Re: equity - ez for you to answer
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Lets say I am in a hand against 2 opponents.
I have the first guy beat 54% of the time.
I have the 2nd guy beat 38% of the time.
What is my equity against both players? (please explain)
ty in advance
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It depends on player how often player 1 will beat player 2. If player 1 will never beat player 2 your equity is 38%. This could happen if players 1 and 2 had the same ranking cards but player 2 had a flush draw. You could also be in a situation where player 1 will beat you almost everytime player 2 doesn't. You have AQ, player 1 has AK, and player 2 has QQ. If you end up beating player 2 you'll almost surely lose to player 1. This isn't a perfect example because you can win, but it's possible to have no equity in this spot.
let A = beating player 1
let B = beating player 2
P(A & B) = P(A)*P(B|A) = P(B)*P(A|B)
If A and B are independant events P(B|A) = P(B) and P(A|B) = P(A).
The problem is neither P(B|A) or P(A|B) are specified or can be determined from the given info.
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