Panama Red said:
sure about that?
Three teams tie for two Wild Card spots
If the A's, Indians and Rays were all tied, the three teams would choose/receive A, B and C designations. Club A would host Club B, and the winner would advance to the Wild Card Game. Club C would then host the loser of the game between Club A and Club B to decide the other Wild Card team.
Again, the three designations are decided by head-to-head records. The A's went 9-4 against the Rays and Indians, the Rays went 9-5 against the A's and Indians, and the Indians went 2-11 against the A's and Rays. So the A's would have first pick of designation and the Rays second.
This would present a fun dilemma for the club with the first choice. Would the A's prefer two shots at advancement (one at home, and, if that doesn't go well, one on the road as Club A), despite the impact that might have on their pitching with yet another sudden-death game looming in the Wild Card Game proper? Or would they prefer to take their chances on a single home game (as Club C) for a chance to get to the Wild Card Game?
It likely depends on pitching matchups, but selecting C likely matches up your ace with someone else's #2. But, since the As don't really have an Ace, I think they choose A and give themselves two chances.
That would leave the Rays to choose between two chances on the road or one shot at home. I think they would choose one shot at home (C), which means Cleveland would potentially have to travel coast to coast for two games, and possibly back to the other coast for the play in game.
Please let this three-way tie happen just for the ABC selection and possible scenarios.
