[BBD]

Hi again DJ. No problem at all. Ok, this is a bit more difficult and I hope I am not giving you the wrong info. But I think the math to figure you AND your opponent holding a suited hand(of the same suit) would look like:
13C4 / 52C4 that is 13 of your suit in the deck before any cards are dealt, of which 4 will get dealt out(2 to you and 2 to your opponent) / the entire deck of 52 cards, drawn from 4 times(2 hole cards to you and 2 hole cards to your opponent)
13C4 = (13)(12)(11)(10) / (4)(3)(2) reduces to (13)(11)(5) = 715
52C4 = (52)(51)(50)(49) / (4)(3)(2) reduces to (13)(17)(25)(49) = 270725
715 / 270725 = 0.00264 or 0.26%

thus you and your opponent will hold a suited hand(of the same suit) 0.26% of the time.

Now, with reguards to your question about you and 2 of your opponents all holding PPs.
To calculate:
4C2 * 4C2 * 4C2 / 52C6 that is 4 of any given rank in the deck with 2 of that rank dealt out to a player * 4 of another given rank with 2 of that rank dealt out to another player * 4 of another given rank with 2 of that rank dealt out to the other player / the entire 52 card deck drawn from 6 times(2 hole cards to each of the three players involved)
4C2 = (4)(3) / (2) reduces to (2)(3) = 6
52C6 = (52)(51)(50)(49)(48)(47) / (6)(5)(4)(3)(2) reduces to (4)(10)(13)(17)(49)(47) = 20358520
6 * 6 * 6 / 20358520 = 216 / 20358520 = 0.0000106 or 0.001%

thus 3 players will be dealt different PPs 0.001% of the time.

again, I certainly hope my math is on the up and up. I'm not really a math guy, I just dabble. hope this helps out!

cheers,
Ben