$10/20... Marginal 44 in MP...

[Catt]

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FWIW you only have to make up 5BBs (10SB) in the scenario you quoted, which with that flop action isn't actually too unreasonable.

Surf

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If he needs 22:1 (implied) to make the call profitable, but he has to put in 2 sbs, doesn't that mean he needs to win ~44 sbs in bets to make the call good? (44sbs total - immediate 21 sbs in pot = 23 sbs implied bets = ~11 BBs).

[MarkD]

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One thing to realize is that the fact I have to call if I don't raise, means that the raise is actually a lot more appealing. I'm really only putting in one extra SB with 18 already in there. It doesn't have to work much for it to be +EV.

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Except for the 62% of the time that one of your opponents has a queen or a 7 on this flop. Add to this probability the probability that someone could have AA, KK, ..., 55 and your raise could end up putting in significantly more than 1 small bet. Also, if you raise this flop I find it hard to believe that you won't often be putting in another two small bets on the turn.

I think you have been way oversimplifying this hand throughout the thread because you knew the result.

[Surfbullet]

Hey Catt,

You're right. My mental arithmetic was quite shoddy. [img]/images/graemlins/smile.gif[/img] thx.

Surf

[Surfbullet]

Hey MarkD,

How did you arrive at this 62% figure?

Thanks,

Surf

[MarkD]

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Hey MarkD,

How did you arrive at this 62% figure?

Thanks,

Surf

[/ QUOTE ]

1 - C(42,8)/C(47,8)

[oreogod]

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[ QUOTE ]
Hey MarkD,

How did you arrive at this 62% figure?

Thanks,

Surf

[/ QUOTE ]

1 - C(42,8)/C(47,8)

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I have no idea how to put work that equation into 62%. Its been a long time since I did math in a classroom.

[zephed]

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Every blue moon you win this hand unimproved all the other times you get shown a better hand or get sucked out on.

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I'm waiting for someone to actually say how often this is and show me some math behind it being incorrect. Then, I'll agree.

For now, I've seen nothing but people's gut reaction which is different than mine (I instantly raised) and that's not much of a discussion.

Someone prove me wrong please, it's very possible I am, but it is at least very close.

[/ QUOTE ]
It's not very hard to do. Even I can do it. Just line up all the possible hands you think they may have, figure out how many combos of each there are, and what your equity is against each. It's really just time consuming work.

Include hands that are on the edge of that range too, just to see if that it affects it.

Against multiple opponents, it's much harder to do. I usually just do this in HU situations because I can never tell what the donkeys are calling with. They could have two undercards and you can't be sure which 2.

[MarkD]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Hey MarkD,

How did you arrive at this 62% figure?

Thanks,

Surf

[/ QUOTE ]

1 - C(42,8)/C(47,8)

[/ QUOTE ]

I have no idea how to put work that equation into 62%. Its been a long time since I did math in a classroom.

[/ QUOTE ]

just go to google and type in:
1 - (42 choose 8)/(47 choose 8)

the theory of the equation is this:
There are 47 unkown cards after we have seen the flop and we have 4 opponents who each have 2 cards. Therefore, there are 47 choose 8 C(47,8) total possible combinations of cards that they hold. Now, if we want to know how many combos do not contain a queen or a 7 then that means 5 cards are missing from the total group so the numberator is C(42,8). This fraction, C(42,8)/C(47,8) is the probability that none of our opponents has a queen or a 7. I subtract that value from one and that is the probability that at least one of our opponent has a queen or a 7.