Glad to see that we were not misunderstanding, and that two posters had it right off the bat.

Bravo!

In the spirit of exactly answering the puzzle, the rake structure was quoted in dollars, and the chance of a tie (around 2 percent) does change the EV.

The lowest dollar amount in rake that can be dark shoved is 55 dollars, and the caller will call with 230 combos (17.35 percent VPIP). The caller must not call and lose money due to ties.

This is the calling range:



and this range has this equity:



At 55 dollars rake (around 26 percent):

The EV of the caller is (110-55)(.6363)-(100)(.3439)-(.0198)(22.50)=0.1545

The EV of the shover is 10(.8265)-(100)(.6363)(.1735)+(110-55)(.3439)(.1735)-(.0198)(22.50)=0.061

At the next lower dollar in rake, the dark shover goes negative (-.33)

Honestly, I don't know how David comes up with these puzzles. IMO, there should be an appendix of them in the book, unsolved, and the readers can discuss and solve them here on 2+2 in the Books and Publications forum!

He got a similar result (20.6%), even though I don't know how exactly he calculated it (I don't use CREV).
I'm confident that my result (20,46%) is 100% accurate though.

Correct, except the caller will only win around 10 of those 15, since the equity of his range vs any two cards is ~66%.
His worst calling hands (A8o) will win about 68% against the weakest opposing pushes (27o).

I agree with all of that. It's a very interesting concept.

It's worth mentioning that the small stack isn't able to then reverse attack the deep stack like this. Though the deep stack also wins only ~4000 EV when he wins, he will also only lose ~4000 EV when he loses, which makes him able to call pushes with less than 50% equity (like in cash games). This is the advantage of deep stacks in tournaments.

I PMed you my real name.

The results shouldn't be in dollars, since the thread is actually about tournament ICM situations.

The chance of a tie is already included in the equilabs equity results.

Also, the range that you used for calling the dark shove with doesn't make much sense to me. For example, why would you call with QTo that has only 57% equity and fold something like A9o/55... that has 60%+ equity?

Yea I think this is where you and I got confused.

Zkesic was evaluating the hand 72o vs another range that is just the top X% of hands in which case the highcard power of 72o vs everything lower than a 7 is moot.

Vs a random range we start including a large swath of hands that are lower than a 7 which causes 23o to be the worse.

Hmmm. Well something is awry.

In your solution the actual expectation value to each player is:

At 20.5 percent rake:

The EV of the caller is (110-43.05)(.6485)-(100)(.3331)-(.0184)(16.52)=9.803

The EV of the shover is 10(.8492)-(100)(.6485)(.1508)+(110-43.05)(.3331)(.1508)-(.0184)(16.52)=1.77


I don't know what is off in your excel spreadsheet, but that is why it is a good idea to include your calculations in your answers.

Whoops. Let me try that again.

At 20.5 percent rake:

The EV of the caller is (110-43.05)(.6485)-(100)(.3331)-(.0184)(16.52)=9.80

The EV of the shover is 10(.8492)-(100)(.6485)(.1508)+(110-43.05)(.3331)(.1508)-(.0184)(16.52)(.0184)=2.07

I'm getting different answers than everybody. I don't know if I'm just having a brain fart here, or if I'm still misunderstanding this poorly worded question.

At 20.5% rake the callers range is easy to verify as {55+, A5s+, A9o+, K9s+, KJo+, QJs}.

When we hold 72o as the shover, this range represents 196 combos out of 1225, and our equity when called is .2655 in a pot of 210*.795, which we pay 100 into. The shover's EV is:

1029/1225*10+196/1225*(.2655*.795*210-100) = -$.5080, which does not match Zkesic.

I believe the correct answer is 20.99% rake:

(1225-184)/1225*10+184/1225*(.2617*((1-.2099)*210)-100) = -$.0004 EV for shoving 72o, and the caller's range is {55+,A7s+,KTs+,QJs,A9o+,KJo+}

At this rake level, the shover's overall EV is:

(1326-188)/1326*10+188/1326*(.3383*(1-.2099)*210-100) = $2.362

The caller's overall EV is:

188/1326*(.6617*(1-.2099)*210-100) = $1.388

Robert. I haven't looked for what is wrong with your calculation, but it is impossible for the sums of the EVs of each player to be greater than $10. Where is that money coming from?

We agree on all of this:



I don't have the exact formula that I used in my calculation, since I made that excel sheet years ago and the formula would be quite difficult to recreate from all that. I am 100% confident that the results are correct though, since I've been using it for years and the results were always 100% like they should be.

Anyways, I will create an EV formula that I think makes sense.

EV for pushers 27o pushes:

When the caller folds + When the caller calls and the pusher gets lucky - When the caller calls and the pusher doesn't get lucky:

1029/1225*10+196/1225*(.2655*.795*110)-196/1225*(.7345*100)

EV = 8,4 + 3,7149 - 11,752 = 0,36

The EV result of this calculation is exactly the same as the result that I got in the excel sheet above.

Your mistake I believe, was at the red part of the formula (you should've multiplied 100 with .7345, which is the chance that the event happens).

I have never really used "formulas" for poker EV calculations. I just think it through. I do not multiply the 100 by .7345 because that term is not supposed to be [loss amount]*[loss frequency]. It is just an adjustment for our contribution to the pot.

Your blue term is incorrect because the size of the raked pot is calculated incorrectly. %Rake is applied to the final pot, not the amount we would win. The raked pot size is 210*.795. We contributed 100 to that pot, so we will win 210*.795-100 = $66.95.

When I fix the blue term our answers match:

1029/1225*10+196/1225*(.2655*(.795*210-100))-196/1225*(.7345*100) = -$.5080

if we are using card removal (72o blocks 4 call combos), then we would have to enumerate every possible starting hand and its removal. The actual strategy is that the shover always shoves (1326 combos) and the caller calls with the combos assigned in range. Enumeration and calculation of every possible scenario would sum up to this.

Oh, you're right. The rake for when the 27o gets called and wins was calculated incorrectly. The correct answer should then really be around 20.99% rake.

Thanks for correcting me.

For 27o I was using card removal when calculating it's fold equity.

Since the pusher shoves 100% of his range, the card removal from those hands evens out to 0.

Then the chance of a flop happening is callrange/1326.

Yes, that's correct.

The chance of flop happening in this case is 85%.

However, the fold equity that 27o has is only 84% (because of bad blockers). Therefore the "1029/1225*10+196/1225*(.2655*(.795*210-100))-196/1225*(.7345*100)" formula is correct, since it's calculating the EV of 27o pushes.

The real poker situation is that the shover is always shoving, and the caller calls with a frequency of 200/1326. The puzzle is when this stops being +EV.

Blind shoving stops being + EV when the shove stops being + EV for 27o, since that's the bottom of the pushers range.

That's why I've been focusing on calculating the point at which 27o becomes 0 EV.

72o may be the first combo that the shover chooses to remove, but the decision to do so is when the entire range of ATC becomes -EV.

More importantly, why are we narrowing the callers range at an EV ~10. The high rake is the ICM penalty that narrows the caller's range. The free 10 bucks is the antes, which widens the callers range. The caller will stop calling when all of this is EV~0, not 10.

20.99% isn't right either. At the resolution of hundredths-of-percents, the pusher can no longer push 72o 100% of the time starting at 20.61% rake.

The caller's equation is
EQ*(210 - R*(2S + P)) >= S
where EQ is the equity of the hand's equity vs. ATC, S is effective stack size in $, P is the pot size in $, and R is the rake[0%-100%].

The pusher's equation is
F*(S + P) + (1 - F)*EQ*(2S + P - R*(2S + P)) >= S
where F is the caller's fold frequency and EQ is the hand's equity vs. the caller's optimal calling range.

At 20.99% rake, the caller's cutoff equity is
EQ*(210 - .2099*210) >= 100
EQ >= 60.27%

His calling range is then 55+, A7s+, KTs+, A9o+, KJo+ (how did you get QJs in that range, browni? it has only 60.259% equity).

Then for the pusher, we have
EV(push 72o) = F*(S + P) + (1 - F)*EQ*(2S + P - R*(2S + P))
= (1045/1225)*110 + (180/1225)*.26094*(210 - .2099*210)
= $100.20 (+$0.20)

Try 20.62% rake now. For the caller:
EQ*(210 - .2062*210) >= 100
EQ >= 59.99%

His calling range is then 55+, A7s+, KTs+, QJs, A9o+, KJo+.

For the pusher, we have
EV(push 72o) = F*(S + P) + (1 - F)*EQ*(2S + P - R*(2S + P))
= (1041/1225)*110 + (184/1225)*.26172*(210 - .2062*210)
= $100.03 ($0.03) <- still 3 cents above 0 EV

Finally at 20.61% rake. For the caller:
EQ*(210 - .2061*210) >= 100
EQ >= 59.98%

His calling range is 55+, A7s+, K9s+, QJs, A9o+, KJo+.

For the pusher:
EV(push 72o) = F*(S + P) + (1 - F)*EQ*(2S + P - R*(2S + P))
= (1037/1225)*110 + (188/1225)*.26264*(210 - .2061*210)
= $99.84 ($-0.16)

Once the caller can start calling K9s, the EV of pushing 72o (as well as 32o and 42o) slips below 0. All values agree with CREV sims.

What actually happens at equilibrium though is that from 20.61%-20.50% rake, 72o for the pusher and K9s for the caller each play mixed strategies. 72o sometimes pushes and sometimes folds, and K9s sometimes calls and sometimes folds. This lets both players profit more than if they were forced to play pure strategies. At 20.49% rake, 72o becomes -EV and always folds.

Yes, that's how we were calculating it. 27o becomes break even when its EV is 0 not 10.

If we average the BE equilibriums we get ~20.555%, which we could say is the final solution to this problem, right?

We can end the thread with that imo.

Hmmm. Vegas is built on getting these sort of calculations exactly right. Imagine this is a trial new bartop videogame, and the house needs each deal to be as close to exactly EV neutral as possible with a deck of 52 cards.

If we use the exact specs y'all have provided:

Pot = $10
Mandatory Dark Bet = $100
Rake percentage = 20.62%
Rake amount = $43.30
Rake deducted from each player if there is a chop = $16.65
Call range as provided: 55+, A7s+, KTs+, QJs, A9o+, KJo+
Combos in call range: 188
Combos in Any Two Cards:1326

Pot after a call minus rake = 210(1-.2062) = 166.70

Odds of a call happening = 188/1326 = .1418
Odds of a fold happening = (1326-188)/1326 = .8582

When the game runs with these specs, (200 million trials) this is what happens:

Caller will call with frequency .1418
Caller will fold with frequency .8582
The call will win with frequency .6527
The call will lose with frequency .3297
The call will chop with frequency .0176

The expectation value for the potential caller is thus:

(.1418)(.6527)(66.70)-(.1418)(100)(.3297)-(.1418)(16.65)(.0176)

6.1733 - 4.6751 - 0.0415 = 1.4565

The expectation value for the dark shover is thus:

10(.8582) + (66.70)(.1418)(.3297) - (100)(.1418)(.6527) - (16.65)(.1418)(.0176)

8.582 + 3.118 - 9.255 - 0.0415 = 2.4035

House is losing money.

[This just slightly expands on MonkeyT1lt's calcs to find the exact cut-off.]

The cut-off is when K9s becomes a profitable call around 20.6% rake, increasing the call range from (14.2%, 55+ A7s+ A9o+ KTs+ KJo+ QJs) to (14.5%, 55+ A7s+ A9o+ K9s+ KJo+ QJs), at which point pushing becomes unprofitable for 72o, 32o, 42o(?).

K9s is approx. 59.9885% equity against ATC.

So the lowest rake where pushing ATC is still optimal is at slightly above the break-even point for K9s:
x = 1 - 100$ / (210$ * 0.599885) = 20.6197%

(Of course, the ATC pusher is +EV for much longer, but at lower rake the pusher's EV can be increased by slightly tightening the range.)

At what rake does the dark pusher lose money

He loses money with the bottom of his range (27o) once the rake gets below 20.50%.

The push is going to be 0 EV for 27o everywhere between 20.61%-20.50% rake.

I am asking how low the rake has to be for the overall EV of the blind pushing strategy to be negative.