#1
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Hold \'em Flop Probability
I'm sure this is right out of Probability 101, but I'm clueless on statistical formulas, so...
In a hold 'em game, if you have two unsuited unconnected cards, what are the odds of flopping two pair, a set, quads or a fullhouse. I've read flopping two pair is 49 to 1, but I'm curious how the odds improve with the other (unlikely) additional favorable flops. Thanks. Sparks |
#2
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Re: Hold \'em Flop Probability
Given starting hole cards X and Y where X > Y, the probability of flopping:
2 Pair using both hole cards: Flops of XYZ = 3*3*44 = 396 Probability = 396/19600 = 2.02% Trips using one hole card: Flops of XXZ = ((3*2)/2)*44 = 132 Flops of YYZ = ((3*2)/2)*44 = 132 Probability = 264/19600 = 1.35% Full House: Flops of XXY = ((3*2)/2)*3 = 9 Flops of XYY = ((3*2)/2)*3 = 9 Probability = 18/19600 = 0.0918% Quads: Flops of XXX = 1 Flops of YYY = 1 Probability = 2/19600 = .01% Total probability = 3.47% = 28:1 against Lost Wages |
#3
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Re: Hold \'em Flop Probability
Good stuff. Is there a book (or website, or other resource) that details this sort of common math calculation for hold em? I've read lots of good books, but even the really great ones seldom get into specific probabilities like this one.
Thanks. |
#4
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Re: Hold \'em Flop Probability
The best book is Holdem's Odds Book available through Gambler's Book Shop.
Best website is Alspach's Mathematics & Poker Page. Lost Wages |
#5
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Re: Hold \'em Flop Probability
Thanks, I just ordered the book. (I noticed it was fifty bucks at Amazon, twenty five from your link.) I appreciate the tip.
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