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05-26-2010 , 01:00 PM
Quote:
Originally Posted by snex
http://en.wikipedia.org/wiki/Nash_eq...er.27s_dilemma

please note the last sentence in this section. "The globally optimal strategy is unstable; it is not an equilibrium." the same statement holds for poker and i challenge you to prove otherwise.
1) Tom Cruise can eat fifty lasagnes in ten minutes and I challenge you to prove otherwise.

2) Are you blind? Please note the third for last sentence in the section. "The Prisoner's Dilemma thus has a single Nash Equilibrium: both players choosing strategy #2 ("betraying")." The sentence you posted is referring to the strategy where neither player defects, which is globally optimal, i.e. the best possible outcome for both players. This is unstable because either player can gain by defecting (which would lead to both players defecting).

3) John Nash proved in 1951 that at least one Nash Equilibrium exists for a game with a finite set of actions, zero-sum or otherwise. Two proofs of this are on the very page you posted.

Now really go away.
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05-26-2010 , 01:20 PM
Quote:
Originally Posted by macthemost
ive been hearing this more and moar.how the fudge can poker be solved? ok games like razz or 5 card draw or fagdooshy could be solved. but stud? holdem? OMAHA???? solved?? i say no way. and wut does solved mean anyway? mathmatically solved? .. u cant solve reading people. u cant solve bluffing. u cant solve floating..so wut do ppl mean when they say poker is getting closer and closer to being a solved game? kinda depressing to think about..
poker = poke + her - he = dildo.
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05-26-2010 , 02:45 PM
Quote:
Originally Posted by Pointland32
2) Are you blind? Please note the third for last sentence in the section. "The Prisoner's Dilemma thus has a single Nash Equilibrium: both players choosing strategy #2 ("betraying")." The sentence you posted is referring to the strategy where neither player defects, which is globally optimal, i.e. the best possible outcome for both players. This is unstable because either player can gain by defecting (which would lead to both players defecting).
and yet, always-defectors cannot out-compete tit-for-tat when you change the starting population. why is this? do you even have an explanation? are you even aware of this fact? you can point to math that neither one of us truly understands all youd like, but simple empirical results prove that you are wrong.
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05-26-2010 , 04:26 PM
Quote:
Originally Posted by jjgoldy5

My impression was that both Chris and Bill believed that you can come close to determining perfect play for simpler games such as heads up limit hold'em and lowball, but that "solving" more advanced poker games would be extremely difficult. Chen is of the belief that you can solve the more advanced games (using a lot of very powerful computers), but I got the impression that Chris was not on board with this thinking.
Wonder what Caro thinks:

http://books.google.com/books?id=yC4...d%20em&f=false
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05-26-2010 , 04:40 PM
Quote:
Originally Posted by FBandit
Sure you do, but math disagrees with you. Optimal strategy exists for every zero-sum game with finite number of actions.
You need a life sir. Perhaps you can figure out an optimal strategy where you can lose your virginity. Picking up a girl from a bar is technically a zero-sum game with a finite number of actions. Your poker hand is your peni.s (You most likely are holding 27os), preflop action is choosing a girl, the flop is your pick up line (finite number of things that could be said), the turn is the conversation, and the river is closing the deal...Pretty sure no one is going to stake you for this endeavour.
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05-26-2010 , 04:41 PM
Here's an easy example of optimal strategy against an optimal opponent not being all that great against a terrible opponent:

Imagine two people each writing a rock paper scissor program. It's easier if it's a program than having 2 people play because then you remove any tells/people not wanting to throw the same thing x times in a row, etc. If I knew my opponent was writing his program to throw every option randomly, I would write my program exactly the same, and we'd push over the long run, but we'd be playing optimally against each other. Now if some dummy comes along and makes a program that only throws rock because that's his favorite one to throw in real life, my program is still going to win 33% of the time, lose 33% of the time, push 33% of the time, and so we'll still push over the long run. My program does no better at playing against someone who is playing optimally vs someone who is using the worst strategy possible. Obviously, my program should be throwing paper every time vs the 2nd guy. So even though my program is written to play optimally vs an optimal opponent, it's absolute garbage vs someone who uses the worst strategy possible.
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05-26-2010 , 04:50 PM
Quote:
Originally Posted by styx2000
can you expand on that, i dont understand how it could be true.



the gto-bot would play exactly break-even, it wouldnt win. in fact, you couldnt find a strategy that loses against gto bot if im not mistaken.
youre right, i guess i just mean that the gto would break even, but the human who adjusts do a non-gto strategy would end up winning.
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05-26-2010 , 04:52 PM
Quote:
Originally Posted by waow
youre right, i guess i just mean that the gto would break even, but the human who adjusts do a non-gto strategy would end up winning.
Adjusting to GTO, are you serious. Do you understand what GTO means?
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05-26-2010 , 04:52 PM
Quote:
Originally Posted by iPLOwgirls
You need a life sir. Perhaps you can figure out an optimal strategy where you can lose your virginity. Picking up a girl from a bar is technically a zero-sum game with a finite number of actions. Your poker hand is your peni.s (You most likely are holding 27os), preflop action is choosing a girl, the flop is your pick up line (finite number of things that could be said), the turn is the conversation, and the river is closing the deal...Pretty sure no one is going to stake you for this endeavour.
rofl

you can say penis you know its not censored

penis penis penis
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05-26-2010 , 04:53 PM
Quote:
Originally Posted by Demonic
Adjusting to GTO, are you serious. Do you understand what GTO means?
youre misreading my post. against a player who only throws rock, the gto bot would breakeven, but the human who adjusts to play only paper (a non gto strategy) would end up winning.

i dont mean to adjust to gto, but to adjust to the player who throws only rock.
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05-26-2010 , 05:13 PM
aejones said it. NL holdem is solved. not a level
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05-26-2010 , 05:52 PM
Quote:
Originally Posted by snex
and yet, always-defectors cannot out-compete tit-for-tat when you change the starting population. why is this? do you even have an explanation? are you even aware of this fact? you can point to math that neither one of us truly understands all youd like, but simple empirical results prove that you are wrong.
I'm starting to think you're just a huge level. Do you actually know what the game theoretical definition of an optimal strategy actually is? The optimal strategy for the Prisoner's Dilemma is for both players to defect and simple mathematics prove I am right. Results of any experiments have absolutely nothing to do with it.

Look up the definition of an optimal strategy. Look at the Prisoner's Dilemma. Then go the **** away.
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05-26-2010 , 06:09 PM
hsnl just had a debate about something similar to this. the intelligence discrepancy between people that post there and people that post in nvg is astounding.
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05-26-2010 , 06:42 PM
I just read this thread and solved life
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05-26-2010 , 06:50 PM
Quote:
Originally Posted by ActionStan
I thought the last CPRG project switched between several strategies that were more or less aggressive than the Nash model as it collected player data. Not a big leap to think that it could beat the hell out of bad players pretty quickly in HULHE.
The computer could run into the problem of a human setting him up. An expert would spot that easier than the comuter would for a variety of reasons.
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05-26-2010 , 07:21 PM
Quote:
Originally Posted by snex
and yet, always-defectors cannot out-compete tit-for-tat when you change the starting population. why is this? do you even have an explanation? are you even aware of this fact? you can point to math that neither one of us truly understands all youd like, but simple empirical results prove that you are wrong.
I wrote a paper about this in undergrad. I'm in my 3rd year of a philosophy of economics Ph D. I know this stuff pretty well.

The rake is irrelevant to whether poker is solvable. This zero-sum issue is whether any gain by player A is matched by a loss by player B. In poker there is no cooperative "win-win" scenario. Prisoner's Dilemma is not a proper analogy to poker, and conclusions drawn from Prisoner's Dilemma are not relevant to whether poker is solvable. [As an aside there are two and only two strategies in PD - cooperate and defect. Tit-for-tat is a strategy for *Iterated PD*. Iterated PD is even less relevant to the question at hand.]

You simply do not know what you're talking about, sorry.

It has been proven that games like poker, which are finite, zero-sum, and representable by game trees all have optimal strategies --- they can be "solved".

The only credible argument is over poker WILL be solved, not over whether a solution does exist. The game tree for non-heads-up NLHE is enormous. I doubt so large that you'd need more atoms than there are in the universe to represent it, but pretty large nonetheless.
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05-26-2010 , 07:22 PM
some of you simply cannot be serious. Something that has incomplete information is impossible to solve.
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05-26-2010 , 08:05 PM
Quote:
Originally Posted by phils08
some of you simply cannot be serious. Something that has incomplete information is impossible to solve.
Rock-paper-scissors is a game of incomplete information. It has been solved.

Claiming that something is impossible when you clearly know nothing about it is rather bold.
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05-26-2010 , 09:36 PM
Gawd — so many idiots itt. I'd do details, but pointland is doing fine (though he's being ignored). So I'll sum up:

Those of you claiming that poker can't be "solved" — to the extent you mean it is impossible to find a strategy that will not at least break even against everyone (ignoring the rake) and not lose against anyone, you're just wrong. Imperfect information, complexity, multiple players, non zero sum... it's all handled in the proof, for Nash received his quasi-Nobel. There are a finite (though large, in NL) number of actions available to the player at each stage; mixed strategies are allowed. Thus, there is at least one Nash equilibrium. That's just the way it is, regardless how much you may want it to be otherwise. Basically, all finite games can be solved; poker is a finite game.

And yes, this means that if and when the computer guys find this solution, or something close to, it, you will not be able to beat them. Tough.
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05-26-2010 , 09:46 PM
Quote:
Originally Posted by AlwaysWrong
Quote:
Originally Posted by phils08
some of you simply cannot be serious. Something that has incomplete information is impossible to solve.
Rock-paper-scissors is a game of incomplete information. It has been solved.
Correct. But phil will probably respond that he means imperfect information about something other than opponents' plays.

But of course those are solvable too. The simplest one I can easily come up with is to start with a variation of the cooperation game (with payoff matrix:
5,51,3
3,13,3
or the prisoner's dilemma
3,30,5
5,01,1
and just change the payoff so that there is some random element, unknowable by the players when they make their decisions, but leaves all the average payoffs the same. In that case it's pretty easy to see that the equilibria of the games are unaffected. Clearly these games are solvable; clearly phils08 has no idea what he's talking about when he claims otherwise.
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05-26-2010 , 09:50 PM
Only read title:

Yes and no.
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05-26-2010 , 09:54 PM
Quote:
Originally Posted by atakdog
Gawd — so many idiots itt. I'd do details, but pointland is doing fine (though he's being ignored). So I'll sum up:

Those of you claiming that poker can't be "solved" — to the extent you mean it is impossible to find a strategy that will not at least break even against everyone (ignoring the rake) and not lose against anyone, you're just wrong. Imperfect information, complexity, multiple players, non zero sum... it's all handled in the proof, for Nash received his quasi-Nobel. There are a finite (though large, in NL) number of actions available to the player at each stage; mixed strategies are allowed. Thus, there is at least one Nash equilibrium. That's just the way it is, regardless how much you may want it to be otherwise. Basically, all finite games can be solved; poker is a finite game.

And yes, this means that if and when the computer guys find this solution, or something close to, it, you will not be able to beat them. Tough.
http://academicearth.org/lectures/nash-equilibrium
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05-26-2010 , 10:19 PM
Of course poker is theoretically capable of being solved.

But to do so requires mathematically quantifying something that is a long, long way from any kind of mathematical quantification: the playing styles and ranges of your opponents. This is part of the intrinsic nature of poker - all strategies have counter strategies. The key to optimal success is correct adaptation to the factors in play AT THAT MOMENT.

Identifying those factors will, it seems to me, always require an evaluation of your opponent at that time which is not readily quantifiable in advance. So I do not see how solving poker will not include assumptions about other players (this one plays tight; this one always c-bets; his range under these circumstances is "X"; and so on). Making those assessments cannot, by definition, be done in advance of the moment, i.e., in advance of "knowing" your opponent(s).

"Solved" implies to most folks the idea that you can plan every move in advance and prevail. In poker my humble opinion is that at best, probably after only a few more years, an understanding will develop about what is the best strategy to employ under each of a huge number of circumstances. But the "circumstances" will still involve subjective assessments about your opponent's play. And. all other skills being equal, the player who is better at those subjective assessments will prevail, over the long run, against the player who is weaker.

So until we are able to instantly quantify and mathematically assess our fellow players, poker will not be "solved" in the sense that you will know what to do in advance of being in the particular situation, and your success then will depend on your ability to correctly recognize the situation.

In other words, the science of game theory should ultimately get to the point of knowing the perfect strategy for each situation; but it will still be an art for any particular player to know and understand exactly what situation they are in at any particular moment so as to be able to apply that "perfect" strategy.

Which should keep the games interesting for as far into the future as I can see.

Skallagrim
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05-26-2010 , 10:22 PM
Quote:
Originally Posted by macthemost
ive been hearing this more and moar.how the fudge can poker be solved? ok games like razz or 5 card draw or fagdooshy could be solved. but stud? holdem? OMAHA???? solved?? i say no way. and wut does solved mean anyway? mathmatically solved? .. u cant solve reading people. u cant solve bluffing. u cant solve floating..so wut do ppl mean when they say poker is getting closer and closer to being a solved game? kinda depressing to think about..
It's obviously possible to do, but pretty hard.
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05-26-2010 , 10:35 PM
Quote:
Originally Posted by Seriously??
Russ Hamilton has solved poker


easy game
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