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Courchevel Question Courchevel Question

09-19-2018 , 10:33 PM
This is 5 card omaha where one of the community cards are exposed before preflop betting.

In a six handed game what is probability someone has a set? Thanks
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09-20-2018 , 01:29 AM
I think you are asking what is the prob someone has a set using only the "spit" card (first community card, dealt face up "pre-flop"), but you could be asking something else.

Could you clarify your question?
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09-20-2018 , 01:56 AM
Guessing that is what you mean, here is what I got:

Clearly only 1 player can make a set using the "Spit" card and two of the same rank from their 5 hole cards. So P6=6*P1 in this case.

P1 = [C(3,2)*C(48,3) + C(3,3)*C(48,2)] / C(51,5)

= 53,016 / 2,349,060

= 2.2569%

So P6

= 6 * 53,016 / 2,349,060

= 318,096 / 2,349,060

= 13.5414%

If that is the question you were asking, someone will probably confirm or deny this answer.

If you are not asking that question, let us know what you are asking.
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09-20-2018 , 06:19 PM
Yes thanks just using the spit card

Wow that seems higher than I would have thought.

Although it seems to make sense. Without removal everyone has 3.25% to have any of the pairs. But the removal makes it harder, so number seems reasonable.
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09-20-2018 , 07:54 PM
Using PIE: 6[C(5,2)*3/C(51,2) - 2*C(5,3)/C(51,3)] = same answer
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09-21-2018 , 12:55 AM
Thanks gents!
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09-21-2018 , 01:37 AM
[(3/51 x 2/50 x 48/49 x 47/48 x 46/47 x 10) + (3/51 x 2/50 x 1/49 x 48/48 x 47/47 x 10)] x 6
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