Hitting the Flop

[tubbyspencer]

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]

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]

Is there a website or some (free) tool I can use to calculate those C(x,y) values?

[Lost Wages]

Try here. Click start and then use the combinations tab on the calculator.

You can also use the COMBIN function in Microsoft Excel.

Lost Wages

[jtr]

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.