Theorem of expected stack sizes

[PrayingMantis]

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you think we're saying M=4 > M=8.

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I do not think that this is what you're saying, certainly not as a generalization. However, if you'll read through Betgo's posts in one of his threads there, he's specifically talking about the advantages of leaving the "orange zone" for the "red zone", or intentionally dropping down from the "orange zone" to the "red zone" (by playing super-tight at the orange zone, for instance), since being in the "red zone" might be more advantagous at times. This is in complete contradiction to the theorem in the OP.

This is with regard to Betgo. I won't get into details regarding what you were saying (the above was enough) - but you can certainly present it again here, or put it in other words, or elaborate on it. If you think that people do not understand it, you should do ALL you can in order to make it clear.

[Proofrock]

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Given a player with a stack size of S1 or S2, seated at a table with certain same conditions for both cases(1), there is no single identical hand that is played by him in both cases to maximize profit(2) and that will yield expected stack sizes of S1' or S2'(3) respectively, for which S1>S2 AND S1'<S2'.

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(1) i.e, same players, stacks, positions and "reads".
(2) or minimze loss, which is the same.
(3) i.e, stack sizes when the relevant hand is over.


....

Note that this theorem isn't based upon any kind of assumed model, and as such it is simply a logical structure (that could probably be dismissed as trivial by some). Moreover, it shouldn't be very hard to refute, since all we need is to think of and describe just one "counter example hand". However, until this example if found, there is no "red zone" theory that can stand to reason.

Comments? refutations? flames?

[/ QUOTE ]

I'm not sure I understand the relevance of the theorem as stated. It seems that, for it to be relevant, it would be Sx' - Sx that you are comparing.

For example, you're in the BB. let S1 = t10000 and S2 = t1000, and another opponent pushes from the small blind with t7000. Regardless of the decision (optimal or not), S2' can never be greater than S1', so the theorem says nothing interesting.

What is important is the net increase in chips, yes? In this case, there are plenty of examples for which S2' - S2 > S1' - S1. Here's one: You are in SB, Blinds t1000/2000.
S1 = t20000, S2 = t1000 (after posting). BB is sitting out. Button pushes for t20000 and you have a read that he has AK or AQ. You have 8s9s.

S1 has to fold, EV = 0 (S1'-S1 = 0).
S2 has to call, EV > 0 (S2'-S2 > 0) based on 5:1 pot odds.

If I've misunderstood the theorem, let me know. Otherwise, I'm not sure it's particularly useful.

-J.A.

[Jason Strasser]

PM.

I'm not going to lie but it took me 10 mins to figure out this simple statement (maybe due to slight hangover).

But I had a thought. Say S1 is 20bb and S2 is 8bb. There is a limped pot and you have Ts9s in the bb. Say you flop a draw (and say only 1 person limped). So 3bb in pot. You decide with both S1 and S2 you will check and evaluate... Which likely means check and raise for both S1 and S2.

Lets say you check after the SB checks, and the limped player moves all in and the sb calls (both 20bb stacks). If you are 20bb deep you fold T9s, but with 8bb you have an easy call. So its possible now for this theorem to be violated, right?

-Jason

[Jason Strasser]

You are saying one stack is easier to play more correctly.

Which is true, but easy does not mean its a better spot.

-Jason

[MLG]

first,

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(maybe due to slight hangover).

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Its 7:15 pm and you're still hungover? WTF?

second,

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Lets say you check after the SB checks, and the limped player moves all in and the sb calls (both 20bb stacks). If you are 20bb deep you fold T9s, but with 8bb you have an easy call. So its possible now for this theorem to be violated, right?


[/ QUOTE ]

Your EV with the draw in this spot is not more than 20x, its probably around 16-18x depending on all factors. So no, its not violated here, although I wonder if you could play around with the numbers to get it to work.

[Jason Strasser]

It involved a barn a sorrority with lots of fat girls and franzia.

Anyway, what does EV have to do with this theorem?

-Jason

[MLG]

ok, then you need to find yourself a better hangover cure.

And the theorem says blah blah blah, lead to an expected stack size, which to me means EV right?

[Jason Strasser]

Oh.

I may have misread.

Hmm I'm pretty sure there is a situation even with EV that may lead to a situation that violates this theorem. Maybe you could add more limpers?

[ QUOTE ]
Oh.

I may have misread.

Hmm I'm pretty sure there is a situation even with EV that may lead to a situation that violates this theorem. Maybe you could add more limpers?

[/ QUOTE ]

Meh.. i was making it complicated.

If you take your hand and just put the two stacks on the border between when calling is correct and when folding is correct then you're good.

[MLG]

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Maybe you could add more limpers?

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The question is then, can you get enough money in the pot to get the EV of the 8x stack over 20x, while still keeping it correct for the 20x stack to fold. There's probably a spot if you fiddled with the stack sizes and limpers and such where it happens.