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Old 07-12-2005, 03:06 PM
stokken stokken is offline
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Join Date: Mar 2005
Posts: 37
Default Looking back at a hand, please help( a little long)

I hope this is the appropriate forum, the situation is small stakes where some tend to show their hands down to often, are to optimistic with raises and some are super agressive. Please read it trough, the question is at the end

I am in ep with QQ, raise, folded round to lp who reraises
( TAG, with gear changes, who has done some rather neat tricky play as well,
seems to have good pre and postflopp play.
He entered the table with the line" Hello fish, get ready to pay up!"
Folded back to me and we go from here on betting as far as it goes

Flopp 9h,7c,3s.

The range I would put him on is basically AA,KK,QQ,JJ,TT,AK, but he plays them all as if
they where AA( If there is 3 or less in the pot)
I bet, he raises, I reraise etc I have long gone decided to see it trough

Turn 5d

OK- 6 ways to AA, 6 to KK, 1 to QQ, 6 to JJ, and 6 to TT
16 to AK
I am ahead to 28 of these 68,3%
I am behind to 12 29,3%
I tie 1 2,4%
If ahead AK-has 6 outs to beat me ie 13%
If ahead JJ and TT has respectively 2 outs each, 4,5%
If behind I have 2 outs 4,5%

Thus is it correct to put it like this?
The times I am up against a-k I will win 87% of the time
Against JJ-TT 95,5% of the time

I tie 100% of the time with QQ

I loose 95,5 % of the time against AA-KK

I then tried to put it like this:
Win 68,3-13-4,5% of the time for the given range 50,8
Loose 29,3-4,5 of the time for that given range 24,8
Neglecting a tie since that is absolute break even
Can I then look at these two as a ratio, giving I am 2:1 favorite if my opponents action is the same for that range?

Is there an easier way to rapidly determine range and profit relation to a given hand?
Also with respect to more then 1 card remaining?
Will it be equally applicable to more then 1 opponent if u define a given range for all the possible hands u are up against?

Someone who knows math please enlighten me( with explanation if possible)


BR Stokken
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