4th-place schedule

I had a conversation with someone the other day about the NFL, and the topic of the NFL's scheduling based on results from the previous season came up. After our talk, I decided that maybe a lot of people, even knowledgable fans, don't quite know how this impact's a team's slate, so I thought I'd clear it up.

The number of games out of the NFL's 16-game season affected by a team's finish the prior year is: two. Consider, that out of the 16 games, 14 are assigned as such (using the 2010 Browns as a model):

6 in-division games (home-and-away against Pittsburgh, Cincy, and Baltimore)
4 games against one AFC division (we play all the AFC East teams this year)
4 games against one NFC division (NFC South)

The other two are decided by 2009's order of finish. By virtue of finishing last, the Browns drew the 4th-place teams from each of the two in-Conference divisions not listed yet (AFC West and South). This year that means matchups with Kansas City and Jacksonville. Both teams are in first place in their respective divisions, KC at 8-4 and the Jags at 7-5. By contrast, last year's AFC North victor (Cincy) has the first-place schedule, earning them 2010 dates with Indy and San Diago, both currently sitting in second place at 6-6.

My point is that the value of playing a "last-place schedule" is very low in the NFL. It's only two games out of sixteen, and because of the churn in the standings each year in the NFL, one's supposedly-favorable opponents are often not so, as the Browns have seen in 2010.