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Here's my solution: We sum over x. The ways of choosing x(>=6) out of 13 cards(of a suit) and ways of choosing remaining (13-x) cards out of (52-x) cards i.e. Prob= ($\sum$[4 * $13\choose x$* $52-x\choose13-x$]/$52\choose13$. The $\sum$ goes from x=6 to 13.

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    FYI, the usual term for "combinations" would be (for instance) "52 **choose** 13." In mathjax it's `$52 \choose 13$` which produces $52 \choose 13$. Meaning that you *choose* 13 objects out of 52 distinct objects. – 2017-01-21
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    Regarding your approach, it's definitely wrong, because you are double counting. For instance, every method by which you can choose 6 hearts and 7 diamonds is counted twice. I didn't check your actual formulas because your basic approach is wrong. – 2017-01-21
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    It's an interesting problem. I think offhand that solving it may involve [stars and bars](https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)) as well as combinations, but not sure without actually reasoning it out fully. – 2017-01-21

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You are double counting.

How many ways are there to get at least six hearts? That's $6$ up to $13$ hearst and then the rest are non-hearts, i.e., $${13\choose 6}{39\choose 7}+{13\choose 7}{39\choose 6}+{13\choose 8}{39\choose 5}+{13\choose 9}{39\choose 4}+{13\choose 10}{39\choose 3}+{13\choose 11}{39\choose 2}+{13\choose 12}{39\choose 1}+{13\choose 13}{39\choose 0}$$ The same goes for the other three suits. However, this double-counts all possibilities where more than one suit has at least six cards! Fortunately, it is not possible that three suits have six or more carsd among 13, so we only count the ways to have at least six hearts and at least six spades, say: $${13\choose 6}{13\choose 6}{26\choose 1}+{13\choose 6}{13\choose 7}{26\choose 0}+{13\choose 6}{13\choose 7}{26\choose 0} $$ Now to arrive at the correct count for the problem, multiply the result from the first formula by four (for the four suits), multiply the result from the second formula by six (for the six possible pairs of distinct suits) and subtract the latter from the former. Of course, ultimately divide by $52\choose 13$ to arrive at a probability.

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Generating functions offer a solution which is conceptually simple but requires a computer algebra system if we are to avoid a tedious calculation.

Consider the complementary problem, generalized slightly: What is the probability that $r$ cards drawn from a deck do not include at least 6 cards from one suite, i.e. all suites include no more than 5 cards each? There are $\binom{52}{13}$ possible hands, all of which we assume are equally likely. The number of hands with $r$ cards in which all suites have no more than 5 cards is $$a_r = \sum \binom{13}{x_1} \binom{13}{x_2} \binom{13}{x_3} \binom{13}{x_4} $$ where the sum is taken over all integers $x_1,x_2,x_3$ and $x_4$ with $0 \le x_i \le 5$ for all $i$ and $x_1+x_2+x_3+x_4 = r$. Let $f(z)$ be the ordinary power series generating function for $a_r$, i.e. $$f(z)=\sum_{r=0}^{\infty} a_r z^r$$ It is easy to see (with some experience working with generating functions, at least) that $$f(z) = \left(1 + \binom{13}{1}z +\binom{13}{2} z^2+\binom{13}{3} z^3+\binom{13}{4} z^4+\binom{13}{5} z^5 \right)^4$$ Expanding this polynomial with a computer algebra system, we find the cofficient of $z^{13}$ is $a_{13} = 504,328,942,832$. So the probability that a hand of 13 cards does not include any suite with more than 5 cards is $$a_{13} / \binom{52}{13} \approx 0.794202$$ and the probability that there is at least one suite with at least 6 cards is $$1- a_{13} / \binom{52}{13} \approx 0.205798$$