Tough, Important, General Case, Game Theory Problem

[Copernicus]

The probability in question is the probability that everyone ELSE folds given that you fold, ie the probability of you folding is 1, and there are n-1 others with a choice.

[George Rice]

I should have explained my confusion. I understand how MBE got the c^(n-1) and that he was figuring that the player's probability of folding is 1. But he won't be folding all the time. And he will win back his ante only those times he folds and the other players fold too. All the other players folding is c^(n-1) but I'm having trouble understand why this shouldn't be multiplied by c, as the player will be betting 1-c times.

The way I look at the problem is that the player will win all the antes (na) when he bets (1-c) and the other players all fold (c^(n-1)); He will lose (b) when he bets (1-c) and one or more of the other players bet too (1-c^(n-1)); and he will win his ante back (a) when he folds and the other players fold too c * c^(n-1) = c^n.

Another way of saying it is that when all the other players fold, hero will win the amount (na), (1-c) times and he will win the amount (a), (c) times. This is wrong? If so, why?

I could understand it if hero only "wins" (a*(n-1)) when he bets (1-c) times. But I don't remember it being written that way. Is this where I went astray?

Thanks.

[George Rice]

Looking closer at MBE's solution is see that he's saying that hero's betting strategy should result in an amount equal to his folding equity (to be optimal), which is c^(n-1). Sometimes when involved in a problem we can forget even a basic principle. So then,

c^(n-1)na - (1-c^(n-1))b = c^(n-1)a

not

(1-c)(c^(n-1))na - (1-c)(1-c^(n-1))b + c^n = 0

The probability of betting can be removed from the left side because betting is a given, and the probability of not betting can be removed from the right side becasue not betting is also a given.

Thank goodness. Try reducing the second equation. For a=1,b=2,n=3 it becomes a cubic equation. For n>3 I don't even want to think about it. :-0

[Copernicus]

its not like you have to solve the equations in general terms, you just plug in the knowns and do the arithmetic.

[Bokonon]

[ QUOTE ]
Lets see what everyone thinks of this answer:
Assumptions:
1. Probability of a victory:
Given you have a hand X, what is the probability that your hand is better than every other players?


[/ QUOTE ]

Ah, but isn't the trick that your hand doesn't have to be better than every other players in order to win? If all the hands are horrible, and you bet, then your 0.3 could end up winning the pot because the guy with 0.4 and 0.5 didn't bet.

This is a fascinating question, and Dr. Slansky is right -- this really gets to the heart of poker theory. Wow.

[Bokonon]

[ QUOTE ]
In fact, if your opponent is playing optimally, you MUST bet if your card is X > c , but if it is less it doesn't matter if you bet or fold. c has been set so that the EV is indifferent to whether you bet or fold when X < c.


[/ QUOTE ]

I just went through a bunch o' math and . . . well, I am in complete agreement, at least for n=2. In this example, you can play mega-loose . . . if your opponent is playing tight, it don't matter, you'll still be even in EV. And if he plays too tight, you'll be ahead.

Which means that to the extent that this example is similar to heads-up, it doesn't matter how tight your opponent plays, as long as you don't play tighter than him (assuming he plays an optimal amount of starting hands) you have no disadvantage.

Which is just bizarre, so I'm sure that it's *not* similar to heads-up [img]/images/graemlins/smile.gif[/img].

[George Rice]

[ QUOTE ]
In this example, you can play mega-loose . . . if your opponent is playing tight, it don't matter, you'll still be even in EV. And if he plays too tight, you'll be ahead.



[/ QUOTE ]

Sure, but if he catches on and plays looser, but still tighter than you, he will destroy you!

[George Rice]

The whole point of an optimal strategy is that it doesn't matter what you do. If you loosen up and bluff more, his EV will not go down because you will be loosing bets to him you would have never made when he is strong enough to bet, and this will offset the bets you win by bluffing. Moreover, if your opponents discover that you are bluffing too much they will loosen up a bit and increase their EV which lowers yours.

It is true that you can bet every hand if your opponent is playing optimally, but it doesn't matter. You are risking losing EV (if your opponent catches on) and have ho chance of gaining EV.

[Cosimo]

If you play at optimum:
* you win money from those who play tighter
* you tie with looser opponents

If you play tighter than optimum:
* you lose money to anyone looser
* you win money from even tighter players

If you play looser than optimum:
* you tie with optimum players
* you win money from too-tight players
* you win money from players looser than you

[George Rice]

You left out that if you play loose you lose money to tighter players who play looser than optimum.