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$50 SnG bubble play.


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#101 Actuary

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Posted 26 June 2006 - 07:08 AM

View PostRocketwadster, on Monday, June 26th, 2006, 4:18 AM, said:

Sorry, I had a ton of other things to do this weekend that didnt involve plagiurism...lol
your little dig at me earlier sounds so silly now.

#102 Rocketwadster

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Posted 26 June 2006 - 07:18 AM

View PostActuary, on Monday, June 26th, 2006, 7:08 AM, said:

your little dig at me earlier sounds so silly now.
I forgot all about it. When I get home tonight, I will try to remember to type it up.

#103 Rocketwadster

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Posted 27 June 2006 - 05:26 AM

View PostSmasharoo, on Friday, January 7th, 2005, 10:59 PM, said:

I rarely play these, but was motivated by all the posts here about how soft they are to give it a shot.Stack sizes are approximate but to scale. Blinds are 300/600.I'm on the BB in 2nd with T4000SB has 6000Button and UTG have ~2500 each.Folded to SB who pushes. I have QQ.What's my play?
After having reviewed Harrington on Hold'em Volume III, I will try to follow Harrington's logic to determine our course of action:Tried. Got stuck when figuring out the third place probabilities. Anyone know how to figure that out? Here is what I have so far:Player Stack First Second UTG 2500 0.167 0.205 Button 2500 0.167 0.205 SB 6000 0.400 0.305 BB 4000 0.267 0.284 15000 1 1 Got mixed up when trying to figure out the third place...

#104 tallytownFSU

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Posted 27 June 2006 - 05:54 AM

Why do you think folding is such a strong option? Id much rather him run 1 card against my made 2.IMO.Might be my live poker talking.
*drags a pot*

"Figjam baby, figjam."

"Hey JJ, whats Figjam mean?"

"Fcuk I'm good, just ask me...."

#105 Rocketwadster

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Posted 27 June 2006 - 10:49 AM

Does anybody around here know how to figure out the probabilities of placing third? First and Second I was able to do fairly easy, but I keep getting the wrong values for third for some reason, and I don't have HOH I, II, or III with me. If memory serves, Action Dan says to take the chip stack of the person you are calculating, and divide it by the total number of chips remaining of each person left (which is only two people for the calculations, as we are assigning two people as coming in first and second). There are 12 combos for first and second with 4 people (I think), BUT, when looking at the chances that SB finishes third for example, isn't there only 6 combinations we need to be worried about (A-B, A-C, B-A, B-C, C-A, C-B)? This is where I am getting confused. My first attempt at doing these calculations, so bear with my probably stupidity...




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