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Playoff tiebreakers change in AFC

Monday - 12/23/2013, 5:59pm  ET

Baltimore Ravens quarterback Joe Flacco watches action in the final moments of an NFL football game against the New England Patriots, Sunday, Dec. 22, 2013, in Baltimore. New England won 41-7. (AP Photo/Nick Wass)

NEW YORK (AP) -- The AFC wild-card race has four teams still in the running: the Ravens, Dolphins, Chargers and Steelers. None can earn the spot outright with a victory next Sunday.

Miami (8-7) must beat the Jets (7-8) and have Baltimore (8-7) lose or tie at Cincinnati (10-5); or win and have San Diego (8-7) win against Kansas City (11-4); or tie and have Baltimore lose and San Diego lose or tie; or tie while the Ravens tie and Chargers tie.

Baltimore must win and have San Diego lose or tie; or win and have Miami lose or tie; or tie and have Miami and San Diego lose; or tie and have Miami tie and San Diego lose; or have Miami, San Diego and Pittsburgh (7-8) lose.

The Chargers must win and get losses or ties by Miami and Baltimore, or tie, with losses by Miami and Baltimore.

Pittsburgh must win at home against Cleveland (4-11) and have the other three lose.

The division titles already are decided, with Denver (12-3) taking the AFC West, New England (11-4) the East, Indianapolis (10-5) the South, and Cincinnati (10-5) the North.

In the NFC, no division is clinched. Seattle (12-3) leads the West, Carolina (11-4) the South, Philadelphia (9-6) the East, and Chicago (8-7) the North. The Seahawks and Panthers own playoff berths, and both will take their divisions with wins Sunday: Seattle at home against St. Louis (7-8) and Carolina at Atlanta.

Philadelphia plays at Dallas (8-7) for the East crown on Sunday. Chicago hosts Green Bay (7-7-1) for the North title.

San Francisco and New Orleans are the front-runners for the wild cards.

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AP NFL website: www.pro32.ap.org


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