Theorem of expected stack sizes
 

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?

Re: Theorem of expected stack sizes
 

I'm sure there's a counterexample. Make S1 and S2 really close together and on the borderline of some implied odds thing (i.e. give someone exactly correct implied odds to call with a small pair preflop off of the BB and play optimally against us with one stack, and then take 1 chip away from us in the other stack).

However, the gist is correct, I think.

Re: Theorem of expected stack sizes
 

[ QUOTE ]
I'm sure there's a counterexample. Make S1 and S2 really close together and on the borderline of some implied odds thing (i.e. give someone exactly correct implied odds to call with a small pair preflop off of the BB and play optimally against us with one stack, and then take 1 chip away from us in the other stack).

[/ QUOTE ]

I don't see how this is a counter example. Please elaborate and be as specific as you can with the details of the hand/s you describe and the details of the theorem.

As a side note - I'll be happy and excited to see this theorem refuted. As for me, I still haven't found a counter example.

Re: Theorem of expected stack sizes
 

Are we assuming the actions of other players at the table do not take our stack into consideration?

Re: Theorem of expected stack sizes
 

this theory makes it very clear that you still don't understand what betgo and i were trying to say.

Re: Theorem of expected stack sizes
 

[ QUOTE ]
this theory makes it very clear that you still don't understand what betgo and i were trying to say.

[/ QUOTE ]

True. You guys were saying S1'- S1 is often bigger than S2' - S2 when S1 < S2.

Re: Theorem of expected stack sizes
 

[ QUOTE ]
this theory makes it very clear that you still don't understand what betgo and i were trying to say.

[/ QUOTE ]

1. It's not a theory.

2. I understand perfectcly well what you and betgo were "trying to say". Moreover, many times during those discussions it was pretty clear to almost everyone that both of you are very confused about your own arguments (see, for instance, your comments with regard to one of your own posts' title.)

3. I thought you already said that you give up and that we "won", didn't you? there was a big post about it. So what exactly is your new perspective on the matter?

Regardless of all this, I'll be happy to read any relevant comments coming from you with regard to this. I won't reply to comments that might turn this into a flame-war, not because I don't like flame wars (I like) but because I don't want LLoyd to lock up this thread. [img]/images/graemlins/grin.gif[/img]

I guess I made a mistake then when specifically "inviting flames" in the OP... In any case, any criticism, harsh as can, is most welcome.

Re: Theorem of expected stack sizes
 

[ QUOTE ]
3. I thought you already said that you give up and that we "won", didn't you? there was a big post about it. So what exactly is your new perspective on the matter?

[/ QUOTE ]

yes. i did give up trying to explain it in detail. i just wanted to let you know one final time that you're still increadibly misunderstanding what we were saying. you think we're saying M=4 > M=8. we're not. i'm certainly not at least, and i'm pretty sure betgo isn't.

Re: Theorem of expected stack sizes
 

I think their theory was that

$EV(X/2)*2 > $EV(X)
-for values of X around 10-15xBB

Since, half of X is more than half as valuable as X, it is correct to take -CEV gambles when the result is endign up in the red zone, b/c conventional EV models underestimate the $EV of those particular stack sizes.

Please tell me if I am wrong about my interpretation.


I do not agree with any of it, I just think I know what you are trying to say.

Re: Theorem of expected stack sizes
 

[ QUOTE ]
[ QUOTE ]
this theory makes it very clear that you still don't understand what betgo and i were trying to say.

[/ QUOTE ] True. You guys were saying S1'- S1 is often bigger than S2' - S2 when S1 < S2.

[/ QUOTE ]

That indeed was a part of what they were saying, but a very uninteresting and trivial part, and certainly not one that qualifies as a reason for why their theory has any kind of merit, or as a "proof" for it. As I said myself in one of those discussions, I completely agree with this particular element of their theory, and it's not new or surprising at all. I was even happy to give my own examples for it being true.

The ONLY interesting and relevant aspect of their theory, and the only element of it that might might make their theory valid, is the possibility that the theorem in the OP is not true. In a very deep and essential way, the negation of this theorem is the most important and "revolutionary" consequence of their theory (or from a different perspective: its most prominent assumption). They might not agree or even understand it themselves, but that doesn't change much, of course.