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  #21  
Old 08-28-2006, 12:00 PM
Aaron W. Aaron W. is offline
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Default Re: 4-flush and two villains

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Why are you 95% sure of a flush?

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It's just an estimate. It's very hard to come up with many situations where netiher player has a diamond in his hand.

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How low would the chance of somebody holding a [img]/images/graemlins/diamond.gif[/img] have to be for bet-fold to be the correct line?

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If you understand the EV calculations, you should be able to work this out. It's a good exercise if you haven't done many of these. (It's actually quite easy.)
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  #22  
Old 08-28-2006, 01:52 PM
gostros gostros is offline
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Default Re: 4-flush and two villains

My first posted EV calculation...

The Bet-Fold Threshold:

EV = 0 = x * (-1) + (1-x) * (6.25)

x = 86.2%

One would have to be less than 86.2% sure that one of the two opponents has a diamond to bet-fold the turn. I think that makes it pretty close, no?
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  #23  
Old 08-28-2006, 02:15 PM
threads13 threads13 is offline
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Default Re: 4-flush and two villains

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[ QUOTE ]
I don't mind the turn check, you aren't getting anyone to fold that's ahead of you. I doubt you are good 1 time in 8 on the river.

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I disagree.

I think the b/f line is much better on the turn. Here's why - basically, on a 3 flush flop, villain will lead out with many non-flush or non-FD made hands like 2 pair or TP or a flopped str8 - he even sometimes bets out with his flopped OESD. A flop of 9TJ will hit a ton of hands that someone limps with UTG. Basically, the line is b/f because there's such a good chance the other two are not liking the diamonds too much and you may even get a better hand to fold. And it's a pretty easy muck when you get c/r.

And to the poster who thinks BB is obviously on the Nut flush draw after calling two cold: He is the BB. He did call a raise PF but that doesn't mean he did it with the Ad or even the Kd or Qd.

[/ QUOTE ]

On this flop, what did the BB call 2 cold with?
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  #24  
Old 08-28-2006, 02:17 PM
Aaron W. Aaron W. is offline
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Default Re: 4-flush and two villains

[ QUOTE ]
My first posted EV calculation...

The Bet-Fold Threshold:

EV = 0 = x * (-1) + (1-x) * (6.25)

x = 86.2%

One would have to be less than 86.2% sure that one of the two opponents has a diamond to bet-fold the turn. I think that makes it pretty close, no?

[/ QUOTE ]

I would not consider it close. The difference between 19:1 and 6:1 is gigantic.

The 10% is deceptive if you don't think of it in light of being relative to a number on the order of 5%. There's an old gem about this sort of thing: Ed Miller would just love this one: KK in MP. Ironically, it's about folding an overpair. I've linked it directly to a post by GuyOnTilt, but it's worth reading the thread, too.
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  #25  
Old 08-28-2006, 02:43 PM
mwg47x mwg47x is offline
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Default Re: 4-flush and two villains

[ QUOTE ]
Why are you 95% sure of a flush?

An earlier poster noted that the flop action could be due to strong(ish) non-flush-draw holdings trying to protect their hands.

How low would the chance of somebody holding a [img]/images/graemlins/diamond.gif[/img] have to be for bet-fold to be the correct line?

[/ QUOTE ]

I think 95% is spot on. I think of it this way:
2 villians w/ 4 cards the chance of one of those cards being a [img]/images/graemlins/diamond.gif[/img] is 100% on average.
I am just learning the math associated with this stuff so bear with me here.
I am correct in thinking that in order for the call on the river to be +EV there has to be a least 21BB in the pot, AND we have to damn sure that the 5% of the time they don't have a [img]/images/graemlins/diamond.gif[/img] that our HIG?
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  #26  
Old 08-28-2006, 04:13 PM
Aaron W. Aaron W. is offline
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Default Re: 4-flush and two villains

[ QUOTE ]
I think of it this way:
2 villians w/ 4 cards the chance of one of those cards being a [img]/images/graemlins/diamond.gif[/img] is 100% on average.
I am just learning the math associated with this stuff so bear with me here.

[/ QUOTE ]

That's not true. Given four random cards out of a full deck, the most common number of diamonds is 1. This is much difference from saying there is one out there 100% of the time on average. (Think of it this way: The average sits between other values. Can you have a diamond out there 110% of the time?)

What information do you have? You know that there are 4 diamonds on the board and I don't have one, so there are 9 left in the deck. You also know that there are 2 non-diamonds out (in my hand) so that there are 37 non-diamonds left. The probability of there being some diamond in someone's hand is 1 minus the probability of there being no diamonds out.

(37/46)*(36/45)*(35/44)*(34/43) = 40.5%

So there's a 59.5% chance that someone has a diamond if you only know what cards are out there.

But we know more. We know that there was a flop donk and coldcaller. That makes this number unreliable because the calculation doesn't take the action into account.
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  #27  
Old 08-28-2006, 04:21 PM
mwg47x mwg47x is offline
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Default Re: 4-flush and two villains

I'm glad I posted. I was all wrapped around the axle and you set me straight.
Thanks Aaron!
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  #28  
Old 08-28-2006, 04:49 PM
gostros gostros is offline
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Default Re: 4-flush and two villains

[ QUOTE ]
AND we have to damn sure that the 5% of the time they don't have a [img]/images/graemlins/diamond.gif[/img] that our HIG?

[/ QUOTE ]
Actually, I don't think we have to be good if neither opp. has a [img]/images/graemlins/diamond.gif[/img]. In fact, I don't think we have to beat anything if nobody has a flush, because both opponents will almost surely fold.

Aaron- does the possibility that a low [img]/images/graemlins/diamond.gif[/img] flush will fold on the turn move the numbers around in any significant manner?

Also, thanks for clarifying the difference between 95% and 86%. I still am a little unclear as to why being 95% sure of something is very different than being 86% sure, but I haven't read that thread yet.
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