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  #31  
Old 03-13-2003, 07:39 AM
Tommy Angelo Tommy Angelo is offline
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Default Re: Proof of Sklansky\'s theorem?

In science, a law is a description, a theory is an explanation.
----------------------------

I had to start with that just because it took me 30 years of squabbling to find it.



By the dictionary, The FTOP is neither fundamental (basic), or a theorem (a statement or formula that can be deduced from the axioms of a formal system by means of its rules of inference.) But that's okay, because we allowed to play with our words. That's how we come up with them in the first place.

What struck me, after reading the FTOP (below) for the first time in a decade, was, huh? And that was odd, because when I first came across the FTOP, I thought it was brilliant. And no matter what sort of grandness the words "fundamental theorem" were meant to convey, that was fine by me, because it *was* grand. This theorem was on par with E=MC2 and "Give peace a chance."

Now when I look at the FTOP, I see "quantifiable mistake" presented and defined. And because that is a concept that slides right off my brain as unusable and irrelevant, so too does the FTOP.

Tommy

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  #32  
Old 03-13-2003, 10:49 AM
gilly gilly is offline
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Default Theorems

All theorems are based on a set of conditions.
A theorem is posed by saying given X and Y then Z. And this must be proven for it to be a theorem.
This is what a theorem is. You cannot try and implement a theorem without first checking that the conditions
are met. This is a classic mistake in a lot of academic literature. You cannot just say that Z holds due to
this theorem unless you know that X and Y are satisfied.

Example: A continuous function on a compact set achieves its maximum

You cannot just say that a function F, on a compact set, will achieve its maximum unless you first check
to make sure F is continuos.
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  #33  
Old 03-13-2003, 02:19 PM
Kim Lee Kim Lee is offline
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Default Here\'s the Darn Proof!

Here's the proof:

A heads-up situation with no rake is a zero-sum game - anything lost by your opponent must be won by you.

That's all. The best play by your opponent is always the worst play for you. Opponent's mistakes make you money.

This means you can safely presume your opponent will make the best play. Suppose you make more money betting than checking when your opponent correctly folds. Then you should bet. But what if your opponent calls without sufficient pot odds? Then your opponent loses and you gain more. You should bet!

The exceptions occur with multiplay pots where drawing hands subsidize each other. Consider a weird hold'em game with 8 cards on the flop. Suppose you have a made straight versus 4 opponents with different 4 flushes. With a large bet they might all correctly fold. But if they all call then you will surely lose.

Sklansky called it a theorem because his father is a mathematician. A theorem is something that has been proved. Fermat's Last Theorem was called a "theorem" in deference to the late mathematician who claimed to have a proof. The Four Color Map Theorem was considered proven long ago until errors were identified in the published proof. We don't go around changing the name from "theorem" to "conjecture" if we will probably need to change it back when a correct proof is established. Tradition dies hard and nomenclature tends to stick after a couple hundred years.
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  #34  
Old 03-13-2003, 02:33 PM
tewall tewall is offline
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Default Re: Proof of Sklansky\'s theorem?

The proof is a formal proof. It just used a computer to help. Just because a computer is used, doesn't mean the proof is not a formal proof.
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  #35  
Old 03-13-2003, 02:59 PM
tewall tewall is offline
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Default From the master\'s mouth

I called it the Fundamental Theorem of Poker in a tongue in cheek way because I knew the FTO of arithmetic, calculus, and algebra were icons though almost self evident, and I thought it would be kind of cool to have something similar for poker coined by me. Furthermore the chapter was meant for advanced beginners so I wanted to accentuate the importance of the concept with a flashy name.
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  #36  
Old 03-13-2003, 03:09 PM
Fishy Fishy is offline
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Default What is the four color map theorem??

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  #37  
Old 03-13-2003, 03:45 PM
BruceZ BruceZ is offline
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Default Re: Gödel

It may be a surprise to a lot of people to learn that not all things are either true or false. There are some things that are neither true nor false, but in a third state of "undecideable", and you can prove that.

Consider the set of all sets that do not contain themselves. Now note that this set contains itself if and only if it does not contain itself. This is a breakdown of so-called "naive set theory" which motivated changes to both set theory and to logic, and indicated the inherent inconsistency of all logical systems.

In the development of mathematics, mathematicians have at times had to make a choice as to which axioms to believe are true, and which are to remain as unproveable. These choices are arbitrary, though they are made in the interest of retaining the most powerful and elegant set of axioms and theorems.
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  #38  
Old 03-13-2003, 03:45 PM
tewall tewall is offline
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Default [url]http://www.math.gatech.edu/~thomas/FC/fourcolor.html[/url]

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  #39  
Old 03-15-2003, 09:43 PM
M.B.E. M.B.E. is offline
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Default Re: Proof of Sklansky\'s theorem?

Ralle is correct about what a "theorem" is.

"Fermat's last theorem" was something of a misnomer, and recognized as such in serious discussions. Some mathematicians did insist on labelling it "Fermat's conjecture".
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  #40  
Old 03-16-2003, 11:46 AM
Kim Lee Kim Lee is offline
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Default How is FTO Algebra \"almost self-evident\"?

The FTO Algebra says every polynomial equation has a (possibly complex) root. Sklansky wrote "I knew the FTO ... algebra ... almost self evident." Can somebody please explain this? Can anybody explain it without complex numbers? Apparently Sklansky has a high threshold for self-evident, higher than Gauss and all his predecessors. At least he considers it beyond obvious.
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