prob of 2 pair on Omaha flop

[gergery]

If I have, say, A234 in Omaha, what is the probability I will get 2-pair on the flop?

If the flop comes 468 what is the chance an opponent has two pair?

--Greg

[BruceZ]

[ QUOTE ]
If I have, say, A234 in Omaha, what is the probability I will get 2-pair on the flop?

[/ QUOTE ]

[C(4,2)*3*3*36 + C(4,3)*3^3 + 2*C(4,2)*C(3,2)*C(3,1)] / C(48,3)

=~ 12.5%

The first term is for only 2 flop cards to be A,2,3, or 4. There are C(4,2) = 6 ways to choose the 2 ranks, times 3 ways to pick each card, times 36 other cards that are not an A,2,3,4. The second term is for matching 3 of your 4 cards, so there are C(4,3) = 4 ways to choose the ranks, times 3 ways to pick each card. The last term is for a pair on the flop. There are C(4,2) = 6 ways to choose the 2 ranks, times 2 ways to choose the rank that pairs, times C(3,2) = 3 ways to choose the 2 cards that pair, times C(3,1) = 3 ways to choose the card that does not pair.

EDIT: The 3rd term is actually called a full-house, so if you didn't want to include this, remove the 3rd term to get 11.9%.

[BruceZ]

My earlier attempt at the 2nd question forgot that we were still playing Omaha.

[ QUOTE ]
If the flop comes 468 what is the chance an opponent has two pair?

[/ QUOTE ]

This is a pain since you hold a 4. Did you intend that? Anyway, assuming you have A234 (or any hand with exactly 1 card in common with 4,6, or 8) and assuming random holdings:

68 + two non-4,6,8: 3*3*C(37,2) = 5994

46 or 48 + two non-4,6,8: 2*2*3*C(37,2) = 7992

668 or 886 + one non-4,6,8: 2*C(3,2)*C(3,1)*37 = 666

664 or 884 + one non-4,6,8: 2*C(3,2)*C(2,1)*37 = 444

446 or 448 + one non-4,6,8: 2*C(2,2)*C(3,1)*37 = 222

6688: C(3,2)*C(3,2) = 9

4466 or 4488: 2*C(2,2)*C(3,2) = 6

6888 or 6668: 2*C(3,1)*C(3,3) = 6

4666 or 4888: 2*C(2,1)*C(3,3) = 4

468 + one non-4,6,8: C(2,1)*C(3,1)*C(3,1)*37 = 666

4668 or 4688: 2*C(2,1)*C(3,2)*C(3,1) = 36

4468: C(2,2)*C(3,1)*C(3,1) = 9

If I didn't miss any, that's a total of 16,054 out of C(45,4) possible hands, or about 10.8%.

If we ignore your hole cards, the number of ways a player can have 2-pair on a flop with 3 different ranks is:

a,b,c mean some flop card
x means non-flop card

abxx: C(3,2)*3*3*C(40,2) = 21,060
aabx: C(3,2)*2*C(3,2)*C(3,1)*40 = 2160
aabb: C(3,2)*C(3,2)*C(3,2) = 27
aaab: C(3,2)*2*C(3,3)*C(3,1) = 18
abcx: C(3,1)*C(3,1)*C(3,1)*40 = 1080
abcc: 3*C(3,1)*C(3,1)*C(3,2) = 81
------------------------------------------------
24426 / C(49,4) = 11.5% .

[BruceZ]

I changed all of the 40s to 37s to take into account our hole cards (4 was already excluded).

[gergery]

[ QUOTE ]

This is a pain since you hold a 4. Did you intend that?

[/ QUOTE ]

No, sorry!

Thanks much for the info.