Will backgammon ever get to this.........?
chuckles_the...@budweiser.com
Walfinho
> http://www.startribune.com/484/story/1313734.html
backgammon can't be "solved", cause it is not deterministic (because
of the dice). Only for deterministic games with no element of luck
like checkers and (theoretically) chess a solution can exist. But iirc
at least one bound for bg exists, i.e. that "Backgammon Misere" must
be over in at most 4000-something moves, although i didn't see a prove
for that, yet.
Michael Petch
skilled and I can say I would challenge 4000 rolls as any limit. With a bit
of bad luck (Good luck in this case) - avoiding an ill timed double is all
you need.
Calkat (A host for Gammonzone - www.gammonzone.com) use to play this format
very well. We called a draw after 1500 moves each (Total 3000). The board at
3000 rolls looked like what it was after 1500.
A better way of disproving the "4000" limit is probably easier done by
exceeding it in play. I don't think you can easily set a bound for this
format.
On 7/21/07 10:38 AM, in article
1185035914....@n2g2000hse.googlegroups.com, "Walfinho"
Michael Petch
Assume PlayerA and PlayerB ar playing regular backgammon. PlayerA rolls
3-1, PlayerB roll Dub5, PlayerA rolls Dub5, and from then on all subsequent
rolls continue to be dub 5. Pretty quickly neither player will be moving.
Is it possible to roll X number of double 5's in a row. Assume X to be some
outrageously high number - Lets say 4billion). It is of course *HIGHLY*
UNLIKELY to but not impossible to do. However - given an infinite number of
rolls, and and an infinite amount of time - it is a certainty that it will
happen.
For me its obvious that their is no theoretical maximum number of rolls to
backgammon. However in practicality - there is one (Bounded by the number of
BG players on the planet, their luck, and the length of time intelligent
life continues to play the game).
On 7/21/07 11:52 AM, in article C2C7A403.42658%mpe...@capp-sysware.com,
Michael Petch
Queue up those people who declare that rolling 4billion double 5's in a row
impossible :). Probably the same people who think the dice on servers aren't
random lol.
On 7/21/07 12:09 PM, in article C2C7A80B.4265B%mpe...@capp-sysware.com,
Walfinho
> > exceeding it in play. I don't think you can easily set a bound for this
> > format.
Disproving the existence of a given bound is always "easy" by finding
a counter-excample, that's right. I must admit, i forgot, where i came
across the result of the 4xxx bound (and i don't think that a prove
was given), and have never played Misere seriously myself.
> Assume PlayerA and PlayerB ar playing regular backgammon. PlayerA rolls
> 3-1, PlayerB roll Dub5, PlayerA rolls Dub5, and from then on all subsequent
> rolls continue to be dub 5. Pretty quickly neither player will be moving.
That's right, there are many examples, where both players cannot move
when the dice always show the same number. But you can add a
constraint, that only "moveable" moves count to the bound and try to
find a bound for this.
Gregg Cattanach
Just because a game has a luck or non-deterministic element, it doesn't mean
that it can't be solved. Hypergammon, (3 checkers per side) has been solved
completely. You can play the Hyperbots at GamesGrid and they play a perfect
game of Hypergammon.
The number of possible positions made this possible as there are only 6
total checkers to worry about. Hugh Sconyers gets the credit for doing the
programming to solve this game.
--
Gregg C.
bob
>
> That's right, there are many examples, where both players cannot move
> when the dice always show the same number. But you can add a
> constraint, that only "moveable" moves count to the bound and try to
> find a bound for this.
Still no bound. Positions could just keep repeating. As a simple
example
suppose each has one checker left. A is on his 5 point and B is on the
bar and on turn.
4001 55s in a row could be rolled.
Bob Koca
Walfinho
> that it can't be solved. Hypergammon, (3 checkers per side) has been solved
> completely. You can play the Hyperbots at GamesGrid and they play a perfect
> game of Hypergammon.
What happens, when the Hyperbots play themselves? What bot does win?
Or is it a draw? (In Backgammon a eternal loop where nobody ever wins
- it would be hard to show a draw in Backgammon) I think, a Hyperbot,
that can choose the numbers on the dice will win all the time - so the
luck element would be the overwhelming factor between two perfect
players in Hypergammon and thus it cannot be "solved", i.e. that there
is a predefined outcome of the game (starting player wins / second
player wins / draw).
> Still no bound. Positions could just keep repeating. As a simple
> example
> suppose each has one checker left. A is on his 5 point and B is on the
> bar and on turn.
> 4001 55s in a row could be rolled.
That's right ... hmm, i really don't remember, where i came across the
Misere bound. If anyone knows, i'd be happy to see, if there is a
prove or if it was just assumed.
David C. Ullrich
>On 21 Jul., 17:30, chuckles_the_scary_cl...@budweiser.com wrote:
>> http://www.startribune.com/484/story/1313734.html
>
>backgammon can't be "solved", cause it is not deterministic (because
>of the dice). Only for deterministic games with no element of luck
>like checkers and (theoretically) chess a solution can exist.
No. Game theory deals with games of chance very nicely; in fact
VonNeumann gave an analysis of poker (or at least a poker-like
game) when he invented game theory.
It's actually _not_ clear to me that there _is_ an optimal
strategy in backgammon, but the problem is not that there are
dice involved, the problem is that there's no bound on the
size of the cube or the length of a game, so it could be
that certain series diverge.
>But iirc
>at least one bound for bg exists, i.e. that "Backgammon Misere" must
>be over in at most 4000-something moves, although i didn't see a prove
>for that, yet.
************************
David C. Ullrich
David C. Ullrich
Is that Hypergammon with a doubling cube? If so I find this
very interesting (with a doubling cube it's not obvious to
me that there _is_ an optimal strategy...)
************************
David C. Ullrich
Gregg Cattanach
>> Just because a game has a luck or non-deterministic element, it
>> doesn't mean that it can't be solved. Hypergammon, (3 checkers per
>> side) has been solved completely. You can play the Hyperbots at
>> GamesGrid and they play a perfect game of Hypergammon.
>
> What happens, when the Hyperbots play themselves? What bot does win?
Of course not, draws are impossible in Backgammon and Hypergammon. A
mathematical proof of this is available somewhere on the web.
When Hyperbot plays Hyperbot whichever side is luckier will win the game,
obviously. It's just that every move by both players would result in the
maximum equity available at each play for the dice rolled.
--
Gregg C.
Gregg Cattanach
The checker play will always be optimal. However, some assumptions have to
be made when devising a cube strategy. I believe you are correct that you
can't prove that one particular doubling strategy is perfect. (i.e. how far
into the window you need to be to double, or exact number of market losing
sequences before doubling, stuff like that.)
--
Gregg C.
bob
>
> - Show quoted text -
In money play with a limit on the cube or in match play the
complete theoretically
optimal strategy (both checker play and cube play) is calculable.
There are a finite number
of positions.
Suppose we now consider money play with no limit on the cube. The
cube could be on infinitely many levels
so there are no longer a finite number of situations. However,
assuming that players wish
to maximize thier expeceted winnings it is tempting to think that only
the position of the cube and not its
value matters. As an example when you are asked if a position is a
theoretical take, the value of the
cube doesn't matter. This is very iikely to be fine, and doing so
leads to equations that are theoretically solvable
(though too many for backgammon, unlike hypergammon, to be practically
solvable). The problem though is that it only works
if we know in advance that the equities actually do exist. If equities
do not exist the equations give numbers that don't mean what
we want them to.
To understand what it means for a position to have an undefined
equity suggest looking at two Douglas Zare's articles on
Gammonvillage with title undefined equity parts 1 and 2.
Bob Koca
David C. Ullrich
I didn't say it couldn't be done, just that it wasn't clear to me that
it could be done...
>(i.e. how far
>into the window you need to be to double, or exact number of market losing
>sequences before doubling, stuff like that.)
************************
David C. Ullrich
David C. Ullrich
_If_ we define "optimal strategy" to be maximizing expectation
then in fact that follows. But:
>As an example when you are asked if a position is a
>theoretical take, the value of the
>cube doesn't matter. This is very iikely to be fine, and doing so
>leads to equations that are theoretically solvable
>(though too many for backgammon, unlike hypergammon, to be practically
>solvable). The problem though is that it only works
>if we know in advance that the equities actually do exist. If equities
>do not exist the equations give numbers that don't mean what
>we want them to.
Precisely. You say "theoretically solvable". Of course not every
system of equations _has_ a solution, but I found a proof some
years ago that a system of equations that I imagine is the same
as the one you're talking about does have a solution. That is,
there _is_ a function E(position) that has all the properties
equity should have (If p is a position where X has a choice
to make then E(p) = max(E(q)) where q runs over all positions
that result from legal choices, if p is a position where the
dice are about to be rolled then E(p) is the appropriate
average over possible rolls, etc). But I never saw how to
show that E _is_ the actual equity.
> To understand what it means for a position to have an undefined
>equity suggest looking at two Douglas Zare's articles on
>Gammonvillage with title undefined equity parts 1 and 2.
>
>Bob Koca
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************************
David C. Ullrich