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  #1  
Old 03-07-2006, 06:22 PM
donkeyradish donkeyradish is offline
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Default When it doesn\'t matter what you do


Often you get a close-call situation where you could call or fold and your EV would be about the same.

But could there ever be a situation when the expected value of:

folding = calling = raising
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  #2  
Old 03-07-2006, 07:18 PM
JaredL JaredL is offline
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Default Re: When it doesn\'t matter what you do

Here's a quick shot. I'm only considering heads up limit poker here.

Folding always nets zero.
EV of calling is zero if and only if the odds against winning are the same as the odds offered by the pot.

Suppose the odds against winning are X:1 and the pot is offering X:1. In other words he is X times as likely to have a better hand than he is to have a worse hand than you. The EV of raising is the EV of calling plus the expected winnings from the action coming as a result of the raise. Since you are getting 1:1 on the raise and the odds against winning are X:1 it cannot be the same EV if he will always call.

The villain here must fold often enough and with the right hands so that the EV of the raise is also zero.

Suppose she calls with probability p when she is ahead of you and q when she is behind. Let Y be the probability that you are ahead when she bets and r be the probability she calls at all. Notice that Y = 1/(X+1) and r = pY + q(1-Y). The EV of raising is (1-r)X + (X+1)r(q(1-Y)/r) - 2r(pY/r). Replace r with pY+q(1-Y), set that bad boy equal to zero and solve for p with respect to q. If your opponent calls your raises with these proportions then they are all the same.

Jared
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  #3  
Old 03-08-2006, 02:43 AM
Aaron_C Aaron_C is offline
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Default Re: When it doesn\'t matter what you do

i got caught up on where your expected value is, your implied odds come into play if you are looking at a drawing hand. But the real problem is that the odds are based off imperfect knowledge of your opponents, and we base these values off a ton of guestimated sums for a situation.... in other words, the estimation will range over a possible set of EVs... we use the average EV since if we have estimated the range correctly the average will be the true EV.
since we could pay more close attention to one varriable than another, a little more thought will tend to lead us to call or fold when we come to a truly "marginal" situation.
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  #4  
Old 03-08-2006, 05:29 AM
bigpooch bigpooch is offline
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Default Re: When it doesn\'t matter what you do

You are facing an opponent who is bluffing with the exact
game theoretical bluffing frequency and will fold against a
raise with the exact game theoretical folding frequency in
accordance to the size of your raise. You have a hand that
can beat a bluff but will lose to all legitimate betting
hands and your opponent will fold to a raise with precisely
those according to game theory.
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  #5  
Old 03-08-2006, 05:50 AM
Aaron_C Aaron_C is offline
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Default Re: When it doesn\'t matter what you do

my question to this response is how precise is your game theory in a limited information game?
Its the same question for econ, we believe an individual acts when marginal benifit exceedes marginal cost. an individual will not take that action when MB < MC. what an individual does when MB=MC is inexplicible because we are dealing with an infinate number of possible values the odds that they are ever equal would be zero.
In an individual pot even if we perfectly balance the equation for EV in terms of dollars for that pot, long term table image, bankroll considerations, personal utitlity of wealth in question etc all matters.
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  #6  
Old 03-08-2006, 12:47 PM
phish phish is offline
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Default Re: When it doesn\'t matter what you do

BicycleKick posted a hand awhile back with 44 against Twinkie's A5 that from Twinkie's perspective, fold, call or raise on the river had very close EV's, I thought.
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  #7  
Old 03-08-2006, 02:41 PM
Aaron_C Aaron_C is offline
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Default Re: When it doesn\'t matter what you do

very close may suggest that other factors than EV may correctly determine your play, but he is asking if it is possible to have exactly equal EV.
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