Re: 10/20 -QQ - 30 to 1 , folding middle set
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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