How Mets can clinch the final NL Wild Card spot

As we enter the final weekend of the regular season, the National League Wild Card race is poised to go down to the wire. Three teams are still in the running. But only one can make it.

The Mets (82-77) currently have possession of that spot, leading the Reds (81-78) by one game and the D-backs (80-79) by two, with three games left to play. New York has a series against the Marlins in Miami on tap, while the Reds square off with the Brewers in Milwaukee and the D-backs face the Padres in San Diego.

The Mets don’t own the tiebreaker advantage against either club. The Reds would get it by virtue of having a better head-to-head record (4-2). The Mets split with the D-backs (3-3) in their six regular-season games, but Arizona would be guaranteed to finish the season with a better division record (the next tiebreaker) than the Mets if the two teams end up tied for a Wild Card spot. New York would also lose a three-team tiebreaker.

The Mets, though, still control their own destiny. Sweep the Marlins and they’re in, no matter what the Reds and D-backs do this weekend.

Due to the tiebreaker math, things get a bit trickier for the Mets if they don’t take all three games from Miami, but they still have multiple paths to get in.

If the Mets go 2-1, which would make them 84-78, they’d clinch if the Reds go 2-1 or worse. (The D-backs would be eliminated in this scenario no matter what, as the best they can do is 83-79). If they go 1-2, they’d need the Reds to go 1-2 or worse and the D-backs to win no more than two games in their series against the Padres. Even if the Mets get swept by the Marlins, they’d still sneak in if the Reds also go 0-3 and the D-backs go 1-2 or worse.

In short, the Mets have the inside track, but every outcome is still very much in play as three clubs fight for one last ticket to October.