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!

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

On Mar 26, 9:13*pm, wrote:
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- Hide quoted text -

- Show quoted text -

Torben wrote in this thread how a set of equations could be written
and solved to find the equity of any
backgammon position. The solutions only make sense as equities though
if the equities are all defined.
In my coin example one can give an equation and solve it but that does
not mean it gives the expected value of the game. If double take is
correct then the player on turn either wins 2 points or gives his
opponent the exact same situation but with the cube doubled. It is
very tempting though wrong to think that E(X) = (1/2)(2) + (1/2)
(-2E(X))
whose solution gives E(X) = 1/2.

Bob Koca

On Mar 27, 12:43*pm, bob wrote:
On Mar 26, 9:13*pm, wrote:





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- Hide quoted text -

- Show quoted text -

* Torben wrote in this thread how a set of equations could be written
and solved to find the equity of any
backgammon position. The solutions only make sense as equities though
if the equities are all defined.
In my coin example one can give an equation and solve it but that does
not mean it gives the expected value of the game. If double take is
correct then the player on turn either wins 2 points or gives his
opponent the exact same situation but with the cube doubled. It is
very tempting though wrong to think that E(X) = (1/2)(2) + (1/2)
(-2E(X))
whose solution gives E(X) = 1/2.

Bob Koca- Hide quoted text -

- Show quoted text -

Ok but you did say that your paradox makes the potential number of
positions infinite. I still don't see how. You have never explained,
why, even assuming your paradoxical scenarios exist, a 32 cube should
be regarded differently to a 64 cube. Both lead to the same
conclusion -- equity undefined.

You said this:

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....

(caps added)

Yet you never explain why your paradox would lead to an infinity of
positions.

Paul

On Mar 27, 1:13*am, wrote:
On Mar 27, 12:43*pm, bob wrote:





On Mar 26, 9:13*pm, wrote:

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- Hide quoted text -

- Show quoted text -

* Torben wrote in this thread how a set of equations could be written
and solved to find the equity of any
backgammon position. The solutions only make sense as equities though
if the equities are all defined.
In my coin example one can give an equation and solve it but that does
not mean it gives the expected value of the game. If double take is
correct then the player on turn either wins 2 points or gives his
opponent the exact same situation but with the cube doubled. It is
very tempting though wrong to think that E(X) = (1/2)(2) + (1/2)
(-2E(X))
whose solution gives E(X) = 1/2.

Bob Koca- Hide quoted text -

- Show quoted text -

Ok but you did say that your paradox makes the potential number of
positions infinite. *I still don't see how. *You have never explained,
why, even assuming your paradoxical scenarios exist, a 32 cube should
be regarded differently to a 64 cube. *Both lead to the same
conclusion -- equity undefined.

You said this:

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....

(caps added)

Yet you never explain why your paradox would lead to an infinity of
positions.

Paul- Hide quoted text -


I clearly said that considering the position and value gives an
infinite number of positions. Ignoring the value does give a finite
number of positions. One must be careful though since equations can be
made but there is no guarantee that the solutions actually give
equities as one might expect.

Bob Koca

On Mar 29, 4:56 pm, wrote:

I'm sure Bob realised 1.e4 is probably a draw. Bob was conducting a thought
experiment which _assumes_ e4 is a win so your reply misses the point.

well i'm aware the discussion/thread was about checkers,
and in the latest postings about specific openings used, or
left out by Schaeffer in his billion numbers database crunching.
but from what i've read in some articles (June 2007) it's
believed that these openings are irrelevant so they
really think checkers is solved, ie a draw.

Now i made a jump back to chess again as
rec.games.chess etc. is about chess, isnt it.
And with similar reasoning, lets say that. after 10 or 20
years or so someone would bother to do a similar
exercise as Schaeffer, but now for chess, starting with
openings d4, e4, Nf3, c4, and then would claim
its a draw. Then of course some people might
say: well you haven't tested e.g. 1.a3! or 1. h3 yet,
so it's not 'proven' yet that chess is a draw.
This then would not be a strong argument, as
after 1.a3 d5! 2.d4 we would get similar positions
as d4 opening lines, whereas after 1.h3 e5 2.e4 we would
get similar opening lines as e4 e5 by transposition.

Only relevant reasoning seems to be to investigate
what opening lines are more double-edged, like eg Sicilian
rather than drawish. If in certain tournaments one would
like to stimulate exciting play, than eg. as response
to e4 they could make eg. c5 obligatory.
But this would change such tournaments into
specific theme-tournaments, as eg. often
is done already in correspondence chess
(eg. d4 games with Benko gambit only or so).
Not the way to go forward imho.

Better extend chess to chess960 i would say,
probably (indeed) much more exciting


best regards
jef