Reading the definition of a Nash Equilibrium, it seems he is correct.

Your next 3 paragraphs are a waste of time. That's not how you go about trying to refute such a claim. What you need to do is to construct a toy game stripped to the bare essentials to illustrate your point unequivocally and not pitch a series of nebulous examples. Hand waving arguments are for the birds.

What you are talking about might be applicable in an exploitive sense, but not in the Nash equilibrium sense. The simple fact that a supposedly optimal action in particular situation is -EV makes the situation inherently unstable i.e. not an equilibrium.

If your opponent is playing an equilibrium strategy you can't gain equity by switching to one different from equilibrium (Nash equilibrium definition).
But, obviously, if you are playing anything -ev you can switch to a different strategy to gain equity, which means the initial strategy you were playing wasn't an equilibrium one.

It's also true, to give you credit, that it's possible you are playing an equilibrium strat and your opponent isn't, so that some of your plays become now -ev.
This is why I said we should assume opponent is playing either a GTO strategy or a perfect counter strategy to our strategy, otherwise it's possible for any X that "limping" atc is right (for example against an opponent who will always fold).

Seriously guys you are not getting what i say or there is a deeper fundamental game theory result i miss that you are doing a poor job to explain and defend as it deserves against someone like me that can understand you if you use math instead of insulting tone (R Gilbert) .

Lets say poker is made of only 2 hands AA and T2o for me as SB and there are no antes! Thats all i ever get dealt in relative ratios of natural frequency 0.45:0.9

My opponent gets normal hands (subject to card removal from mine of course) and knows i get only T2o and AA in the stated frequencies.

I can only move all in or fold by the way and have 10bb only say!

If i move all in with AA only and always fold T2o he will fold every time he doesnt have AA!!!

In order to avoid that i have to also push T2o with some to be calculated frequency if not necessarily 100% (needs to be solved) . Then he has to figure also his equilibrium range to call that mix. Now inst my push with T2o every time i do it locally negative idea vs his calling range and folding probability?


It secures however overall success when mixed with AA that AA alone wouldnt achieve. Obviously random hand vs T2o,AA is 43.8%. We do not know of course if equilibrium is natural mix yet.

All i have to figure now for the equilibrium is how often to push T2o and how often AA (always) and also what range he has to call us with.

Can you show here that my T2o push is not negative locally?

I don't think I can explain it any better than what I did in my previous post: if your opponent's strategy is GTO and fixed and if your strategy is doing something which is -ev (like bluffing hands when you will get called very often) you can improve your equity by changing it. Since you can improve it by changing it, by definition of Nash equilibrium, you are not in a Nash equilibrium, ie your strategy is not GTO.

As for your example, if you bluff T2o few times, bluffing it is going to be +ev against an opponent who responds perfectly to your strategy, because he will always fold, so you are winning the pot both with AA and T2o. If you bluff too often it's going to be -ev because you will often be called. It can be shown (and it has been shown a lot of times on this forum) that when you play a GTO vs GTO strategy pushing T2o is the same equity as folding it (ie pushing it is 0 equity).

Here is my answer to David's initial question, far from being a final one

Let's suppose X is such you can "limp" (ie put in the minimum $1 required) atc.
By theory experience it's pretty obvious you should also have a raising range (you put in more than the minimum $1) and it's also obvious your raising range is going to be stronger than your "limping" one.
Now, if you can "limp" with the worst holdings in your range (may or may not be 32o, not important), it's obvious the button will never fold to a limp, because he's getting better odds than you to put in the one dollar and your range is weaker than atc, while his range was atc when you decided to put the $1 in (ie with the worst holdings he can have he's in a better situation than the one you were in when you decided to at least "limp", so he's going to do the same).

Now, since our "completing" range is pretty weak (weaker than a random hand), button is going to raise you a lot and you are going to fold with your worst holdings (unless X>=110 obviously, in which case we already know it will be a preflop allin atc).
We don't know how often the button is going to raise, but probably somewhere along (at least) 70/80% of his range, given what we know about HU play.
The remaining times we are going to see a flop and our equity is around 35%, but given the fact our hand is very weak and will mostly flop bluff catchers out of position, it's unlikely we will win our fair % of what's in the pot. I'd guess our true equity will be somewhere around 20-25%.

So, to sum it up (using 80% raise for the button and 25% equity for the cutoff postflop), if we put $1 in with our worst hand, we lose it 80% of the times to a raise and we win [(X+2)*0.25-1] the remaining 20% of the times.
Our win expectation is thus
-0.8+(X+2)*0.05-0.2
thus to be +ev we need
(X+2)/20 >= 1
X >= 18

Using 70% raise and 25% postflop equity (my most optimistic guesses), we'd get X>=11
Using 80% raise and 20% postflop equity (most pessimistic guesses), we'd get X>=23

So I would go with something like X>17 for my best guess
X > 8 seems a little too optimistic in my opinion