Nerdy math question here

So lets say you're preflop or on the flop or whatever and your opponent puts you all in and hand ranges says you should have about 40% equity if you call and you're getting the right pot odds to call off. I understand this is +EV for you and logic would tell me that this hand would therefore be -EV in the same amount minus rake of course for the other player but the other player has gotten all the money in middle with 60% equity which is obviously a +EV play long term.

Is it possible for 2 players to be +EV in the same hand?

At certain decision points in certain cases, it's possible for everyone in the hand to be +EV due to the money currently in the pot and odds they are getting.

Toy example: All 10 players have $200 stacks. There's a raise to $199 preflop, and everyone calls. One guy then shoves $1 on the flop into the $1990 pot. It's very likely that all 10 players have a +EV call at this point.

There's of course more realistic examples. This happens all the time in Limit, especially on the flop, where the best hand and a good draw (who are both +EV) raise and reraise each other and drag along a few other players (who are all -EV).

GcluelessEVnoobG

Yes. If there is already money in the pot, it is very easy for two players to both be +EV.

True. But in OP's example, someone made a sizing error.

Sort of depends on how you want to define it, but strictly speaking no, you cannot take an action that is +EV for multiple people.

If we take an action, you calling a bet, we are comparing it to some other alternative (usually folding). In that case, your equity is generally reported as your eventual wins or losses compared to the alternative(s).

So in the case where you have 40% equity as an example, you might expect to make $10 by calling. That means that by calling, on average the other person is losing $10 compared to you folding. Each action and alternative must equal up to $0 total. Becasue the amount of money in the total system does not change.

It's important to note that there is a distinct difference between the questions of:
'Is it possible for two (or more) people to meke +EV decisions in a hand (or on a given street)'

and

'Is it possible for one decision to be +EV for two (or more) people'

The first case is an easy yes, and what people are referring to. The second case is a always no heads up, although it gets weird if you have multiple people in a hand and the person with the most equity folds and raises two other peoples equity as a result. Not common, but it happens.

Not sure I get that IRTM?

Let's say pot is $100 on the turn. Mr. TP bets $20 offering Mr. Flush Draw 6:1 to chase. The single decision (betting by Mr. TP / calling by Mr. Flush Draw) are +EV for both of them.

Gmaybelistanexampletoillustrate?G

The correct way to think about this, IMO, is hand vs range.

So yes, multiple players could be +EV.

Consider V1 moves in not with one hand, but with a range.

V2 calls vs V1 with his hand vs V1's range.

V3 calls vs V1's and V2's combined ranges and can still be +EV.

The sum of player EV’s is always the pot before last bets are made assuming rake=0. If Player A has EV = $10 when he bets into a $40 pot and player B closes the river action with a call, the EV of player B is $30.

Of course, assuming no chopped pot, the actual payout will be the total pot to only one of the players. But EV is an average so both players can expect profit though only one will get it.

It stands to reason that two players can only TRULY be +EV in the same pot if there is money from other players in the pot. In a heads-up game, if you play a scenario out a million times (range vs range), only one of the 2 players will end up on top. (or chop) With other players' money in the pot, you can find scenarios where the 2 players in question can both be ahead after a million iterations of a given scenario.

With limited information and fallible range construction (AKA our lives), it can appear that 2 players heads up are +EV. You may each be correct to form the +EV estimation from where you stand, but the reality is that one of you is less accurate than the other, which will show itself over a million iterations of the same scenario.

I suggest you read the second paragraph of the quoted posting one more time.

Also, recall that EV is a statistical average. If you call a fair coin toss correctly you win $2 or else you lose $1. Your EV is 50 cents. Yet not once in a million tosses will you collect 50 cents.

I don't think that's right. I've always understood EV = (W%*$W) – (L%*$L). Therefore, the EV in your coin flip scenario is $0. We expect to win $0. If the coin is weighted such that heads wins 75% of the time and tails wins 25%, and we always bet heads, then our EV is $0.50, while the player betting tails has an -EV (-$0.50)

That is correct. A fair coin toss with both players wagering the same amount will have an EV of 0. In statman's example above, he was either talking about unbalance bets, or (more likely) forgot that one of those dollars you'd win was already yours, so you'll actually win $1 from your V, or lose $1 to your V.

Still, he's right that you'll never actually collect your EV number.

IRTM pretty much summed it up. I think the continuing confusion is between is the blurring between multiple sequential actions and one action, combined with equating +EV with maximum +EV.

GG, to go back to your flush example. The bet of $20 is +EV for player 1. He is putting money in good against player 2. For player 2, his maximum EV declined with the bet since his maximum EV would have been generated by a check. Player 2's EV goes up when he calls because he can hit his hand with pot odds. Player 1's EV goes down since the best solution is for player 2 to fold.

However, the maximum EV for player 1 is to bet as much as player 2 would call.

I really know how to calculate EV; Really. For the record, I thought it was clear that this was a bet favorable to one player where if he guesses right he wins $2 and if he guesses wrong, he loses $1.

By fair coin toss, I meant the coin was fair, 50/50, not that the bet was fair.

Sorry. I see people do EV calcs using pot odds wrong by adding their portion of the pot into the potential winnings side of the ratio all the time, so I just assumed it was more likely an oversight.

It's not just plus and minus ev, it's how plus and how minus ev. In limit, just because your opponent *should* call your bet with his draw doesn't mean you shouldn't bet. Charging him x is better for you and worse for him than charging him zero.