Pocket tens post flop.

[Cobra]

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