#1
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prob of 2 pair on Omaha flop
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 |
#2
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Re: prob of 2 pair on Omaha flop
[ 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%. |
#3
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2nd question
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% . |
#4
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another correction
I changed all of the 40s to 37s to take into account our hole cards (4 was already excluded).
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#5
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Re: 2nd question
[ QUOTE ]
This is a pain since you hold a 4. Did you intend that? [/ QUOTE ] No, sorry! Thanks much for the info. |
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