Re: Theorem of expected stack sizes
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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?
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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.
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