Proof of Sklansky's theorem?

[AceHigh]

"If memory serves the FTOP holds true in all situations"

This is not true. Andy Morton has proven that there are situations where you lose when an opponent makes a mistake. See the link below for details or look up Morton's theorem on google.

http://groups.google.com/groups?q=Mo...com&rnum=5

[BruceZ]

The following link shows the history of great minds who have tried to prove this:

http://www-gap.dcs.st-and.ac.uk/~his...f_algebra.html

Note this paragraph:

Viète gave equations of degree n with n roots but the first claim that there are always n solutions was made by a Flemish mathematician Albert Girard in 1629 in L'invention en algèbre . However he does not assert that solutions are of the form a + bi, a, b real, so allows the possibility that solutions come from a larger number field than C. In fact this was to become the whole problem of the FTA for many years since <font color="red">mathematicians accepted Albert Girard's assertion as self-evident.</font color> They believed that a polynomial equation of degree n must have n roots, the problem was, they believed, to show that these roots were of the form a + bi, a, b real.

Amazing this passage even uses the word self-evident. So it may be regarded as self-evident or almost self-evident even though it is far from self-evident how to prove it. Note that Leibniz (no idiot) even thought he proved it to be false! I think it can be proven without complex numbers purely by modern algabraic means by proving that all polynomials have a splitting field.

The Fundamental Theorem of Arithmetic states that all integers &gt;1 can be expressed as a product of prime numbers. I think this is self-evident since any positive integer &gt;1 is either prime, or is a product of other factors which are either prime or the product of other factors, etc.

The Fundamental Theorem of of Calculus states that if f(x) is continuous on [a,b], and G(x) is an antideriviative of f(x), then integral[a,b]f(x) = G(b) - G(a). This is proven by noting that integral[a,x]f(x) is an antiderivative of f(x), so

integral[a,x]f(x) = G(x) + C
G(a) + C = integral[a,a]f(x) = 0
C = -G(a)
integral[a,b]f(x) = G(b) - G(a)

[Jimbo]

I have always thought Mortens Theoeum was interesting but it has it's own problems as well. Carried to it's logical conclusion you would want everyone to fold preflop when you raise with pocket aces. Since this is clearly wrong I will stick with Sklansky's interpretation as it is written and described including his exceptions.

[AceHigh]

No, Morton's theory has more to do with flop and turn play, where others players bad calls make it correct for good players to call.

I believe Morton's Theory has been proven.

[Jimbo]

I believe I have just disproved it. Here is another example for you. The flop comes 222 and you have the same hand as before AA. According to Morten you would want most of the other players to fold because if enough people call someone might draw out on you with a lower pocket pair or an ace and the 4th deuce comes. How absurd is this thinking?

[ACPlayer]

Godel

Godel's Law which says, in effect, that "you can't intellectually know everything."

Any axiomatic system is incomplete and the consistency of an axiomatic system cannot be proven within the system.

[AceHigh]

Morton's Theorem only applies to specific situations. Most times it does not apply. It is not a catch-all theorem like FTOP.

BTW, if you want a disproof of FTOP, it's never correct to call on the river according to FTOP. You should raise or fold, clearly that is a flaw.

FTOP is a little bit like HEFAP's starting hands guidelines, use them but apply them rigidly.

[Jimbo]

BTW, if you want a disproof of FTOP, it's never correct to call on the river according to FTOP. You should raise or fold, clearly that is a flaw. Mind if I ask where you came up with that one?

[AceHigh]

common sense. If you have the best hand you should raise, if you are beat you should fold.

[Jimbo]

common sense. If you have the best hand you should raise, if you are beat you should fold. I thought you might say that. Please give me an example of when you might call when you knew all the other players cards. If the other players are playing incorrectly they will call your raise, if you are beat you should either fold or bluff-raise but it is hard to make a case to just call under these circumstances though I am sure you could contrive a rare example.