Deepest \"possible\" stacks
 

Just out of curiosity, where do you think the theoretical treshold for stack size in terms of BB is in a NLHE game, so that all the chips are virtually never going into the middle? Of course there are maniacs who would gladly call a 50X potbet on the flop, but consider a game with about six good/expert players.

For instance, would a $1-2 blind NL game with $1500 stacks play any different than the same game with $3000 stacks?

Re: Deepest \"possible\" stacks
 

In a NL game where all the chips can go in at any time, there is no threshold.

I suppose wiht a pot-limit game, you'd have a practical threshold, as you describe it. But even there, if I have quads and you have higher quads, it's all going in no matter how much we have.

But, the deeper the stacks, the more implied odds becomes more important than pot odds. So, a $1500 stack game has about half the implied odds on most streets than a $3000 stack game.

Re: Deepest \"possible\" stacks
 

Yes, the thing is, that with ridiculously big stacks, say
100,000BB, how can anyone ever be allin without the nuts? There must be some point where that happens?

Re: Deepest \"possible\" stacks
 

[ QUOTE ]
Yes, the thing is, that with ridiculously big stacks, say
100,000BB, how can anyone ever be allin without the nuts? There must be some point where that happens?

[/ QUOTE ]

Infinite stacks.

But I'm also that, since poker has a set number of hands of discrete value relative to stacks, there must be some point where the second nuts is of equal value to the nuts, and some point where that stops being the case.

Re: Deepest \"possible\" stacks
 

How would the flop action go with infinite stacks with AK v AK v QQ on a QJT board?

In a three way all in the QQ is a money favorite, but heads up it is an underdog. How would each player play their hand correctly? Each of the AK players would be making an "infinitely" big mistake to get all in on the flop, but the QQ would make an infinitely big mistake to get all in unless he was guarenteed to get two callers.

Re: Deepest \"possible\" stacks
 

[ QUOTE ]
How would the flop action go with infinite stacks with AK v AK v QQ on a QJT board?

In a three way all in the QQ is a money favorite, but heads up it is an underdog. How would each player play their hand correctly? Each of the AK players would be making an "infinitely" big mistake to get all in on the flop, but the QQ would make an infinitely big mistake to get all in unless he was guarenteed to get two callers.

[/ QUOTE ]

Easy cop-out answer: with infinite stacks, nobody would get in preflop without AA.

Re: Deepest \"possible\" stacks
 

[ QUOTE ]
[ QUOTE ]
How would the flop action go with infinite stacks with AK v AK v QQ on a QJT board?

In a three way all in the QQ is a money favorite, but heads up it is an underdog. How would each player play their hand correctly? Each of the AK players would be making an "infinitely" big mistake to get all in on the flop, but the QQ would make an infinitely big mistake to get all in unless he was guarenteed to get two callers.

[/ QUOTE ]

Easy cop-out answer: with infinite stacks, nobody would get in preflop without AA.

[/ QUOTE ]
thats not saying much. people can fold KK just 200BB or 300BB deap...KK could be easily folded 1000BB deap or whatever.

Re: Deepest \"possible\" stacks
 

No one would ever lose with infinity stacks.

What is infinity minus infinity?

Re: Deepest \"possible\" stacks
 

[ QUOTE ]
No one would ever lose with infinity stacks.

What is infinity minus infinity?

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Well, that´s not really the point, let me reformulate the question:

Find (roughly )a number X, so that playing with stacks of X BB is the same as playing with bigger stacks.

That is, given expert players, who play "optimally", how many BB:s deep would they need to be to guarantee that there would never be an all-in? Is that even possible? Would it be possible with say 2 players but not 3 or more, or vice versa?

Re: Deepest \"possible\" stacks
 

An interesting question.

We can answer this for some toy games. For example, Chen and Ankenman present a solution for the half-street [0,1] game in terms of two functions x(s) and y(s) which specify for a bet size s (in relation to a pot size 1) how good of a hand you need to call or bet, respectively. (Lower numbers == better hands in this game.)

x(s)=(6/7)/(1+s)
y(s)=(3/7)/(1+s)^2

Obviously these equations permit unbounded bet sizes. But suppose there are only 169 possible hands, as in Texas Hold'em. Then the best possible hand is 1/169 or about 0.006. What bet size can only be called by a hand this good?

1/169 = x(s) = (6/7)/(1+s) = 6/(7*(1+s))
7*(1+s)=6*169
1+s = 6*169/7
s = 6*169/7 - 1 = about 144

What bet size corresponds to betting with the best hand (or a corresponding bluff)?

1/169 = y(s) = (3/7)/(1+s)^2 = 3/(7*(1+s)^2)
(1+s)^2 = 3*169/7
s = sqrt(3*169/7)-1 = about 7.5

This suggests that stack sizes greater than about 7.5x the size of the pot do not change the play of this toy game, since the optimal strategy is never to bet more than that.

(Of course, changing from a continuous to a discrete game creates all sorts of inaccuracy.)

There is not any easy way to turn this estimate into a multiple-full-street game. If there is only one raise per round then 7.5x per round gives us a final bet about 3200x the size of the initial blind. Two raises per round gives an estimate of 10,000,000x the initial blind.

I suspect the true answer for Hold'em is probably somewhat less than that. [img]/images/graemlins/wink.gif[/img]