[Proofrock]

[ QUOTE ]
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.