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Boomer2k10
Joined: 22.09.2010

If you want to know what one of these range building exercisies you can watch any of my videos where I use Combonator (There is a silver one where I explore a JTo river bluff)

And no, you do NOT use your opponent's range when calculating whether something is a bluff.

You use your range which got you to this point.

i.e. preflop you'll have your preflop range (same on flop usually as we c-bet 100% in HU pots) but the action will dictate what our current range is and the decision on the river is essentially a equilibrium calculation where we know we'll be given our opponent certain odds to call us.

Lets have a look at a river bet very simply:

Assuming a simple relationship (i.e. opponent is all in if he calls) If we are giving our opponent 6-1 to call us on the river then in order to make him indifferent to calling us we have to bluff with 14% of our range (1 in 7 bets). That means no matter what stretagy our opponent choose (call% or fold%) he cannot take advantage of us bluffing.

The correct counter-strategy for him is to fold the bottom 17% of his range.

Why 17%?

Becasue we are getting 5-1 to bluff so in order for us to not make a default profit he has to call 5/6ths of the time.

If he adopts this strategy there is no adjustment we can make in our bluffing range to take advantage of it and there's no incentive for him to change so the game is in equilibirum as both players are playing optimally. If we bluff more he takes advantage of us with his calling frequecy and if we bluff less he takes advantage with his folding range.

Now obviously poker is more complex than this but that's the basis of Equilibrium play, to reach a point where your opponent cannot take advantage of what you're doing.


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YohanN7
Joined: 15.06.2009

Originally posted by Boomer2k10

Originally posted by YohanN7
Can someone point me to a rigorous*) statement and proof, preferably detailed, and specialized to poker, of the existence of GTO in FL HUHU and FL multi-handed.

*) Rigorous = Hardcore mathematics by a hardcore mathematician published in a journal of mathematics. (At Wikipedia they would say "sourced".)

/Johan = :f_confused:

Yes it's called Nash Equilibrium which has been around for years and has been proven to be correct for non-cooperative 2-player games and has been explored for multi-player games but is seen as ridiculously hard which is why when you see range building exercises they are almost exclusively for 2-player hands (i.e. postflop) as multi-player Nash is impossible for even a super-computer to calculate in poker (even modern day bots are a factor of almost 1,000,000 short of HU Game Theory Optimal Play)

Yes, I knew this. I'm still asking for a statement and a proof, preferably specialized to poker, but an abstract one will do - and a place to find it. I am going to spend some time to acquire first hand knowledge.

/Johan = :f_confused:


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YohanN7
Joined: 15.06.2009

And no, you do NOT use your opponent's range when calculating whether something is a bluff. You use your range which got you to this point.

This clarifies a lot. Let me try to expand on this.

We do use our opponents range, but we do so implicitly. His range is encoded it the previous action, hence always encoded in our own range. We still need to determine what we consider to be a bluff. This has to be a statistical average of his range. Even if I'm wrong about his range being encoded in our range (logically probably meaning his range is always 100% - what else could it be then, and why?), we need to pin down what constitutes a bluff and what constitutes a value bet. In the end it is a Sklanskian loop over the possible holding of our opponent, averaged, but not weighted. Please don't dispose of this reasoning right away, because I sometimes have trouble expressing myself in an unambiguous way.

You post beautifully describes what GTO is in this one spot (provided we got there legitimately!). I'll watch the vid (once again). Rest assured, i'll have follow up questions.

/Johan = :f_confused:


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redskwerl
Joined: 04.03.2008

nash's theorem shows that for every finite game (a game with a finite number of players, with finite strategic options), there exists at least one nash equilibrium.
all poker games are finite, therefore they must have NE


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Boomer2k10
Joined: 22.09.2010

Originally posted by redskwerl
nash's theorem shows that for every finite game (a game with a finite number of players, with finite strategic options), there exists at least one nash equilibrium.
all poker games are finite, therefore they must have NE

:f_thumbsup:

Pretty Big game though even if it's finite :)


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YohanN7
Joined: 15.06.2009

Thus there is at least one Nash equilibrium in full ring 100 000 000 BB deep NL Holde'm?

/Johan = :f_confused:


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Boomer2k10
Joined: 22.09.2010

Yup

Best of luck to anyone trying to work it out though


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YohanN7
Joined: 15.06.2009

Originally posted by YohanN7
Can someone point me to a rigorous*) statement and proof, preferably detailed, and specialized to poker, of the existence of GTO in FL HUHU and FL multi-handed.

*) Rigorous = Hardcore mathematics by a hardcore mathematician published in a journal of mathematics. (At Wikipedia they would say "sourced".)

/Johan = :f_confused:

Hey Yohan, here ye go: Non-Cooperative Games, John Nash.

Excerpt from the introduction: "As an example of the application of our theory we include a solution of a simplified three person poker game."

Always go to the sources! :)

/Johan = :f_confused:


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YohanN7
Joined: 15.06.2009

Hi again!

The original theorem proved by Nash is probably slightly more generally valid than it is usually presented as being. "In every game with a finite set of strategies..." can probably be generalized to "In every game with a compact set of strategies...", where "compact" has a certain technical meaning*).

As far as No Limit Hold'em goes, the number of strategies is (in theory) infinite, since any amount can bet, up to, and including, your, or your opponents, stack size. In practice, the set of strategies is finite, since bet sizes are limited to e.g. two decimal places in the currency used.

I think that the Nash theorem is true for infinite sets of strategies under certain conditions, in certain games, but not for all types of conditions.

Consider the following, very very slightly modified, game of No Limit Hold'em: You may bet any amount except your whole stack. That is, the limitation is that you may not go all in, or put your opponent all in. There are no GTO strategies in this game. Can you see why :evil:?

Spoiler

Consider first any GTO strategy (Nash equilibrium) in regular No Limit Hold'em. Call it strategy Z. This strategy will always have situations where you move all in. This strategy is, by the rules, not allowed in the modified game. You can come arbitrarily close to it in the modified game, but will never really achieve it. Suppose that you bet very close to all in where the strategy Z in the regular game dictates all in. Call this strategy X. There is a strategy, strategy Y, that will exploit strategy X by betting an amount even closer to all in in the spots in question, and thus always winning against strategy X. QED.

In other words, there are sequences (or, for the nerd, nets) of strategies converging to a GTO stragegy, strategy Z, but the strategy Z is not an allowed strategy :).

*) I haven't done this rigorously, but the original proof relies on theorems that in turn rely on what I here call compactness.

/Johan = :f_confused:


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taavi1337
Joined: 29.05.2009

60/3


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