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I've read that given 2 suited cards your chances of improving to a 4 flush by the flop is 10.9 percent. I'm trying to verify this for myself mathematically. Given that C(n,r) is the number of r-combinations of a set with n-elements does the following make sense?

P(x) = ( C(11,2) * (50 - 11) ) / C(50,3)

Where P(x) is the prob. if drawing 3 cards having 2 of the same suit from the 50 remaining cards.

Thx

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I don't think it's a combination problem as much as it's a pure probability problem:

To get a four flush after the flop, having started with two suited cards would go as follows.

There are three ways the flop can produce two same suited cards: 1st & 2nd, 1st & 3rd, 2nd & 3rd. Add those probabilities and subtract the probability of all three being the same suit, and you should have your answer.