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  #1  
Old 10-17-2005, 06:19 PM
Fillamoore Fillamoore is offline
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Default Question about the fundamental theorem of poker

I was rereading sklansky's Theory of Poker last night and was reading his fundamental theorem. In doing so i came across something that puzzled me and perhaps you gentelmen (and ladies) can help me out.

Lets look at an example from Texas Holdem. Say that you're in the Big Blind with JTo. One limper and its folded around to you and you check. Heads up to the flop. The flop comes giving you a gutshot straight draw on a board of:
Q - 8 - 2.

Lets also say that we check the flop and make an incorrect call and see the turn, which brings an ace. (as some of you have probably already noticed this is from a very simial example in the book). Lets also say that we KNOW our opponent has KQ, he's passive, and that we KNOW if we bet he will call. Does this make it correct to bet? I ask this because in his fundamental theorem he says any time an opponent plays differently than he would if he could see your cards, you gain. If we bet here, our opponent will just call. Therefore we gain 1 bet because he should have raised. However, our mistake of betting with the worst hand cost us 1 bet MINUS the roughly 9% that we improve to the best hand...so therefore our mistake isn't as great as our opponents, and we therefore gain over him. So back to the original question, is it correct to bet even though we KNOW our opponent will call and NOT RAISE?
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  #2  
Old 10-17-2005, 07:13 PM
TaintedRogue TaintedRogue is offline
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Default Re: Question about the fundamental theorem of poker

Only if he'll call the river when you make your hand. That will give you a fraction of +EV long term.
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  #3  
Old 10-17-2005, 09:39 PM
Xhad Xhad is offline
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Default Re: Question about the fundamental theorem of poker

You're taking this too literally.

Consider this: If you know he will call if you bet, then betting is mathematically the same thing as checking, then calling a bet.
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  #4  
Old 10-17-2005, 11:39 PM
Fillamoore Fillamoore is offline
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Default Re: Question about the fundamental theorem of poker

Thats my problem x. The thing is, although its the same mathematically, its a different situation. If i check, he is correct in betting while i am incorrect in calling, and this would reflect if we played this out many times. However, if i bet out and he just calls, he makes a mistake according to the fundamental theorem of poker and therefore i gain what he misses, that is 1 BB.

This is why i posted this. I can see that it is obviously the same mathematically, but the reason for the inquiry is because it doesnt make sense to me thta betting out even though we know he will call is correct when check calling is incorrect.

However, after pondering this i believe i came up with the answer to my question. Post your thoughts please.

I think the reason it is "correct" to bet is because, in theory of course, we all face the EXACT SAME situations WITH THE EXACT SAME CARDS if we were to play out infinitely long. So while its true that he misses a bet and makes a mistake, i gain because when IM faced with a bet and hold the same KQ against JT, and I raise, i dont lose anything and therefore and 1 BB ahead of my opponent. If i were to just call as well both of our mistakes cancel eachother out in a sense.

* this post was actually supposed to follow X's but whenever i clicked his link for some reason i was logged out...not sure why. Anyways, you get the idea
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  #5  
Old 10-18-2005, 12:06 AM
Xhad Xhad is offline
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Default Re: Question about the fundamental theorem of poker

[ QUOTE ]
Thats my problem x. The thing is, although its the same mathematically, its a different situation.

[/ QUOTE ]

No it is not. If you KNOW he will call and you KNOW his hand is better than yours and you do not have pot odds to call a bet from him, you are still risking the same amount of money to win a pot of the same size. It is the same thing.

[ QUOTE ]
If i check, he is correct in betting while i am incorrect in calling, and this would reflect if we played this out many times. However, if i bet out and he just calls, he makes a mistake according to the fundamental theorem of poker and therefore i gain what he misses, that is 1 BB.

[/ QUOTE ]

The problem is that you were incorrect to bet the worst hand in the first place under all the assumptions you're making. Look at it this way: If he would bet if you check, but only call if you bet, then betting or check/calling both result in you paying 1BB to see the next card. Thinking the EV in these two situations is any different shows a fundamental misunderstanding of the entire concept. There is no Poker God out to punish people for making a "mistake".

Put another way, you lose less when he calls vs. raising; you do not gain anything vs. checking.

[ QUOTE ]
I think the reason it is "correct" to bet is because, in theory of course, we all face the EXACT SAME situations WITH THE EXACT SAME CARDS if we were to play out infinitely long. So while its true that he misses a bet and makes a mistake, i gain because when IM faced with a bet and hold the same KQ against JT, and I raise, i dont lose anything and therefore and 1 BB ahead of my opponent. If i were to just call as well both of our mistakes cancel eachother out in a sense.

[/ QUOTE ]

If he would call when you bet, but bet if you check, then betting is just doing his dirty work for him, regardless of whether or not he philosophically made a "mistake". If he would bet if you check, or raise if you bet, then you're intentionally putting money in with the worst hand which obviously can't be right.

The game does not "know" why money goes in, only that it does go in. If you bet when you know you're going to be called, it's the same thing as calling even if your opponent "should" raise.
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  #6  
Old 10-18-2005, 02:41 AM
Shandrax Shandrax is offline
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Default Re: Question about the fundamental theorem of poker

I don't get the Fundal Theorem of Poker anyways. If your opponent makes three incorrect calls in a row and then draws out on the river you have gained nothing. Card distribution of the next hand is independent and the cards don't know that you lost a big pot on the last hand so they can't do you any justice. If you are a 9:1 favorite and lost once already it doesn't guarantee you 9 wins in a row.

I would change the whole theory to whenever your opponent makes an incorrect play he put himself at risk, so your chances of gaining something from it increase.
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  #7  
Old 10-18-2005, 07:24 AM
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Default Re: Question about the fundamental theorem of poker

I like to think of it this way.

On any given hand, there is a maximum you can lose and a maximum you can gain. The same applies to your opponent.
It's a sliding scale and a fight to swing the balance in your favor.

In your example, you are willing to make a losing bet and a losing call.

In both scenarios you (wrongly) push the balance in favor of your opponent.

Where you gain, is that given the opportunity, Villian didn't take his opportunity to swing the balance EVEN FURTHER in his favor.

So as explained by Xhad. Whether you bet and your opponent calls. Or you check\call. Still has exactly the same result on the sliding scale.

EXCEPT the bet is worse because you give your opponent a chance to swing the balance even further in his favor.

Saying that this is your gain is slightly misleading. It's villian's loss. You gain by default but the balance doesn't shift in your favor.

Another example.
Kwazzie's fundamental theorum of Murder.

Choice 1. Having my balls cut off then shot dead OR
Choice 2. Being shot dead then having my balls cut off.

I take Option 2 and I can say I'm better for it. But I haven't gained.

What your suggesting to do in your example, is to cut your own balls off then give your murderer the choice.
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  #8  
Old 10-18-2005, 11:06 AM
Tilt Tilt is offline
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Default Re: Question about the fundamental theorem of poker

But aren't YOU doing something different than you would do if you could see his cards?
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  #9  
Old 10-18-2005, 11:13 AM
Fillamoore Fillamoore is offline
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Default Re: Question about the fundamental theorem of poker

wow, this is getting annoying, why cant i post after this? When i click the other links it just logs me out of the servers, anyone know why?

In any event, thats a funny take on it krazzie, the only problem that both you and X have is that you dont seem to realize PRECISELY what im saying. I understand everything you say about it being no different. This is why im posting this. Because according to The Fundamental theorem, they ARE DIFFERENT...which im on your side and agree that they arent...and this is my questio, WHAT makes them different. The problem, which you say is that although he was incorrect in not raising, i was incorrect in betting. However, the difference being is that i have outs to the best hand which will get there about 9% of the time, so while it was incorrect to bet out, my mistake has not cost me 1 full BB while his not raising has cost 1 full BB. Does this make sense now? perhaps i shouldve said that i belive the fundamental theorem is flawed in this sense. However i belive i came up with the reason for this in my last post. Comments are appreciated. If you disagree me at all, please let me know why.
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  #10  
Old 10-18-2005, 12:44 PM
RoundTower RoundTower is offline
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Default Re: Question about the fundamental theorem of poker

His failure to raise does not cost him 1 full BB, but about .91 of a BB.
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