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You People Suck at Playing Multi-Way Hands

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  1. #1
    spoonitnow's Avatar
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    Lightbulb You People Suck at Playing Multi-Way Hands


    Go read this week's column and figure out how to stop sucking when you're playing against multiple opponents in the same hand.
  2. #2
    Really good post.
    Erín Go Bragh
  3. #3
    Quote Originally Posted by seven-deuce View Post
    Really good post.
    +1 yep, nice one!
    Quote Originally Posted by Jay-Z
    I'm a couple hands down and I'm tryin' to get back
    I gave the other grip, I lost a flip for five stacks
  4. #4
    spoonitnow's Avatar
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    Thanks guys.
  5. #5
    Enjoyed the article, attempted to solve the question in OP.

    Assuming villains can't raise & can only call.

    If the pot is P & our bet size is aP, the two players are called A & B.

    Options
    1) A & B fold
    2) A calls B folds
    3) A folds B calls
    4) A & B call

    EV of each outcome
    1) P
    2) (0.55)(a+1)P - (0.45)aP
    3) (0.55)(a+1)P - (0.45)aP
    4) (0.35)(2a+1)P - (0.65)aP

    If we say 1 happens W%, 2 X%, 3 Y%, 4 Z% and we know W + X + Y + Z = 1 and each is >= 0

    W * P + X * [(0.55)(a+1)P - (0.45)aP] + Y * [(0.55)(a+1)P - (0.45)aP] + Z[(0.35)(2a+1)P - (0.65)aP]

    WP + [X+Y][(0.55)(a+1)P - (0.45)aP] + Z[(0.35)(2a+1)P - (0.65)aP]

    WP + [X+Y][0.1aP + 0.55P] + Z[0.05aP + 0.35P]

    As every variable is > 0 by definition this bet will always be profitable.
    Last edited by Savy; 03-16-2015 at 10:27 PM.
  6. #6
    spoonitnow's Avatar
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    Quote Originally Posted by ImSavy View Post
    Enjoyed the article, attempted to solve the question in OP.

    Assuming villains can't raise & can only call.

    If the pot is P & our bet size is aP, the two players are called A & B.

    Options
    1) A & B fold
    2) A calls B folds
    3) A folds B calls
    4) A & B call

    EV of each outcome
    1) P
    2) (0.55)(a+1)P - (0.45)aP
    3) (0.55)(a+1)P - (0.45)aP
    4) (0.35)(2a+1)P - (0.65)aP

    If we say 1 happens W%, 2 X%, 3 Y%, 4 Z% and we know W + X + Y + Z = 1 and each is >= 0

    W * P + X * [(0.55)(a+1)P - (0.45)aP] + Y * [(0.55)(a+1)P - (0.45)aP] + Z[(0.35)(2a+1)P - (0.65)aP]

    WP + [X+Y][(0.55)(a+1)P - (0.45)aP] + Z[(0.35)(2a+1)P - (0.65)aP]

    WP + [X+Y][0.1aP + 0.55P] + Z[0.05aP + 0.35P]

    As every variable is > 0 by definition this bet will always be profitable.
    I can very much appreciate your enthusiasm, but for the sake of other players who aren't as inclined for math, I'll say that this is largely overkill. We just need >50 percent against one call and >33 percent against two calls to prove it, etc., like you learn in the article.

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