So each division ends in a five-way tie.
So you have to do the five-way tiebreaker scenario, with the question being who gets first pick at roles A, B, C, D, or E
I think you'd have to assume it goes the same as a three-team tiebreaker:
Determining A, B, C Designations in Three-Team Tiebreakers
1. All Three Clubs Have Identical Records Against One Another
• Club with highest winning percentage among three tied clubs in intradivision games chooses its designation, followed by the team with the next highest winning percentage in intradivision games. If any two of the clubs have identical winning percentages, then then the two-Club tiebreak rules will break that tie. If all three Clubs have identical winning percentages, then;
• Club with the highest winning percentage in the last half of intraleague games chooses its designation, followed by the team with the next highest winning percentage in intraleague games. If any two of the clubs have identical winning percentages, then then the two-Club tiebreak rules will break that tie. If all three Clubs have identical winning percentages, then;
• Club with the highest winning percentage in the last half plus one intraleague game, provided that such additional game was not between any of the tied Clubs, chooses its designation, followed by the team with the next highest winning percentage in the last half plus one intraleague game. Continue to go back one intraleague game at a time until the tie has been broken.
All teams should have identical intradivision records, taking us to bullet point 2
Teams would not necessarily have identical winning percentages in the last half of intraleague games, creating the separation needed. Let's look at the AL East to see how that would go.
Looks like the Rays would be 38-32, to start...