12-30-2007, 01:06 AM
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#8
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Experienced Starter w/First Big Contract
Join Date: Oct 2006
Posts: 6,424
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Re: 19-0 can be beat under unusual circumstances
Quote:
Originally Posted by spacecrime
Cannot be done.
There are four divisions, so there can only be 2 16-0 teams.
For instance, if a team from AFCE goes 16-0, then no other team from AFCE can be 16-0,obviously.
But the AFCE will play an entire division from AFC, so if the AFCE team is 16-0,then obviously no team from the division they play can be 16-0, either.
This year, it means that no team from AFCN could be 16-0.
That means the other two 16-0 teams would have to come from the other two conferences, but because they play each other, that is impossible for two teams to be 16-0.
Try it yourself for this year or any year. Find three teams that could be 16-0. You cannot because it is simply impossible.
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You're absolutely right, except that you had a typo calling divisions "conferences". (But I had worse typos than that until I edited this note ...)
If there are two 16-0 teams in the same conference, then there are two divisions in which no team is better than 15-1. And teams in the same division as the 16-0 squads can be no better than 14-2. So no other team in the same conference can be better than 15-1.
And by the way, the only way there can even be two 16-0 teams is if at least one of them didn't win its division the prior year. 
Last edited by Fencer; 12-30-2007 at 04:53 AM..
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