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stokken
09-20-2005, 02:50 AM
There has been alot of posts about AA vs KK and the likes.
I have searched, but not found quite what I am looking for.

And what I am looking for is maybe a bit extensive?

The probability for a spesific PP is 1/52 x 3/51 and if 9 handed you multiply by 9 for the likelyhood that one hand contains this spesific pair.Since there are 9 occasions for this particular event to take place.
How do u go about calculatng the probability of running into an overpair given your hand is XX and n is the number of possible overpairs( with this I am thinking along this line I have JJ, n=3) and m is the number of opponents yet to act?

BruceZ
09-20-2005, 03:08 AM
[ QUOTE ]
The probability for a spesific PP is 1/52 x 3/51 and if 9 handed you multiply by 9 for the likelyhood that one hand contains this spesific pair.Since there are 9 occasions for this particular event to take place.

[/ QUOTE ]

Multiplying by 9 is an approximation, and this double counts the times that 2 players have this pair. For the exact answer, you need 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=), which is also what you need to answer your question, but in this case that just involves subtracting the times that 2 players have this pair, which is C(9,2)/C(50,4).

[ QUOTE ]
How do u go about calculatng the probability of running into an overpair given your hand is XX and n is the number of possible overpairs( with this I am thinking along this line I have JJ, n=3) and m is the number of opponents yet to act?

[/ QUOTE ]

Use the inclusion-exclusion principle. The first term will be 9*n*6/C(50,2), since there are n*6 overpairs, the probability that a particular opponent has an overpair is n*6/C(50,2), and there are 9 opponents. This term alone is usually quite accurate, and you can usually get an accuracy better than 0.1% with 2 or 3 terms.