what are odds on these (2) scenarios - please help!

Anyone know what the odds are that two players could make a flush on the flop (10 handed)

And does anyone know the odds that three players could each hit three different sets on the flop (10 handed)

Thank you!

[BBD]

ok, not 100% sure about this but I think the math for all three players flopping a set would look like this:
2C1 * 2C1 * 2C1 / 46C3 that is 2 outs in the deck for player 1's PP with 1 on the flop * 2 outs in the deck for player 2's PP with 1 on the flop * 2 outs in the deck for player 3's PP with 1 on the flop / the total 46 card deck to be drawn from 3 times.

2C1 = 2
46C3 = (46)(45)(44) / (3)(2) reduces to (46)(15)(22)= 15180
2 * 2 * 2 / 15180 = 8 / 15180 = 0.00052 or 0.052%

thus all three players will make a set on the flop 0.05% of the time.

As to your question about 2 players flopping a flush. I think the math would look like:
9C3 / 48C3 that is 9 of you and your opponent's suit remaining in the deck (as you both hold a suited hand) three of which will be on the flop / the total 48 card deck drawn from 3 times
9C3 = (9)(8)(7) / (3)(2) reduces to (3)(4)(7) = 84
48C3 = (48)(47)(46) / (3)(2) reduces to (16)(47)(23) = 17296
84/ 17296 = 0.00485 or 0.485%

thus you and your opponent will both flop a flush 0.48% of the time you both hold a hand of the same suit.
As i said, I'm not really sure this is the correct way to solve the problem. If I have this all wrong, does anyone here know the correct way to do this? Thanks!

[LetYouDown]

Assuming you have K-K vs. J-J vs. 8-8 and you want all three to flop a set:

8/C(46,3) = .0527% or 1896.5 to 1.

For two people to hit a flush, assuming both hold the same suit (obviously):

C(9,3)/C(48,3) = .4857% or about 204.9 to 1.