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Tycobbler has posted a portion of the relevant MLB rules. Here's the plain English version of what it says and how those actual rules would have applied to your three-way tie scenario:
Since Oakland had a better head-to-head record against both other tied clubs, it gets first choice of which designation it wants to be (A, B, or C). Boston, having a better head-to-head record against one opponent but not the other, would get second choice of which designation it wants to be. Minnesota, having a losing head-to-head record against both other clubs, gets whatever designation is left.
The games are then played as follows:
Day 1: B at A
Day 2: C at A/B
The winner of the game on day 2 is the winner of the tiebreaker and gets the wildcard spot.
Now, as for how OOTP handles it, I'm not sure. Obviously it is not applying a true three-way tiebreaker. I've pointed out several times how a three-way tie should be resolved, but it seems certain situations are still not correctly coded into the game (likely due, at least in part, to how rare a three-way tie is. It's never happened yet in all of MLB's history. The fact that under certain scenarios the three-way tie can be complicated to resolve probably contributed as well).
Even if OOTP disregarded the head-to-head record part of the equation and simply designated the three tied teams as A, B, and C randomly, that would be fine with me as long as the sequence of the games in the tiebreaker were properly played.
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