What's the chance one of my opponents has a flush draw?

[BruceZ]

Without taking into account likely calling hands, the random chance of at least 1 of 4 opponents having two cards that match the suited part of a two-tone board is only about 23%.

No it's not. It is less than 11/47 * 10/46 * 4 = 20.3%.

In fact, it is exactly:

C(4,1)*C(11,2) / C(47,2) -
C(4,2)*C(11,4) / C(47,4) +
C(4,3)*C(11,6) / C(47,6) -
C(4,4)*C(11,8) / C(47,8) +
= 19.3%

Note: The first line is the same as 11/47 * 10/46 * 4.

If you want to do this for a particular group of starting hands, just replace the 11's with the actual number of possible flush cards your opponents could hold, and replace 47 with the total number of possible cards your opponents could hold. You can do this for different raising situations too. In fact, there is no reason everyone shouldn't have a good idea what these numbers are for different situations. Make a spreadsheet.