Heading into Thursday night's games, the Rays sat atop the American League wild-card race, one game ahead of the Indians and two of the Rangers with four to play. The Rays controlled their destiny: keep winning and they host a one-game wild-card playoff on Wednesday night at the Trop.
But if they stumble along the way, there is the possibility of a two- or three-way tie.
In short, if there is a tie to win a spot in the playoffs, it will be broken on the field. If there is a tie to determine where the wild-card game would be played, there is a tiebreaker plan in place. And in this case, it would simply be the first criteria, head-to-head records.
Here are the various scenarios:
Rays and Indians tie for the two wild-card spots: Both are in, game is in St. Petersburg (based on Rays winning season series 4-2).
Rays and Rangers tie for the two wild-card spots: Both are in, game is in Arlington (based on Rangers winning season series 4-3).
If the Indians and Rangers tie: The Indians would host based on their 5-1 edge in the season series.
Now if the top wild card is claimed and there is a tie for the second, there would be a one-game playoff, presumably on Monday, and the same tiebreakers would be used: Rays would host Indians, Rangers would host Rays, Indians would host Rangers.
It gets a little trickier if there is a three-way tie for the two spots. In that case, there would be two playoff games. The teams are ranked based on their records against each other, which looks like this: 1. Indians (7-5), 2. Rays (7-6), 3. Rangers (5-8).
As the top team, the Indians get to pick whether they want to be Team A, Team B or Team C (with the Rays picking second) under the following format:
Team A: Hosts Team B on Monday. If it wins, advances to the wild card. If it loses, plays Team C on Tuesday for the second wild-card spot.
Team B: Plays at Team A on Monday. The winner advances to the wild card. The loser plays at Team C on Tuesday.
Team C: Hosts the loser of Monday's game on Tuesday, and the winner is the second wild card.