An Expression of the Value of Position

[punter11235]

I am very pleasantly surprised by this article as this was a problem I was trying to solve for some time (unfortunately I didnt get there). I have a request for the author : could you please refer me (or give a summary here which would be really great) to a place when the Nash equilibrium solution is proved ? Does it imply that this solutions is better than any mixed stragy possible ?

I hope for more interesting (and maybe not practical) mathematical articles in the future /images/graemlins/smile.gif

Best wishes

[a rebours]

[ QUOTE ]
I have a request for the author : could you please refer me (or give a summary here which would be really great) to a place when the Nash equilibrium solution is proved ?

[/ QUOTE ]

I'd like to see that too.

good article, I liked it.

[JaredL]

Thanks for the feedback. I'm going to warn you that this post is going to be more technical than the article.

While many people have solved it, probably the best link you can find for this is an article on [0,1] poker (http://www.math.ucla.edu/~tom/papers/poker2.pdf) by Chris Ferguson (yes, that Chris Ferguson) and Tom Ferguson. They review work by von Neumann and others and go through some versions of this, more general than the one I've done. I used a set bet size (pot), they solve the arbitrary case, which is fairly simple to do, they also solve the version that allows raises, including the check-raise.

They also demonstrate that it's an equilibrium. I wanted to avoid all discussion of proving it was an equilibrium as that would distract from the point - why position is valuable with no cards to come. Loosely speaking you can simply take the partitions made by the points where the action changes (for example the value at which you switch from bluff betting to check-folding) and for each player explicitly writing down their expected payoff from each pure strategy given the opponent's strategy in the equilibrium. If you do this, you will see that the strategy given has taken the best pure strategy for each partition and thus the whole space.

As for the other question "Does it imply that this solutions is better than any mixed stragy possible ?" To answer this question one simply needs to use the properties of 2-person zero-sum (or in this case, fixed-sum depending on your interpretation of the antes) games. In 2-person zero-sum games if you have a Nash Equilibrium, it must be that whatever value the players get out of the game will hold for EVERY Nash Equilibrium. So if there were to be a Nash Equilibrium involving mixing on more than 1 point then this Nash Equilibrium will also have the same value for each player. Note that the equilibrium I've given DOES allow for mixing at the indifference points, for example at 1/12 player 1 can either bluff bet, check-fold, or mix between these using any ratio of one to the other she wishes.

To see that this must be the case consider this game. Player 2 wins himself a long run value of 1/12 from repeating the game (assuming he stays in the same position) in this equilibrium. It is an equilibrium because player 1 is acting optimally given what he is doing. So if 1 plays some other strategy then it must be worse for her (or she gets the same payoff). Taking this into account we can see that player 2 can guarantee himself a long run average payoff of 1/12 by using the given strategy. So if there is an equilibrium, player 2 must get 1/12 or more. Doing the same analysis but switching players tells us that player 1 must get a payoff of at least -1/12 in any other equilibrium. The only way these two equations can hold if the game is zero-sum is for 1 to get -1/12 and 2 to get 1/12 in any equilibrium.

I hope that made sense, if not feel free to ask any and all follow up questions you may have.

Jared

[a rebours]

makes perfect sense /images/graemlins/smile.gif ty for the explanation.

just out of curiosity: "In 2-person zero-sum games if you have a Nash Equilibrium, it must be that whatever value the players get out of the game will hold for EVERY Nash Equilibrium." this must be true for every 2-person game, even if it is not zero-sum, right?

assume that there is a game with two different Nash Equilibriums which have different values. if player 1 is getting value A in equilibrium 1 and value B in eq. 2, and B>A, then he wld do better to switch from strategy 1 to strategy 2. then player 2 wld also have to switch to strategy 2, because that's his best response (by definition). but that wld mean that strategy 1 is not Nash Equilibrium.

am I missing something here? cannot think of an example of a game having Nash Equilibriums with different values for more than two players either, but it probably does exist.

[JaredL]

The reason I suspect that you're not coming up with an example is that you're still thinking more in competitive (from a real world type definition of competetitive not game theoretical) type situations which tend to be zero-sum.

Here's a simple example. You and I are playing a game where we both say 1 or 2. If we say the same number we each get that number. If we say different numbers we get nothing. In this game there are two equilibria in pure strategies. One what we call pareto dominates the other, meaning that everybody gets more (technically it means that somebody gets more and everybody gets at least as much).

(1,1) and (2,2) are both equilibria. To see this suppose I'm putting 1 and you know that. If you put 1 you get 1 while if you put 2 you get nothing. Hence playing 1 is a best response. The same argument will hold for me so (1,1) is an equilibrium. Similarly (2,2) is an equilibrium.

To see why this is possible in this case and not the zero-sum case, follow the argument I made in the post above. I said that if player 2 were to change strategies, she would be worse off, or get the same payoff. If this happens that must give player 1 a better payoff (or the same) due to the game being zero-sum - player 2's loss is player 1's gain.

Notice that that didn't happen in this 1 or 2 game. Suppose you put 2. 2 is the equilibrium strategy for me. If I put 1 I get a lower payoff, but now you also get a lower payoff. Furthermore, you get a payoff that is so much lower that if you could redo your action based on my action you would rather change.

Again, the reason it doesn't hold for all two-player games is that you cannot guarantee equilibrium payoff if you play an equilibrium strategy, and playing the strategy for equilibrium A could give you a payoff lower than equilibrium B if the other player is playing her equilibrium B strategy.