Theorem of expected stack sizes

[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.

[PrayingMantis]

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

[/ QUOTE ]

Jason, I agree that it's essentially simple, but it's a bit tricky to put in specific words. If you have a better wording for it, please go ahead, but after I read few of the other posts between you and MLG, I think maybe you didn't read as it was meant to be read, and that's why it looked so simple to you...

Anyway gl in refuting it, it's getting late now here and I'm going to sleep - so I'll check it again tomorrow.

[PrayingMantis]

[ QUOTE ]
and the theorem says blah blah blah, lead to an expected stack size, which to me means EV right?

[/ QUOTE ]

You got it right, yes, it's about the EV, that is, the _expected_ stack sizes.

[Proofrock]

[ 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 ]

I don't think so.

If we call all-in,
S' = (prob of winning)*(amount in pot) (assume a loss knocks you out).

If we fold, S' = S - (x = amount we put into pot).

prob of winning = p, amount in pot = (D + nS) where D = dead $, n = the number of players all-in that match your stack size.

we fold if p < (S-x) / (D + nS).

If we call, S' = p*(D + nS). We're trying to find two stack sizes S1 and S2 s.t. S2' > S1' but S1 > S2. We need S1 to fold, but S2 to call, so we'll have:

S1' = S1 - x, p < (S1 - x)/(D + nS1), p > (S2-x)/(D+nS2), and S2' = p*(D + nS2).

define y = S1-x. Then S1' = y, (S2 - x)/(D+nS2) < p < y/(D+nS1), and S2' = p*(D+nS2) > (S2 - x) and
S2' < y*(D+nS2)/(D+nS1) = S1' (D+nS2)/(D+nS1).

But D + nS2 < D + nS1 b/c n*S2 < n*S1, so S2' < S1'.

I did this very quickly, so it's not unlikely that I made a mistake. If this is correct, though, then increasing the number of limpers won't work to provide a counter example.

-J.A.

[gambelero2]

Let's take 88 against a 50% of average normal player (a maniac gets played with, a really tight player gets a release). The BB is 10% of the other player's stack.

You're 3% to 6% of the players from the bubble. The bubble payout is 1.5x.

Hero has S1 other player has S2.

Situation 1 S1 = S2*1.25
Situation 2 S1 = S2*.75

In situation 1 a reraise allin is a much better play than situation 2. You are more likely to get a release against Aj Aq, hands over which you have a minimal edge. If you get beat by either a drawout or a bigger pair, you will still probably make the bubble.

Note the extra EV comes from a greater probability of a fold by the other player and because there is an overlay to retaining a small stack into a payout category.

[PrayingMantis]

[ QUOTE ]
Let's take 88 against a 50% of average normal player (a maniac gets played with, a really tight player gets a release). The BB is 10% of the other player's stack.

You're 3% to 6% of the players from the bubble. The bubble payout is 1.5x.

Hero has S1 other player has S2.

Situation 1 S1 = S2*1.25
Situation 2 S1 = S2*.75

In situation 1 a reraise allin is a much better play than situation 2. You are more likely to get a release against Aj Aq, hands over which you have a minimal edge. If you get beat by either a drawout or a bigger pair, you will still probably make the bubble.

Note the extra EV comes from a greater probability of a fold by the other player and because there is an overlay to retaining a small stack into a payout category.

[/ QUOTE ]

It looks like you have misinterpreted the theorem. It is about a player having 2 possible stacks, not 2 different players at the table with those 2 stacks.

[PrayingMantis]

I don't see the theorem being refuted up until now. I didn't find any counterexample either.

Regardless, I was thinking about some interesting and very relevant IMO conseqences of it:

Even if there are 2 stack S1 and S2 (S1>S2), that for given expected stacks of S1' and S2' respectively, S2'-S2>S1'-S1 for _each and every hand played_ (!), still, according to the the theorem that still stands, it is always more advantagous to have stack S1 than having S2, for the simple reason that any MTT model that assigns a higher $EV value (or in other words, share in prize-pool) to a shorter stack (that is, a model that says that there could be cases in which S1>S2 AND (share of prize pool of S2) > (share of prize pool of S1) for same player) is absurd.

The above might look complicated but it is very simple. It means in simple words that no matter how much more chips you can make with each hand when you have a shorter stack, you should still prefer a bigger stack.

However, this is true ONLY for a MTTs. When we have this rare spot where S2'-S2>S1'-S1 for every hand played, and it's a _cash game_, we obviously better have the shorter stack, because we only care about the profit we make by playing the hand, and not where our stack is/was/will be.

I find it to be a rather interesting distinction.