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Old 03-07-2007, 04:09 PM
Phat Mack Phat Mack is offline
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Join Date: Sep 2002
Location: People\'s Republic of Texas
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Default Re: How many callers does Buzz need?

[ QUOTE ]

[ QUOTE ]
But what if the missed clubs pair the board?

[/ QUOTE ]Only one club pairs the board on the turn (the ten).

[/ QUOTE ]

Buzz,

I'm sorry, I did a terrible job on my previous post. I'll try to correct it. "Missed club" was a poor choice of words. I should have written "non club."

My concern wasn't with the way the odds for hitting 4th street (8/44) were calculated, it was with the way the odds for hitting 5th street were calculated by multiplying the complement of 8/44 (== 36/44) by the P of a good card coming (8/43).

My problem with 36/44 is that the 36 contains 8 cards that pair the board. If any of these 8 cards come on 4th street, you will drop the hand. Therefore, I think it best to subtract these 8 cards, and calculate the 5th street odds as 28/44 * 8/43.

jmo

Mack

edit:
[ QUOTE ]
You think Hero should have four opponents to call the bet. I certainly like four better than two, but four is not what I was thinking at the time. In retrospect and to be on the safe side, from the viewpoint of getting proper odds to initiate fresh money into the pot, maybe Hero needed three opponents instead of just two.

[/ QUOTE ]

Three opponents is what I was thinking, but I don't have a problem with the bet. I think being the aggressor makes up for some of the bet equity...
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