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krishanleong
01-13-2005, 02:43 PM
Assume your opponent plays 4% of his hands. What hand is the breaking point for having 50% equity against his likely hands. How about 8%, 12% and 16%.

Also, if a tight raiser will only raise with AA-JJ and AKos, what is the probability that he has a pair?

Krishan

Cobra
01-13-2005, 03:37 PM
I don't know the first answer to the first question, but the second is easy. There are 6 ways of getting any pair and 4 pair possibilities. There are 16 ways of combining AK into a hand. Thats a total of 40 combinations with 24 being pairs. Therefore if he raises the probability of a pair is 24/40 = 60%.

Lost Wages
01-13-2005, 03:41 PM
Assume your opponent plays 4% of his hands. What hand is the breaking point for having 50% equity against his likely hands. How about 8%, 12% and 16%.

You would have to put him on a range of hands. Then use PokerStove.

Also, if a tight raiser will only raise with AA-JJ and AKos, what is the probability that he has a pair?

You can eaisly figure these problems. There are 4*4 = 16 ways to make any given non-pair and (4*3)/2 = 6 ways to make any given pair. So:

AA - 6 ways
KK - 6 ways
QQ - 6 ways
JJ - 6 ways
-----------
Total 24 ways

AK - 16 ways

So it's 3:2 in favor of him having a pair.

In practice, you would take the cards you were holding into account. For example, if you were holding AK then the numbers would change to:

AA - 3 ways (3*2/2)
KK - 3 ways
QQ - 6 ways
JJ - 6 ways
-----------
Total 18 ways

AK - 9 ways

Now it's 2:1 he has a pair.

Lost Wages