[mosta]

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Omaha hi/lo, 9 handed game.

The board reads:

K [img]/images/graemlins/diamond.gif[/img] 5 [img]/images/graemlins/heart.gif[/img] 3 [img]/images/graemlins/spade.gif[/img] Q [img]/images/graemlins/diamond.gif[/img]

Your hand :

A [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/diamond.gif[/img] 4 [img]/images/graemlins/diamond.gif[/img] 6 [img]/images/graemlins/club.gif[/img]


Assuming that the villain holds an A2 combination in his four card hand, what is the probability that he was dealt a higher flush draw?

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[deleted]

let me try it this way: he has an ace, a deuce, and two cards. if any 2 or them are suited in diamonds, he has a higher flush draw than yours. his deuce can not be a diamond. so the question is of his ace and two other cards, how many ways can they be suited in diamonds, versus not.

his total number of possible hands is 1 of 3 aces, 1 of 3 deuces, and 2 from the 38 other unknowns (if I've counted the knowns right: 4 on board, 4 my hand, 3 aces, 3 deuces). so his # of possible hands is
3C1 * 3C1 * 38C2 = 6,327. (or 3*3*703)

I think we have to proceed by cases (can't do the one-minus trick) because we have to consider 2 or 3 diamonds versus 0 or 1 diamonds. I'll count the suited ways b/c 3 diamonds is easy:

he can have 3 diamonds w/ Ad & non-d deuce:
1 * 3 * 8 * 7 = 168
he can have 2 diamonds w/ Ad and non-d deuce:
1 * 3 * 8 * 30 = 720
or by non-d ace and deuce and 2 other diamonds:
2 * 3 * 8 * 7= 336

so the chance he is suited in diamonds is:

(168+720+336)/6327= 19.34%

good problem.