Probability in Texas Hold'em

[Malakor]

I've been working on this question for quite some time, and would like to know if you guys could tell me if I'm on the right track or if I'm missing something:

"If the flop has 3 of the same suit, what are the odds that AT LEAST one of the 5 players (2 cards each) has a flush?"

I believe that I need to find out the probability of nobody having a flush and then calculate "1 - result".
An idea I have is to calculate:

Nobody has any of the flop suit:
$\displaystyle C(39,2) * C(37,2) * C(35,2) * C(33,2) * C(31,2) = 72093527906400$
+
One person has one of the flop suit:
$\displaystyle C(10,1) * C(39,1) * C(38,2) * C(36,2) * C(34,2) * C(32,2) = 48062351937600$
+
Two people have one of the flop suit:
$\displaystyle C(10,2) * C(39,1) * C(38,1) * C(37,2) * C(35,2) * C(33,2) = 27907172092800$
+
Three people have one of the flop suit:
$\displaystyle C(10,3) * C(39,1) * C(38,1) * C(37,1) * C(36,2) * C(34,2) = 13953586046400$
+
Four people have one of the flop suit:
$\displaystyle C(10,4) * C(39,1) * C(38,1) * C(37,1) * C(36,1) * C(35,2) = 5919703171200$
+
Five people have one of the flop suit:
$\displaystyle C(10,5) * C(39,1) * C(38,1) * C(37,1) * C(36,1) * C(35,1) = 2089307001600$

All possible deals:
$\displaystyle C(49,2) * C(47,2) * C(45,2) * C(43,2) * C(41,2) = 931901075582400$

Which gives:
$\displaystyle 2089307001600 + 5919703171200 + 13953586046400 + 27907172092800 + 48062351937600 + 72093527906400 = 170025648156000$

and divide the sum by all possible deals:
$\displaystyle \frac{170025648156000}{931901075582400} \approx 18.25\%$

And inverting gives:

$\displaystyle \frac{761875427426400}{931901075582400} \approx 81.75\%$

But this fraction seems a bit too high, am I doing something wrong or is the probability really that high?