View Full Version : ace duece in omaha/8

01-28-2003, 04:22 AM
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.

01-31-2003, 10:23 AM
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.