[SwissPoker]

I understand this ...

If I throw a dice 11 times, what is the probability that 8 times I throw a six?

C(11,8) * (1/6)^8 * (5/6)^3 = 0.00006 = 0.006%

Now, I thought I can apply this formula to the following question:

The scenario is draw poker. If I make 11 flush draws, what is the probability that I make my flush two times?

C(11,2) * (.1915)^2 * (.8085)^9 = 0.297 = 29.8%

But if I make 40 flush draws, what is the probability that I make my flush two times?

C(40,2) * (.1915)^2 * (.8085)^38 = 0.00887 = 0.88%

Should the probability not be more than 100% that I make my flush two times out of 40 times?

What am I doing wrong? Thanks for any help.