The only one having a definition problem is you.

No it isn't

Player A has a nut flush draw on the turn. Pot is 300. He goes all in for 50. If he doesn’t hit one of 9 outs he will lose

EVa = Fb 100 + (1-Fb)* (.18*350 – 0.82*50) , where Fb is B’s fold equity.

Now it is easy to show that no matter what B’s fold equity is, A’s decision to go all in has positive EV.

Now, B faces a decision, call or fold. He decides to call.

EVb = 0.82*350-0.18*50= 319, plus EV.

NOW, what is A’s EV after B called. It is

EVa after B called =0.18*350- 0.82*50 =31, plus EV

So, we have A having +EV when he made his decision, B having +EV when he made his decision in response to A’s bet and A still having +EV after B responded.

How many more examples do you want that shows two players can have +EV, even under your special condition?

true if based on ranges and not actual hands.

If player B's call is +EV for player B (i.e. better han folding) then it is -EV for player A. This doesn't necessarily make player A -EV in the hand. I haven't referred to any "special condition" - I've tied to answer the OP's question and at the same time pointed out the simultaneity fallacy. Am I missing something here, statmanhal?

1) Without rake or fees, poker is a zero sum game. The amount of money in the game doesn't change.

2) Plays can be made where all players involved can make a +ev decision. Example. Both players have 5BB. One player posts the SB, the other player the BB. SB receives 99 as his hand. BB receives AKs. SB goes all-in (+ev because he is 52% favorite to win the hand). BB calls (+ev because he has the right odds to call).

3) Point 1 and 2 have nothing to do with eachother.

Really? Why is that iyo?

I think your problem is that when B correctly calls, it reduces A's EV. But that doesn't make A's EV negative. A still has positive EV. I and many others have shown through numerous examples that both players can have positive EV but that doesn't mean that the game is not zero-sum if rake free. .

After nearly 60 postings of similar arguments on both sides, I doubt very much that you are going to change your mind, so if you want to insist that something called simultaneity fallacy exists then be my guest.

Please stop with the "heroes" and "villains" nonsense; neither exist in poker, only in Survivor Season 20.

It IS possible for both players to make +EV plays; and poker is not zero-sum in general, only in tournaments where there is a finite number of total chips among all players to start and end the game. But even in the tournament case, each HAND is not zero-sum, only the tourney as a whole. So "zero sum" is never RELEVANT to poker and I can't WAIT for the day that people stop applying the term "zero sum" to poker.

Nearly everything you said above is completely false. Each hand of poker is zero sum, provided there is no rake. Poker IS zero sum in general. If you think either of these things isn't true, then I suspect you don't know what zero sum means.

Also, you're kind of taking a scattershot approach to this forum, answering a dozen posts in a short period of time, with 1 or 2 sentence replies, that are largely non-sequiters or nonsense. I recommend taking a more considered approach.

I recommend you think about what I said instead of writing it off as nonsense. Failing that, quit poker, I guess.

two particular hands have a certain % of equity against each other. ranges are wider. getting 77 in against 88 on a 378 two tone flop (100bb stacks) is -EV. however, it's +EV to stack off with 77 against most villains ranges in this spot... when you consider 33, 78, FDs, SDs and air. that's a HUGE difference in equity.

That's what I said. Looks like we agree after all. I'll stick with calling it a definition problem and lock myself up.

OK, how about you define "zero sum" for me and give me one example of how a hand of poker isn't zero sum, given no rake.

Zero Sum means one player's gain must be equivalent to other player's losses. So in a hand of poker, my maximum possible winnings are what other players are prepared to put into the pot. Therefore poker is the absolute definition of Zero Sum.

In regards to your comment of "How about you read my post rather than dismissing what I'm saying". I'd like to adivse you to try saying something. Your post is a series of generalizations without ever explaining how poker is not a zero sum game, or why a tourney would be as a whole, but not the individual hands.

Personally, I'm suspecting you're just trolling at this point.

Hey, guys, I just started doing some EV calculations today and ran into the same problem, so I read your posts and decided I wanted to contribute, so I made an account in the forum


What I came to realize after giving it a bit of thought is that:

1) Poker is definitely a zero sum game
2) "Poker in theory" is also a zero sum game, but the EV's don't add up to one another, they add up to the size of the pot. So if there is a heads up pot and one player has EV of +5$, it doesn't mean that the other player should have an EV of -5$, it means that player 2 should have an EV=(the size of the pot)-(other players' EV)
So if one player has EV=+5$, and pot is 10$, then the other player also has an EV of +5$. Since 5+5=10, and 5+5-10=0, then you can see how even EV is a zero sum game, if you don't forget about the pot
So not only can both players have +EV on a single street, but that can be often be the case in poker.

Let's put it into practice in the example given in the beggining of the thread with the hand 9hTh vs AA.

hero's EV is +1.75$ on the flop,
villain's EV is +3.25$ on the flop. ( it's not 6$ as OP has said, he had made a calculus mistake )
And pot is correctly 5$ (the sum of both players EV's)

Sooo, if both players have +EV, how is anybody losing any money on the hand in the long run?

Well, when we calculate the EV of a hand on a single street, we ignore the dead money one has already put into the pot. If you want to know the TOTAL EV of the hand (I just made this name, don't know if this concept already has a name), you must take into account the dead money. In this case, since pot is heads up, both players has contrubuted equally 2.50$ each in dead money.

So you take the EV of the last known action before the showdown (in this case the flop), and from that you substract the dead money contributed. And that should give you the "TOTAL EV for the hand".

In this case Hero's EV is 1.75$, his dead money are 2.50$, so his total EV for the hand is 1.75-2.50= -0.75$

Soo hero is losing 75 cents on average on that hand. Let's check out how villain is doing 3.25-2.50= 0.75

Oh, villain is winning 75 cents on average. So it is a zero sum game after all ;D

Well, what's the moral here. When calculating your EV in a vaccum for a single street, it's not the point to make it on the plus side. You have to take THE BEST of all EV lines, because that will give you the biggest chance to break even on your dead money investment.

The hand above is just a set up, in which hero ended up having the worst of it, and will go broke if it keeps appearing. But by jamming the flop hero made sure he will at least lose the smallest, if he cannot win. If he decided to fold the flop instead, that would be EV=0 for the flop (because folding is always 0), but his TOTAL EV for the Hand would be 0-2.50= -2.50$
-0.75$ is not great, but it's still better than -2.50$


It's the first time I dig so deep into this, so I would love to hear what you guys think about this and if you think it is correct or not.


P.S. I saw another discussion where people were arguing over if folding the SB has an EV of 0, or an EV of -0.5bb...

Well, you can see from what I've written that it can be put into 2 different concepts, where folding pre would have and EV of 0, and the total EV of the hand would be 0-0.5 bb= -0.5 bb

Hope it makes sense

C ya

In the actual example that got me thinking about this I estimated that a river bet had a +2.32$ EV, while a check/call had a +1.93$ EV. Didn't seem like much of a difference in the beginning, but when I figured out my dead money investment was 2.05$ I came to me that check/calling was actually a losing play, even though so very slightly less +ev!

No, imagine a situation where the two players are playing heads up and with no rake. It is still a zero sum game as even if both players are making +EV decisions, one player will eventually be up money and the other player will be down that same amount in money. Just because players are both making +EV decisions, doesn't mean at the end of their heads up match they'll both somehow magically be up. There's zero extra money added in the game. If there is rake, then it is possible for both players to be down money (if, say, one player has a tiny edge over the other player). Then, the game will not be zero sum.

I agree with the assertion that poker is a zero sum game.

However, when we actually play poker we should be thinking about profiting almost every time we put money in the pot, otherwise we should be folding; bluffcatchers and weak draws come to mind as the exceptions as these hands will earn near 0 ev as calls or raises.

Both players have AA

Each player is going to win the pot ~50% of the time

You guys are making no sense. How is poker with rake not a zero sum game?

Player 1 winnings = Player 2 losses - rake
Player 1 winnings - Player 2 losses + rake = 0

Hello?

The first definition when I google "zero sum game:"

"In game theory and economic theory, a zero-sum game is a mathematical representation of a situation in which each participant's gain (or loss) of utility is exactly balanced by the losses (or gains) of the utility of the other participant(s)."

Don't count the rake in the second equation. It's money going to a non-participant.

Player 1 doesn't win the rake. It's like there is 1 extra player that always takes money from the game so it's not 0 sum.

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You can choose to track money going in and coming out of the pot instead of treating folding as 0 EV but this is not the way you do it.

Also both methods will reach the same conclusion, they are just different methods of accounting.

I don't remember exactly how to do the method where you track all bets but I think involves tracking your stacksize before your decision and comparing it with the average stack size after the decision.

So for example let's say it's on the flop and you have 30% equity the pot is 6bb, your stack is 97bb, and your opponent bets 3bb.

Normally you could calculate your EV like this:
.3*(6+3)-.7 *(3) = .6

If you fold your EV is 0 so calling is .6bb better than folding.

You could also do this:
When you win your stack will be 97-3+12 = 106 and when you lose your stack will be 97-3 = 94.

You end up with 106 30% of the time and 94 70% of the time:

.3*106+.7*94 = 97.6

If you fold your bb stack is 97 so calling is 97.6-97 = .6 bb better than folding. The same result just different accounting methods.

Also for an explanation of why you should ignore the money you previously put into the pot look at the economic idea of sunk cost.

Basically the cost you paid to reach this point to make a decision has nothing to do with the economic decision at this moment.

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Casino poker is a minus sum game, though often beatable.

Poker games where all monies stay in the game or can be cashed out by the players are zero sum. In a zero sum situation, if I lose a dollar, another player (or players collectively) has won a dollar.
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Note that in the casino minus sum game it is possible for all players to lose over time because of the money that is drained off of the table by the house fees. In a true zero sum game this is not possible; total losses are equal to total wins.