How many wins will it take to win the NFC East in 2012?

Giants are currently 6-4. This is their remaining schedule.

Green Bay (6-3)
@ Washington (MNF) (3-6) -- Washington gets extra rest after playing Cowboys on Thanksgiving
New Orleans (4-5)
@ Atlanta (8-1)
@ Baltimore (7-2)
Philadelphia (3-6)

How many of those games are the Giants going to lose? 3 games are very likely losses.

Green Bay will be motivated, hoping to catch the Bears for the NFC North title.
Atlanta will still be fighting for best record in the NFC.
Baltimore will be at home, where it never loses to NFC teams.

2 more games are not gimmes.

Washington will be coming off a kind of mini-bye playing on MNF after playing the previous Thursday.
New Orleans will be fighting for its playoff life.

So, it looks like the Giants are likely going to lose 3-4 more games, and finish at 9-7 or 8-8. They will have at least 2 NFC East losses.


The Cowboys are 4-5. This is our remaining schedule:

Cleveland (2-7)
Washington (3-6) (Thanksgiving)
Philadelphia (3-6)
@ Cincinnati (4-5)
Pittsburgh (5-3)
New Orleans (4-5)
@ Washington (3-6)

How many of these do the Cowboys need to win to win the NFC East? Only one of these teams currently has a winning record -- Pitt. And they have lost to the Raiders and TItans. All 7 of these games are winnable, and if the Cowboys take care of business and win the next three at home, the Cowboys will likely be favored in all of them.

But the question is -- which of these must the Cowboys win to win the division?

If the Cowboys beat Cleveland, Washington, Philly, and @ Washington, they need to only win one of the remaining 3 games -- @ Cincy, Pitt, and New Orleans -- to finish 9-7 and likely take the NFC East. The Giants aren't likely to win more than 9 games, and if Dallas wins it's last 3 NFC East games it will finish 5-1 in the division and win the tiebreaker.

However, this scenario may force Dallas to win the last game @ Washington, which is not the best path, given how Dallas has fared recently in win-and-in/lose-and-out last games..

So, the next best hope is that the Giants lose @ Washington, giving them 3 NFC East losses. If that happens, the Cowboys may need only 8 or 9 wins going into the last week of the season to have clinched.

My bottom line is that the winner of the NFC East will likely win 9 games, just like last season.

If the Cowboys and Giants tie at 9-7, here's the likely breakdown.

If Dallas can go 5-1 in the NFC East, it will win the division b/c the Giants have already lost 2 NFC East games.

If Dallas and the Giants go 4-2 in the NFC East, the next tiebreaker is record in common games. The common games are 4 NFC East games, plus the NFC South and AFC North. Dallas is 2-1 in the NFC South, and 0-1 in the AFC North. NY is 2-0 in the NFC South, and 1-2 in the AFC North. We have NO, Cleve, Cincy, and Pitt left. They have NO, Atlanta, and Baltimore left. For us to get to 9-7 with an NFC East loss, we need to win 3 of our common opponent games, which would leave us 5-3. For the Giants to get to 9-7 with no more NFC East losses, they will need to lose at least 2 games to common opponents, which would leave them 4-4. I know that's pretty complicated. But, the conclusion is that if both teams finish 9-7, Dallas will almost certainly win the NFC East, and Dallas will probably not need to beat Washington the last week of the season.

Dallas likely needs to: beat Cleveland, Washington, Philly, @ Cincy, and one of Pittsburgh/New Orleans for the last game to not matter.

And even if Dallas stumbles late and finishes 8-8, they will be in if NY loses to: Green Bay, @ Washington, @ Atlanta, and @ Baltimore, or beats Washington but loses to New Orleans.

What's your prediction?

Another user-created commentary provided by a BTB reader.

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