The Mathematics of Poker

[morgan180]

[ QUOTE ]
Bill or anyone else who is qualified:

I'd like to take a mathematics refresher course before reading your book, can you recommend a good book or software application for an adult who hasn't attended a math course in 20 years, yet is a very fast learner? I think it will be like getting back on a bike for me.

TT [img]/images/graemlins/club.gif[/img]

[/ QUOTE ]

I'm certainly not qualified TT but I agree with RVG - the first two chapters do a good job of refreshing, building the basic statistic foundation, and really explain it well. I was particularly impressed with how they used plain language and examples to talk about some of the results presented by the equations.

Plus they reference back to the original equations frequently so you can flip back a couple of pages, re-read the explanation of the underlying formula and jump back to where you were with a clear understanding of the section.

HTH.

edit: If I had to make a recommendation I would say an undergraduate entry stats class would cover most of the material covered in the first few chapters and flipping through the rest it looks like it could be covered in a first year calc class to cover some of the logs stuff, etc.

[RedGladiator]

i knew my BSc maths would come handy one day [img]/images/graemlins/cool.gif[/img]

[sputum]

OK maybe a dumb question but...
If one renders the opponent's choices indifferent, how can he make a mistake?

[thylacine]

[ QUOTE ]
OK maybe a dumb question but...
If one renders the opponent's choices indifferent, how can he make a mistake?

[/ QUOTE ]

Not a dumb question at all. If you play in an unexploitable way, you do tend to equalize the expectation of SOME of your opponent's choices, but not necessarily ALL of them. Mistakes are still possible, except in very simple examples.

Also, your opponent also needs to be careful not to play in an exploitable way.

[felson]

[ QUOTE ]
[ QUOTE ]
OK maybe a dumb question but...
If one renders the opponent's choices indifferent, how can he make a mistake?

[/ QUOTE ]

Not a dumb question at all. If you play in an unexploitable way, you do tend to equalize the expectation of SOME of your opponent's choices, but not necessarily ALL of them. Mistakes are still possible, except in very simple examples.

Also, your opponent also needs to be careful not to play in an exploitable way.

[/ QUOTE ]

also, indifference means 'indifferent when we consider the total range of opponent's possible hands.' it doesn't mean your opponent is indifferent for every possible hand he might have. if you bet and he has the nuts on the river, he had better raise.

for a lot of opponent's hands (maybe most or even all of them in some games), there will be a single correct play. but if the opponent somehow couldn't see his hole cards, but only knew his hand range, then he would be indifferent.

[sputum]

So if we know his range we can make our actions unexploitable against it but he can still misplay his actual hand? Sounds plausible.

[Patrick Sileo]

Game theoretic equilibria have the property that no individual player can change his action and make himself strictly better off (vs the equilibrium strategies of the other players). They come in two forms: pure strategies and mixed strategies. The former consist of a set of non-random actions (one per player). If a player takes an out-of-equilibrium action, he may be worse off. Since poker is a zero-sum game, when your opponent is worse off, you are (generally) better off.

In a mixed-strategy equilibrium, each player is randomizing over a set of non-random actions. Let these actions be say {A,B,C}, selected with probabilities p(A), p(B), p(C), respectively, where p(A)+p(B)+p(C)=1. It is a property of this type of equilibrium that a player's expected payoff is unchanged if he chooses a different randomization over the same strategy set {A,B,C}. In this sense, mixed-strategy equilibria are "mistake-proof". However, should an "incorrect" action "D" be included in the randomization, the player now may well be worse off, as in the pure strategy equilibrium case.

In general, game theory's non-exploitable strategies are most useful against better opponents, since you cannot be exploited. On the other hand, it can be worth playing a theoretically exploitable strategy against an inferior opponent so that you can fully exploit him.

[Djeorge]

I have a question regarding finding nash equilibria. is there necessarily only one nash equilibrium in say HU poker and if so, why? Is this true for all zero-sum games ? I don't know much about game theory but my intuitive guess is that there is only one.

[Patrick Sileo]

For a classic and very readable text on game theory, see "Games and Decisions" by R. Duncan Luce and Howard Raiffa.

From the text:

"In the domain of mixed-strategies, every zero-sum, 2-person game has at least one equilibrium [strategy] pair, and when there are several, they are equivalent and the equivalent strategies are interchangeable."

[Djeorge]

[ QUOTE ]
For a classic and very readable text on game theory, see "Games and Decisions" by R. Duncan Luce and Howard Raiffa.

From the text:

"In the domain of mixed-strategies, every zero-sum, 2-person game has at least one equilibrium [strategy] pair, and when there are several, they are equivalent and the equivalent strategies are interchangeable."

[/ QUOTE ]

Are you saying that if a 2-player zero-sum game has 2 eqiulibria, A=(a_1,a_2) and B=(b_1,b_2), then e.g. C=(a_1, b_2) is also an equilibrium?