It’s pretty simple for the Colts:
4 wins and they are in.
At a 10-6 tie with the Jags, Indianapolis wins the South based on victory verses common opponents.
The first tiebreaker (head to head) would be split. The second tiebreaker (division record) would be split 4-2.
The Colts and Jags play 12 games in common.
4 are AFC South games. In those games, both teams would be 3-1. (Indy lost to Houston, Jax lost to Tennessee). Their other division loss would come at the hands of one another.
The other eight common games are:
NFC East games (4): Jags would be 2-2 (losses to Philly and NYG) Colts are 2-2 (losses to Philly and Dallas)
NFC West games (4): Jags would be 2-2 (losses to KC, SD). Colts would be 3-1 (losing only to San Diego)
That would make Indy 8-4 and Jacksonville 7-5
IF the two teams finished tied at 9-7…
The Colts must win the 3 division games. If they do so, but drop the Oakland game AND the Jags lose to Houston or Oakland in addition to Indianapolis, the Colts still win the tiebreaker for common opponents. This is because all the Jags remaining games are against ‘common opponents’. A loss by the Jags to any of the teams they play other than Indy ensures they lose either the division or the common opponent tiebreaker.
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