Was curious to know whether EV takes into account the pot size prior to an all in and call.

For example.

Player 1 has K K
Player 2 has 9 9

Turn: ($990.00, 2 players) 9 5 Q 2
Player 1 bets $5.00 and is all in, [color=red][b]Player 2/b] calls $5.00 and is all in

River: ($1000.00, 2 players) Q

Player 1 mucks K K (Two Pair, Kings and Queens)
(Turn 5%)
Player 2 shows 9 9 (Full House, Nines full of Queens)
(Turn 95%)
Player 2 wins $1000.00

Will Player 1 EV be -$950 and Player 2 EV be +$950?

Or does it take into account and exclude the pot prior to the all in bet?
Player 1 (turn pot) $10 x 5% = $0.50.
Player 1 (turn pot) $10 x 95% = $9.50.

So

Player 1 total EV is $990 + $0.50 = $990.50
Player 2 total EV is $990 + $9.50 = $999.50

Obviously this would not balance as both players cannot be expected to win more than the actual pot. But Player 1 has positive EV by calling the turn bet of $5 to win a $1000 pot.

The EV equation, in general, for an all-in lead bet that is called, is

EV = Pr(win)*$Win – Pr(lose)$Lose

In this case, for Player 1, Pr(win) = 5%, $Win = $995, $Lose = $5, so

EV1 = 0.05*995 – 0.95*5 = 45

Clearly, for a chance to win a huge pot with a minimal investment, you always play unless you know you are drawing dead.

In any pot, the sum of all player EV's is equal to the pot before final bets are made. Therefore Player 2 EV = 990-45=945. This can be seen from his EV equation:

EV2 =0.95*995-0.05*5 = 945

Thanks for your response.

Why does the total EV for both players total $990 when there is $1000 to be won?

Also, must we subtract the $ invested in in order to calculate net EV?

EV1 = -$455
So, by making this call OTT, Player 1 is -$455 EV rather than -$495

EV2 = +$445
If Player 1 folds OTT, Player 2 is +$495 EV

*****not counting rake.

I take it that you either do not understand my EV analysis or do not accept it. The above response makes no sense IMO.

You can define EV at a decision point as the expected or average change in stack size as a result of the decision. In your example, the decision point for Player 1 is going all-in for $5 when the pot is $990 and his winning chance is 5%. So, his baseline stack size at the decision point is $5. If he were faced with the identical situation 100 times, he would win 5 times and lose 95 times. For each of the 5 winners, his ending stack would be $1000, an increase of $995 from his baseline stack of $5. For each of the 95 losses, his ending stack would be 0 for a -$5 change from his baseline stack. Thus, the average change in stack size as a result of the all-in bet, or EV, is

(5*995 +95*(-5))/100 = $45.

Brokenstars comment on including rake as part of the EV analysis has some rationale, but since rake varies by game and site, in a Poker Theory forum, it is difficult to conceptualize except in a general way. For low stakes, the rake is significant so I guess one could say that there are cases when EV is done that you might make a different decision with including rake than that with not including it. Perhaps, some posters can provide specific examples.

Now understood. Grateful for you taking your time to write detailed responses..

The only amount of money we are losing is our $5 call. Sure, we’ve already put money in to the pot, but don’t think of that as “belonging” to us anymore. It now belongs to the pot, so it’s not something we can lose.

Back when I was at 5nl guy limped in btn and it checked to me in BB. We're just heads up for a 12c pot or whatever and board runs out AKQJT and he ships it for 100bb+. So.... I folded... cuz... rake...

Ha ha. How to lose by winning.