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10-06-2005, 08:13 AM
Hey guys,
I am holding pocket tens. I raise 4x BB. 2 reasonably aggressive players come along for the ride.
The flop comes 9c 9s 2h.
Seeing as my poocket tens are an overpair to the flop, what is the probability I have the best hand at this point. e.g. no one has overpairs to me, a nine or pocket twos.

my situation was fairly desperate (should have probably moved all in with pocket tens but felt i needed a caller)
After flop I was second to act, first guy checked, If i was going to make a move it had to be all in, so I did. I then fell to the man on my left who had A 9.
Was I justified in my actions?
Any math or reply is greatly appreciated.

Cobra
10-06-2005, 10:28 AM
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

Paizzon
10-06-2005, 10:42 PM
Cobra, could you please explain the different variables in your formula that came up to 16.2%? Also, if you have the time, how you came up with .008%.

I love math and I am just starting to study the different advanced math for poker. Thanks.

Paizzon

BruceZ
10-07-2005, 12:25 AM
[ QUOTE ]
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

[/ QUOTE ]

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 (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=&amp;Number=417383&amp;page=&amp;v iew=&amp;sb=5&amp;o=&amp;fpart=), 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.

Cobra
10-07-2005, 10:44 AM
Sloppy work by me Bruce. Thanks for pointing out the mistakes.

Cobra