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  #31  
Old 10-26-2007, 07:55 PM
Heisenb3rg Heisenb3rg is offline
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Default Re: 10/20 -QQ - 30 to 1 , folding middle set

Wolfram his math is wrong.
Quote:

"if the chances of winning against ANY of the callers is 1/3 we should have called"

This is not true... its
p[beating all 3] = P[beating A] * P[beating B] * P[beating C]

they are all indepdent variables..
If B is > 1/3, the term on the leftside can still be < 1/30.
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  #32  
Old 10-26-2007, 08:46 PM
Wolfram Wolfram is offline
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Default Re: 10/20 -QQ - 30 to 1 , folding middle set

[ QUOTE ]
Wolfram his math is wrong.
Quote:

"if the chances of winning against ANY of the callers is 1/3 we should have called"

This is not true... its
p[beating all 3] = P[beating A] * P[beating B] * P[beating C]



[/ QUOTE ]
That is what he said? [img]/images/graemlins/confused.gif[/img]

Well, the way he worded it is confusing, but his mathematical formula is the same as yours I think.
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  #33  
Old 10-26-2007, 09:06 PM
MicroBob MicroBob is offline
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Default Re: 10/20 -QQ - 30 to 1 , folding middle set

[ QUOTE ]

Beat: I lost a 600$ pot

[/ QUOTE ]

[ QUOTE ]

also I never claimed I folded the best hand on the river for ppl assuming so...

[/ QUOTE ]


fwiw - you kind of led people into that assumption.
I too thought you had folded the best hand until you clarified later in the thread that you weren't trying to say that.

Given the maniac-crazy involved in the hand and the idea that the TAG is likely aware that he will call down with A-high on weird boards and other such nonsense I think you can call....but it's close-ish.

I've lost a few of those 'big pot 2 opponents so I can fold the river' type hands lately though which obviously sting so that may be swaying my judgement somewhat.

I've also won a couple of those 'there is no way I'm good here against these two opponents but I'm calling anyway because the pot is huge' type hands and those are always fun, pleasant surprises.
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