|
|
#22
05-17-2005, 11:38 AM
|
|||
|
|||
Re: This was a Clarkmeister, was it?
[ QUOTE ]
[ QUOTE ] despite the fact is -EV any way you play it, betting out is less -EV then just check-calling because you have some fold equity.[/b] [/ QUOTE ] Obviously betting out is better than check-calling. But if they're both -EV, check/fold. [/ QUOTE ] I don't think this is right, as it would invalidate his theorem. Open-folding is 0 EV too. Should you do that every time a 4-flush hits on the river? 
|
|
#23
05-17-2005, 11:40 AM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
[ QUOTE ]
Say I'm sandwiched between 2 players with a nut flush draw. They cap the flop and turn. I miss. Should I bet the river because it's the only chance I have to pick up the pot. [/ QUOTE ] This I would fold. But heads up with a four flush I think there is often value in a river bet.
|
|
#24
05-17-2005, 11:47 AM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
[ QUOTE ]
[ QUOTE ] [ QUOTE ] despite the fact is -EV any way you play it, betting out is less -EV then just check-calling because you have some fold equity.[/b] [/ QUOTE ] Obviously betting out is better than check-calling. But if they're both -EV, check/fold. [/ QUOTE ] I don't think this is right, as it would invalidate his theorem. Open-folding is 0 EV too. Should you do that every time a 4-flush hits on the river? [/ QUOTE ] Can you tell me why you're following a theorem you think tells you to take -EV action?
|
|
#25
05-17-2005, 11:50 AM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
Since you asked [img]/images/graemlins/smile.gif[/img]
I believe to calculate the odds an opponent has a club (since the two cards are not mutually exclusive) you need to determine the odds that your opponent does not have a club and subtract that from 100%. 1-36/45*35/44=.3636 or around 36.4%. Even if the opponent always calls with any club the river bluff will be successful nearly 2/3 of the time. Most opponents would call with more hands but still it doesn't have to be successful that often (and is easily folded to a raise).
|
|
#26
05-17-2005, 11:52 AM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
[ QUOTE ]
Since you asked [img]/images/graemlins/smile.gif[/img] I believe to calculate the odds an opponent has a club (since the two cards are not mutually exclusive) you need to determine the odds that your opponent does not have a club and subtract that from 100%. 1-36/45*35/44=.3636 or around 36.4%. Even if the opponent always calls with any club the river bluff will be successful nearly 2/3 of the time. Most opponents would call with more hands but still it doesn't have to be successful that often (and is easily folded to a raise). [/ QUOTE ] You correctly calculated the chance that a random hand has a club, but the chance he has a club or any other hand he'll call with is drastically increased by the fact that he hasn't misseed a single bet postflop.
|
|
#27
05-17-2005, 12:04 PM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
[ QUOTE ]
[ QUOTE ] [ QUOTE ] [ QUOTE ] despite the fact is -EV any way you play it, betting out is less -EV then just check-calling because you have some fold equity.[/b] [/ QUOTE ] Obviously betting out is better than check-calling. But if they're both -EV, check/fold. [/ QUOTE ] I don't think this is right, as it would invalidate his theorem. Open-folding is 0 EV too. Should you do that every time a 4-flush hits on the river? [/ QUOTE ] Can you tell me why you're following a theorem you think tells you to take -EV action? [/ QUOTE ] If I understand Clarkmeister's idea correctly, then no matter what you do, it will be -EV. Betting just happens to have an EV that is the least negative. This all assumes that the 3 conditions for the theorem are met. Namely: 1. you're HU 2. You're in first position 3. a 4th flush card hits on the river. I also think it is assumed that you actually have a hand worth showing down. So, if clarkmeister is correct, then check/folding must be -EV as well.
|
|
#28
05-17-2005, 12:05 PM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
folding always has an EV of 0.
Krishan
|
|
#29
05-17-2005, 12:06 PM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
[ QUOTE ]
[ QUOTE ] [ QUOTE ] [ QUOTE ] [ QUOTE ] despite the fact is -EV any way you play it, betting out is less -EV then just check-calling because you have some fold equity.[/b] [/ QUOTE ] Obviously betting out is better than check-calling. But if they're both -EV, check/fold. [/ QUOTE ] I don't think this is right, as it would invalidate his theorem. Open-folding is 0 EV too. Should you do that every time a 4-flush hits on the river? [/ QUOTE ] Can you tell me why you're following a theorem you think tells you to take -EV action? [/ QUOTE ] If I understand Clarkmeister's idea correctly, then no matter what you do, it will be -EV. Betting just happens to have an EV that is the least negative. This all assumes that the 3 conditions for the theorem are met. Namely: 1. you're HU 2. You're in first position 3. a 4th flush card hits on the river. I also think it is assumed that you actually have a hand worth showing down. So, if clarkmeister is correct, then check/folding must be -EV as well. [/ QUOTE ] you are confusing -EV on a single street with a -EV play. given a non-zero pot on the river, a -EV for one street play can become +EV (as in calling one bet in a 20 BB pot with only a 10% chance to win). check-folding is *always* zero EV.
|
|
#30
05-17-2005, 12:16 PM
|
|||
|
|||
|
Re: This was a Clarkmeister, was it?
[ QUOTE ]
you are confusing -EV on a single street with a -EV play. given a non-zero pot on the river, a -EV for one street play can become +EV (as in calling one bet in a 20 BB pot with only a 10% chance to win). check-folding is *always* zero EV. [/ QUOTE ] Allright, so in clarkmeister's theorem, betting is -EV, but the whole street is +EV? Is that what youre saying?
|
 |
|
|
Linear Mode
