AL wild-card race reset: Breaking down A's chances of making playoffs

Share

OAKLAND -- With just nine games left in the regular season, the A's are in a great position to lock up a playoff berth for the second consecutive year.

Oakland sits atop the AL wild-card standings at 92-61, two games ahead of Tampa Bay and 2 1/2 games in front of Cleveland. The A's also have the easiest remaining schedule of the three clubs.

Here's a breakdown of each team's final three series and their odds to make the playoffs:

A's: 92-61 (9 games remaining)

3 vs. Rangers (74-79)
2 at Angels (69-83)
4 at Mariners (64-88)

Oakland will play its final nine games of the regular season against sub-.500 AL West opponents. With a magic number of eight to clinch the top-wild card position, the A's likely only need to win five of the nine games. A 6-3 record would just about guarantee them the top spot.

According to FanGraphs, the A's have a 96.3 percent chance of making the playoffs.

Rays: 90-63 (9 games remaining)

4 vs. Red Sox (79-72)
2 vs. Yankees (99-54)
3 at Blue Jays (61-91)

The Rays have the most difficult remaining schedule of the three wild-card contenders, with six games against the Red Sox and Yankees. New York still has something to play for as they try to beat out Houston for home-field advantage, while Boston's lineup is always dangerous.

FanGraphs gives Tampa Bay a 59.9 percent chance of making the playoffs.

Indians: 89-63 (10 games remaining)

1 vs. Tigers (45-106)
3 vs. Phillies (78-72)
3 at White Sox (66-86)
3 at Nationals (83-68)

The Indians have two tough series remaining as they battle the Phillies and Nationals from the NL East. Washington currently leads the NL wild-card race, while Philadelphia is three games out of the second spot.

According to FanGraphs, Cleveland has a 44.0 percent chance of making the playoffs.

Tiebreakers

The A's own tiebreakers against both the Rays and Indians, having won the season series against each club. That means, in the case of a two-team tie between the A's and either Tampa Bay or Cleveland, Oakland would still host the Wild Card Game.

[RELATED: Red-hot A's have work to do to attract fans]

It gets more complicated if all three teams tie for the two wild-card positions. The A's still own the tiebreaker, so they would host the Rays, with the winner earning the top wild-card spot. The Indians would then host the loser, with the winner of that game claiming the second wild-card position.

Of course, if the A's handle their business, it won't come down to tiebreakers.

Contact Us