The Hitch said:
I know its unfair to give GB 3rd seed becuase they never played Dallas but it makes more sense than giving Dallas 3rd seed because of conference record. GB played the NFC South for christ sake while dallas the NFC best, of course they got a better conference record.
I pointed out in my post that teams that got to play the NFC South had a big advantage. Sure, that’s unfair. But there isn’t much one can do about that. We don’t know ahead of time which division is going to suck, and in fact, no one saw what was coming in the NFC South. They had two strong teams in 2013, and a few years ago, they were the strongest division in the NFL. Other than weighting wins and losses according to a strength of schedule factor, this is unavoidable.
+ its stupid imo to have a tiebreaker system change depending on number of teams.
Use one tiebreaker system the whole way (points scored, SOS, h2h, whatever).
The whole point of the system is that if one rule doesn’t separate, you have to go the second rule, and so on. But if you’re going to use head-to-head—and again, I’m against it—on three teams, you have to stipulate they all played each other. It doesn’t make any sense otherwise.
The problem is 2012 to 2013. 2012 New England played Denver because the AFC E winner played the AFC W winner. New England got home
2013 New England played Denver because the AFC E winner played the AFC W winner. New England got home
How does that not rotate
Its very simple and doesn't require much change in other teams schedules.
Each year every team plays 1 other division in their conference and 1 team from each of the other 2 divisions (depending on finishing position).
In these matchups, each year each division has 4 home games vs 1 division and 4 away games vs the other.
So for example this year the entire NFC East played all 4 of their NFC North matchups away.
The Eagles went to GB
Cowboys went to Chicago
Giants to Detroit
Redskins to Minnesota.
In 2012
Denver went to NE
SD to New York
KC to Buffalo
Oakland to Miami.
Why did this not rotate for 2013. To make it fair those Winner vs winner, 2nd place vs 2nd place, 3rd place vs 3rd place games should rotate.
Here are the AFC inter-divisional schedules for 2012 and 2013:
2012:
West vs. all of North, East vs. all of South
West at East
South at West
East at North
North at South
2013:
West vs. all of South, East vs. all of North
West at East
East at South
North at West
South at North
Yes, West played at East in both 2012 and 2013. Suppose there had been a rotation, and East played at West in 2013. Then, to balance, South would play at East in 2013, and to balance this, the other two home/road relationships would also have to be reversed, giving us:
East at West
South at East
West at North
North at South
But now North and South are not rotating, North is playing at South two years in a row. When you have all the teams in one division playing all the teams in another division, you are left with teams in each division playing an all home or all road series against teams in two other divisions. There are four divisions combining to make four series, with each division involved in two of these series.
It turns out that you can’t rotate all of those series. All you can do is make sure that the same home-road relationship doesn’t occur three times in a row. In fact, in 2014, East—which got to play West at home two years in a row—had to play South on the road for the second year in a row. What goes around comes around.
Of course, Denver also played NE on the road this year, but that was different. Each team in the West played each team in the East this year, with two on the road and two at home. In this case, other factors would have made it difficult for Denver to host NE.
Edit: Here’s a simple mathematical proof. Call the four divisions A,B,C,D. When all the teams in one division play all the teams in another division, indicate with a dash, e.g., A-B. When each team in one division plays one team in another division, indicate with >, with the second team the home team, e.g., A>C.
So suppose A-B and C-D in one year. A possible schedule for the other interdivisional games is:
A>C
D>A
B>D
C>B
The only other possibility is:
A>D
C>A
B>C
D>B
Now suppose in the following year we have A-C and B-D. One possible schedule is for the other games is:
A>B
D>A
C>D
B>C
Notice that per this schedule, D>A will occur for the second year in a row if the first possibility for the first year is used. OTOH, if the second possibility is used, B>C repeats. And if the other possible schedule is used for year 2:
A>D
B>A
D>C
C>B
Either C>B or A>D repeats.
The basic problem is that in year 1, A and B must have opposite home-away schedules vs. C and D, such that if D>A, then C>B. But in year 2, when A and B are not playing all each other’s teams, D>A must go with B>C.