[PrayingMantis]

[ 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.

[/ QUOTE ]

You understand the theorem. However, you seem to miss its context and relevent role in the discussions with regard to what you might call "the red zone" strategy.

Your example about usefullness of thinking in terms of Sx'-Sx is very clear, but does not have any particular importance, and it's trivial in many senses, as was demonstrated in the past. There are certainly case where the net win for a hand could be bigger as the stack is shorter. However, the "red zone" strategy for MTTS specifically implies that there might be advantages in very specific cases for having a smaller stack, over having a bigger stack _in terms of over-all EV_.

However, the only meaning of any kind of an advantage in this context, is in the ability to _grow your stack to a higher point than in the other case_. That is, the theory basically claims (this is unavoidable) that there are stacks sizes S1 and S2 (let's say now that S2>S1), for which the "road" to a "higher stack" (call it S3, and S3>S2>S1) is "shorter" for S1 than for it is for S2, even when S2>S1.

This is in contradiction to the theorem in the OP, that is: if the consequences of their theory is true, then the theorem is not. But as long as the theorem isn't refuted (by at least one clear counter example) their theory cannot be true.

BTW This important point was discussed numerous times in the threads about those theories, but not in a such a completely theoretical was. That's what I am trying to do here.