#1
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ace duece in omaha/8
Can anybody help me with this probability problem. I am trying to figure out the probability that if you hold a2xx you will end up with the nut low. i know there is no low about 42.5% of the time and i think you get counterfieted 42.5% of the time. And 1/3 of the time somebody else will also have a2 (assuming low limit loose game with 8 players). My question is if you only call the flop if you flop a low or 4 to a low, what percentage of the time will you end up with the nut low? i have no simulator and was wondering how playing only hands that include a2 would fair in a loose 3/6 game.
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#2
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Re: ace duece in omaha/8
I think you will be eaten up by the blinds if you wait for A2 to play a hand.
The chance of not getting at least one Ace is: 48/52 * 47/51 * 46/50 * 45/49 = 0.72 So 28% of the time at least one of your cards is an Ace. The chance of not getting at least one 2 in your other three cards is: 47/51 * 46/50 * 45/49 = 0.78 So 22% of the time that you got an Ace you will also get a deuce. That works out to getting an A2 about 6.2% of the time or 1 out of every 15 hands. |
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