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  #21  
Old 02-20-2003, 10:44 AM
Ralle Ralle is offline
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Default Re: Proof of Sklansky\'s theorem?

> You say "can" be proven, are you saying they havent been
> proven yet? Because if they have, then you wouldn't use the
> word "can", you would use the word "have", no? Somebody
> else used the word "can" as well, that's why I'm wondering.

This is a valid question. For a theorem to be accepted as such
a proof would normally be required. Until the proof is presented
the "theorem" would usually be considered a conjecture.

However, in general we don't examine the proofs of all theorems
we encounter. Instead we take it for granted that a proof exists,
and that it is commonly accepted to be correct. It's just not
worth the bother to go through all proofs ourselves. And in many
instances, we are not even able to understand the proof. That is
why in practice it is enough that the theorem "can" be proved. But
like I said, that "can" is only a "can" for us, if you understand
what I mean.

Anyway, this is more a question of semantics. Maybe it's better to
simply use the word "have" like you suggest.
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  #22  
Old 02-20-2003, 05:55 PM
BenD BenD is offline
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Default Re: This doesn\'t seem that complex

My point, and the reason I put down premises, is that the FTOP is mathematically sound in heads-up situation. That is, it is provable.

Every play has a mathematical expectation. Playing perfectly is choosing the play that has the highest mathematical expectation. Determining mathematical expectation of a hand requires that you know both your hand and your opponents.

The first post I made was intended as a general proof, but let's take a specific case.

You are playing no limit hold-em against one opponent. Game structure: You each have 10 chips. Blinds are 1-2. You are in the big blind (small blind is on the button). You have KK. Small blind simply calls the 1. What is the perfect play?

Most people will say raise (or perhaps to trap), but the real answer is, you don't have enough information.

Take different possibilities. Your opponent has:

AA -- clearly you should check here and fold if (a) you don't improve and (b) you cannot get your opponent to fold by betting.

KK -- check again, and attempt to outplay your opponent when a "scary" flop comes.

QQ -- Raising when you get 4-1 and are giving 2-1. Your opponent will likely call. However, a small raise (to make the pot larger, and then and then a check call (or raise) on the flop when no queen flops, is the best way to get the chips

etc...

If your opponent has a hand that will not call a big raise, and doesn't have a good shot at KK, but may call a flop bet when they hit something (a hand like KJ, when they flop either a K or a J), then raising is clearly a terrible play. But if your opponent has a hand with a better shot, but one they will still fold, like with 67s, then a raise is a better play.

As a mathematical model, in heads-up situations, the FTOP is both useful and provable...in practice, knowledge of your opponent's hand is not always available.

In fact, you can calculate just how large a FTOP mistake is if you know both hands. EV of a play, and choosing the play with the best possible EV is how you win at poker in the long run.

A useful thing for newer players to do is to calculate how bad a mistake before the flop is...that is, to calculate the EV of a play you make. When doing this, you are essentially determining how big of a FTOP mistake you made.
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  #23  
Old 02-22-2003, 06:48 AM
felson felson is offline
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Default Multiway exception to FTOP

FTOP depends on two assumptions.

1. Your opponent will always act in such a way as to maximize his expectation.
2. Your opponent's expectation plus your expectation equals the pot.

Therefore, if your opponent plays perfectly, his profit is at your expense.

In a multiway pot, assumption 2 is violated, since at least one other person has an interest in the pot. Thus, Opponent A's profit may come at the expense of Opponent B and not you. In fact, Opponent A's correct play can even be profitable for you.

Here's an example. It's the turn in holdem. The board is

A [img]/forums/images/icons/spade.gif[/img] T [img]/forums/images/icons/diamond.gif[/img] 9 [img]/forums/images/icons/heart.gif[/img] 3 [img]/forums/images/icons/club.gif[/img]

EP has A [img]/forums/images/icons/diamond.gif[/img] K [img]/forums/images/icons/heart.gif[/img] . You have 8 [img]/forums/images/icons/club.gif[/img] 7 [img]/forums/images/icons/club.gif[/img] . LP has 5 [img]/forums/images/icons/heart.gif[/img] 3 [img]/forums/images/icons/heart.gif[/img]

EP bets. You call, with an open-ended straight draw. Now LP has to make a decision. Do you want him to see your cards?

Well, if the pot is big enough, LP is correct to call in the hopes of rivering 2 pair or trips. If he hits his hand, though, it doesn't affect you at all. Because if he hits, it means you missed your straight draw anyway and cannot win the pot. When LP calls on the turn, he's actually taking money from EP.

You should be happy when LP calls. Because when you make your straight, that money will be yours. If it were allowed, you could even show him your cards on the turn to encourage him to call (if you wanted to reassure him that you don't have 9 [img]/forums/images/icons/spade.gif[/img] 9 [img]/forums/images/icons/club.gif[/img] , for example).

So, LP's optimal play actually profits you. You don't mind if he sees your cards. This is an exception to FTOP.

Note that I made some assumptions here which are pretty reasonable (in particular, that you can't bluff out EP, and that LP would fold on river if he doesn't improve).

This example is precisely why S&M emphasize that it's important to play suited cards in their "loose games" section. In loose games, the big pots will cause people to chase two pair or gutshot straights. When they hit, they'll be taking profit from the guy with top pair top kicker. As they chase, however, they're effecting handing money to the people with flush draws. That's who you want to be.
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  #24  
Old 02-22-2003, 06:58 AM
felson felson is offline
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Default Re: Proof of Sklansky\'s theorem?

I think that mathematicians would agree that Fermat's Last Theorem was not a true theorem until Andrew Wiles proved it. It was (whimsically) called a "theorem" because Fermat claimed to have proved it (and was almost certainly erroneous in his claim). But it wasn't a theorem in the strict sense of the word, and mathematicians would not accept it to be true until it had been proved.

Really, Fermat's Last Theorem and the four-color theorem were conjectures until proved, not theorems. The term "theorem" is reserved for statements which can be proved from postulates (i.e. fundamental mathematical assumptions) or other theorems. The proofs may be extremely long or be done by computer. But a proposition is not a theorem until it is proved.

In addition, I don't think that I need to read the entire proof of a theorem before considering it to be a theorem. If a large part of the mathematical community has reviewed and accepted the theorem's proof, that's sufficient for me.
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  #25  
Old 02-27-2003, 11:48 AM
MCS MCS is offline
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Default Re: Proof of Sklansky\'s theorem?

The four-color conjecture HAS been proven (making it a theorem), and it USES computers to expedite the proof. They wrote a program since the proof relies on checking a large number of cases.


I know this because three semesters ago I took graph theory with Robin Thomas, one of the authors.


And the use of the term "theorem" without a proof bothers me as well.

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  #26  
Old 02-28-2003, 11:42 AM
ResidentParanoid ResidentParanoid is offline
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Default Re: This doesn\'t seem that complex

You *can* compute a mathematical expectation (in theory) even if you can't see your opponents cards. You now just factor in the distribution of starting hands that he could have before he calls. After he calls (if he calls) you factor in the distribution of starting hands that he would call your raise with.

In any case, you can compute a mathematical expectation without seeing your opponents cards. You just have a much higher level of uncertainty when you don't see the cards.

And there is nothing in the FTOP that is going to stop me from raising KK in holdem when I can't see my opponents cards.
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  #27  
Old 03-12-2003, 11:51 AM
Al Mirpuri Al Mirpuri is offline
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Default Re: Proof of Sklansky\'s theorem?

A theory once proven becomes a law.

Creationists are always citing the fact that Evolution is a theory not a law.
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  #28  
Old 03-12-2003, 12:03 PM
Al Mirpuri Al Mirpuri is offline
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Default Re: Fermat\'s Last Theorem

Fermat's Last Theorem states that xn + yn = zn has no non-zero integer solutions for x, y and z when n > 2.

Concerning it, Fermat wrote "I have discovered a truly remarkable proof which this margin [Fermat is writing this remark in the margin of his copy of Diophantus's 'Arithmetica'] is too small to contain."

Fermat's proof was never found amongst his mathematical papers. Posterity labelled the Conjecture (because that is all it is without a proof) a Theorem because to do otherwise would have been to label Fermat a liar.

Andrew Wiles, the Princeton academic (an Englishman), who eventually solved it has expressed the fear that Fermat might well have had a simpler more beautiful proof that is now lost to history.

We will never know.
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  #29  
Old 03-12-2003, 02:13 PM
Carl_William Carl_William is offline
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Default Re: Proof of Sklansky\'s theorem?

My 2 cents:
In the book "Sklansky on POKER THEORY", GBC Press 1978, David presented "THE FUNDAMENTAL THEOREM OF POKER" on page 27. Using an analogy from science (engineering whatever), I propose that it would be more appropriate to call it "David Sklansky's FUNDAMENTAL LAW OF POKER." If it is a law -- than you don't have to prove it. A law can be considered valid until somebody’s disproves it or tosses it out. For instance, in the engineering-science field: “The second law of thermodynamics is a major building block in the engineering science field.” It has never been proved to be true but more important it has (repeating myself) never been disproved.

Also getting back to theorems: True: Theorems are proved, but behind every theorem there is at least one assumption whether we know it or not (there is always an assumption associated with every theorem). A German mathematician (what was his name?) proved (or tried to prove) that it is essentially impossible to prove anything. If we go down deep enough it may be almost impossible to prove anything. Some people think that the “Second Law of Thermodynamics” might be violated in deep space.
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  #30  
Old 03-13-2003, 04:07 AM
Ralle Ralle is offline
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Default Gödel

First, I agree that "Sklansky's Fundamental Law of Poker" would be an appropriate label. Good suggestion Carl!

You are probably thinking of "Gödel's incompleteness theorem", which states that:
1) All consistent axiomatic formulations of number theory include undecidable propositions.
2) If number theory is consistent, then a proof of this fact does not exist using the methods of first-order predicate calculus.
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