Pot odds taken holistically doesn't make sense.

A

aliceabusmus

Rising Star
You sit at a full table playing $5/$10 NLHE.

The dealer takes $5 from the pot for the rake (drop). Folds to Lojack who raises to $30. Folds to YOU on the Big Blind. You have to call $20 into a pot of $40 or 2-1 immediate pot odds. You call and the pot is $60.

Flop - you check, the PFR bets pot ($60). Once again, you have 2-1 pot odds. You call and the pot is $180.

Turn - you check, the PFR bets pot ($180), you have 2-1 pot odds and you call. You call and the pot is $540.

River - you check, the PFR bets pot ($540), you have 2-1 pot odds and you call. You call and the pot is $1620.

So, in each of four betting rounds, you had 2-1 odds to call and needed only to be best more than 1/3 of the time for these calls to be profitable.

BUT, the total pot is now $1620. If you win, your profit since sitting down is $810. If you lose, you lose $810. If you're best only 1/3 of the time, you do NOT break even, you lose on average $270! You have to win 50% of the time just to break even.

What am I getting wrong? How can I make four consecutive calls that are profitable in a vacuum that add up to a grossly unprofitable result?
 
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H

Hermus

Rock Star
You have to accept that the pot build on prior streets is not your money anymore. If you regard the pot as dead money (which it is) in each situation the maths actually does check out. So while you're indeed correct that the river is a win/lose scenario, the statement that you either win or lose $810 is incorrect. You either risk 540 to win 1620 by calling or you risk nothing to win nothing by folding (remember the money in the pot is not yours anymore).
 
NWPatriot

NWPatriot

Rock Star
...

So, in each of four betting rounds, you had 2-1 odds to call and needed only to be best more than 1/3 of the time for these calls to be profitable.

BUT, the total pot is now $1620. If you win, your profit since sitting down is $810. If you lose, you lose $810. If you're best only 1/3 of the time, you do NOT break even, you lose on average $270! You have to win 50% of the time just to break even.

What am I getting wrong? How can I make four consecutive calls that are profitable in a vacuum that add up to a grossly unprofitable result?

You aren't doing anything wrong. The poker community has called something profitable that really isn't. Pot odds equity (POE) tells us the minimum equity we need to call a bet to break-even on the calling chips only (the play). It DOES NOT tell us the minimum equity needed to break-even for the entire hand (the line), which would need to include the chips we have previously invested.

I call this the "pot odds equity myth" in a book I am writing, that is near completion. I do have a solution for this, and I simply call it MCE (minimum call equity) and it tells us exactly what you want to know: what equity do I need to break-even for the hand?

To have your stack be larger at the end of this hand, on average, you would have needed 50% equity in order to accomplish this. This is all about EV (expected value), so we need an understanding of EV first, in order for the solution to make sense.

I can share an excerpt of the book with you if you are interested. Private message me and I will get it to you. The formula adjustment is very simple, but describing the entire concept is too lengthy for a post like this. I hesitate to share the formula without its supporting dialog, as I fear it may be too confusing to many readers.


Great question. Good luck and God bless.
 
H

Hermus

Rock Star
It DOES NOT tell us the minimum equity needed to break-even for the entire hand (the line), which would need to include the chips we have previously invested.


That's because equity of a line is irrelevant if you're facing a bet on any street. Money spent in a hand are sunk costs and the only thing that matters is the EV of either folding (which is always zero btw because the pot is not yours) calling or raising. It doesn't make sense to calculate break-even equity for an entire hand and it doesn't aid you in making good poker decisions. Assigning value to sunk costs is purely emotional and should be avoided in all areas of life, but especially poker.
 
A

aliceabusmus

Rising Star
NWPATRIOT: Thanks for your reply. I don't know why I thought of this last week but I've been unable to think of much else since! I'd love to see the exerpt, however, I'm new here and cannot send a private message until I've posted 15 times. Not sure if you can PM me though or what other method is copacetic.

HERMUS: While I mostly agree with you, that still leaves the paradox -- a player can make 4 consecutive EV+ decisions that when added together are EV-. Four positive numbers cannot sum to a negative one. At least ONE of those numbers MUST be negative. When the BB calls preflop, it isn't really meaingfull that she's getting 2-1 UNLESS the bet puts one of the players all-in.
 
H

Hermus

Rock Star
Okay I'll give it one last try. If two people decide at the start of the hand that the OOP player will bet pot every street and the IP will call every time of course the player with more than 50% equity will take more than half of the total pot. We don't need a book or special calculations for that.

Poker doesn't work that way though. We get three decisions facing a bet. Fold, call or raise. If we regard the money we invested in the hand as sunk costs, as we should, each decision stands on its own. We get 2:1 on each bet, therefore we need 33% equity to continue (assuming implied odds, reversed implied odds and fold equity don't play a role in this hand). There is no paradox here. If you get the right price to chase your draws you do it. If you don't you fold.
HERMUS: While I mostly agree with you, that still leaves the paradox -- a player can make 4 consecutive EV+ decisions that when added together are EV-. Four positive numbers cannot sum to a negative one. At least ONE of those numbers MUST be negative. When the BB calls preflop, it isn't really meaingfull that she's getting 2-1 UNLESS the bet puts one of the players all-in.


This also doesn't really make any sense either. Hypothetically, if you're good 40% of the time on each street and on each street you face a pot bet, chasing the draw is profitable even if you're the equity underdog. Each street doesn't add up to a -EV play because that doesn't make any sense. Like I said the money in the pot is dead money. Therefore, it only makes sense to look at each decision in isolation. If you really want to add the EV of this 40% equity hand for this exact line we get:

EVpre + EVflop + EVturn + EV River =

(60 * 0.4) + (-20 * 0.6) + (180 * 0.4) + (-60 * 0.6) + (540 * 0.4) + (-180 * 0.6) + (1620 * 0.4) + (-540 * 0.6) = 480

So because we get the right price for chasing the draw on each street the EV of chasing the draw as the equity underdog is still a positive number. There is no paradox because each +EV call adds up to a +EV line.
 
NWPatriot

NWPatriot

Rock Star
That's because equity of a line is irrelevant if you're facing a bet on any street. Money spent in a hand are sunk costs and the only thing that matters is the EV of either folding (which is always zero btw because the pot is not yours) calling or raising. It doesn't make sense to calculate break-even equity for an entire hand and it doesn't aid you in making good poker decisions. Assigning value to sunk costs is purely emotional and should be avoided in all areas of life, but especially poker.

We may have to agree to disagree. I get the idea of sunk costs, but where I get sideways is when we start calling things profitable, when at the end of the hand our stack is less than it was when we started the hand. This is what 0EV plays do for us. Yes, we are breakeven on the bet or call we are making, but if our entire strategy is built upon making 4 0EV plays each hand (one at each street), then we are probably a losing poker player.

I am only suggesting that if want to call something profitable, then the only measure of true profitability will be if our stack is larger at the end of the hand than at the beginning of hand. Anything else starts to feel like self-deception.

We have heard it said that the BE% (immediately profitable) for a 0% equity bluff is to get 33% folds on a 50% pot size bet. This is 0EV for the better. If it is 0EV for the bettor, than the caller has all of the EV of the current pot. If playerA and I were to play and playerA did this 9 times, I would be quite happy to oblige and fold 3 times and call 6 times and I would be ahead after those 9 hands. PlayerA played a 0EV strategy and lost chips along the way. Assuming a 100 chip pot on the flop with a 50 chip 0% equity bluff:
  • PlayerA EV bet = .333*100 + .667*((0%*150) - (100%*50)) = 0EV
  • PlayerB EV call = (100%*150 - 0.0%*50) = 150EV every third hand for 50EV avg
  • Both of these EV answers imply the same thing: PlayerB gains 50 chips each hand, and PlayerA loses 50 chips each hand on average (the so called sunk costs).
There is more to 0EV than meets the eye, in my opinion. Of course there is really no such thing as 0% equity on the flop, so if a 0EV strategy is making money, it is not because the strategy is flawless, it is because there are suckouts and over-folding taking place to counteract the flaw in the strategy.

If we want to win we must do better than 0EV, otherwise we cannot make up for our losses, which are coming no matter how hard we try.

Let me know your thoughts, I am sure I am missing something, but I have spent a lot of time on this, and I think I am seeing something worth pursuing here.

Good luck and God Bless
 
H

Hermus

Rock Star
We may have to agree to disagree. I get the idea of sunk costs, but where I get sideways is when we start calling things profitable, when at the end of the hand our stack is less than it was when we started the hand. This is what 0EV plays do for us. Yes, we are breakeven on the bet or call we are making, but if our entire strategy is built upon making 4 0EV plays each hand (one at each street), then we are probably a losing poker player.

I am only suggesting that if want to call something profitable, then the only measure of true profitability will be if our stack is larger at the end of the hand than at the beginning of hand. Anything else starts to feel like self-deception.

We have heard it said that the BE% (immediately profitable) for a 0% equity bluff is to get 33% folds on a 50% pot size bet. This is 0EV for the better. If it is 0EV for the bettor, than the caller has all of the EV of the current pot. If playerA and I were to play and playerA did this 9 times, I would be quite happy to oblige and fold 3 times and call 6 times and I would be ahead after those 9 hands. PlayerA played a 0EV strategy and lost chips along the way. Assuming a 100 chip pot on the flop with a 50 chip 0% equity bluff:
  • PlayerA EV bet = .333*100 + .667*((0%*150) - (100%*50)) = 0EV
  • PlayerB EV call = (100%*150 - 0.0%*50) = 150EV every third hand for 50EV avg
  • Both of these EV answers imply the same thing: PlayerB gains 50 chips each hand, and PlayerA loses 50 chips each hand on average (the so called sunk costs).
There is more to 0EV than meets the eye, in my opinion. Of course there is really no such thing as 0% equity on the flop, so if a 0EV strategy is making money, it is not because the strategy is flawless, it is because there are suckouts and over-folding taking place to counteract the flaw in the strategy.

If we want to win we must do better than 0EV, otherwise we cannot make up for our losses, which are coming no matter how hard we try.

Let me know your thoughts, I am sure I am missing something, but I have spent a lot of time on this, and I think I am seeing something worth pursuing here.

Good luck and God Bless

I'm happy to agree to disagree. Like I said and also showed, +EV calls add up to a +EV line. Both of you seem to make a jump from logical thinking to "oh no my stack is smaller there must be something going on" while in reality there is nothing going on. Were you to play the exact same situation a million times each player would just walk away with their equity share and depending on if you made a profitable call that number would be higher or lower than your starting stack.

Not sure what you're trying to say with the maths example as I don't really see the practical application and it's also not related to pot odds. But lets say that your assumptions are correct, player A has a pure bluff and player B is folding 33% of his bluff catchers (see where this gets a bit weird because that is just not how ranges work). Player A is always breaking even (i.e. player A is indifferent to betting and checking pure bluffs) and player B walks away with the dead money. Player A doesn't lose $50 because player A is not entitled to the pot. And sure player B is quite happy to pick up $50 by defending the correct frequency, but that's not really saying much. The only conclusion you can draw here is that if player B always has the best hand, defending at the correct frequency just wins the pot. That's not really a conclusion that's just common sense.
 
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