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'.

--------------------------------------------------
(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.

Re: Theorem of expected stack sizes
 

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

[/ QUOTE ]

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.

Re: Theorem of expected stack sizes
 

[ 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'.

--------------------------------------------------
(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.

Re: Theorem of expected stack sizes
 

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

Re: Theorem of expected stack sizes
 

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

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

-Jason

Re: Theorem of expected stack sizes
 

first,

[ QUOTE ]
(maybe due to slight hangover).

[/ QUOTE ]

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

second,

[ QUOTE ]
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.

Re: Theorem of expected stack sizes
 

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

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

-Jason

Re: Theorem of expected stack sizes
 

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?

Re: Theorem of expected stack sizes
 

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?

Re: Theorem of expected stack sizes
 

[ 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.

Re: Theorem of expected stack sizes
 

[ QUOTE ]
Maybe you could add more limpers?

[/ QUOTE ]

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.

Re: Theorem of expected stack sizes
 

[ 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.

Re: Theorem of expected stack sizes
 

[ 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.

Re: Theorem of expected stack sizes
 

[ 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.

Re: Theorem of expected stack sizes
 

[ 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.

Re: Theorem of expected stack sizes
 

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.

Re: Theorem of expected stack sizes
 

[ 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.

Re: Theorem of expected stack sizes
 

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.