"triple set" is my new favorite drink!

[tampapoker 0]

I was playing in an annual two table NL tourney this past weekend. THe buy-in was pretty substantial and we had 20 players. The event only paid the top three places 1st = 50%; 2nd = 30% and 3rd = 20%. I made it into the top three and I had about 40% of the chips, the guy I will call "A" had about 50% of the chips and the guy I will call "B" was shortstacked with 10% of the chips. Immediately after we got to three players, I was in the small blind, A was in the big blind, and C was on the button. I was dealt pocket ks. C moved all in pre-flop, I only called his pre-flop all in (it was about 4.5 times the amount of the big blind) and A also calls. The flop comes: K-8-3 in that order (and a rainbow flop)I check and A bets the pot. I raise him double his bet and he pushes all in. I call.As it turns out, A had pocket 8-8 and B had pocket 3-3. WE ALL FLOPED SETS.I ended up winning the pot, but I was wondering if any of you math wizards can calculate the chances that 1) three players would have a pocket pair) and 2) all three players would flop a set. Also, has anyone else ever seen anything quite like this before. We were all shocked.

[....Ian.... 0]

i dunno how to figure the chances of all of yall having pairs......but the chances (provided that 3 players have pairs) of all 3 flopping a set are 1 in 1732.5 (or 0.06%)

[BetItAll33 0]

i dunno how to figure the chances of all of yall having pairs......but the chances (provided that 3 players have pairs) of all 3 flopping a set are 1 in 1732.5  (or 0.06%)

By my math, it's about 4869 to 1 against all three of you being dealt pocket pairs at a 3 handed table. As for flopping a set, I think those numbers are too high Ian. Since order isn't important, the chances of all three flopping a set should be:(2/46) * (2/45) * (2/44) this is 1/11,385 or .000205.