Assuming a completely random holding (say 1 opponent in big blind getting free flop) 3 of the suit flopped, you don't have any of the suit, so 47 cards left with 10 of the suit. Percentage of the one person having two of the suit would be (10/47)*(9/46) or about 4.16% which is about 23:1 against. Using the 4.16% against multiple random hands (less realistic) calculate the odds that noone has flopped it and then subtract that from 1. So, for two people use 1-(.9584)^2 = 8.15 % or about 11.3:1 against,

for 3 people 1-(.9584)^3 = 11.98% or 7.35:1 against and for 4 people about 15.6% or 5.4:1 against. This is against completely random hands, so in real play you would probably want to adjust the odds down according to the texture of the game.