Re: raising with draws - how much FE needed?
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I don’t know if this is any easier but this is what I came up with a while back to calculate required fold equity.
Called = total pot if called.
W = Stack if (called * Equity) - Initial Stack
Fold % required = (W) / (W - Current Pot Size)
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I'm not sure I understand. It seems to me that the numerator is bigger than the denominator here, so you would require that the opponent folds more than 100% of the time????
I might just be misinterpreting your formula. Can you give an example of how you would use this formula to calculate required FE?
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Sure.
You’re stack size is 150 (opponent has you covered).
Pot size is 50.
You’re equity if called is 20%.
W = ((150 + 50 + 150) * .20) – 150
W = (350 * .2) – 150
W = -80
-80 / (-80 ) – 50 = -80 / -130 = 0.6153846
You need him to fold 62% of the time for this to be chip EV neutral.
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