There are $6$ single-Heart cards and $1$ double-Heart card, so to get at least $n$ successes in $x$ draws, we must either
- Draw at least $n$ single-Heart cards and no double-Heart card, or
- Draw the double-Heart card and get at least $n-2$ single-Heart cards in the other $x-1$ draws.
In the first case, if we are to draw $k$ single-Heart cards, there are $\binom 6k$ ways to choose them, and there are $\binom 9{x-k}$ ways to choose the remaining $x-k$ cards from among the $9$ non-Heart cards, so there are $$\sum_{k=n}^x{\binom 6k\binom 9{x-k}}$$ ways in all.
We can do the second case in much the same way. The double-Heart cards must be chosen, and then we have $\binom 6k\binom 9{x-1-k}$ ways to choose $k$ single-Hearts and $x-1-k$ non-Hearts, giving $$\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}$$
There are $\binom{16}{x}$ ways to draw $x$ cards, so the probability is
$$\frac1{\binom{16}{x}}\left(\sum_{k=n}^x{\binom 6k\binom 9{x-k}}+\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}\right)$$
There is no need to approach by breaking into cases. Case work will just become too messy and does not generalize nicely to other sized hands.
Consider answering the question instead as: "How many ways are there to form a six-card hand such that at least one of the suits is missing?"
Let $A$ be the event that clubs are missing, $B$ that diamonds are missing, $C$ that hearts are missing, and $D$ that spades are missing respectively. You are tasked with finding $\Pr(A\cup B\cup C\cup D)$
This expands by inclusion exclusion to be:
$$\Pr(A\cup B\cup C\cup D) = \Pr(A)+\Pr(B)+\Pr(C)+\Pr(D)-\Pr(A\cap B)-\Pr(A\cap C)-\dots - \Pr(C\cap D) + \Pr(A\cap B\cap C)+\dots + \Pr(B\cap C\cap D) - \Pr(A\cap B\cap C\cap D)$$
By symmetry, this simplifies to be:
$$ = 4\cdot \Pr(A) - 6\cdot \Pr(A\cap B) + 4\cdot \Pr(A\cap B\cap C)$$
noting that we could leave off the intersection of all four since it is impossible for all suits to be missing simultaneously.
The probability that clubs are missing? That is answered simply by a hypergeometric distribution argument and is $\Pr(A) = \dfrac{\binom{39}{6}}{\binom{52}{6}}$. Similarly for the probability that both clubs and diamonds are missing as $\dfrac{\binom{26}{6}}{\binom{52}{6}}$ and so on.
Your probability is then:
$$\dfrac{4\cdot\binom{39}{6} - 6\cdot\binom{26}{6}+4\cdot\binom{13}{6}}{\binom{52}{6}}$$
Best Answer
The probability of four different suits is the probability that the second card is not the same suit as the first and the third card is a different suit than the first two and the fourth card is the last suit: $$\frac{39}{52}\cdot\frac{26}{52}\cdot\frac{13}{52}=\frac{3}{4}\cdot\frac{2}{4}\cdot\frac{1}{4}=\frac{3}{32}$$