[Closed] EV diff

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Last Post by beefpuff 15 years ago

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_Anonymous_

04.03.2011, 17:37 0

Had this hand in my last session and it sais that the EV difference is -4.26. Does that mean that I am expected to loose 4.26$ on average if I play same hand like this everytime? :f_confused:

Grabbed by Holdem Manager
NL Holdem $0.10(BB) Replayer
SB ($22.20)
BB ($9.58)
UTG ($11.63)
UTG+1 ($4.05)
Hero ($10)
BTN ($2.76)

Dealt to Hero T♦ J♦

fold, fold, Hero raises to $0.40, fold, SB calls $0.35, fold

FLOP ($0.90) 6♠ Q♠ 9♥

SB checks, Hero bets $0.59, SB calls $0.59

TURN ($2.08) 6♠ Q♠ 9♥ 8♣

SB checks, Hero bets $1.37, SB raises to $6.19, Hero raises to $9.01 (AI), SB calls $2.82

RIVER ($20.10) 6♠ Q♠ 9♥ 8♣ T♥

SB shows 6♥ 6♦
(Pre 47%, Flop 74.1%, Turn 22.7%)

Hero shows T♦ J♦
(Pre 53%, Flop 25.9%, Turn 77.3%)

Hero wins $18.76

2 replies

_Anonymous_

04.03.2011, 17:45 0

I think I got it, so nwm... It means that this time I won 20$ (-rake), but on average I would only win around 16$ (because he will hit his FH). Am I correct?

beefpuff

Joined: 15.03.2009

Posts: 73

04.03.2011, 17:59 0

Yes

I'm pretty sure the calculation is:

EV Difference = (Amount won) - (equity*Pot Size)

So unless someone has 100% equity when going in before the river (other guy(s) are drawing dead), the winner of the pot will always win more than EV expected. If you lose a pot and weren't drawing dead before going all-in you will have lost more than EV expected.