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Rams are one win away from having a stranglehold on the NFC

Green Bay Packers v Los Angeles Rams

LOS ANGELES, CA - OCTOBER 28: Running back Todd Gurley #30 of the Los Angeles Rams high fives fans after beating the Green Bay Packers at Los Angeles Memorial Coliseum on October 28, 2018 in Los Angeles, California. (Photo by Joe Robbins/Getty Images)

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The 2018 season isn’t even half over, but the Rams are close to having an overwhelming lead for home-field advantage throughout the playoffs.

At 8-0, the Rams have the best record in football. The second-best record in the NFC belongs to the 6-1 Saints, and the Rams visit New Orleans this week. With a win, the Rams would have a 2.5-game lead and the head-to-head tiebreaker advantage, and it would be very, very tough to envision anyone catching them.

Then attention would turn less to whether the Rams can get the No. 1 seed in the playoffs, and more to whether the Rams can go 16-0. The Rams’ Monday night game against the Chiefs on November 19 in Mexico City looms as potentially the biggest regular-season game of the season, not because it would have much of an impact on the Rams’ playoff position but because it could be the signature game of the Rams’ run to 16-0.

Of course, the Rams may not even try to go 16-0. Sean McVay has made it clear, both last year after the Rams clinched the NFC West and this year in the preseason, that he prioritizes keeping his players healthy. The Rams might rest their starters late in the season even if they’re undefeated.

But we’re getting ahead of ourselves. If the Saints beat the Rams, suddenly they both have one loss, and the Saints have the head-to-head tiebreaker. The Rams would go from the overwhelming favorites to having a real threat to their status as the top team in the conference. (And Washington, currently 5-2 and atop the NFC East, may find itself in contention for a bye week as well.)

So this week’s game between the Rams and the Saints is a very big one. It may shape what the rest of the year looks like in the NFC.