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

(Why did you post this to rec.games.chess.politics?
It has nothing to do with FIDE or USCF politics.)

wrote:

If we agree that owning a 2-cube has the same theoretical
meaning at owning a 4-cube or 16-cube (or whatever level)

Are we allowed to factor in human psychology, or are we
assuming a perfectly rational player? To a real human,
there is an intangable value to seeing which way the
game goes, and a variable cost to making a bet that
approaches his net worth.

Even assuming a perfectly rational player, he has a finite
net worth, and thus a point comes where he *must* resign
rather than accept the double even though he has a 90% plus
chance of winning, because he cannot cover the bet.

That being said, none of the above assumptions leads to an
infinite number. A bet that is twice what he can afford to
cover is the same as a bet that is four or eight times what
he can afford to cover -- none are attanable states. Thus
the doubling cube itself does not make backgammon infinitely
long / unsolvable.


--
misc.business.product-dev: a Usenet newsgroup
about the Business of Product Development.
-- Guy Macon http://www.guymacon.com/

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

"jefk" wrote in message
...
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.

That's a practical, not absolute, solution. We were talking about working
out EVERY combination for every opening, and how the solved openings can be
avoided by force of rule, leaving the unsolved remainder for tournaments.
The point was that it's possible for a GAME to be solved but for rules to be
enacted that change the game to something not solved.

Then again, my name isn't Bob, so....