Poker Probability Problem I can't even begin to grasp how to approach such a statistics problem.

Given 9 different cards: two cards for 2 players, and 5 community cards.
Find n, where n is the number of same suited cards out of 9; so that 50% of the time at least one player has a flush.


For example:
If all 9 cards are the same suit, a flush will be present 100% of the time. Conversely, if only 4 cards are the same it drops to 0%.
n = 5 will rarely have a flush, and finding such a percentage is where I'm having trouble.

Thanks in advance to whoever broadens my view.
If you think about it you can easily remove $n=8$ as a possibility. 

For $n=7$, this is the case where two cards are not from the suit. Fix one of these cards and notice that in most hands ($5$ out of every $9$) this card will not be in any of the players' hands (sorry for using the word hands with different meanings), and this of course corresponds to the case where at least one of the players has a flush.
If you think about it you can easily remove $n=8$ as a possibility. 

For $n=7$, this is the case where two cards are not from the suit. Fix one of these cards and notice that in most hands ($5$ out of every $9$) this card will not be in any of the players' hands (sorry for using the word hands with different meanings), and this of course corresponds to the case where at least one of the players has a flush.