New to this 6-max thing, no peel right

[rzk]

hm, perhaps we are both right and are just talking about different things. let me make my statement even clearer:

suppose an optimal villain bluffs 24% of the time whatever that means. our EV of playing against an optimal villain is 0.

suppose villain A bluffs 40%, then he's exploitable and our EV against him is >0.

suppose we perfomed action X, as a result of which he decided to reduce his bluffs. suppose he reduced them to 26%. then he became closer to optimal and our EV against him became closer to 0.

isn't that completely obvious?

[Oink]

[ QUOTE ]
suppose an optimal villain bluffs 24% of the time whatever that means. our EV of playing against an optimal villain is 0.


[/ QUOTE ]

With a given hand or with a range of hands?

I understand you are talking about AT specifically right?

Well if what about the hands that cant call when he bluffs more than 24%?

Say he either bluffs 40% or 24%

Then a hand that cant call even when he bluffs 40% will make more money when he starts bluffing less as it will see more SD's and win


Also a hand that can call in both cases but wont valuebet will make less money when he bluffs less.

EDIT: a hand that beats the hands he stops bluffing

I cant make it any clearer with my english skills and with my state of mind.

BUT YOU CANT JUST CONSIDER THE EV OF AT WHEN HE BLUFFS LESS: WHAT ABOUT ALL THE OTHER HANDS IN OUR RANGE!

Sorry cant make myself clearer. 99.9% sure you are wrong now that the wine is loosing its effect and if you dont understand what I am talking about I can only advise for you to rtalk to someone who can explain it in your words

[rzk]

[ QUOTE ]
[ QUOTE ]
suppose an optimal villain bluffs 24% of the time whatever that means. our EV of playing against an optimal villain is 0.


[/ QUOTE ]

With a given hand or with a range of hands?

I understand you are talking about AT specifically right?


[/ QUOTE ]

no, of course not. i think that's where the misunderstanding lies. i'm talking about the EV of playing this villain as a whole, on all possible future hands.

the AT hand is the same as "action X" in my previous post. if we call down with it that will reduce the villain's bluffing frequency on future hands. in my example it got reduced from 40% to 26%.

my statement is that if on _future hands_ it will be 26% instead of 40% then on these future hands the villain will be less profitable to play against since he became closer to optimal.

[Oink]

[ QUOTE ]
my statement is that if on _future hands_ it will be 26% instead of 40% then on these future hands the villain will be less profitable to play against since he became closer to optimal.

[/ QUOTE ]

Thats the conclusion you cant make.

See my previous post

[rzk]

[ QUOTE ]
[ QUOTE ]
my statement is that if on _future hands_ it will be 26% instead of 40% then on these future hands the villain will be less profitable to play against since he became closer to optimal.

[/ QUOTE ]

Thats the conclusion you cant make.

See my previous post

[/ QUOTE ]

i thought your previous post assumed i was talking about only the AT hand, which i'm not.

if he starts bluffing closer to optimal then yes, as you mentioned in your previous post, some hands gain in EV and some hands lose in EV.
but the question is: has the overall EV increased or decreased?

if his strategy is now closer to optimal then it is less exploitable and less profitable to play against. i really don't see any room for disagreement here, the statement is true almost by definition.

the only situation when his making his bluff frequency closer to optimal is better for us is if he replaces his "bluffing-too-much" errors with some kind of other errors. in other words, he can start to bluff optimally but make other elements of his strategy so much worse that his new strategy is more exploitable than the old one. it doesn't sound like that's what you meant, but just in case, is it?

btw, sorry to argue for so long - i'm really surprised we are disagreeing on what i thought to be a simple issue.

[Oink]

[ QUOTE ]
btw, sorry to argue for so long - i'm really surprised we are disagreeing on what i thought to be a simple issue.


[/ QUOTE ]

No problem. I was drunk.

Closer to sober now



I think I have found the problem

[ QUOTE ]
if his strategy is now closer to optimal then it is less exploitable and less profitable to play against. i really don't see any room for disagreement here, the statement is true almost by definition.


[/ QUOTE ]

there you go

There is plenty of room for disagreement there!!!!

The fact that a player plays suboptimal will not necessarily increase the overall payoffs for other players!

I dont know why you would think that?

In a zero sum game that would be the case. But river play with a pot > 0 is not a zero sum game

I cant come up with any examples right now, but I remember reading several scientific articles on the subject presenting examples in which player A enjoys a lower (higher) payoff when player B plays less (more) optimal

Also, what definition are you talking about?

[rzk]

[ QUOTE ]
The fact that a player plays suboptimal will not necessarily increase the overall payoffs for other players!

I dont know why you would think that?

In a zero sum game that would be the case. But river play with a pot > 0 is not a zero sum game

I cant come up with any examples right now, but I remember reading several scientific articles on the subject presenting examples in which player A enjoys a lower (higher) payoff when player B plays less (more) optimal



[/ QUOTE ]

ok, if that's what you mean then i'm 99.9% sure i'm right [img]/images/graemlins/laugh.gif[/img] all those examples are not applicable here, but i need some time to figure out how to explain why. you can enjoy another glass of wine while i collect my thoughts [img]/images/graemlins/tongue.gif[/img]

[rzk]

ok, so all i have to do is prove that the river play with pot>0 is equivalent to a zero-sum game and you'll agree with me right?

well, here goes. before the start of the betting steal an amount equal to pot/2 from each player's wallet. their strategy for playing will of course remain the same, but now

winning of player A + winning of player B = 0.

so the new game is a zero-sum game with the same strategy as the original game.

[Oink]

You cant do that as the pot is distributed between the players depending on strategies chosen and strength of hands

But yeah I would have agreed [img]/images/graemlins/smile.gif[/img]

More formally: the transformation of the game you are suggesting is not neutral in the sense that the players payoffs are altered. So its a whole new game.

[rzk]

[ QUOTE ]
You cant do that as the pot is distributed between the players depending on strategies chosen and strength of hands

But yeah I would have agreed [img]/images/graemlins/smile.gif[/img]


[/ QUOTE ]

of course i can do that. but since you don't believe me i found this link for you:

http://ockenfels.uni-koeln.de/downlo..._leininger.pdf

on page 4 it says:

"A constant-sum game is equivalent to a zero-sum game". you can search for that text for an explanation.

an even better explanation is here (omg, why aren't long links posting correctly? so annoying):

<font class="small">Code:</font><hr /><pre>
http://www.google.com/search?hl=en&amp;safe=off&amp;q="the+strategic+equ ivalence+becomes+obvious+as+soon+as+one+realizes+t hat"
</pre><hr />