Odds of 3 out of 4 getting flush

[aloiz]

In my home game before starting we always deal an open face hold'em hand to determine dealer. On one occasion there was four of us playing, and after dealing the hand three of us were suited in clubs. The flop comes down with two clubs, and the river brought the third giving three of us the flush. We wondered what the odds were so I took a stab.

Assuming the following. Four hands dealt and everyone in to the river. Only three of the four can make the flush (not all four). The board may only contain three of the flush cards.

I took a shot at it a came up with something, but I'm sure there is a better way to do it.

Case 1: Fourth player's hole cards contains one of the flush cards)
C(4,3)*4*C(13,6)/C(52,6) * C(7,1)*C(39,1)/C(46,2) [hole cards] * C(6,3)*C(38,2)/C(44,5) [board]

Case 2: Fourth player has none of the flush cards
C(4,3)*4*C(13,6)/C(52,6) * C(39,2)/C(46,2) [hole cards] * C(7,3)*C(37,2)/C(44,5) [board]

Adding both cases I get 2.533E-5

Is this right? Can anyone show me a better way to do it?

Thanks,
aloiz