another easy q?...

[ryan1986]

I was working out the total number of flops possible and came up with 50*49*48=117600, this seems to be all the permutations possible since you have 50 outcomes for the first card and so on...but most people seem to post that 19600 is the number of possible flops, I understand this is a combination of 50, 3 ways but I cant seem to understand how to reach this value except dividing my value by a factor of 6...and if this is the correct way to get that value what is the reasoning behind dividing by six? any help appreciated thanks!

[Siegmund]

Yes, dividing by 3! = 6.

All three cards are turned over at once... so it makes no difference if the flop is A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, or C-B-A.

In general, for combinations, the 'top' (50*49*48 here) is the number of ways of selecting the objects one by one, and the 'bottom' is number of different sequences the elements of a particular set can be selected.

[Stephen H]

The answer is, do you care about the difference in flops:
AsKsQs
AsQsKs
KsQsAs
and so on? Or are they the same? For poker purposes, they're completely the same, so use the combinatorial value instead of the permutation. The reason it works out to dividing by 6 is that there are 6 permutations to the order the 3 flop cards can come: AKQ,AQK,KAQ,KQA,QAK,QKA.
The formula for N choose R: N!/((N-R)! * R!)
or in this case 50! / ((50-3)! * 3!) = 50! / (47! * 3!) = (50*49*48)/(3*2) = 19600

[ryan1986]

finally understand...much thanks
I was having trouble figuring out that formula, much more difficult than just dividing by 6 [img]/images/graemlins/smile.gif[/img]