Don't remember whether I ever asked this before.

Two perfect players (under these rules) are playing 1 2 blind nlh. Player 1 knows what player 2 is holding. Player 2 realizes that his cards are being seen but player 1 doesn't realize that his cheating is known. Player 2 also knows that player 1 is unaware he was found out. Assuming Player 1 has amnesia after every hand (in real life he would figure out his cheating was discovered pretty quickly). Who has the edge? Does it depend on stack size? If so at what approximate stack size, does the edge switch to the other player.

I think Player 2 has an advantage if they're both perfect strategists.

If P1 knows Nash Equilibrium heads up strategy, and P1 assumes his knowledge of P2's hole cards is unknown, and assumes P2 will also be playing in Nash equilibrium, P1 would know the postflop betsize percentages that would force P2 to fold or mostly fold his currently holding.

Immediately, then, I think P1's strategy would involve few to no egregiously large preflop raises as P1 wants to keep SPR within the range where shoving is rarely a correct postflop option for an opponent playing in regular Nash Equilibrium.

When holding a losing hand, P1 would want to bet the largest sizes postflop that would force a naive P2 to continually call to the river, and then on the river bet a sizing where a naive P2 would have to fold or mostly fold.

When holding a winning hand, P1 would do the same on the flop and turn, and then on the river bet a sizing where a naive P2 would have to call or mostly call.

If P2 knows his hand is a fold to the river bet in regular Nash equilibrium, he would call, and vice versa. So in a way both players are exposing their hands, but only one player knows that they are.

edit: This is for ~100bb play.

You can show an arbitrarily low stack size where P1 regains the advantage. At starting stacks of 1bb, the BB is blinded in and the SB has a choice to commit .5BB for a pot of 2BB or fold. Since any starting hand has 25% or more equity vs 100% range, P2 would have to call with everything in the SB, whereas P1 would be able to make a few insane folds when holding undercards vs a pair.

Very interesting question!

I think player 1 has the advantage at all stack sizes and it isn't even close. Though player 2 has a leveling advantage, his cards are exposed and that is just too much of an edge to give up. On most boards player 2 is going to be capped and player 1 can have a massive value range. In some spots he'll be overbluffing, but more often than not, player 2 is going to have to fold due to being capped. Also, when player 2 hits the nuts or a really strong hand, he's never getting paid off.

Imagine player 1 getting dealt T7o on the button. Player 2 can 3-bet so many hands here for value.

After thinking about this for 30 seconds, I completely agree. It would be a blood-bath.

P2 is never (or always if you prefer) capped because P2 never has a range in this game.

If P2 is put into a spot where he's "supposed" to fold, he would call. P1 would not balance any of his bet sizes / bet ranges because he's only ever playing against P2's hand who he assumes P2 will take the regular Nash Equilibrium action with.

Toy scenario: P2 holds a bluffcatcher on the river. P1 holds a hand that will not win on showdown.

In regular (i.e. non-exposed) Nash Equilibrium P2 would call (or mostly call) with this hand facing a bet size of 50% or less and fold (or mostly fold) with this hand facing a bet size over 50%. P1 therefore bets more than 50%. P1 and P2, both being perfect players as per OP's premise, know the Nash Equilibrium strategy. P2 knows he is supposed to fold his hand to this action and knows that P1 doesn't know that P2 knows that P1 knows P2's cards; therefore P1 would never hold a value hand; therefore P2 should call.

I assume you reversed the player numbers in this statement.

Player 2 (with hand exposed) would obviously not open T7o if he could not profitably defend it vs a 3-bet / postflop aggression if he is a perfect strategist.

It may be in this game that P2 plays very few buttons and gains all of his EV in the BB. I think this part may be dependent on stack sizes and what preflop strategy is used, which is quite hard to say.

Also if I'm right then this game is one form of a type of game.

In a heads up poker game where P1 knows P2's hole cards and P2 does not know P1's hole cards, P1 has an advantage.

In a heads up poker game where P1 knows P2's hole cards, P2 does not know P1's hole cards, and only P2 knows that P2 knows that P1 knows P2's hole cards, P2 has an advantage.

In a heads up poker game where P1 knows P2's hole cards, P2 does not know P1's hole cards, and P1 knows that P2 thinks that only P2 knows that P1 knows P2's hole cards, P1 has an advantage.

In a heads up poker game where P1 knows P2's hole cards, P2 does not know P1's hole cards, and P2 knows that P1 thinks that P2 thinks that only P2 knows that P1 knows P2's hole cards, P2 has an advantage.

(with the amnesia and perfect players premise in the OP)

And so on.

If he could name a hand to have on the button what would it be (with large stacks)?

P2 can play rivers perfectly. Because p1 knows p2's theoretical correct move with the specific cards p2 is holding, p1 will go for max exploit and massively overbluff or underbluff. P2 is aware of this and can counter exploit.

It's in P2s best interest to draw the hand out to the river and play as passively as possible on all earlier streets.

So it appears in a deep game p2 will win. In a short stacked game, p1 has an advantage because short stack play is about pure equity and not as much leveling and range defending. Pure equity calculations will be trivial for p1 because he has massive information advantage and denying equity is valuable in short stack matchups. So p2 can't decifer p1's intentions that accurately.

At around 50bb, 99-TT
At around 100bb, JJ-QQ
250bb+, AA

Okay, I see what your saying. If player 1 knows player 2 has a weak/mediocre hand he'll overbet with his bluffs and bet small with his value hands. I would consider that exploitative play though which can be argued as perfect play. However, I think an opponent playing GTO would play in a way where no matter how his opponent plays he is going to lose the least amount possible. So that could also be argued as perfect play. If player 2 can exploit player 1, then player 1 is not playing a GTO strategy.

A problem that I can see occurring assuming that player 1 takes the exploitative route:

100BB effective stacks, 1/2 NL
player 1 min-raises to $4 with J T
player 2 calls with 8 5

Flop($8) 5 6 A
player 2 checks
player 1 bets $24 as a bluff
player 2 calls

Player 1 would now be tipped off that player 2 is not playing the Nash equilibrium strategy that he is assuming player 2 is playing. Since the amnesia is taking place after every hand and not every street, player 1 can now change his strategy.

Player 1 would not bet $24 on the flop. He'd bet an amount he knows player 2 will call down until the river, and then "over"bet because that would yield the highest payoff from his perspective.

Homework assignment:

Toy game. Player 1 and 2 both are dealt a real number between 0 and 1. High number wins.

Both players ante half the pot.

Player 1 bets the pot or checks. Player 2 calls or folds if bet into but cannot raise or even bet if checked to. Obviously Player 1 has positive EV and player 2 has negative EV. What is the strategy and what is the EV?

If player 1 can see 2's hand it increases. What is the strategy and what is the EV?

If player 2 knows of 1's cheating but 1 is unaware that brings 1's EV down somewhat. Does it go below the EV of a normal game? Does it actually become negative?

If it goes below the EV of the normal game it comes close to proving that Player 2 has the edge with at least some stacks in my original question.

This homework assignment is actually not that hard. Do it.

I suspect that in the undiscovered cheating variant, P1 will have infinitely many co-optimal strategies because of betting and checking having the same payoff when P1 holds a better hand than P2 and P2's hand is in his folding range, which would then affect P2's play and thus his EV in the discovered cheating variant because P2 cannot know which of these strategies P1 would be using.

The strategy for player 1 is to bet the top 25% and bottom 12.5% since player 2 must call 50% of the time. Player 2 will obviously call with the top 50% and fold the rest. Player 1 will check with the middling numbers (.125-.75).

To find the EV of player 1 we can break up the EV into 3 categories: Value Bets, Checked Numbers, and Bluffs. After finding the EV of each of these we can add them together after multiplying them by the frequency in which they occur.

Value Bets
Player 1's EV for his best number (1) will be 1.5 EV on calls and .5 EV on folds. Since player 2 is calling and folding exactly half the time those numbers can be averaged. (1.5 + .5) / 2 = 1. For player 1's worst number he is break even on calls for an EV of 0 and wins from folds for .5 EV. (0 + .5) / 2 = .25. We can average player 1's best betting number and worst betting number to get the entire EV for value bets. (Intuitively this seems right since the EV progression should be linear as the number gets better.) (1 + .25) / 2 = .75.
EV = +.75

Checked Numbers
The best checking number (.75) has an EV of (.75 * .5) - (.25 * .5) = .25. The worst checking number (.125) has an EV of (.125 * .5) - (.875 * .5) = -.375. Average those both and we get (.25 + -.375) / 2 = -.0625
EV = -.0625

Bluffs
The bluffs break even since they win and lose exactly half the time.
EV = 0

(.75 * .25) + (-.0625 * .625) + (.125 * 0) = .2265625

So in this game player 1 is going to be making about +.23 EV on average. Player 2 will be the opposite of that and lose -.23 EV on average.

I hope I'm right. I may do the other stuff later if no one beats me to it.

Okay for the cheating variants. Player 1 will bet 100% of his numbers when he sees player 2 has a bottom 50% number (okay, he is indifferent between checking and betting his better numbers, but betting is better in a real world scenario since it allows for mistakes to happen). When player 2 has a top 50% number, player 1 will only bet if he is ahead.

Calculating the EV from player 2's perspective here may be easier since the strategy is entirely based off of what player 2 is holding.

Top 50%
Player 2 wins +.5 EV with his best holding (since it gets checked down) and loses -1.5 EV with his worst holding. On average he loses (.5 + -1.5) / 2 = -.5.
EV = -.5

Bottom 50%
Player 2 folds and loses -.5 EV.
EV = -.5

Since the EV for the top 50 and bottom 50 percent of numbers are the same we can assume that is the EV for player 2. Player 1 is the opposite of that and wins +.5 EV on average.


If player 2 knows about player 1's cheating then player 2's strategy should be to fold whenever player 1 bets. His weakest numbers won't be strong enough to call and his best numbers are never winning since player 1 would have checked instead.

Top 50%
The best number is winning +.5 EV. The worst number is break even. (.5 + 0) / 2 = .25
EV = +.25

Bottom 50%
Player 1 always bets and player 2 always folds resulting in -.5 EV
EV = -.5

Since both groups happen at equal frequencies we can average them.
(.25 + -.5) / 2 = -.125 EV for player 2.


To sum things up, these are the EVs for player 1 in each scenario:
Regular game ---> EV = +.23
P1's cheating ---> EV = +.5
P2's awareness of the cheating ---> EV = +.125

If player 2 calls 50% of the time, the EV for player 1 of betting [0,.125) is 0 while the EV of checking any of those hands is the number itself. So player 1 would never bluff and so this is not a Nash equilibrium.

On my phone while at work, so for now I will contribute the first model so others can add in the versions with exposed hole cards.

This is the Von Neumann model. P1 and P2 contribute an ante of 1 unit in the pot. P1 may bet 2 (which is actually the exact optimal bet size for this model), and P2 may call or fold. If P1 checks, the hands are compared. If P2 calls, the hands are compared. If P2 folds, P1 wins the pot.

From memory, the solution is thus:

P1 bets all values above 7/9 and below 1/9.

P2 calls all values above 5/9.

Value of the game to P1 is 1/9.

Deriving these values using indifference is pretty basic, 3 equations in 3 unknowns.

Doing without indifference is a little longer, 6 equations in 6 unknowns.

Now, the fun part is how will P1 exploit the exposed cards of P2, and what will P2 respond with, knowing this subterfuge?

Nope. Player 1 expects to be called by the top half of Player 2's hands. So if player 2 is in the top half Player 1 will never bet unless he has the winner and 2 should always fold.

BUT

If player 2 is in the bottom half, Player 1 will bet all hands (either as a bluff he expects to succeed, or a value bet that where he hopes Player 2 misclicks).But since Player 2 is getting 2-1 odds he must call if there is a 1/3 chance he has the best hand. Which means he must call with the upper third of his normally folding hands. (Because with those hands there is somewhere between a 1/3 and 1/2 chance his hand is best when facing a bet.)

In other words in this strange game Player 2 should call with 1/6 of his hands but NOT the top 1/6.

While the fractions are slightly off, TheGodson is basically correct, and you should understand why betting the worst of your hands as a bluff is part of an optimal betting strategy.

This is quite an effort, but incorrect. Your strategy is good, but the math is wrong.

A conjectured strategy is solved for the values of the variables, not the other way around.

I haven’t seen a correct answer yet, so here goes.

Standard game:

P1 will bet the range [0, x], [(1-2x), 1]. P2 will call at a frequency which makes x indifferent between betting and checking. Therefore the EV of betting with 0 equity when called must be x.

x = f+(1-f)*-1
x = 2f-1

The other indifference point occurs when we have 50% equity when called. f must therefore be twice as far from 1 as our worst value bet.
f = 1-4x
f = 1-4(2f-1)
f = 5/9
x = 1/9

P1’s betting range is [0, 1/9], [7/9, 1]
P2’s calling range is [5/9, 1]

P1’s EV of [0, 1/9] is 1/9.
P1’s EV of [1/9, 7/9] is his number, with average 4/9.
P1’s EV of [7/9, 1] is 5/9+4/9*(.75*3-1) = 10/9
P1’s overall EV is 1/9*1/9+6/9*4/9+2/9*10/9-1/2 = 1/18 =~ .0556 units.

P1 cheating, P2 unaware:

If P2 is weaker than 5/9, P1 bets and wins. If P2 is stronger than 5/9, P1 bets only if they have a better hand, and they get called and win.

P1’s EV when cheating is 5/9+4/9*(2/9*2)-1/2 = 41/162 =~ .253 units.

P1 cheating, P2 aware, P1 unaware of P2’s awareness:
P2 won’t win when facing a bet with a number > 5/9 so he should fold. With a hand weaker than 5/9, P1 May be bluffing, so he should call if he has at least 1/3 chance to win.
P2’s calling range is [1/3, 5/9]

P2’s EV of [0, 1/3] is 0.
P2’s EV of [1/3, 5/9] is (1/3+5/9)/2*3-1 = 1/3
P2’s EV of [5/9, 1] is their number, with an average of 7/9
P1’s overall EV is -(2/9*1/3+7/9*4/9-1/2) = 13/162 =~ .0802

Seems right. In my post I assumed a 50% calling rate because I forgot that some of P1s bad hands would still win a showdown.

So this example doesn't prove that there are cases where the cheater catcher does better than the fair game. But it doesn't disprove it either. What do you think?

Solving all three versions of the toy game for EV in terms of Bet size will determine if there is any value of Bet which is +EV for P2.

Once this is complete, is there any scenario of real poker with multiple bets and raises that could improve the EV of P2?

It’s going to be a bloodbath for P2 if anything more than a shove/call is an option.

I believe the original scenario will be a bloodbath. P1 gets destroyed because P2 will effectively have complete information on the river in NL, since P1 will bluff with a sizing that causes P2 to fold 100% of the time assuming P2 is unaware of cheating. Since P2 is aware, he can bluffcatch in situations he’s supposed to fold and fold against sizings he’s supposed to call against. It is also generally in P1’s best interest to make it to the river where P2 has a very clear and large advantage, since in the cheating/unaware version P1 has 100% pot share against most hands on the river due to being able to bluff so effectively.

I’m not sure at which stack sizes P1 has the advantage. Surely just very short stacks, but I’m not sure how to estimate more precisely.

Change the toy game to NL and P2 will clean up.

P1 leaks too much information in a NL format. He is only winning the toy game because it’s a limit format.