2nd question
My earlier attempt at the 2nd question forgot that we were still playing Omaha.
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If the flop comes 468 what is the chance an opponent has two pair?
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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
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24426 / C(49,4) = 11.5% .
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