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View Full Version : Help w/ probability calculation regarding flopping the nuts

J.Copperthite
04-14-2005, 07:19 PM
I play a lot of Omaha H/L, and have come to grips with the volatile nature of the nuts in that game. In Hold 'em, if you flop the nut straight, while it won't be the technical nuts by the river a good % of the time, it will hold up quite a few times even when it isn't the nuts. In Omaha, when the nuts change, usually you are drawing dead if you don't have a redraw.

I'm interested in calculating the odds that the best hand on the flop will stay the best hand by the river. Say I flop Broadway in Omaha but I have no draw to a flush or full house on the flop. If the flop is Q /images/graemlins/club.gifJ /images/graemlins/club.gifT /images/graemlins/diamond.gif, and I have the A /images/graemlins/spade.gifK /images/graemlins/heart.gif in my hand, there are 9 cards that pair the board on the turn, and 9 clubs (or is it 11? I'm wondering because in order for a flush to be possible, at least two clubs must be in someone elses hand) to make a possible flush, that means I am looking at dodging 18 cards twice (assuming my two side cards do not make pairs or are clubs).

p = (18/45) + (18/44) = .81

Does this mean that, if my calculations are fundamentally correct, that my nut straight will not be the nuts by the river 81% of the time, given the 2 of a suit nature of the flop? Also, how do I account for a runner-runner situation, such as two straight diamonds coming out on the turn and river. I'm new to this kind of probability, and I would be interested to see if I my mathematical thinking is correct so far. Thank you for all your help in advance.

- Jeff

gonepunting
04-14-2005, 07:43 PM
The chance that you have the nuts on the turn means dodging 9 pairing cards and 10 clubs (not counting the club ten). Whether someone else holds two clubs or not is not part of the problem. That would chance of your having second nuts but still being in the lead.
On the river you need to avoid 12 pairing cards and 9 clubs.