ANAHEIM -- The A's remain in the hunt for the best record in the American League, though it's very much a possibility they wind up sharing that honor with the Red Sox.
Oakland enters Wednesday one game behind Boston with four games to play. But the A's hold the advantage in a tiebreaker scenario, according to research done by the team's media relations staff.
The first tiebreaker is head to head, and the A's and Red Sox split their six games. The second tiebreaker is based on intradivision success, and if the clubs finish with identical records on the season, the best Boston can do is tie Oakland in that category. That also holds true for the third tiebreaker, decided by their records against American League clubs.
The fourth tiebreaker is success in the last half of intraleague games, so the final 71 American League games. Oakland is currently 42-25 in that stretch, compared to Boston's 40-29 ledger. There is no scenario where the Red Sox can overcome that deficit without finishing the season with a better overall record than the A's.
Thus, the A's would be awarded the No. 1 seed, not only giving them home-field advantage throughout all playoff rounds but also a matchup with the winner of the Wild Card Game in the AL Division Series.
This is a significant edge, though if Boston were to hang on to that first seed, the A's entered Tuesday with a magic number of three over Detroit to finish with the second-best record in the AL, which would ensure them of home-field advantage in ALDS.