Re: This was a Clarkmeister, was it?
 

[ QUOTE ]
Clarkmeister correct applied?

[/ QUOTE ]

No - this is just a river bluff.

Sarge[img]/images/graemlins/diamond.gif[/img]

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?

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.

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?

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).

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.

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.

Re: This was a Clarkmeister, was it?
 

folding always has an EV of 0.

Krishan

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.

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?