NEW YORK -- The New York Yankees finally won -- a coin flip.

Unable to clinch the AL East despite a big lead, the Yankees won a coin flip Friday and would be the host if a tiebreaker game is needed against the Boston Red Sox on Tuesday.

On Sept. 13, New York held a nine-game lead in the AL East. But the two-time defending World Series champions lost 12 of their next 15 games, getting outscored 108-50 and allowing 10 or more runs six times.

New York began Friday with a 3½-game lead over the second-place Red Sox.

In order to force a tiebreaker, Boston would have to sweep a three-game series at Tampa Bay, and the Yankees would have to lose three games at Baltimore and then lose a makeup game against Florida on Monday at Yankee Stadium. That would give New York an eight-game losing streak heading into a tiebreaker.

Other coin flips were conducted Sept. 12. If the Athletics and Mariners tie for the AL West and Cleveland wins the wild card, the A's and Mariners would play a division tiebreaker at Seattle.

If Cleveland ties the Athletics for the wild card, the tiebreaker would be in Oakland, and if the Indians tie the Mariners, the tiebreaker would be in Seattle.