Inspired by this answer I tried computing the probability of getting at least a $k$ card flush being handed $n$ cards from a deck of $52$. Even for $k = 5$ it says that it's difficult to find a general formula, though I seem to have gotten it rather too easily.
The way I see it the possible cases are $13 \choose k$ equal cards per suit, times $4$ suits, times (and here's the different bit) $52-k \choose n-k$ other possible choises for the $n-k$ empty spots on your hand filled by the $52 - k$ cards left in the deck.
I'll make myself more clear just in case: for each of the $4$$13 \choose k$ possible flushes you have $n-k$ spots that can be filled with any card from the deck but the $k$ ones you already have been dealt to score the flush itself.
Hence the formula $$\frac{ 4{13 \choose k}{52-k \choose n-k}}{52 \choose n}$$
Note: of course $1 \leq k \leq 13$.
The mistake here is that you are counting some combination twice.
Indeed we have same sets of cards in (13,k) and (52-k,n-k).
Note indeed that for n=52 and k=13 we obtain 4 since we are overcounting.