therrinn
Sunday, July 30th, 2006, 4:27 PM
QUOTE (Bizzle @ Sunday, July 30th, 2006, 7:14 PM)

If zimmer wins, TC is the 1 seed, TZ is the 2 seed, and DA is the 3 seed.
If zimmer loses, DA is the 1 seed, TC is the 2 seed, and Zimmer is the 4 seed.
It's fair, it's correcct, and it's logical.
End. Of. Frickin. Story.
Let me explain to you why this is inherently illogical. I'll start off with your interpretation, and then show you why even with your interpretation, your conclusion is still wrong. (all of this is assuming Zim wins tmw)
Deciding the number 1 seed Conf record, blah blah, we have a 3-way tie between DA, TC, and TZ.
Heads up Tiebreaker - Fine, say you need a sweep, this doesn't get us anywhere.
Common Opponents - Depending how you interpret this, either DA wins outright, or DA is eliminated from the number one seed. IMO, since DA holds the lead over both teams individually, DA sweeps this tiebreaker. IF you take the approach of Bizz/Zim, then we look at the subset group of teams that all 3 of us have played. There, DA is eliminated from contention for the number one seed.
So then, you return back to the start of the tiebreak process, and TC becomes the number one seed.
HOWEVER, THAT DOES NOT MAKE TZ THE NUMBER 2 SEED.
After you determine the number one seed, you do not take the runner up and make them the number two seed. You then go through the whole process again to get the number 2 seed.
Look at record, conference record, you get a tie between TZ and DA.
Headsup tiebreaker -> no match played
Common Opponents tiebreaker -> DA > TC
=> DA gets the number two seed.
Now I'm not saying I agree with that second interpretation of the 'common opponents tiebreaker', but even if that is the interpretation you go with, you still get
1. TC
2. DA
3. TZ
as opposed to
1. TC
2. TZ
3. DA