Re: Why is game in Indianapolis
posted at 11/10/2009 9:01 PM EST
OK. Here is how it goes.
Most people on this board that each NFL team plays an en entire division from their own conference every three years. That means in 2002, the Pats played the AFC West, 2003 - the South, 2004 - the North, and so on. Of course they play two games against that division at home and two games on the road and switch the home games every three years. So if the Pats played the AFC West in 2002 and had the Chargers in SD, then in 2005 when the played the AFC West again, they played the Chargers in NE. So that much we are all good with.
As to how this pertains to the original question of why the Pats are once again in Indy, they are playing the entire AFC South division this year and it happens to be Indy's turn.
OK, so why were the last two years in Indy as well?
As each AFC team plays another AFC division each year, this leaves two other divisions in the AFC. So for two games, each team plays once against the other teams in its conference that finished in the same place in their own divisions.
E.G. If the Pats are playing the AFC West in 2002 and they finished second in the AFC East in 2001, then they play the second place teams from the AFC South and North in 2002. (I will refer to these games as "same place" going forward.)
Great, but how do they determine which team plays at home - or why did the Pats play in Indy the last few years.
Every time they play an entire AFC division for the next two years, they play the same place team from that division on the road.
E.G. They played the AFC West in 2002. In 2003 and 2004, when they played the same place team from the AFC West, it was on the road.
So, in 2006, they played the entire AFC South. This meant that for 2007 and 2008, they play the same place AFC South team game in that team's city. Of course, the same place team from the South each time was Indy and the game was in their stadium.
I hope that helps clear things up a bit for people. If you have any questions I will try to answer them.