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Old 07-03-2007, 04:14 PM
jackflashdrive jackflashdrive is offline
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Default A logical proof of the AEJONES theorem

A logical proof of the AEJONES theorem.

1) If the immediate EV of a decision in poker is close, then one should select the action that is best for meta-game purposes. For example, action at a 6-handed 3-6 NL table could proceed such that the EV of calling an allin PF with AKs or folding is close to 0.

2) For the purposes of meta-game, in 0 EV situations calling an all-in is better than folding (because it keeps people from taking +EV shots at you, which is +EV for you) and pushing is better than folding (because it generates action on other hands, which is also +EV).

3) Similarly, there will be many decisions that have an immediate -EV but neutral EV when meta-game is factored in. In these situations one is always pushing and calling rather than folding (see 2 above). Since we are ambivalent w/r/t money in these circumstances (unless variance is a concern -- but we'll assume sufficient bankroll), the only other concerns are personal preferences. Everyone prefers to push and call rather than fold, because one who pushes and calls has balls, and one who folds is a pussy. Thus, these decisions that have neutral EV w/r/t metagame actually have positive EV when personal preferences to have balls and not be a pussy are included.

3b) An exception to (3) above is that calling a push with a hand that cannot beat anything makes one an idiot who gives insane value. In this circumstance, no advantage is obtained w/r/t balls and the EV w/ metagame is even more negative.

4) At the extreme, decisions with very negative immediate EV, and somewhat negative metagame, will nevertheless be neutral with respect to having balls or being a pussy. In these situations we are ambivalent about pushing/calling vs. folding.

5) Thus, a fully rational poker player should be completely ambivalent about pusing with any hand (regardless of how negative the EV) vs. folding, and he will also be ambivalent about calling with any hand that has the potential to beat *something* vs. folding (see condition 3b).

6) Thus, a rational player will always be pushing with a hand range of [random], and will call a push with any hand that beats a hand range of [random] so long as their opponent is regarded as rational. This is another way of phrasing the AEJONES theorem that "nobody has anything" and additionally extends his theorem to include the following: "Good players will call a push with any hand that beats something".
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  #2  
Old 07-03-2007, 04:28 PM
Ship Ship McGipp Ship Ship McGipp is offline
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Default Re: A logical proof of the AEJONES theorem

[ QUOTE ]
" Calling Stations will call a push with any hand that beats something".

[/ QUOTE ]

Yea, pretty much. But some calling stations are good.

Basically, I am not smart enough to comprehend this post, so I'll tell you what I understand: Lots of decisions are really close, better players recognize what side of those close decesions they should err on. Those good players will often choose to call if it's close to neutral (I can count MANY situations that are posted or that I find myself in everyday where I think to myself, 'calling is definitely okay, so is folding, I don't think one is even definitively better' and by that thought process, we will find outselves often making decisions based on style.

I am kind of an aggro-trapping type of player, depending on the opponent, and I definitely err on the side of aggro or loose, in general, "this is close, I can justify a call."

I will say that TONS of situations come down to: this guy is a pussy, he folds; this guy makes hero calls, he calls. Just a function of style, in a lot of situations, and the EV will be very, very negligible.
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Old 07-03-2007, 05:23 PM
Fight Club Fight Club is offline
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Default Re: A logical proof of the AEJONES theorem

[ QUOTE ]
2) For the purposes of meta-game, in 0 EV situations calling an all-in is better than folding (because it keeps people from taking +EV shots at you, which is +EV for you) and pushing is better than folding (because it generates action on other hands, which is also +EV).


[/ QUOTE ]
how exactly is someone going to take '+EV shots' at me because I didn't take a coinflip in the past
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  #4  
Old 07-03-2007, 05:46 PM
gir gir is offline
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Default Re: A logical proof of the AEJONES theorem

[ QUOTE ]
Basically, I am not smart enough to comprehend this post

[/ QUOTE ]

Aren't you AE Jones?
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  #5  
Old 07-03-2007, 05:54 PM
jackflashdrive jackflashdrive is offline
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Default Re: A logical proof of the AEJONES theorem

[ QUOTE ]
[ QUOTE ]
2) For the purposes of meta-game, in 0 EV situations calling an all-in is better than folding (because it keeps people from taking +EV shots at you, which is +EV for you) and pushing is better than folding (because it generates action on other hands, which is also +EV).


[/ QUOTE ]
how exactly is someone going to take '+EV shots' at me because I didn't take a coinflip in the past

[/ QUOTE ]

To the extent that you fold in marginal situations (and to the extent that you fold in general), you expand the range of hands with which others can profitably bluff, semi-bluff, and press 'coinflips'.

E.g., If I think you will call/push-over my semi-bluff non-nut flush draw with any hand as good as A high, and if I think your range consists in large part of A-high and better, then I obviously can't push my flush draw because I will almost always get called and at best it is a coin flip. Whereas if I can fold out a good portion of your coin-flipping range then I may end up with a +EV semi-bluff.
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