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Game Theory and Poker
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Lets see if I've understood this... apologies if I'm a bit too verbose here, RoundTower, want to see if I understood it properly.
I also have a question or two about the applying this to games in general:
With an Ace, the payoff/profit for a check is 1. No matter what player 2 has or does, in response to a raise, the minimum payoff is still 1, and could be as high as 2, if the other player calls the raise. So rationally, you'd always bet.RoundTower wrote: »here's how you solve this problem:
there is nothing to be gained from checking with an Ace, so you should always bet.
If you call with a king, the EV of the profit is 0.RoundTower wrote: »there is nothing to be gained from betting with a King, so you should always check.
But, if you raise, the EV of the profit falls to -.5 (half the time player2 will have an ace, and you'll lose 2, half the time player2 will have a queen, and you'll only gain 1, because they will rationally fold, knowing they are beaten, hence -.5 EV)
-.5 EV being less than 0 EV, it's rational to always check, with a King.RoundTower wrote: »with a Queen you should adpot a mixed strategy, either check or bet. let's say you bet with probability x, 0 <= x <= 1
So, this is where it gets a little trickier.
When you have a queen, and player2 has an ace, you are going to lose money. Nothing can be done about that. But half the time player1 has a queen, player2 has a King, and not an Ace.
When player1 has a queen, and player2 has that king, player2 doesn't know if player1 has a queen or an ace. Therefore there's a chance of being able to bluff player2 out of the pot, by raising.
Hence, you can't say that player1 should either always check or always bet here.
The game theoretic thing to do will be to adopt a mixed strategy, which means that you don't always take a single action, instead you choose which action according to a probability distribution. As you can only either check or bet, the distribution here can be represented by a single number - essentially the ratio of checks to bets.
So, the only thing to do to figure out optimal play for player1 is to figure out what that number is.
Simple enough - he's ahead with an ace, and behind with a queen. The only time he'll have an ace and have a decision to make is when you are trying to bluff him with your queen, so he obviously calls. The only time he has a queen and has a decision to make is when you have raised with your ace, so he obviously folds.RoundTower wrote: »for him, he should always call with an Ace and fold with a Queen.
So, thats what player2 does when player2 has an ace or a queen.RoundTower wrote: »He should call with a King some of the time - enough that you don't get to win the pot every time with a Queen, but he also doesn't want to pay you off every time you have an ace. So say he calls with probability y.
There are two situations in which player2 can have a King, and have a decision to make about whether to call or fold.
One is when player1 has an Ace, and has made a bet, because it made no sense to just check with an ace, and the other is when player1 has a queen and has made a bet as a bluff (with frequency X).
So, when player2 has the King, and sees a raise, he has to call a certain amount of the time - ie, as you say, with probability Y.
What the Y is to maximise player2's EV obviously depends on what X player 1 is using, given that player1 moves first each time.
And also, if player1 wants to maximise player1s winnings, then player1s X should be chosen mindful of player2s Y. For example, if Y was 1, player1 would make X to be 0 - if player2 always calls with the King, player1 will always check the queen, (and maximise profits by getting paid for his Ace).
But if Y was 0, player 1 would make X to be 1 - if player2 never calls with the King, player1 will always bet with the queen, and while player1 won't get paid for his ace, he'll take down the pots when he only has a queen instead.
But we're trying to find the equilibrium strategy, which is the best X given that player2 will chose the best Y...
So, the expectation are:RoundTower wrote: »Then there are 6 possibilities for how the cards get dealt, all equrobable. When you have an Ace and he has a Queen, for example, you make €1. When you have a Queen and he has an Ace, you lose €2 x of the time and lose €1 (1-x) of the time, so on average you make €(-2x -1(1-x)) which is €(-x-1). You can also find the outcomes for the other four possibilities too, and average them together to get the expectation of the game. You should get (x + y - 3xy)/6.
[player1card,player2card]:[expectation]
AQ:1
AK:y+1
KA:-1
KQ:1
QA:-x-1
QK:2x-3xy-1
And these are all equally likely, the expectation of the game is the sum of them all divided by their number, so as you said, (x+y-3xy)/6, again, where x is the probability of player1 to bluff with a queen, and y is the probability of palyer2 to call with a king.RoundTower wrote: »Now to find the optimal x, you need to know that you should choose x such that it you are indifferent to his choice of y, in other words, the terms containing y should sum to 0.
So we now have the expectation of the game (from the perspective of player1, and remembering the game is zero sum), expressed as a function of X and Y.
What player1 wants to do is choose X such that the value of the expectation, given the worst value of Y player2 can choose for that X, is at a maximum.
This happens when player1 chooses 1/3 as the value. (see pic)
I'm not sure I understand what you say that the player 'should choose x such that you are indifferent to his choice of y, in other words, the terms containing y should sum to 0' - can you explain this part in more detail?
I understand that 1/3,1/3 is the equilibrium from the function, and theres plenty of ways to calculate that, but I don't fully see how you calculated that?RoundTower wrote: »So y - 3xy = 0 for all y, therefore x = 1/3. Similarly you should get y = 1/3, and the expectation of the game should be that you, on average, make one eighteenth of a euro. He expects to lose one eighteenth of a euro on average
Now you could see how the game changes if you have a different choice, for example if the bet amount was €2 instead of €1, or more interestingly, if you had a choice to check, bet €1 or bet €2.
I'm pretty sure I understand the analysis of this game that you gave.
I'm wondering though, how well does the method used generalise? Working this stuff out by hand, like you did there, could get pretty complex as the game complexity grew. How would you go about analysing more complex games? Is it possible to work these things out algorithmically, or to represent the game rules in some standardised format, and apply something to calculate the equilibrium positions in a general way? (or do some sort of stochastic approximation?)
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A perfect example is the guess 2/3's of the average puzzle.
In this, players attempt to guess a number that is 2/3s the average all of all players guesses. The solution is simple, yet hugely flawed.
Applying this to the AKQ problem, I'd assume that player B calls with a K 50% of the time. Now what is our EV, and how often should we bluff.
I understood the solution like this, That B calls 100% of the time with a K b/c A's bets with Q's 33% of the time. This is where A's edge lies in the puzzle, and it's b/c B never gets a chance to raise etc.
GT is flawed so, jeepers. Just wondering if you could quantify that and add it to a puzzle, like this-
A is on the river with the 2nd nuts and bets,
B has the third nuts and is thinking something along these lines,
1, A bets a hand that is beating me x% of the time
2, A bets a hand that is losing to me x% of the time
mmm, stuck here, b/c A is considering the times in which he's beat, which would/could take into consideration A's tilting tendencies as well as when he's playing solidly.
Maybe you have to come up with 2 different numbers. . .
I found this article very intersting and is totally relevant to this thread
http://www.twoplustwo.com/magazine/issue60/Christenson-commentary-approximating-game-theoretic-optimal-strategies-full-scale-poker.php
C
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This makes sense to me, in many real world scenarios, especially where it isn't a fully correct model of the real world, and where it isn't a zero sum game, and where the other players are acting intuitively.I'm pretty sure his father is (or was) a leading GT professor, in UCLA IIRC
When I look at game theory problems, I tend to approach it as if I was playing a reasonably rational opponent , but not one that plays with perfect rationality
For example, the prisoners dilemma as mentioned earlier.Example, Nash's say that the answer to the travelers dilemma is 2, yet we all know that picking say 94 would have a much higher EV against most rational people.
This is the huge flaw in nash's equilibrium, it assumes everyone has perfect rationality, which they don't.
A perfect example is the guess 2/3's of the average puzzle.
In this, players attempt to guess a number that is 2/3s the average all of all players guesses. The solution is simple, yet hugely flawed.
Applying this to the AKQ problem, I'd assume that player B calls with a K 50% of the time. Now what is our EV, and how often should we bluff.
I think the equilibrium analysis here is pretty useful though.
Like, as player1 I now know the correct way to play this game, and if we player and you are player2, I'll always beat you - in the end, I'll always come away profitable (over the long run, expected value, etc).
Without the game theory analysis, its possible that as I played I might wander either 1) make a silly mistake (not betting with my Ace) or 2) without making silly mistakes, still wander into that blue corner of the surface where I have -EV.
If B calls with a K 50% of the time, we should never bluff, and our EV is about .08 - which means the game is worth more to player1 than it was before. In other words, B is now losing more money from deviating from the best strategy.
So yes, if you have an opponent thats deviating from the best strategy like this, then it does make sense to change from the game theory best strategy in order to maximise how much money you take off your opponent.
However, we must remember that the deviation the opponent did didn't improve the game from their perspecitive - by deviating from .333 player2 did not improve his chances to take money from player1 - only gave player1 an opportunity to take more money from him.
If player1 still just played .3333 then he'd still beat player2 always.
Where as if player1 instead deviates to never bluffing, in an attempt to maximise the amount taken from a player2 who we think is calling half the time, and player2 starts to never call, suddenly the EV for player1 goes to zero - no longer making any money at all.
You get into this cycle of trying to figure out what your opponent is actually playing, having to second guess them, and repeat that again and again - whereas with the GT optimum player1 can just sit there and take their earnings.
Its not a perfect model of the real world and how the real world works.I understood the solution like this, That B calls 100% of the time with a K b/c A's bets with Q's 33% of the time. This is where A's edge lies in the puzzle, and it's b/c B never gets a chance to raise etc.
GT is flawed so, jeepers.
But the analysis of this poker game, as done here, isn't flawed as such - you need to be careful about what claims are made, but the claims that are made are accurate - .333333 is the way to go if you are player1, and you'll take home the best average profit over time, even against an infinitely rational and smart opponent.Just wondering if you could quantify that and add it to a puzzle, like this-
A is on the river with the 2nd nuts and bets,
B has the third nuts and is thinking something along these lines,
1, A bets a hand that is beating me x% of the time
2, A bets a hand that is losing to me x% of the time
mmm, stuck here, b/c A is considering the times in which he's beat, which would/could take into consideration A's tilting tendencies as well as when he's playing solidly.
Maybe you have to come up with 2 different numbers. . .
I found this article very intersting and is totally relevant to this thread
http://www.twoplustwo.com/magazine/issue60/Christenson-commentary-approximating-game-theoretic-optimal-strategies-full-scale-poker.php
C
Will have to look at that in more detail later, be interested to read more about this.RoundTower wrote: »I'm not sure what the proof is that this works as a method to find the equilibrium of the game, I just know it does.
I should also read more about mechanically trying to solve bigger games that better approximate the real world, to trying to find partial solutions to them. Interesting stuff...RoundTower wrote: »As for the argument that "game theory is flawed because it assumes rational players" - this isn't really true. Examples like the Travellers Dilemma are interesting because they are atypical and provide unexpected results. If you could find optimal play for poker against rational players, firstly, you would probably be able to crush almost any game in the world, and secondly, it would then be a relatively small step to move towards a strategy that exploits weaker players (for example, in the example game, if the guy always called with a King you would never bluff with a Queen).
You can't solve any "real" poker game by breaking down all the trillions of possibilities, like we did for the one-card game, but you can still apply some of the conclusions from it. For example, you can generalise the result to say that your optimum bluffing frequency should be such that your bluffs:value bets are in the ratio bet size:pot size+bet size (which was 1:3 in the example).
Thanks for the example, and the solutions/discussion about it, RoundTower.0 -
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I was thinking about player 2's calling frequency and was wondering why his optimal number wasn't 33% if player 1's betting frequency with a Q was known to be 33%. To my mind, this would save player 2 money or am I getting into second level thinking here, where 2 is adjusting already?
A more complete poker puzzle would be a JQKA game with player 2 being able to bet and raise.
Thanks y'all for the thread,
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Not sure what you are saying here.I was thinking about player 2's calling frequency and was wondering why his optimal number wasn't 33% if player 1's betting frequency with a Q was known to be 33%. To my mind, this would save player 2 money or am I getting into second level thinking here, where 2 is adjusting already?
The poker problem, as stated by RoundTower here, has equilibrium at 1/3,1/3, as he said, and we worked out - so player2s calling frequency is 33.33..%.
Maybe you misread something there?
That'd be a more interesting problem all right, although could be a lot harder.A more complete poker puzzle would be a JQKA game with player 2 being able to bet and raise.
Thanks y'all for the thread,
C
I'll take a look at the version with the reraise at some stage, and then maybe an extra card, but each extra decision grows the tree a lot.0 -
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