I don\'t get the Bernoulli Equation right
 

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.

Re: I don\'t get the Bernoulli Equation right
 

Your math is perfect, your interpretation slightly off. You are computing the probability of making exactly two of your flush draws.

If you want the probability of making at least two of your flush draws, you have to sum the results of your equation from two to eleven or forty (you'll save time by computing for zero and one, then subtracting that total from 1). There's also a Normal approximation that doesn't require summations.

If you want the expected number of successful draws, just multiply the probability by the number of attempts.

Re: I don\'t get the Bernoulli Equation right
 

In addition to the above post, the probability of something occurring is never going to be more than 100%. Even the probability of flipping a coin a million times and getting at least one heads is less than 100%. Same with your situation.

Re: I don\'t get the Bernoulli Equation right
 

Thanks for all the answers.