EV diff

    • TiciBoy
      Joined: 13.01.2010 Posts: 1,235
      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:diamond: J:diamond:

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

      FLOP ($0.90) 6:spade: Q:spade: 9:heart:

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

      TURN ($2.08) 6:spade: Q:spade: 9:heart: 8:club:

      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:spade: Q:spade: 9:heart: 8:club: T:heart:

      SB shows 6:heart: 6:diamond:
      (Pre 47%, Flop 74.1%, Turn 22.7%)

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

      Hero wins $18.76
  • 2 replies
    • TiciBoy
      Joined: 13.01.2010 Posts: 1,235
      I think I got it, so nwm... It means that this time I won 20$ (-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
      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.