[Insty]

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Here's what I know so far.

1. Assume you have the option to call (but not raise) the final bet by the
only other player still in the pot. Imagine a bet putting one of you
all-in or a bet on the river.
2. There may be dead money in the pot from blinds or earlier bets.
However, assume the bet is "fair," i.e. the pot odds equal the odds against
your winning the bet.
3. And so far, unfortunately, assume the tournament pays only two places with at least three still in it.

The ICM then implies that calling the bet has negative EV.


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I disagree.
With dead money in the pot it should be possible to construct a situation where calling has a positive ev. Unless the stuff about it being "fair" means this isn't possible, in which case just simplify the problem by assuming 0 dead chips.

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Sketch of the proof. Your total EV is comprised of several terms.

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Ok.

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It is fairly easy to check that your expected value for finishing first or for
finishing second when anyone except the bettor finishes first is the same
whether or not you call the bet.


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I don't see how this works, can you explain this further?

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The only remaining term is when the
bettor finishes first and you finish second. Here your EV decreases if you
call the bet. In a nutshell, this is because f(z)=(y+z)(x-z)/(1-y-z) is
concave down (x represents your percentage of the total chips, y the
bettor's, and z your loss, which could be negative).


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Where does this function come from?
I agree that it is -ev if you might be knocked out, but am I mistaken when I say that it is never -ev to call if you have more chips than the villain?

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This same proof shows that raising on the river if you are certain your
opponent will call is worse than calling.

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I'll have to think about this.