Sklansky and odds on bluffing
Hi, everyone, I am fairly new to poker and have begun reading Sklansky's theory of poker. I was wondering if anyone can translate this paragraph for me:
"When using game theory to decide whether to call a possible bluff--assuming your hand can beat only a bluff and assuming your judgement doesn't give you a hint--you must determine the odds your opponent is getting on a bluff. Make the ratio of the calls to your folds the same as those odds. If your opponent is getting 4 to 1 odds on a bluff, you must call randomly four out of five times to make that bluffing unprofitable."
All I need to know is what it means to "get 4 to 1 odds on a bluff" Does that mean 4 times it isn't a bluff and one time it is, or vice versa? Or does it mean neither of those two things. Much help appreciated.
Denouement
Re: Sklansky and odds on bluffing
Quote:
Originally Posted by denouement
All I need to know is what it means to "get 4 to 1 odds on a bluff" Does that mean 4 times it isn't a bluff and one time it is, or vice versa? Or does it mean neither of those two things. Much help appreciated.
Denouement
"4 to 1 odds on a bluff" means that he must be successful in his bluffing attempt at least 4 out of 5 attempts to break even.
Re: Sklansky and odds on bluffing
Quote:
Originally Posted by Aceofone
Quote:
Originally Posted by denouement
All I need to know is what it means to "get 4 to 1 odds on a bluff" Does that mean 4 times it isn't a bluff and one time it is, or vice versa? Or does it mean neither of those two things. Much help appreciated.
Denouement
"4 to 1 odds on a bluff" means that he must be successful in his bluffing attempt at least 4 out of 5 attempts to break even.
Don't you mean unsuccessful? I don't usually bluff at pots with 4x pot bets.
4:1 is the ratio between his bet and the pot he will win if it is successful. So it only needs to work 2 out of 5 times to be profitable and 1 out of 5 times to break even.
-'rilla