play with username "Bob" at GetClub? I was
searching for Bob for long time. You left playing long before. Play a
few games at GetClub and see how well it plays now.

Play Chess at:http://www.GetClub.com/Chess.html

Help Bot used to say that you use Computer's help while playing
against GetClub is that True?

Your games are remarkable. Only Zebediah matches your game style.

Bye
Sanny

Play Chess at:http://www.GetClub.com/Chess.html

* *A different Bob than me.

Bob Koca- Hide quoted text -

- Show quoted text -

Huge coincidence since Bob is an extremely uncommon name.

Paul Epstein

My recollection is slightly different. I remember looking at results
from the highest level of checkers play and a player would win a match
by 1 game to 0 with 20 draws. However, that's an impression from
memory only -- I haven't been able to check it. Can you back up your
claim with hard stats? If the world no. 1 plays the world no. 2,
would they draw less than 95% of their games? (I doubt it.) I think
it's a pretty dead game at the highest level.

I could get exact stats from the American Checker Federation. I am
pretty sure though that, for instance world championship matches, are
something above 80% draws, if maybe not 95%, even with 3-move
restriction. Whether that means it's a "dead" game is a matter of
definition. Certainly at the NON-championship level, the percentage
of draws is very much lower (you can verify this by looking at on-line
play sites). High-level tournaments, when they achieve under 75%
draws, are considered "lively."

But my point was that, while computer analysis shows an absolute draw
for unrestricted play and very likely will show the same for 3-move
play, humans still win and lose. As you point out, it is at a
relatively low percentage, but still, the game continues to be played.

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

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, 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

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!

On Mar 25, 9:11*pm, wrote:

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.


I doubt that there is a siutation like that in backgammon. Note
that you would need the 18 non-monster rolls to lead to repeating
positions. There has been one proposed but it only can arise as the
result of an illegal checker play.

Here is why such a situation will affect theoretical money play:

Theoretical money play means making the move that maximizes equity,
assuming perfect play from the opponent. Equity means the expected
value of the position. Take my coin flipping example again. If both
players use the always double/take strategy then the player on turn
will win $2 with probability 1/2, will lose $4 with probability 1/4,
will win $8 with probability 1/8, will lose $16 with probability
1/16 ... The expected value is 2(1/2) + (-4)(1/4) + (8)(1/8) + (-16)
(1/16) + ... which does not converge. The expected value is not even
defined.

Bob Koca

On Mar 26, 11:14*pm, bob wrote:
On Mar 25, 9:11*pm, wrote:

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.

* *I doubt that there is a siutation like that in backgammon. Note
that you would need the 18 non-monster rolls to lead to repeating
positions. There has been one proposed but it only can arise as the
result of an illegal checker play.

*Here is why such a situation will affect theoretical money play:

*Theoretical money play means making the move that maximizes equity,
assuming perfect play from the opponent. Equity means the expected
value of the position. Take my coin flipping example again. If both
players use the always double/take strategy then the player on turn
will win $2 with probability 1/2, will lose $4 with probability 1/4,
will win $8 with probability 1/8, will lose $16 with probability
1/16 ... *The expected value is 2(1/2) + (-4)(1/4) + (8)(1/8) + (-16)
(1/16) + ... which does not converge. The expected value is not even
defined.

Bob Koca

Yes, indeed. I realised that. However, you appeared to claim that
this no-equity position implies that the value of the cube needs to be
taken into account when enumerating positions. You don't show this.
In your example, whether the cube is on 2 or 4 or whatever, it's still
the same no-equity verdict.

Paul Epstein