[Wolfram]

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i'm trying not to be results oriented, but if the chance we are winning against any of the callers is 1/3, we should have called.

e.g. P[beating all 3] = P[beating one]^3 = (1/3)^3 = 1/27, and we are getting 26.5 to 1.

just thought i'd put that out there.

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thats so wrong heh.
If we are beating one caler 100% of the time and another 0% of the time, we still lose the pot.

Results?

Donker had the 5s2s
UTG(maniac) had KdTh
Precitable TAG had KhQs

I was in fourth place on the river.

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His math is right, but his premise is very unlikely (i.e. we're never 1/3 against all the callers respectively in this spot).

Edit: Bah, I post late.