It shouldn't be hard if you program commuters. I will acknowledge your help in the book. (Probably titled The Theory of Poker Applied To No Limit).

There is ten dollars in the pot. Me and you get a two card starting holdem hand.

I bet 100 all in or fold. You call or fold.

Th house rakes a giant amount from the 210 dollar pot if you call. Nothing if you don't. If it the rake is high enough you can't call with two aces. If it is somewhat lower, say fifty bucks, and I bet in the dark every hand, there are a few hands you would call with, namely any hand that is more than a 100-60 favorite over a random hand. But even though you would call with these hands I should still bet every time because I will steal the ten so often and sometimes win when I don't.

At what rake should I stop pushing everything in the dark? (I say "in the dark" because your I want your strategy to be based on my hand being random.)

(I want to know because it analogizes to certain tournament bubble situations)

To be clear, where is the 10 in the pot coming from? Because if it's not from our stacks, we only play AA as long as rake is less than 44.2% of the total pot. And if it is any higher, then we walk away from the tables.

If the caller has any combos that can call, then at MDF the worst call will break even, and the whole range of calls will certainly be +EV. This is actually *because* your range is any two cards. If you have a more nuanced shoving strategy, then maybe you can shove with a very high frequency, but still not ATC.

I'm just here looking for the commuter programmers.

We should stop pushing when the EV of the caller's range is > $10.

At zero rake 726 combos are +EV calls and the caller's EV is 762/1326*(.5679*210-100) = $11.07. I could program something to find the answer, but I think trial and error will be enough.

This didn't take long.

At $3 rake or $4 rake the caller's range is the same:
706/1326*(.5749*207-100) = $10.12
706/1326*(.5749*206-100) = $9.81

With rake in increments of $1, villain can dark shove profitably if the rake is >= $4. Since the caller's range doesn't change for more precise rake simple algebra could calculate the exact amount of rake.

I used Equilab's feature which calculates +EV hand ranges based on pot odds (hand range calculator).

Edit: Alright, I won't be lazy. Here is the answer to four significant figures because this is the most Equiliab presents.
Rake >= 210-(10*1326/706+100)/.5749
Rake >= $3.389

I think David wants to know at what rake we have to fold AA. Part of a Sklansky thread is deciphering the Sklansky post.

I’m not sure what exactly there is to decipher. Sklansky asked a question in clear English, and I gave an answer. If that turns out to not be what he meant, then I’ll happily answer him when he asks a new question. In the meantime I’m not going to guess about what he wants when it seems pretty clear to me.

As for the amount of rake for which the caller breaks even with AA, that can be figured out by setting EV to $0 and solving for the size of the pot, knowing AAs equity against random cards.

For any hand with equity ‘e’ it should be this:

Rake <= 210-100/e

It would be best to start by drawing a decision tree with all the known and unknown variables.

No. I am essentially asking at what rake it is unprofitable to push with 32. The answer should be at about the point where the other guy can call with about the top 12 or 13% of his combos.

You’re going to have to clarify because that’s a totally different question from what you asked. Do you want the range we can profitably shove assuming the opponent plays optimally under the assumption we are shoving any two cards? Do you want Nash ranges? Do you want to know the rake level where we can profitably shove dark?

I ran it through CREV and it seems the answer is around 20.6% rake.

I feel like I am missing something here because I am not sure how you arrived at this simplification.

I had assumed he was asking at what rake is is shove less than EV neutral and I am not sure if the two things are the same or if you're solving for something else (i.e. a solution that has higher EV than shoving blindly, even when shoving blindly might still be +EV).

I'm still in the deciphering Sklansky phase.

So far this is what I gather:

1) There is ten bucks free money on the table.

2) David is betting 100 bucks with any two cards.

3) This is because the house charges a very high rake on calls, but no rake on folds.

4) We get to look at our hand, and decide on a range worth calling and paying the high rake.

5) This is profitable for David to continue shoving any two cards, until the rake drops to a point that it stops being profitable.

^^^One of these seems like it does not belong with the others.^^^

According to my calculations, ~44,1% is the postflop rake level at which calling/folding AA has the same EV.

Yeah I wrote that 10 hours ago but it's not what OP wants it seems

I thought it was a binary choice of shove dark or fold/don’t play. “What is the rake level for which a dark shove profits?”

That's what I am asking except it's not where dark pushing becomes unprofitable but rather when is it even more profitable to refrain with some hands.

Ok, I did the math/sims and the results were

For the "pusher" holding 27o when the rake is 20.4%:


For the "pusher" holding 27o when the rake is 20.5%:


Fot the caller when the rake is 20.46%:


This is the callers calling range when the rake is 20.5%:


When the rake drops to 20.4%, the caller starts calling A8o also, which makes the pushers jamms -EV.

Conclusion:
20.46% rake is the point at which A8o becomes a 0 EV call and is the point at which 27o becomes a 0 EV jam.

How is it a dark push if we can make a decision to refrain based on our hand?

Do you actually want to know the minimum rake level such that looking at our cards will never change our strategy?

That's good stuff but wouldn't 23o be better for a heads up situation? I thought 72o was worse with multiple players but 23 was worse for heads up due to high cards being more important.

No, I've tested all hands and 27o has the lowest EV vs that calling range (also vs pretty much any other range).

Great. Thanks again.

I thought exhaustive montecarlo vs any two cards has 23o at the bottom. This was the combo Sklansky uses.

So, there is a rake at which dark pushing is optimal, and maximizes the return to the pusher? And at a slightly lower rake, the dark pusher will remove a combo to maintain equilibrium?

OR....

The dark pusher (cool villain name) simply pushes whenever it is +EV to do so.

Which is it?

That seems right. Thanks.

Did Monkey T1lt get this also?

Do you or he want to PM me your real name?

I don't get why some people didn't realize what I was asking.

I would like to point out that the question becomes a little harder if it involves rakes low enough where the pusher shouldn't push all hands.

If I were to approximate the answer I believe your result means I can say that "if the rake is 20% or higher the pusher can push all his hands in the dark (laying 10-1 odds) and at that 20% figure the other player will call about 15% of the time and win almost 12 of those 15. His worst calling hands are about 60% against a random hand and will win about two thirds against the weakest opposing pushes."

Do you agree with all that?

Assuming you do, do you agree that I can analogize that into saying that a very large stack can push against a small stack of about five buy ins or less in a tournament where there are a few tiny microstacks and only one player needs to be eliminated to make the bubble? I come to that conclusion thusly:

100 players buy in 1000 in a nlh tournament with only one big blind and no antes. Ten players make the bubble prize of 2000. After which they shoot for the first prize of 80,000. No intermediate prizes. If you make the bubble with 5000 your EV is 2000 plus 5% of 80,000 or about 6000. (In real life slightly more). If you make the bubble with 10,000 your EV is about 2000 plus 8000 or about 10,000. So if the blind is about 500 and you are in it and a big stack moves you all in, you are risking 6000 EV to win another 4000 EV. If he moves in in the dark you need a 60% chance against a random hand to call.

Since you need a 60% chance to call, the situation is essentially the same as the rake problem. So the big stack dark pusher can indeed do his evil against a thinking opponent who is trying to maximize their EV when that opponents stack is up to about 5 buy ins.