With the AL Wild Card three-way tie the most likely massive tie available, it’s interesting to note that the tie-breaker for this eventuality is a bit unusual.

Three-Club Tie for Two Wild Card Spots:
After Clubs have been assigned their A, B and C designations, Club A would host Club B on Monday, Sept. 30 (tentatively). The winner of the game would be declared one Wild Card winner. Club C would then host the loser of the game between Club A and Club B on Tuesday, Oct. 1 (tentatively) to determine the second Wild Card Club.

Most three-way tie breakers are designed to eliminate two teams, so one loss and the team is out. In this case, however, only one team gets eliminated, so it’s winner advances.

Note that the A,B,C designations are not automatic. The clubs are ranked based on winning percentage against each other, and other tie breaker rules. The team with the best rank gets to pick which of the three designations they prefer, then the second team picks from the remaining two.

The three-way head-to-head, which I believe would be used here, has Cleveland 7-5, Tampa Bay 7-6, and Texas 5-8. So Cleveland would need to decide if they want to play the first game at home, and if they lose go on the road, or take their chances with a single game at home on day two with an extra day of rest. Right now, Danny Salazar would be in line to pitch on Monday, but with a day off, Zach McAllister could pitch on Tueday instead. The only way to get Ubaldo Jimenez the start would be to give him Sunday off, but if the tie is in play, that’s not going to happen. I think the Indians would be better off taking the two shots.

Then Tampa Bay would need to decide if they want the possibility of two road games, or the day off and playing at home. No doubt they take the day off. They have David Price lined up for the game either way, but the Rays are 51-30 at home. So my best guess is that Cleveland would host Texas on Monday, and the loser would travel to Tampa Bay on Tuesday.