American League Wild-Card: Six Teams Tied For Two Spots
In this scenario, no wild-card spots have yet been claimed after 162 games.
Team A: Yankees
Team B: Twins
Team C: Angels
Team D: Royals
Team E: Rangers
Team F: Mariners
Game 1, Oct. 2: Team B (Twins) at Team A (Yankees); loser is eliminated
Game 2, Oct. 2: Team D (Royals) at Team C (Angels); loser is eliminated
Game 3, Oct. 2: Team F (Mariners) at Team E (Rangers); loser is eliminated
A reminder: Based on the existing rule about not re-designating that is applied to a four-teams-for-one-spot tiebreaker, it seems unlikely that MLB would add that additional layer of complication. If that is the case we would have three teams remaining and MLB's tiebreaker scenario for this situation (three teams for two spots) is now in play.
Game 4, Oct. 3: Game 2 winner (C or D) at Game 1 winner (A or B); winner gets first wild-card
Game 5, Oct. 4: Game 4 loser (A or B or C or D) at Game 3 winner (E or F): winner gets second wild-card
To Reach The Wild-Card Game (no re-designating):
Team A has to win two games, both at home.
Team B has to win two games, the first on the road and the second at home.
Team C has to win two games, the first at home and the second on the road.
Team D has to win two games, both on the road.
Team E has to win two games, both at home.
Team F has to win two games, the first on the road and the second at home.
(If MLB does re-designate, it would look like this: Game 4: New Team B at New Team A, winner gets first wild-card; Game 5: Game 4 loser at New Team C, winner gets second wild-card)