[Insty]

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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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I have dusted off my algebra and managed to derive this result.
(basically the +chips and the -chips cancel each other out.)

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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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I also managed to derive a similar, although less elegant formula. Which shows the same thing.
Your mathematics is obviously much better than mine.

I assert that for tournaments that pay out more than 2 places further placings may slightly alter the curve they will not fundamentally alter it's shape.
Therefore the result above holds for all (regular) payout structures.
I don't have the mathematical ability to prove this, so I challenge anyone to disprove it. [img]/images/graemlins/smile.gif[/img]

In summary:

I now agree with NoSelfControl and believe his proofs and mathematics to be good.

It is always -ev (or at best breakeven) to call.
Even if you have the bigger stack..

Insty
who needs to study some more mathematics.