How many callers does Buzz need?

[Phat Mack]

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But what if the missed clubs pair the board?

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

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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

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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.

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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...

[Phat Mack]

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Isn't the main reason to bet the flop so that we fold out the tight players behind us whom hero says are very likely to have A2 so that we win if the backdoor low comes in?

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Yeah, we want to fold some hands, A2 for example, but not others: We want to fold the nut flush draw, but let the Q-high flush draw draw; If were against a set and two 2-pairs, we want them all in there. The trouble is, we often drop the chasers we want (the dead money), and get stuck with those we don't.

In my opinion, this might be a consideration if we were contemplating a raise, but here, there's not much we can do. It might be a valid argument that the only ones we will drop will be the ones we want to stay, but I still think it's best to bet here, for the reasons Buzz mentions.

jmo

[Buzz]

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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,

[/ QUOTE ]Hi Mack. You are right. That is a better way to do it.

Thanks.

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and calculate the 5th street odds as 28/44 * 8/43.

[/ QUOTE ]I have to think about that part. Haven't had my coffee yet.

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I think being the aggressor makes up for some of the bet equity.

[/ QUOTE ]Agreed. And actually that was on the fringe of my thinking at the time.

Buzz