Estimating Chance You Are Beat On The Flop

[parappa]

(3/46)+(3/45)=.065+.067=13.2% chance your lone opponent was dealt a queen, right?

[srt19170]

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(3/46)+(3/45)=.065+.067=13.2% chance your lone opponent was dealt a queen, right?

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No, he's counting the number of starting hands involving a Q in a particular pre-flop strategy.

-- Scott

[gaming_mouse]

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Hi, Mouse. How did you get 17% of holding a Q?

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Total # of Sklansky pre-flop hand containing a queen divided by total # of Sklansky pre-flop hand. Using HEPFAP.

I'll emphasize again that this approximation does not take position into account and will only work for tight players. But my idea can be extended for variables like position and looseness.

gm

[gaming_mouse]

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Each permutation in the range has equal chance taking into account the cards the other plaers have. Just download them and see for your self they are free.


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Thanks. You are right, pokerstove can do the same calculation that I have done, plus give an exact estimate for a specific board.

But my question is really: Would it be worthwhile to memorize what these stats are for tight and loose players, and know the stats for each position?

I think it might be a starting point in some situations (especially agains unknown players), as long as it doesn't distract you from learning individual player idiosyncrasies.

Just curious for more thoughts.

gm