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  #11  
Old 10-11-2005, 10:29 AM
UMTerp UMTerp is offline
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Default Re: Hypothetical Question

[ QUOTE ]
You're kidding, right?? He'll call with Ax and pocket pairs and no more.

[/ QUOTE ]

Why would calling with a pocket pair be any more profitable for the BB than calling with a suited connector? Remember, aces are "outs" for him too.
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  #12  
Old 10-11-2005, 10:30 AM
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Default Re: Hypothetical Question

I'd be making $500.00/hr.
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  #13  
Old 10-11-2005, 10:30 AM
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Default Re: Hypothetical Question

I don't think the kings can beat any decent game at all.
At the very least the situation is not nearly
as clear-cut as the first few posters claim it to be.

It's all about the FTP.

If the other players are willing to make 'negative' EV
calls (raises?) preflop to set themselves up for the huge
EV of playing against face-up cards.

Poker is more about uncertainty and deception than cold
odds. And the kings give up way more than they get.
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  #14  
Old 10-11-2005, 10:31 AM
ALL1N ALL1N is offline
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Posts: 156
Default Re: Hypothetical Question

[ QUOTE ]
Definitely I agree if we are raised on any street, we should just call down. You don't reraise someone when your cards are face up. Furthermore, frequently on the turn, and always on the river, we should refrain from even *betting*.

The problem is that to answer David's question, we need to know what that "game theory optimal amount" for folding is. And that's very difficult, especially considering that it's board dependent.

[/ QUOTE ]

I don't think David was after an exact answer, and the game theory folds aren't going to save us much.

But if you want to work it out, the optimal amount to fold is 1/(1 + Pot), and this works fine on the river, but as you work back each street the 2 K outs have to be mixed in. Eh, goodnight!
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  #15  
Old 10-11-2005, 10:34 AM
ALL1N ALL1N is offline
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Default Re: Hypothetical Question

In a typical ten handed 30-60 game where you normally make $50 an hour, what would your win rate be if you were dealt two face up kings every hand and your opponents didn't collude?
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  #16  
Old 10-11-2005, 10:36 AM
ALL1N ALL1N is offline
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Posts: 156
Default Re: Hypothetical Question

[ QUOTE ]
Why would calling with a pocket pair be any more profitable for the BB than calling with a suited connector? Remember, aces are "outs" for him too.

[/ QUOTE ]

Aces are not outs. We aren't folding if an ace comes up (any more than the game theory optimum, that is).

Suited connectors flop draws, which suck in a 2BB pot when they've got no fold equity, or underpairs, which suck full stop.
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  #17  
Old 10-11-2005, 10:37 AM
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Default Re: Hypothetical Question

I don't think I suggested collusion, except maybe between
one's preflop self and postflop self. I do believe that
the more players who play correctly against the kings the
worse it will be for the kings, but thats not collusion either.
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  #18  
Old 10-11-2005, 10:39 AM
ALL1N ALL1N is offline
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Default Re: Hypothetical Question

[ QUOTE ]
I don't think I suggested collusion, except maybe between
one's preflop self and postflop self. I do believe that
the more players who play correctly against the kings the
worse it will be for the kings, but thats not collusion either.

[/ QUOTE ]

Ok, sorry, I misread. Anyway, give me an example of a situation where you think it would be profitable apart from AA or Ax/PP in the BB?
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  #19  
Old 10-11-2005, 10:43 AM
jason_t jason_t is offline
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Default Re: Hypothetical Question

[ QUOTE ]


1) Someone has AA. They will reraise preflop, and we will lose 4BB total calling down.

This will happen 9/221, so 9 * (-4.75) / 221 = -0.19 BB

[/ QUOTE ]

9/221 is not correct for two reasons.

Once Hero has been dealt KK at most two people can be dealt AA. Fixing on one player, the probability that he his dealt AA is (4 choose 2) / (50 choose 2). The sum of these probabilities for 9 people is 9 * (4 choose 2) / (50 choose 2) and is close to the right answer but it's wrong because it double counts the times that two players hold AA. The probability that two particular players hold AA is (4 choose ) / (50 choose 4) and multiply this by (9 choose 2), the number of ways to choose 2 players from 9 players. Hence the probability that our Hero ends up against AA is

9 * (4 choose 2) / (50 choose 2) - (9 choose 2) * (4 choose 4) / (50 choose 4) =.0439.
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  #20  
Old 10-11-2005, 10:48 AM
ALL1N ALL1N is offline
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Default Re: Hypothetical Question

Yeah you're right.. I was just trying to get an approximation down. Anyway, my error seems to amount to an overestimate of about $36.

edit - I've made lots of other assumptions/crude maths too in getting the approximation.
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