(3) players hit (3) different sets on flop (10 handed)

[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.

like I said though, you might want to post this in probabiltiy as my math might be incorrect.