View Full Version : Hitting the Flop
tubbyspencer
08-12-2004, 12:20 PM
Hypothetical: 3 Players see a flop with different unpaired cards (AQ and K9 and 78, for eg.)
What is the chance that the flop will pair none of them?
How about 2 Players or 4 Players?
Lost Wages
08-12-2004, 12:54 PM
For 3 players:
There are 46 unseen cards so C(46,3) = 15180 possible flops.
There are 7 ranks that will pair no one for a total of 28 cards.
C(28,3) = 3276 flops that pair no one.
Probability = 3276/15180 = 21.6%
Lost Wages
tubbyspencer
08-12-2004, 12:58 PM
Is there a website or some (free) tool I can use to calculate those C(x,y) values?
Lost Wages
08-12-2004, 01:58 PM
Try here (http://www.wiley.com/college/mat/gilbert139343/java/java05_s.html). Click start and then use the combinations tab on the calculator.
You can also use the COMBIN function in Microsoft Excel.
Lost Wages
That C(x,y) stuff is not hard to do with a standard calculator or even a pen and paper. If take the notation "N!" to mean the number N x N-1 x N-2 ... x 3 x 2 x 1, then C(A,B) is just A! / ( (A-B)! x B! ). So the number of distinct hole card combinations, for example, is C(52,2) which is 52! / 50! x 2!. Which is easy, as most of 50! can be factored out of 52! to give a final answer of 52 x 51 / 2 = 1326.
vBulletin® v3.8.11, Copyright ©2000-2024, vBulletin Solutions Inc.