I understand now. It would be ok to use a -Pot in the EV equation?
W%*W$ - L%*L$ - Pot? To make it easier to see exactly how much EV I have (or I'm giving) when I make the bet?
e.g. Played this hand few days ago
MP 0,20 9
7
($6,97)
CO Fold
Hero Call 8
8
($10)
SB Fold
BB Call
Flop K
T
8
BB check, MP bet 0,47, Hero raises to 1,55, BB fold, MP call
Turn 3
(3,64)
MP check, Hero bets 2,75, MP calls (leaving 2,47 behind)
He has 30% pot odds and 18,18% equity. His implied odds are 5,99 (2,75/(2,75+2,75+3,64+X)=0,1818
So, his EV, having 2,47 left, is: EV = -0,64 $, correct? (0,1818*(2,75+3,64+2,47))-(0,8182*2,75)
EV for the bettor is EV = 4,28$
If his remaining stack was exactly 5,99, his EV would be exactly 0.
If his remaining stack was exactly 5,99, the EV for the bettor would be exactly the pot, like
statmanhal showed:
EV = (0,8182*(2,75+3,64))-(0,1818*(2,75+5,99))
EV = 3,64$ = The pot OTT before hero making the bet.
So this is the equilibrium point, right?
But can I add the -Pot on the Bettor EV equation?
EV for his actual remaining stack of 2,47
EV = (0,8182*(2,75+3,64))-(0,1818*(2,75+2,47))
EV = 4,28
If I subtract the pot before making the bet
EV = 4,28 - 3,64
EV = 0,64.
So the EV for the bettor is
+0,64 and EV for the caller is
-0,64. The EV numbers are always equal, just that one is positive and other is negative, unless both are exactly zero. That's beautiful lol.
But my EV after I bet on the turn is only 0,64 cents, thought it would be higher than this (I know that EV is higher if included pot before betting)... Still, a better result than showing OTT and never getting called, since by betting and getting called I'm winning +0,64
Now I want to go deeper on it, will try to add reverse implied odds to it.