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Assuming you open raised to 4XBB and there are six people left to act I will determine the probability that one or more of those six people were dealt a problem hand for you. The probability that they played that problem hand I will leave up to you.
Problem Hands and Combinations
AA - 6
KK - 6
QQ - 6
JJ - 6
99 - 1
22 - 6
A9 - 8
T9 - 4
98 - 8
I assume they would not play K9, J9, or 97.
I also assumed that they would have raised with AA, KK, QQ. So I eliminated them from the calculation.
You sum the hand combinations and divide by the total possible hand combinations.
=33/(50c2)*6 = 16.2%
The second term is difficult to explain but it comes out to .008%. So the probability that one or more of the six remaining opponents was dealt a problem hand is
16.2 - .008 = 15.3% or 1 in 6.5 times
Cobra
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Some comments:
1. Since there is a 2 on the flop, there are only 3 ways to hold 22, and so there are only 30 problem hands, not 33. This would change the second term calculations, as well as the first term.
2. Since we know the flop cards, there are only C(47,2) possible hands, not C(50,2).
3. There are actually much less than C(47,2) possible hands, since only a fraction of these will be played. It doesn't make sense to carefully consider what problem hands your opponents might play, and then assume that they are holding random hands. The inclusion-exclusion analysis makes sense pre-flop before we have any other information, but not post-flop.
To Paizzon: He is using the
inclusion-exclusion principle, so you must thoroughly understand that first. I had written out an explanation of the terms before I realized that the method was used inappropriately in this case due to point 3 above.