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Yankees would host Red Sox in tiebreaker

Boston holds 3-game lead, but N.Y. wins crucial coin flip

NEW YORK - The New York Yankees won a coin flip Monday and would host the Boston Red Sox if a one-game playoff is needed to break a tie for the AL East title.

The Yankees, who trail the Red Sox by 3½ games, end the regular season with three games at Fenway Park from Sept. 30-Oct. 2. If the teams finish with identical records and the wild-card club comes from another division, the Yankees and Red Sox would shift to New York for the tiebreaker game.

New York and Boston last finished the regular season in a tie for first in 1978, when Bucky Dent’s three-run homer boosted the Yankees to the division title.

If game is needed to break a tie between Cleveland and the Chicago White Sox in the AL Central, the Indians would host, and if the Los Angeles Angels and Oakland need a tiebreaker in the AL West, the game would be at Anaheim.

In the event a tiebreaker is needed for the wild card, New York and Oakland would play at Cleveland, and Oakland would play at New York.

If three teams finish with the same record and a tiebreaker is needed, the teams in the same division play a tiebreaker for first place, and the loser meets the team from the other division in a one-game wild-card tiebreaker.

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If a tiebreaker game is necessary in the NL wild-card race, Philadelphia would host Houston and Florida, the Marlins would host Houston, the Astros would host Washington, and the Nationals would host the Phillies and Marlins.

© 2010 The Associated Press. All rights reserved. This material may not be published, broadcast, rewritten or redistributed.

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