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

[Oink]

No its not.

Hero is not good 3% or whatever is enough. Maybe Hero will win this 1 in a thousand. But certainly not 1 in 30

The skalansky stuff you quote are for people who cant read hands. If they always call here they wont make a mistake in another spot where they fail to see why they might be ahead.

In this pot Heisenberg isnt winning. EVER

[numbnuts007]

I would have lost a 620 pot. I've got to pay off this river. I absolutely understand the fold, can't say it's wrong, i support the position, but the fact that I will be good sometimes combined with how bad I will tilt if I fold the winner, i've got to call. Maybe this makes me a sd monkey, but so be it.

In reality, I think alot more people call in this spot than it seems from this thread. Whether or not it's the right thing to do is obviously a different issue.

Tough hand Heisenberg, I'm tilting just thinking about being in that spot.

[numbnuts007]

I'm revising my previous post. A fold is correct. I still call, I don't think I can back out of that, but a fold is correct. I can't think of three hands that could withstand that much action, that you beat (unless there are 5 or 6 Qs in the deck). I can't say a call is THAT bad though. If not from a mathematical perspective, at least from a pshycological one.

[waffle]

Results plz

[yellowjack]

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.

[Enervate]

Sick hand

[Heisenb3rg]

[ QUOTE ]
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.

[/ QUOTE ]

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.

[yellowjack]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

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.

[/ QUOTE ]

it was a figure i threw out there
in addition, i just took chance of beating one person as 1/3, i didn't use the 0% chance anywhere at all

[Heisenb3rg]

just saying the math is very misleading..

Id estimate the chances im ahead of SB are about 1/100, the chances im ahead of maniac are like 5/6 of the time and the chances im ahead of the overcalling TAG are like 1/1000.

Multiply those fractions together, and thats the chance im winning the hand.

Even though my chances against one of hte callers is very good (how u phrased it)

[Wolfram]

[ QUOTE ]
[ QUOTE ]
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.

[/ QUOTE ]

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.

[/ QUOTE ]
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.