If there's no 10$ in the starting pot then the Nash Equilibrium ranges for this game are: 0% push, 0% call. And that neatly results in a 0EV for both players and a total of 200$ chips on the table.

That's my last interaction with you before we can agree whether this is a zero sum game or not. If you keep dodging topics and never admit you might have been wrong then this is a waste of time.

The jamming player jams 100% of hands. For every amount of rake, the calling player adjusts what equity he needs to call, which maps to some percentage of hands. Thus, for any given rake amount, we can determine the percentage time the calling player will fold, and knowing his calling range, what the jamming player's average equaity against the calling range is. Therefore, we can determine the overall EV for the jam.

The question OP is asking is at what rake amount should the jamming player stop jamming 100% of hands, given that the calling player will call with the correct range for any given rake amount. As 32o is the worst heads up hand, the question can be narrowed to be: at what rake is folding 32o preferable to jamming, assuming the calling player knows that the jamming player is folding 32o and adjusts accordingly.

At a $60 rake, the calling player risks $100 to win $50. Thus he calls with any hand that has over 66.7% equity against a random hand (jamming player is still jamming 32o). This range is 88+,AKs which is 3.5% of hands. This means, when the jamming player happens to have 32o, he wins $10 96.5% of the time, and loses an average of $67.07 3.5% of the time (32o vs 88+,AKs has 15.68% equity), for a net EV of $7.03.

At a $50 rake, the calling player risks $100 to win $60. Thus he calls with any hand that has over 62.5% equity against a random hand (jamming player is still jamming 32o). This range is 66+,KJs+,A9s+,ATo+ which is 9.8% of hands. This means the jamming player wins $10 90.2% of the time, and when he happens to have 32o, his equity is 25.3%, which makes his net EV on the jam with 32o be 10*0.902+(110*0.253-100*0.747)*0.098 = $4.43

At $40 rake, the calling player calls with 55+,QTs+,KTo+,K9s+,A4s+,A7o+ which is 18.4%. 32o against that range is 28.426%. Net EV of 32o jam is 10*0.816+(110*0.28426-100*0.71574)*0.184 = $0.74

At $35, call range is 55+,JTs,QTo+,Q9s+,K9o+,K7s+,A2s+,A5o+, EV of 32o is -$1.90

At $38 the EV is $0.08.

It would therefore appear that, using whole dollar amounts, the rake at which the jamming player should fold 32o instead of blindly jamming all hands is $37.

The question is fairly basic and has been settled after a few posts, check MonkeyTilts and ZKesic posts on the first page.

(Seriously, if you want to contribute or learn something then avoid this thread like the plague, it's riddled with misconceptions and it won't go anywhere because participants can't even agree on the most basic definitions. This is more of a social experiment than anything else.)

Except, they got the answer wrong because they were doing calculations using 72o for a heads up situation. It wasn't until page 3 when ZKesic posted the correct answer of 17.48%.

And I thought it would be useful to lay out the problem in a more clear way than had been done previously in the thread.

It's a mistake to narrow this down to 32o. The ranking of hands depends on the calling range and the calling range changes as you change the rake. (32o & 72o happen to become unprofitable blind shoves at the exact rake% in this specific game though, so the difference does not matter here.)

The correct answer to the original question is 20.6% (When is it no longer optimal for the pusher to simply shove any two cards?), the correct answer to the follow up question (At which point does showing ATC actually lose money?) is 17.5%.

EDIT: Nevermind, I'm wrong. 72o was the correct hand to use. Which makes 20.46% the correct answer.


That answer is incorrect because they used jam hand of 72o. The correct answer is right around 18%.

I may have been mistaken about 17.48% being correct, since as you say, it's an answer for a different question. But 20.6% is absolutely not right. You can do the math yourself. Look at my post and see that $38 rake is still profitable for 32o, which is 18.1% rake.

~18% is the correct answer to the original question. 17.48% is the answer to the second question.

At 20.6197% rake, with the calling range of 14.5%, 55+ A7s+ A9o+ K9s+ KJo+ QJs. That's the rake point where K9s just barely moves into the calling range to make the bottom of the pusher's range (72o, 32o and 42o) unprofitable:

Lets check 32o:
32o equity against the calling range is: 27.1901%
The calling range has 192 combos, none of them blocked by 32o.

Caller calls against 32o with 192/1225 combos, or 15.67347% of the time, and folds 1033/1225 combos, or 84.32653%. (Note the card removal here, 32o is getting called more often than 14.5%.)

EV for 32o:
1033/1225 * 10$ +
192/1225 * (210$ * (1 - rake) * equity - 100$)
= -0.1367$

Your EV formula isn't right since the rake% isn't mentioned in it at any point. The rake affects both the caller and the pusher, but you only applied it to the caller.

Also, the fold equity that 23o has here is 80.08% (244/1225). You wrongly used 81.6% (244/1326) because you weren't considering card removal.

The correct formula for this would therefore be something like: 10*0.801+0.199*(0.2843*(0.81*210-100))-0.199*(0.7157*100)=

My excel sheet gives the exact answer:


23o is a -EV jam at 19,05% ($40) rake according to this.

We shouldn't even be using 23o as the bottom of the pushers range though, since it has more equity and higher EV than 27o, as I've mentioned before:

Yes, quite the bamboozle.

How do we unravel it?

If two billionaires want to shove piles of chips back and fourth in a casino for whatever reason, and the casino charges rake for this, every chip must be able to be counted. No chips are created or destroyed. The only way to do this is below:

First, we check the rake drop box. Maybe our missing chips are in there. After all, this is raked poker.

Contents of rake box:

Shover always shoves, and caller calls with probability 188/1326 and the house collects 200*.2061=41.22
188/1326*41.22 = 5.84416
93.77768 + 100.71132 + 5.84416 = 200.33316


Hmmmmm. That is too much. But that has to be the contents of the rake box. The house is not a player here, and the house gets its fair share of the action always.

Lets divide the rake in half, and add it to their final results. How much rake did each player pay?
5.84416/2 = 2.92208
Add this to each player:
Shover: 93.77768 + 2.92208 = 96.69976
Caller: 100.71132 + 2.92208 = 103.6334

We know the caller should be making a profit, but when we return the rake the players paid, there is still too much money. We must account for this.

Lets change the ranges. Lets go crazy. Lets make our players irrational.
After all, these are billionaires putting 100 bucks at a time on the table.
Lets make the shover shove always and the caller call always.
This results in either player winning with probability (.4978) and players tie with probability (.0404) -----> [.4798 + .4798 + .0404 = 100 percent]
Pot is always 200, and house always collects exactly 41.22 which makes the pot 158.78.
Shover wins 158.78 with probability .4798 and gets a net 76.18264
Caller wins 158.78 with probability .4798 and gets a net 76.18264
41.22 + 76.18264 + 76.18264 = 193.58528

Nope. Now much more money has disappeared.

Arggh!

Whatever the solution is here that can count the chips on ANY poker table, will be THE SOLUTION and work for all cases of poker in ANY format. Raked, not raked, SNG, tournaments, home games. Whatever.

Well, there is only one way to account for the chips.

You have to understand that whatever the shover’s winnings are with the betting strategy, these ARE the caller’s losses with the calling strategy.
Simultaneously, whatever the shover loses by betting, IS the caller’s winnings by the alternate calling strategy.
Whatever way you look at it depends on which chair you are sitting in. In poker, we call this position.

In this thread, the shover wants to know how often to shove at whatever rake.
So, we will treat his net proceeds from betting as +EV. Conversely, we will treat the caller’s net returns in response to our betting strategy when we are in the bettor’s position as -EV.
So, shover has +EV and caller has –EV.
AND (this part is truly divine...)
Specifically because this is raked poker, we can always verify the starting net worth of our players by checking the rake box!

Lets give it a whirl and see if we find the missing chips.

Bettor:

Caller folds with probability (1138/1326) and we get back (1138/1326)(100) = 85.82202
The dealer collects EXACTLY 41.22 and keeps it.
Pot is now precisely 158.78.
(188/1326)(.3446) we win and collect 158.78
(188/1326)(.3446)*158.78 = 7.75756
With probability .0176 there is a tie:
(188/1326)(79.39)(.0176) = 0.19810

Add up every chip:

85.82202 + 7.75756 + 0.19810 = 93.77768

So, we know he started with 100 and finished with 93.7768.

We are in the bettor’s chair, so this is a loss of 6.2232, so EV = -6.2232

Now the caller:
100*(1326-188)/1326 = 85.82202
(188/1326)(.6526)(158.78) = 14.69120
(188/1326)(79.39)(.0176) = 0.19810

Add ‘em up:

85.82202 + 14.69120 + 0.19810 = 100.71132

We know he started at 100, so this is a gain of .71132

This makes the shover’s final stack 100 – 6.2232 + .71132 = 94.48812

Conversely, the caller’s final stack is 100 + 6.2232 - .71132 = 105.51188


Now, there are our chips!!!!

The overall EV of the shover’s betting strategy is to lose 5.51188 bucks.

The overall long term expectation of the callers strategy is to win 5.51188 bucks.


Whatever one player wins, the other player loses.
Whatever one player loses, the other players wins.

ALWAYS.



Now, we can take our method back to regular poker with regular players:

Betting strategy, as we sit in the betting position:

We shove 100 percent, and with probability (1326-188)/1326 the dealer gives us back the whole 100 plus the 10 so 1138/1326(110) = 94.40422

With probability 188/1326 there is a call and the dealer collects exactly 43.281 from the pot and keeps it.
This leaves precisely 166.719 in the middle of the table.
With probability (.3298) we win that pot:
(188/1326)(166.719)(.3298) = 7.79560945
With probability (.0176) there is a tie, and we get back 83.3595
(188/1326)(83.3595)(.0176) = 0.20801

94.40422 + 7.79560945 + 0.20801 = 102.40784

Caller:
We fold with probability (1326-188)/1326 and we keep all 100.
100*(1326-188)/1326 = 85.82202 kept behind the betting line.
With probability 188/1326 we call and push our 100 in the middle.
The dealer collects 43.281 and keeps it.
Pot is now 166.719...
With probability (188/1326)(.6526) we win and collect 166.719
(188/1326)(.6526)(166.719) = 15.4257572
With probability .0176 there is a tie:
(188/1326)(83.3595)(.0176) = 0.20801

85.82202 + 15.4257572 + 0.20801 = 101.4557872

Now that we know we are solving for our betting strategy from the position of the bettor, we see that 102.40784 is larger than 101.4557872, so the bettor got a net increase of

2.40784 - 1.4557872 = .9520528 Nice. A 95bb/100 winrate.

Conversely, the caller:

1.4557872 - 2.40784 = -.9520528 Ugh, a minus 95bb/100 winrate.

But, where did they start?!? ….105 each. They equally owned the antes. This is confirmed by the rake box.

Final stacks to each player:

Shover: 105.952053
Caller: 104.047947



So, to sum it up:



1) Raked poker is zero sum.

2) You have to solve for both players according to position.

3) Only by comparing the +/- EV of both positions will you know your actual expectation of your strategy.



Now... not every resource available to us as we study hold’em takes this into account. Some calculators, solvers, softwares, etc. Do NOT DO THIS.

Alright then.

Are you saying that the caller should be defending wider when the rake is higher?

Also, the caller calling Q9s while folding 55/A8o doesn't make much sense.

Below has corrected final stacks:




So, to sum it up:



1) Raked poker is zero sum.

2) You have to solve for both players according to position.

3) Only by comparing the +/- EV of both positions will you know your actual expectation of your strategy.


The caller does not want a –95 bb/100 winrate, so he immediately switches to a call range of



66+,A5s+,K9s+,QTs+,JTs,A9o+,KTo+,QTo+

This call range wins with probability .6354 against ATC. It loses with probability .3446 and ties with probability .0200

This now gives the caller a final stack of 100.4967 and the shover 99.5032. (50 bb/100 in favor of caller).

Shover requests rake to be lowered to 17.48 percent:



Caller DOES NOT CHANGE RANGE.



Now the caller wins 100.8398 and the shover wins 99.1602 (83 bb/100 in favor of caller.)



Caller says: “My range works for calling a jam of 100 into a pot of 10 at any rake around 25 percent.”

Shover says: “Raise the rake to 25.20 percent”



Caller thinks for a minute, and says: Black is my favorite color. I will add Qs9s and Qc9c to my call range. (now has 240 combos in range)



This results in probabilites of .6349 win and .3451 lose and .0200 tie.



The final stacks are 100.05 for caller and 99.95 for shover. A five-cent duel.



The shover says: “I want to keep shoving, BUT I CAN NOT FIND ANY HANDS TO FOLD.” (doing so only increases the caller's profits.)



Now that is a Nash Equilibrium.

The caller can always do a little better with mixed strategy. Those two combos work.

Are "Game of Chicken" or "Prisoners Dilemma" also zero sum then? Seems that using this process any two player game would satisfy your definition of zero sum.

I'm well aware that they are not considered zero sum in the mainstream literature. I was wondering if they are considered zero sum in your alternative theory. If not, then why? Because if you "only compare the +/- EV of both positions" like you suggest, that's going to be zero sum. (It's interesting that you decide to draw the line here and not just go full alternative universe though.)

I doubt I'll change your mind on your position but noticed what I believe to be mistakes in your math or reasoning so just wanted to point them out.

Where did you get 93.77768 and 100.71132? I checked the post you quoted but couldn't find those numbers. Maybe you caclulated them independently for this scenario where we only consider the $200 pot, but if they are EV payouts from the $210 pot then you are obviously going to have too much money when you add things up with the rake from the $200 pot.





The money disappeared because you omitted when players tie. The equation should be:

Rake+Player1EV+Player2EV

41.22+(.4798*(158.78)+.0404*(.5*(158.78)))+(.4798* (158.78)+.0404*(.5*(158.78)))

Or more simply

41.22 + .5*(158.78) + .5*(158.78)



This would be incorrect in this instance because of the free 10 in the middle that was contributed by no one. We only care about EV after this decision point so even if we did contribute money earlier to reach this point it is a sunk cost. So money gained/earned cannot be considered to exclusively come from the other player.




I am confused why you are omitting the $10 here but whatever.

I also don't think you're calculations are correct as presented.

If you want to use the "track your stack" method I believe you have to use the difference between your final stack and starting stack. Your equity seems to be off as well (.0176+.6526+.3446 = 1.0148). So the calculations would be the following.

Let's assume the caller and tie percent remains fixed and just subtract the extra .0148 equity from the shover.

Caller equity: .6526
Shover equity: .3446 - .0148 = .3298
Ties: .0176

Double check: .6526+.3298+.0176 = 1

Caller folds:

((1326-188)/1326)*(100-100)

Caller calls, we win:
(188/1326)*.3298*(158.78-100)

Caller calls, we lose:
(188/1326)*.6526(0-100)

Caller calls, we tie:

(188/1326)*(.0176)*(79.39-100)

Added up:

0+2.7485+-9.2525+-.0514 = -6.5554

Caller:

Fold:

((1326-188)/1326)*(100-100)

Call, win
(188/1326)*.6526*(158.78-100)

Call, lose

(188/1326)*.3298*(0-100)

Call, tie
(188/1326)*(.0176)*(79.39-100)

Add them up:
0+5.4386+-4.6759+-.0514 = .7113

So why aren't the payouts equal? Because the rake took money out of the pot. If we treat rake as a player with a net positive payout the game should be 0 sum:

Rake:

41.22*(188/1326) = 5.8441

5.8441 + .7113 + -6.5554 = -0.0034

I think we can assume a rounding error as I know I rounded the last digit. Hopefully that convinces you that is indeed the correct method for tracking EV with stacks.

So now we know the EV of the players without the $10. But the shover wins $10 ((1326-188)/1326) of the time:

((1326-188)/1326)*(110-100) = 8.5822

Shover net:

8.5822 - 6.5554 = 2.20268

Now this is only for the rake you provided. But it's clear both parties are still +EV in the game at that rake level, so you would have to test other rake levels using the correct math.

Maybe I am missing something but can someone explain why the game as presented is not constant sum if you treat the rake as an additional player with the payout of the raked amount?

Or is the argument just the game is not 0 sum? Thanks!

Yeah, I actually mentioned earlier in this thread that this can be modeled as a 2-player game with variable sum, or as a 3-player game with constant sum.

Neither helps Robert though. He wants to use the Nash = maximin result, and that only holds for 2-player zero sum games. So he is bent on showing that this is a 2-player zero sum game.

He is clearly not arguing for a 3-player game here:

Thank you. I do remember that now that you mentioned it. Thanks!

Whoops. I should note this is not the correct final EV. The -6.5554 came from considering pot sizes that were only based on $200.

Would have to redo the calcs for $210 and since that post was just to illustarte the correct method I am not going to do that here.

just_grindin,

Thank you for pointing out my algebra mistake(s). Algebra should matter for me as much as anyone else. I suppose I should run my equations by symbolab before posting.

plexiq,

I think ICWUDH. If not, I learned a lot anyway and I do enjoy a spirited discussion.

I still want to get this right:

2-person poker is always constant-sum.
So, strictly competitive.
There will be a number that every pair of strategy profile payoffs adds up to.

2-person poker without free sklansky dollars flying around is always also zero-sum.

But, when there is free money added, it may or may not be zero sum, depending on the selection of strategy profiles.

The way to know you are presently at a Nash Equilibrium, is that it STAYS ZERO SUM when you add more free money, and no player has incentive to change strategy.

The only equilibrium I can see that does this is mine. My players will just get more money. Shover gonna shove, caller gonna call. At the other equilibriums they would have to renegotiate this.

Nope, that's obviously not the case for raked poker. Constant sum means that the payoffs of all players add up to one constant amount for any possible outcome of the game.

Payoffs in this game are the player's stacks at the end of the hand. The sum of stacks clearly remains constant in unraked poker. In raked poker the sum isn't constant, if the caller folds then the pot isn't raked. So sometimes rake gets deducted from the pot, other times it isn't. Therefore the game is not constant sum. (IF you want to treat the house as a player then the sum of payoffs is constant, but it's now a 3-player game and that means Nash doesn't equal maximin.)

That's not the definition of a Nash Equilibrium at all, how did you get to that claim? The NE definition was posted earlier itt.

---

Let's take a different angle:
You realize that the ranges you posted are not actually a maximin solution? (That's because the game isn't zero sum and using maximin doesn't make any sense here, which you probably noticed when calculating the ranges.)

The Nash Equilibrium play against crackpots is to ignore them.