Theorem of expected stack sizes

[PrayingMantis]

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

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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.

[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.

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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.