Re: ICM problems
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This leaves open the interesting possibility that the general statement of the ICM might reduce to a relatively simple formula, through the application of higher mathematics. Does anyone know if this has been attempted? Or if it's been shown to be impossible?
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I've tried and failed so far, but some simplifications could be possible.
One way of ruling out some levels of complexity is to work out examples exactly. Some of the denominators are huge, so any simplification would still have to be able to produce huge denominators, too.
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Does anyone have a good (ICM) approximation? I found one which is computable, but which does not preserve the property that the matrix of finishing probabilities has the property that each row and column sum to 1.
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Is this approximation available anywhere?
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No, at least not from me. Others could have developed it independently, as I just made some crude approximations and renormalized some probabilities to 1.
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Another tractable problem related to the ICM is showing that you always lose E$ when you take an EChip-neutral gamble. As I recall, no one posted a proof on a past thread where this was conjectured. Maybe it would be good to gamble to knock out the short stack on the bubble when you have the second or third stack, but the ICM doesn't seem to say this.
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Isn't this another way of saying that for any case of the ICM, the curve is always convex end-to-end?
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I think the curve you mean is the result of taking chips from all opponents equally or in proportion, but I refer to gambling against each opponent individually. I expect that a counterexample could be constructed easily if you allow sidepots, but perhaps multiway gambles are also ok if there are no sidepots.
I didn't specify the payout structure, and I think it should hold for all decreasing structures. It suffices to prove it for satellites, i.e., that the probability of reaching the nth place or higher always decreases when you take an EChip-neutral gamble.
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