Since poker is a game of decisions based on estimations of ranges and equity, players in the same hand can both make a +ev decision?

Not really. When you make the right move based on ranges etc you have made a long term strategic +ev decision. But the way score is kept is based on the actual hands, not the ranges.

Ya, thats why I said in theory. It's just something I never really thought about.

A zero sum game is one where one player's winnings is equal to the other player's losses. Poker is not a zero sum game because of rake and fees, not because both players can have a +EV decision.

But home games don't have rakes and fees. Those aren't part of poker, they are part of commercial enterprise.

Well, in some of the home games I played, the 'rake' for stale chips and warm beer makes if far from a zero-sum game.

definitely

zero sum just means money gets shifted from one person to the other. Your loss is someones gain and vice versa.

Two players HU can both be +EV (or both -EV) if the decision is based on ranges. Turn the cards face up and clearly one is +EV and one is -EV. This makes it zero sum, if there is no rake or fee. Tasty snacks and good beer makes any social home game +EV imo.

There are several errors in this quote IMO. If it is not showdown, whether cards are turned up or not, both players can be +EV at this point for there are still unknown cards to come. The +EV comes about because of the dead money in the pot - the bets made in previous rounds and by those who have folded. Also, two players cannot be both -EV. If they were, who gets the money?

I'll agree with the good snacks and beer statement.

Cool! Give us one example of both players being +EV with the cards turned up.

This is wrong. If there is no rake or fees, poker is a zero sum game, period.

Where in the quoted statement did I say otherwise? The statement you quoted is only referring to how EV may be 'distributed' and that has nothing to do with zero-sum.

If there are n players in a game, it is zero sum only if the sum of all the winnings and losing is zero, period.

Then give us an example of both players being +EV with the cards turned up.

I'm sure there are lots, as long as it is before the river. In terms of EV if cards are flipped over before the river, both players can both make a +EV decision. Lets say the flop is 8h7h3d, the pot is 5$, both players have 5$ behind. hero asks what villain has, villain shows 9hTh for a OESFD, Hero shoves and shows AA no heart. Villain calls. Hero's equity (winning %) is 45% and villain's equity is 55% on the flop.

Hero's ev = ((pot+bet)*heroequity)-((bet*villainequity))
=((5+5)*.45) - (5*.55)
Hero's EV = 1.75$
So, hero stands to win the 5$ currently in the pot + the 5$ call from villain (for simplicity sake, hero knows he has no FE), hero stands to win 10$ 45% of the time and loses his 5$ bet, 55% of the time

Villain's ev = (pot*villainequity)-(call*hero equity)
= (10*.55) - (5*.45)
Villain's EV = 6$
Villain's stands to win 10$ 55% of the time and loses his 5$ call, 45% of the time

I don't think two players in the same hand can make a -ev decision, with cards face up or down. But as you can see both players can make a +ev decision on a given street, although villain stands to gain alot more, both players are profitable in the long run because of the dead money in the pot along with equity of future streets

What does turning the cards up have to do with it? The above example will work but it can be shown a lot simpler. Think of a pot of $100 and each player has $1 after the flop. So ,assuming neither player is drawing dead, they are each getting 100 to 1 odds to stay in the hand. With at least one out for each at that point, they clearly are both +EV.

It's heads up. That's not free money in the middle. Run the hand over and over and they just take each others money. Whoever is +EV is statistically more likely to win in the long run. The opponent loses the same amount. EV is not just about odds and one player's outs here. They can't both be plus EV because it's zero sum. They might be "pot committed" because it's cheap to call but this not the same as "+EV".

You're right, you didn't really say that, sorry. However, saying that both players can be "+EV" needs a little clarification though. Yes, two players can be +EV for certain bets on later streets, but when the hands are originally dealt, given how each player is going to play those hands, one player is +EV and one is -EV, or they're going to break even.

You are right, Blizzuff, it does need clarification. I'm not disagreeing with statmanhal - of course both players are in for their last $1 with $100 in the middle even if one thinks he might be drawing dead. But the situation here is zero sum given it's heads up and if you run it again an again only one player can ever come out ahead - and the one that is more likely to do so is "+EV".

Let's step back a second. OP started the thread wondering how poker can be a zero-sum game and yet two players can each have a +EV decision.

Facts:

1. If there were no rake or fees, poker is a zero sum game. The sum of all players winnings and losses over a series of games would be zero.

2. In any single hand, heads up or not, it is possible for 2 or more players to have +EV at a specific betting decision point. In heads ups or not, there is dead or free money that players put in the pot on previous rounds. This money is why more than one player can have +EV

1 and 2 are not inconsistent as I tried to say earlier. EV and zero sum are two different concepts. EV is a mathematical expectation. If you are flipping a fair coin and win $1.10 with a head and pay $1.00 if a tail, your EV is +$0.05. Yet, you will never win 5 cents - it will always be win $1.10 or lose $1.00. Zero sum on the other hand is a measure of actual winnings and losses.
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This is another definition problem.

Both players can not have positive expectation at any given point in time. However, both players can be faced with a position where the *relative* EV of each choice means that calling is better than folding (i.e. they have negative expectation either way, but folding is more negative than calling)

Assume a simple poker game of hold-em with just one board card and each player bets $1 or folds prior to the 'flop'. Big Bucks puts in $10 in the pot each hand just to watch two players play. Suppose A gets AA and B gets KK. At this point, deciding whether to fold or not, we have for A and B:

EVa = 0.8*11-0.2*1 > 0
EVb =0.2*11 -0.8*1 > 0

Both players have a +EV decision – not relative to each other. These are the average positive amounts they would win if the same situation came up a large number of times.

Note that the sum of the EV's is 10. If we consider Big Bucks a player, then his EV = -10 so all EV's sum to 0, consistent with a zero-sum game.

Yes it is a definition problem. The confusion is a result of a simultaneaty fallacy. For example, we are on the turn heads up. Player A can bet, check or (theoretically) fold. There is an expectation associated with each decision. Player B acts after player A has made his or her decision, not at the same time. Both players can make a +EV (and it is relative) decision. But if you compare their EV at any give time, it is based on all of their options - not the assumption that they will both perform a predetermined action.

If we consider Big Bucks a player then we have 3 players, so of course 2 can be +EV and 1 -EV. It's a multiway pot.

Zero-sum games are a special case of constant-sum games, in which choices by players can neither increase nor decrease the available resources. In zero-sum games the total benefit to all players in the game, for every combination of strategies, always adds to zero (more informally, a player benefits only at the equal expense of others). Poker exemplifies a zero-sum game (ignoring the possibility of the house's cut), because one wins exactly the amount one's opponents lose. Other zero-sum games include matching pennies and most classical board games including Go and chess.

Many games studied by game theorists (including the famous prisoner's dilemma) are non-zero-sum games, because some outcomes have net results greater or less than zero. Informally, in non-zero-sum games, a gain by one player does not necessarily correspond with a loss by another.

Constant-sum games correspond to activities like theft and gambling, but not to the fundamental economic situation in which there are potential gains from trade. It is possible to transform any game into a (possibly asymmetric) zero-sum game by adding an additional dummy player (often called "the board"), whose losses compensate the players' net winnings.