Since you are not fond of Wikipedia's informal definition, maybe we can agree on this one?
(Roughly: For all players, the strategy of the player has to be in the set of best responses of that player against the strategies of the other players.)

Excellent. I will show that your shover should not be shoving, and that your caller can exploit him at will.

Only at NE, is this not possible.

Only at NE are both players locked together in equilibrium.

Cool. But maybe explain first how your proposed equilibrium is not in blatant conflict with this definition?

plexiq,

It would seem that your carefully calculated rake level of 20.6% (or 20.6197%) and your carefully constructed call range of 55+, A7s+, KTs+, QJs, A9o+, KJo+ do not meet the criteria of the OP when he said specifically:


Here is the EV of your shover, vs your call range, using 1 billion Monte Carlo trials with Equilab:

Rake = 20.6197%
Pot = 10
Stacks = 100
Net win at showdown = (210*(1-.206197)) - 100 = 66.6986
Probability caller wins showdown = .6521
Probability shover wins showdown = .3298
Probability players tie = .0176

So, here is the EV of the shover:

EVshove = EVfolds + EVcallerloses - EVcallerwins - EVties


10*(1138/1326) + (188/1326)(66.6986)(.3298) - (188/1326)(100)(.6526) - (188/1326)(16.6507)(.0176) = $2.41


The EV of the caller is:

EVcallwins - EVcalllose - EVties

(188/1326)(66.6986)(.6526) - (188/1326)(100)(.3298) - (188/1326)(16.6507)(.0176) = $1.45


So, your shover is still positive, and still has an EV greater than the caller.

This does not meet the criteria quoted.

PLease post a rake level and a call range that meets the criteria of OP. Also, I request that you explain why 20.6% did not work out after all.

The 20.6% calculation was not for the question you quoted, the quoted question was a follow-up. (Hence, a different calculation was posted later, with a different result. That's the 17% part, but I didn't check that one.)

The original question in OP was: What's the lowest rake for this game where pushing 100% is still the Nash Equilibrium strategy?

The answer to that question is slightly above 20.6197%.

The quote is what OP always wanted. And he said he never intended otherwise.

Are you now in the 17 percent camp?

It's two separate questions. Maybe try to keep up with the thread a little bit?

This is question #1 (20.6%):
This is question #2 (17%), posted after the first one was answered:

plexiq,

This seems to be where we stand, with regard to the quote above from OP.

My answer to this quoted question has always been 26.21 percent (pure caller).

ZKesic’s answer is 17.48%, and you tentatively agree, but have not reviewed that solution specifically.

The carefully calculated rake and range of 20.6% does not apply to the above quote.

Correct?

The 20.6% rake range applies to the OP, not to the follow-up question you quoted. But since all pushes are profitable against a rational caller above the 20.6% rake threshold, the answer to the second question has to be considerably lower than that.

I didn't check, but 17% looks reasonable. It clearly can't be 26%. But then again, your caller is not acting rationally. We discussed that above, players maximizing their own payoffs etc.

ZKesic,

You were within one cent of the correct calc for your shover. Your solution above to 3 significant digits, with 1 billion Monte Carlo Equilab trials:

Rake = 17.48%
Pot = 10
Stacks = 100
Net win at showdown = (210*(1-.1748)) - 100 = 73.292
Probability caller wins showdown = .6278
Probability shover wins showdown = .3507
Probability players tie = .0216

Here is the EV of the shover:

EVshove = EVfolds + EVcallerloses - EVcallerwins - EVties


10*(1050/1326) + (276/1326)(73.292)(.3507) - (276/1326)(100)(.6278) - (276/1326)(13.354)(.0216) = 0.14123 = 14 cents


The EV of the caller is:

EVcallwins - EVcalllose - EVties

(276/1326)(73.292)(.6278) - (276/1326)(100)(.3507) - (276/1326)(16.6507)(.0176) = 2.21667 = $2.22


However, there is a problem. In the quote from David Sklansky above, it requires that the blind pushing strategy be profitable for the shover to shove with 72o.

In your proposed solution, your shover would increase his +EV by checking for 72o and folding it.

1 billion Monte Carlo trials:

Rake = 17.48%
Pot = 10
Stacks = 100
Net win at showdown = (210*(1-.1748)) - 100 = 73.292
Probability caller wins showdown = .6272
Probability shover wins showdown = .3513
Probability players tie = .0216

Here is the calculation (where 'Prob. Shove' means the probability that the shover checked for 72o and then bet):

EVshove = (Prob. Shove)EVfolds + (Prob. Shove)EVcallerloses - (Prob. Shove)EVcallerwins - (Prob. Shove)EVties

(1314/1326)10*(1050/1326) + (1314/1326)(276/1326)(73.292)(.3513) - (1314/1326)(276/1326)(100)(.6272) - (1314/1326)(276/1326)(13.354)(.0216) = 0.16140 = 16 cents

So, your shover should not be shoving, and would profit by folding 72o at a rake level of 17.48 percent.

Now, I am sure that you can reconfigure some numbers and ranges, to get this problem corrected, so that you can give an answer to the quote above from OP. Of course, you are free to do so.

However, I think you might want to wait just a bit, while I ask a few questions of our friend plexiq.

plexiq,

Help me out, please, to see if I am keeping up in this thread.

At a very high rake level, say 30 percent, the shover always shoves no matter what. To do anything else would diminish his returns and lower his EV.

This holds true, vs ANY +EV calling range for the caller. So long as the caller is not suicidal, the shover should shove.

You can enter any +EV response from the caller, and the shover is unfazed.

It would seem that, from the mind of David Sklansky, a truly perfect GTO full range pure strategy shove all-in has emerged.

Is all of this true, specifically because the shover is shoving at a rake level above Nash Equilibrium?

Thank you.

Wow, we are almost on the same page now. And it only took like 50 extra posts

Clarification on the 17.48% rake state:
We are searching for the rake threshold where the ATC pusher's overall EV against a rational caller is zero. Since the overall range is supposed to be 0EV, the bottom of that range will include -EV hands. So the 17.48% rake state is not a NE, I mentioned that before.

(Little caveat: If the game rules actually force the pusher to play without looking at his cards then this state is indeed a NE. Maybe that's what ZKesic meant here, idk.)

Im not sure what you mean with the bold part. The rake is a game parameter, not as part of any strategy. There exists a Nash Equilibrium for any given rake level. (You could treat the house as a player that sets the rake level at the beginning of the hand, but that'd be going down an entirely different rabbit hole.)

All of this was already stated 60+ post ago, you guys just seem to be running in circles (which is why I wasn't really responding).

Plexiq is right:
The 20.6% rake is the point at which the pusher should stop pushing with any two cards when he can see his cards.
The 17.48% rake is the point at which the pusher should stop pushing with any two cards when he can't see his cards.

That's the difference here.

You've basically confirmed with your calculations here that my 17.48% result was correct.

In this case the pusher can't see his own cards and is playing his whole range the same way.
At 17,48% rake he would push his whole range.
At 17,47% rake he would fold his whole range.

This entire thread has boiled down to what happens at different rake levels, and what rational participants choose to do to maximize their own personal +EV at those rake levels.

When the rake is 30 percent, how does any rational bettor respond, versus any caller, that is not suicidal (his call has to be a +EV call).

He shoves all-in blind, because checking for 72o would be irrelevant, he would shove it if he saw it.

He can do this when the rake is set very high at 30 percent.

He cannot do this when the rake is very low, say 15 percent.


At both rakes, He must serve his own interest, by maximizing his own EV.


At 30 percent rake, he could check for 72o and still shove.


However, at 15 percent he must check for 72o (and some other bad hands) and fold it.


Since there exists a Nash Equilibrium for any rake level, the rational bettor stays in equilibrium by checking for bad hands, if necessary, and shoving otherwise.


Do we at least agree that this bettor is behaving rationally and maximizing his own +EV, against any caller that is not suicidal?

What you wrote in the last post seems correct and mostly obvious, but doesn't really clarify what you meant by "rake level above equilibrium".

Since we have equilibrium strategies for every rake level, what do you mean by "rake above equilibrium"?

Well, at some rake level between 30 and 15 the rational bettor decides to maximize his +EV by checking for 72o and folding it if he sees it.

This maximizes his EV, versus any call range that is not suicidal.

Correct?

Yeah, that rake level where the pusher needs to start checking for 72o etc is at 20.619%. I still don't get your formulation, but that's not important anyway.

These points you guys made above are good points. I do think my replies below are true wrt to poker strategy, but I admit a caller at true NE would not be able to pull off a free 67 cents.


The reason that the caller with 77+, ATs+, AKo was able to squeeze out 67 cents of EV by varying strategy, is because this solution is in pure strategy, and in mixed strategy (down to a single combo) at 26.25 percent rake any variation of overfolding by the caller will only be about 5 cents of EV either way and results in very big profits for the shover.

In some spare time, I will solve for the mixed strategy call range at 26.25 percent rake. A +/- variation of about 10 cents will suffice, for my own peace of mind.

The above is what I consider to be the weakest part of my solution.





Below I have what is the weakest part of your solution:

If your version of GTO shove all-in is guaranteed its maximum EV, no matter what the +EV call range of the caller is, so long as it is a non-suicidal call, then why is it that at 20.6197 percent rake, your shove suddenly loses money and goes in the red because a potential caller decides to make a +EV call with this range:



because he read somewhere online in a poker strategy forum that it was a NE call in a very similar situation?

OR

When a caller uses my call call range from my solution but at your 20.61 percent rake, suddenly the caller is the one with the profit?

Do you really solve your call ranges, to ensure that the bettor playing against you is guaranteed to win more than you in the long term? Is that even poker?

A GTO bettor and caller should be immune to such variations.

My shover is guaranteed a profit, versus any +EV non-suicidal caller. My shover is guaranteed a huge profit versus any overfolder. This is what happens when a GTO bet is placed.

My caller can call with confidence, knowing that the call makes money and requires the bettor to actually be correct in placing that bet. This is what happens when a GTO call is made.

We solve for Nash Equilibrium by using the definition I posted above. The definition simply states that all players need to play a best response against the strategies of the other players. A best response means any strategy that achieves maximum EV for that player, ie it needs to play all +EV hands and fold all -EV hands, 0EV hands may be played mixed. If you find a set of strategies where each player's strategy is a best response against the other players' strategies, then you got a NE. It's really that simple and also quite trivial to test if a set of ranges is a NE.

If the caller could increase his own EV by deviating then that would show that his previous strategy wasn't a best response. In your example the caller loses EV when switching to your proposed range, so whatever happens in that case is not a violation of the NE condition.


I don't know what you calculate, but it's not a Nash Equilibrium.

So, its ok that a perfect GTO shove can lose money to a +EV call?

That does not sound right.

Yup, that's perfectly ok. Note that this can't happen in 2 player constant sum games, so that's probably why you haven't seen it.

I agree that the 10 dollars in the middle is quite a sticky wicket wrt constant-sum with rational participants.

Do you think that we are discussing this now in a more respectful manner?

Do you still think that I am intentionally poisoning a thread?

Well, it seems all is quiet for over a day, so it is time to wrap up this valiant discussion.

Poker is zero-sum. Those chips come from somewhere, and they go to somewhere. They are neither created or destroyed. If there is any calculation that says otherwise, that is a mistake.

Any resource you may use, to study the game of hold'em, including any EV calculator, solver, etc, must observe this or it will lead you astray. Plus, when you enter your ranges into the solver you must do so with the trust that the solver is properly programmed.

I should have proved this many posts ago, but the discussion got sidetracked for reasons that I detail below. It is a short and worthwhile read, IMO. (Plus a last piece of the puzzle to solve at the end.)

In poker there are always blinds, and sometimes antes. Blinds (or straddles) are mandatory bets that belong to the player that placed them.

Antes, on the other hand, are mandatory fees that are put in the middle, that simultaneously belong to all players. Even if a MTT structure has the BB pay all the antes, the BB does not own those.

So, the total of antes, divided by the numbers of players still in the hand, are part of the net worth of those players.

Here, both players started with a net worth of 105. Both players finish with a EXPECTATION total net worth of 210. If the players are GTO, and the rake is affordable around 26.25 percent, they will both finish at EV~105. If one or both players are exploiters, then one of them will finish with on average more than the other but the sum of their literal net worth in Expectation Value is always 210.

Now, every time they play a hand and the caller calls, 98 percent of the time one of them goes to the rail, the other player gets his predicted EV result, and the house collects the rake. The winning player's increased stack, plus the rake to the house, is always 210. (about two percent of the time they tie and get to both keep 105).

So, why does a discussion go on so long without one of the participants pointing this out?




Sometimes, learning takes effort. Sometimes, discussions must continue until both parties can learn to talk in the same language. Only then, will the precise correctness or lack thereof from either party be understood by the other.



I know here that a GTO call is always minimax. Minimax means that your maximum loss is minimized. The caller is guaranteed to break even, and scores big if the bettor is off course.





By my own fault, I did not do this right away. I was not listening since I know K9s works.



Well, yes, I did double down. This is because all calls in poker are minimax. Period. Any format Cash, MTT, STT, SNG.

This is very insulting. (The poisoning part, since I am indeed very stubborn)



Poker is zero-sum.

Now, I am trying to speak my language (minimax) to a fellow forumer who only speaks a language of maximin. In poker, maximin (maximizing your minimum gain) will get you to the right answers eventually, but the calculations are more difficult, and there are important pitfalls to be avoided when doing so.


Modern poker strategy is always ranges, you solve for the range first, see what is in there, and then do accordingly when you get dealt that hand by the dealer.

To the credit of browni3141, he is now trying to speak my minimax language. If I had simply proved the game is zero-sum then we would be on the same page. Kudos to browni3141.

More incivility, and yes, forgetting that poker is always zero-sum was not my mistake. The rest is an apt description of Nash Equilibrium poker. We live in that universe. If the participants in the conversation cannot agree on what universe in which they live, then trying to prove your own argument according to the rules of different universes will not work.



Now I am again trying to restate minimax in a language that works in a universe where maximin does not equal minimax in poker.

Translation: “Alright new plan, let’s talk in terms of EV. The caller is supposed to be optimal, but you show him losing 5.50. Why would he do so? “

Oh no. no. In one of our universes there is magic money where both players are employed to create EV working together. (That is positive sum).


Again, my universe, my language. But, I am trying to use the language of EV that we both seem to understand. However, asking anyone in another universe to evaluate their universe according to your own rules will simply result in them using their own rules, not yours.

Yes, some rereading of many different resources was in order. But we can get there. We are almost there....

My last try. I did not understand that plexiq was missing the zero-sum fabric of our poker universe.

As they say: “Houston, we have a problem.”

If an avid participant in a conversation about poker does not believe that a solution where a caller wins 67 cents and the bettor wins 6.17, shows that the caller lost 5.50, then drastic incremental measures of logic must be employed. I must step into the alternate universe and speak that language entirely if there is any hope of communication.

New strategy: In your universe, your answer does not meet the requirements of the puzzle. IMO, I am nitpicking, but one tiny step at a time.

Translation: So, pick one answer, either 17.48 percent or 20.6197 percent. (we both know 17.48 is not it)

We must agree that in whatever universe we reside, the prize that we seek is indeed a Nash Equilibrium. But now I must use the language of +EV and maximin. Not easy to do.

The term “+EV” in poker strategy usually means one decision is more profitable than the other. But here in this thread, we are talking about literal Expectation Value to either player by using an exact supposed optimal strategy.





Yes indeed. We are now making progress in a maximin universe. And, we are both actually down that rabbit hole. We always have been.

Now, that I am listening, I can restate my position in his language, with the important +EV caveat. This caveat means usually “A Thinking Opponent”:



“Houston, we have communication!”



Bingo. We agree. We are indeed seeking Nash Equilibrium. Major progress has happened.

Now that I am listening and speaking maximin, I can narrow down why plexiq cannot accept my answer. This is because in his universe, maximin poker does not equal minimax poker. In my universe these are equal, since poker is zero-sum.

And, I can now restate the weakness of his position in the language that he understands....

And, finally, I can now speak enough maximin to get my point across.....

Poker is zero-sum. This means that minimax=maximin and NASH EQUILIBRIUM IS ALWAYS MINIMAX IN ALL CASES.

You can get there with maximin too, but as I have demonstrated it is a tricky path.

So. There is one large but quiet elephant in the room. Why does K9s (and QTo) work as a call versus a range of any two cards? Why was the carefully calculated rake of 20.6197 percent wrong? I have proven that my range is minimax. It turns a profit. Simultaneously, every individual combo contained therein must by itself turn a profit when you are dealt it and face a Sklansky GTO jam with any two cards.

How?

As far as I can gather, this has never been proven before on this forum.

This is a lot to take in all at once, and I don't want that part of the puzzle to distract any readers of this post from the importance of what is at the top of this post. Plus, maybe someone else would like to post first, how specifically, K9s works and TURNS A PROFIT in this puzzle.

Hint: The METHOD was shown by someone in this thread, however the correct CALCULATION of that method was shown by someone else.

If no one wants to take the floor for more meaningful discussion, nor post the proof of K9s, then I will do so tomorrow.

-Rob

Raked poker is not zero sum, or even constant sum. A game is constant sum if the sum of player payoffs is constant. Since the pot is not always raked, the sum of player payoffs is not constant. (That's assuming that uncalled bets are not raked, I believe you also used that rule in your calculations?)

You previously said that the 10$ free money in the pot are the tricky part here, but it's actually the rake that changes the dynamics.

since the rake amount never changes, then how do your players manage a payoff that is simultaneously positive. Remember, payoffs are the LONG TERM EXPECTATION of their strategies. You are saying that they are guaranteed a simultaniously positive return, that is also not constant but varies over strategies, while the house is always taking THE SAME amount from the pot. At the very least, your calcs would have to add up to ten bucks.

I am trying to help you. Do you really think it is necessary for me to show that your players can not create magic money?

But the house is not always taking the same sum. No rake is taken when the caller folds.