Permutation Question on drawing cards, Please help

[tbradley77]

I'm trying to work up draw odds for a statistics paper, and having a little trouble. Say you ahve two suited cards. I know from looking online the chance of flopping 4 to flush or a flush is 11.79% and the chance of flopping a flush is .84 leaving the chance of a flush draw at 10.95%. How do I calculate the 10.95% chance of a flush draw.

Another example. You ahve suited connectors, what are chances of flopping a straight draw. I know the logic. There are 8 cards that help you at first, and then 4 cards for the next card, and there are 3 different possibilites of drawing an open ended draw, but I don't know how to factor in that there are three cards drawn but you only need two cards to hit something.

[MickeyHoldem]

The number of possible flops is 50 choose 3 = 19600
(50 choose 3) is sometimes written 50c3 or c(50,3) and meens the ways to choose 3 objects (cards) out of 50.

so when you have 2 suited cards you will flop the flush 11c3 / 50c3 = 0.008418367

to flop a flush draw you need 11c2 * 39 / 50c3 = 0.109438776

For the straight... let's say you have 56.. you can flop a straight 4 ways with 234, 347, 478, or 789... this would be = 4*4*4*4 / 50c3 = 0.013061224

open-ended draws occur 3 ways with a 34x, 47x, or 78x... the remaining card can be any card that doesn't complete the straight (there would be 42 of these cards in each situation) so this = 3*4*4*42 / 50c3 = 0.102857143

[Cobra]

Mickey I would like to clarify something either you or I are doing wrong.

In your example you say that hitting 34x would be 4*4*42 for the 42 remaing cards. I believe that by doing it this way you are double counting the three and four. I think you should exclude the 7,2 they make a straight and the 3,4 for the x. This would leave 34 remaing cards. Then you would add back in the times that you hit two three's or two four's. So it should read for 3,4,x

=4*4*34 + 2*(4 c 2)*4 = 592 boards not the 672 boards you come up with.

We also did not add in the boards leaving a double gut shot, specifically the 9,7,3 and 8,4,2 boards. I believe the answer should be.

=3*(4*4*34 + 2*(4 c 2)*4) + 2*4*4*4= 1904 boards/ (50 c 3)

= 9.7 %

Cobra

[MickeyHoldem]

Damn... you right!... this will double count the 334 and 344 flops!

Also looking at that Omaha problem.. I over counted some hands there too...