Game Theory Problem

David Sklansky · #1 12-23-2006, 12:49 AM

Just made this up. You and your opponent are dealt the proverbial real number between zero and one. You both ante one dollar. You are first and can bet two dollars or fold. If you bet he can call or fold only.

But before you bet you can throw two more dollars into the pot to see his hand.

With what hands would you do this and what is your overall strategy assuming:

A. He knows the rules and turns his hand face up when you toss in the two bucks (before possibly betting two more)?

B. He doesn't know the rules and doesn't understand that you sweetened the pot to gain x-ray vision (but he does know the pot is sweetened)?

Don't even attempt this question if your not very well versed in game theory. I don't know the answer and won't attempt it. But I think I will recognize the right answer if it is properly explained.

Utah · #2 12-23-2006, 02:38 AM

It is too late to try and calculate part A (and I havent started thinking about part B) but, if I understand the problem correctly, I think you are talking about a very large problem (although not very complex) and you end up with a very large densely populated grid.

for player one and for each discrete possible card value you have the following strategies:

Bet, Pay 2 and Bet if opponent has 1, Pay 2 and bet if opponent has 2, Pay 2 and bet if opponent has 3, fold.

Symbolizing it is,

1 - B, PC1, PC2, PC3, F
2 - B, PC1, PC2, PF3, F
3 - B, PC1, PF2, PC3, F
4 - B, PC1, PF2, PF3, F
5 - B, PF1, PF2, PF3, F
6 - B, PF1, PC1, PC2, F
7 - B, PF1, PC2, PF3, F
8 - B, PF1, PC2, PC3, F

Strategies 3, 5, 6, 7, 8 can be removed as they are intuitively dominated.

So, for each possible card value, there are 3 strategies. So, if we assume we can have numbers 1-10, there are 59,000 strategies available before further reduction of dominated strategies. If we assume 5 possible card values then we have 243 strategies.

Player 2 has 4 strategies,

Call bet, Call bet if $2 pay
Call bet, Fold to bet if $2 pay
Fold bet, Fold to bet if $2 pay
Fold bet, Call bet if $2 pay

I would think the forth could also be reduced. So, again, there are 3 options for each card value leading to the same number of strategies.

So, if you assume 5 card values you can solve this quite easily with a 243x243 matrix by using matlab or another matrice calculator. The problem gets harder with 10 card values because a 64k x 64k dense matrix is a mofo to solve and I am not sure there is a calculator that can do it (although I may be wrong).

GiantBuddha · #3 12-23-2006, 11:09 AM

For those of us with precisely zero chance of solving this problem, what's the best book for getting started with game theory?

BluffTHIS! · #4 12-23-2006, 11:47 AM

[ QUOTE ]
For those of us with precisely zero chance of solving this problem, what's the best book for getting started with game theory?

[/ QUOTE ]

Special Thread For Chen-Ankenman Mathematics of Poker

Utah · #5 12-23-2006, 11:53 AM

I just woke up an realized my answer is incorrect. The number of discrete card strategies for player 1 is exponentially based on possible opponent cards and it is also exponentially based on player 1's possible card. The reason is that for each type of card my opponent can have my options grow exponentially. The form on the middle classification - ie, when a pay $2 - of player one is (action types)^(possible cards player could have). In my first post, it was 2^3 because it assumed 3 possible player 2 cards. Then, you have to add +2 for the options of just folding or just raising. So # of player 1 strategy options =

(2 + 2^(# of player 2 cards)) ^ (# of player 1 cards)

So, for only 5 discreet values possible for cards, you are talking 45,435,424 possible strategies. That is a pretty big matrix.

If you wanted to approximate instead of solving discretely, you could probably come up with a very good +EV answer to apply to actual gameplay by reducing the portion of responses when player 1 pays the $2 to a simple more/less than classification. The formula would then be:

{2 + 2^2) ^ (# of player 1 cards)

This is solvable if the # of player 1 card possibilities is not too large.

Utah · #6 12-23-2006, 12:12 PM

[ QUOTE ]
For those of us with precisely zero chance of solving this problem, what's the best book for getting started with game theory?

[/ QUOTE ]
Not to hijack this thread with a book discussion, but the best book as a primer to game theory is Poundstone's Prisoner's Dilemma. It will take you through basic calculations, tit-for-tat, dollar auctions, the game of chicken, etc. It is written for the non-technician and it is a very fun read. Do not start with some complex math book.

I would then recommend The book from the 50's from RAND called "Compleat Strategist". Then, to really get into the details, I would look at the website of Tom Ferguson (Chris 'Jesus' dad) to start learning the math - http://www.math.ucla.edu/~tom/. I started at his site when I first learned tough calculations and I contacted him and he directly helped me through some very tough problems. Of course, what do I know as my answer could be wrong in this thread [img]/images/graemlins/smile.gif[/img]

note - you can also download for free the Gambit software to model simple game theory problems. The software is very nice and easy to use. The only downside is that it cannot handle tough problems as it quickly bogs down to a stop. http://econweb.tamu.edu/gambit/

UpstateMatt · #7 12-23-2006, 01:29 PM

I don't have the time to work this one out right now (maybe I'll try later tonight), but one things is instantly clear:

If we pay to see his hand (and he knows we see his hand), then we're going to bluff (X/3) percent of the time, and fold (100-X-(X/3))% of the time, where X is the percent of the time we have him beat (and thus value-bet). That will make him indifferent to calling.

Ex: We pay to see his hand, he has .300001, we'll value-bet 30% of our hands (.3 or better), bluff with 10% of our hands, and fold the other 60%. That way, 3/4 of the times he calls the 2 dollars, he loses 2 dollars, and the last quarter he wins 6 dollars, making him indifferent.

I'll write more later.

mikechops · #8 12-23-2006, 02:08 PM

I'd guess this isn't going to be an easy problem. Von Neumann and Morgenstern (sp?) analyzed a simple form of poker in "Theory of Games and Economic Behavior". They laid out the equations and you end up with an horrendous integration to solve. They ended up just describing the solution which included bluffing.

BTW, If you are interested in game theory and are prepared to follow the math, then this book is seminal. I was a high school student when I read it and I thought I understood most of what they were saying.

UpstateMatt · #9 12-23-2006, 02:35 PM

[ QUOTE ]
I'd guess this isn't going to be an easy problem. Von Neumann and Morgenstern (sp?) analyzed a simple form of poker in "Theory of Games and Economic Behavior". They laid out the equations and you end up with an horrendous integration to solve. They ended up just describing the solution which included bluffing.

BTW, If you are interested in game theory and are prepared to follow the math, then this book is seminal. I was a high school student when I read it and I thought I understood most of what they were saying.

[/ QUOTE ]

No offense, but that book is a bit dated. It certainly was groundbreaking, but there are better, easier-to-understand, more complete modern books. I thought this was an excellent starting point. And as far as poker problems go, Mathematics of Poker has some excellent lessons on game theory analysis.

Utah · #10 12-23-2006, 02:55 PM

[ QUOTE ]
I don't have the time to work this one out right now (maybe I'll try later tonight), but one things is instantly clear:

If we pay to see his hand (and he knows we see his hand), then we're going to bluff (X/3) percent of the time, and fold (100-X-(X/3))% of the time, where X is the percent of the time we have him beat (and thus value-bet). That will make him indifferent to calling.

Ex: We pay to see his hand, he has .300001, we'll value-bet 30% of our hands (.3 or better), bluff with 10% of our hands, and fold the other 60%. That way, 3/4 of the times he calls the 2 dollars, he loses 2 dollars, and the last quarter he wins 6 dollars, making him indifferent.

I'll write more later.

[/ QUOTE ]

The terminal value of the node is very easy. It is combining it with the previous decisions to form the complete game strategies that is tricky [img]/images/graemlins/smile.gif[/img]

I normally only solve these problems with discrete values as opposed to a continuum of o to 1 and it has been a long time since I have worked on game theory problems so, maybe I am missing something here. But, it seems in this problem I first need to know the terminal node value of .3000001 before I decide to go down that branch. Thus, I need a ".3000001 strategy". Then, what about .4546321? Now, I can find the cutoff point where the value of the node goes to zero. But, that may not help you because it still may be profitable to sometimes choose a node with a negative value so as to "trick" your opponent is the scheme of the larger strategy (however, I am not sure about this)- ie., my opponent can calculate that cutoff point too and if I never paid $2 past a certain point then I have conveyed information about my hand that can be used against me.