Permutation Question, Please help

[tbradley77]

I'm trying to figure out exactly how to calculate % with permutations. Say you have two suited cards. What are the cahnces you slop a flush draw? I know the answer should be 10.95%, because the cahnces of flopping a flush is .84 and the chance of flopping at least 2 of the same suite is 11.79. But every way I calculate it isn't working. Please help. I was just doing 11/50*10/49 but that doesnt factor in the third card, which really doesn't matter what it is

[Cooker]

This is how you calculate it:

percent to flop a flush draw = 3 * (39*11*10)/(50*49*48) = 0.1094 or 10.94%. Note this is to flop exactly a flush draw not including the odds to flop a flush.

How do you get the part on the right? Obviously, the denominator is simply the total number of possible flops given your 2 cards and the numerator is 39 unsuited cards * 11 suited left for the 3 flush * 10 remaining suited cards to give a 4 flush. The factor of 3 comes from the fact that there are three ways the 2 flush cards can be arranged among the 3 cards.

One thing that confuses a lot of people is that the way I calculate it considers the flop 8/images/graemlins/heart.gif K/images/graemlins/diamond.gif T/images/graemlins/heart.gif as different from K/images/graemlins/diamond.gif 8/images/graemlins/heart.gif T/images/graemlins/heart.gif. This isn't a problem, because I do this multiple counting in both the numerator and denominator (that is why I must include the mystery factor of 3 in the front). Basically, you must always multiple count or never multiple count, but be careful not to mix the 2. And also be careful that you understand what numbers someone is giving you.

[jtr]

Think about the flops you're interested in as a proportion of all possible flops.

First: all possible flops after you've seen two hole cards are 50 x 49 x 48 = 117600. (Note that this gives us an ordered list of all the possible flops. Ac Jd 7h is different from Ac 7h Jd. If we wanted to say that order doesn't matter we need to recognize that each ordered flop can be trivially rearranged in 3x2x1 = 6 ways, so we would divide by 6 to get 19,600 flops where order doesn't matter.)

Sticking with the ordered list of flops for a moment, how might a flush draw come out? We assume of course that your two hole cards are suited. Now, if "F" represents your flush cards, and "X" represents some other card, it could be FFX, FXF, or XFF. Can I confirm that you're not interested in FFF, the made flush, right? We're just talking about odds of flopping a flush draw here.

So, for the first possibility, FFX: the first flush card could be any of the remaining 11 cards of the suit. So 11 choices there. The second flush card could be any of the remaining 10. And the final non-flush card can be any of the remaining 39 cards in the deck that are not in your suit. 11 x 10 x 39 = 4290 "FFX" flush draw flops for you.

By similar logic there are 4290 of the other two orderings in which your flush draw could come out. So now we have 4290 x 3 = 12870 (ordered) flops that give you a flush draw.

All we need to do is to compare that number with the total number of ordered flops. 12870 / 117600 = 10.944%, giving you your odds of flopping a flush draw (but not including the chances of flopping a made flush).

[jtr]

Bah, nice work Cooker, you beat me to it. Shall we open a Permutation Academy for Beginners together?

[Cooker]

I think it odd how similar our posts were. Specifically that we both addressed the point of ordered vs. unordered to some extent when it really wasn't really needed. I find your discussion more clear. I think it was because I was in a hurry when I typed mine.