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Rays pick two road games in event of three-way wild card tie

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The Canadian Press
9/27/2013 4:38:40 PM
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NEW YORK -- Tampa Bay decided it would rather have two chances to earn an AL wild-card berth rather than a home game in the event of a three-way tie.

Cleveland would host the Rays in a tiebreaker game Monday if there is a three-way deadlock for the two AL berths, and the loser would play at Texas on Tuesday for the second spot in the post-season, Major League Baseball said Friday.

Rays manager Joe Maddon insisted he's not thinking about the possible outcomes, preferring to leave all decisions to the front office.

"Our guys are going to think it all the way through," Maddon said before facing Toronto on Friday night. "I don't have time to think it all the way through. I have no interest in thinking it all the way through. I'm just concerned about tonight's game and trying to get this right. If we do that, that stuff will take care of itself."

Because each team split season series against the two others, winning percentage in games among all three was used to determine tiebreaker placement. Cleveland (7-5) had the highest at .583, followed by Tampa Bay (7-6) at .538 and Texas (5-8) at .385.

Given the choice, Cleveland opted to be "Team A" and host Monday, and Tampa Bay decided to be "Team B" and play on the road Monday. Texas became "Team C."

If there is a two-way deadlock for one wild-card berth, the tiebreaker game would be Monday. Based on regular-season head-to-head record, Tampa Bay would host Cleveland, Texas would host the Rays and Cleveland would host the Rangers.

Entering play Friday, there were 64 different possible scenarios for the final weekend's results between the Indians, Rays and Rangers, 15 of which would require some sort of tiebreaker.

Matt Moore (Photo: The Canadian Press)

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(Photo: The Canadian Press)
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