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Sparks
11-24-2004, 08:55 PM
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

Lost Wages
11-24-2004, 11:20 PM
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

larrondo
11-25-2004, 03:53 PM
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.

Lost Wages
11-26-2004, 12:31 AM
The best book is Holdem's Odds Book (http://www.gamblersbook.com/weblink.cby/detail/475401.html) available through Gambler's Book Shop.

Best website is Alspach's Mathematics & Poker Page (http://www.math.sfu.ca/~alspach/).

Lost Wages

larrondo
11-26-2004, 02:20 PM
Thanks, I just ordered the book. (I noticed it was fifty bucks at Amazon, twenty five from your link.) I appreciate the tip.