The Importance of Winning a 2-0 vs. 2-0 Matchup in the NFL
The Bills and Patriots hope to start 3-0; only one of them will.
Rick Stewart/Getty Images
This Sunday, the Buffalo Bills (2-0) play host to the New England Patriots (2-0). It’s early in the season, but it’s important to not underestimate the importance of winning a 2-0 matchup.
Since 1990, there have been 30 2-0 vs. 2-0 matchups in the NFL. The winner of such matchups has made the playoffs 80 percent of the time. Meanwhile, the loser of such matchups has made the playoffs only 33 percent of the time.
It appears a loss in a 2-0 vs. 2-0 matchup can be a real momentum killer. Check out the table below for a more complete analysis.
Playoff Odds Based off a Win / Loss in a 2-0 vs. 2-0 Matchup
| Pct of Times in Playoffs Since 1990 | |
| All 3-0 Teams | 75.9% |
| Without a 2-0 vs. 2-0 Matchup | 74.4% |
| With a 2-0 vs. 2-0 Matchup | |
| Winner | 80.0% |
| Home Winner | 84.2% |
| Away Winner | 72.7% |
| Pct of Times in Playoffs Since 1990 | |
| All 2-1 Teams | 54.9% |
| Without a 2-0 vs. 2-0 Matchup | 58.4% |
| With a 2-0 vs. 2-0 Matchup | |
| Loser | 33.3% |
| Home Loser | 27.3% |
| Away Loser | 36.8% |
Since 1990, a team that lost a 2-0 vs. 2-0 matchup has a 25 percent less chance of making it to the playoffs than a team that is 2-1 and didn’t play in a 2-0 vs. 2-0 matchup.
The 2010 Green Bay Packers and 2001 Miami Dolphins are the only teams since 2000 (out of 15 games) to have lost a 2-0 matchup and still make the playoffs.
This article contains information and text taken from my sports research blog SportsDelve.com.
References: Pro-Football-Reference.com, ShrpSports NFL Standings
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