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Red Sox - Athletics tiebreaker still up in the air

The division is clinched, but home field advantage is up-for-grabs, as is the potential tiebreaker.


The playoffs are right around the corner, and with the Red Sox already having wrapped up the AL East, the only question remaining is who the Red Sox will play, and where?

At the moment, the Red Sox hold the best record in the American League. If that holds, they'll face the winner of the Wild Card play-in (very likely to be Cleveland, Tampa Bay, or Texas) with home field advantage locked up throughout the playoffs. But holding that lead is not that simple.

The Oakland Athletics, at 93-63, are just 1.5 games back of the Red Sox--just one behind them in the loss column. Holding a lead like that is difficult, but by no means impossible, particularly if the Baltimore Orioles aren't playing for their playoff lives in that final series of the year (their elimination number is down to just three). That job becomes much easier, though, if the Red Sox happen to have the tiebreaker on their side, meaning the Athletics would have to outperform them by a whole 2.5 games in these final days of the year. A 2-3 record from the Red Sox, for instance, would have to be met by a 5-1 record from Oakland.

But do they? The answer, unfortunately, is that we have no idea.

Home field advantage is, according to's Paul Hagen, decided by the following in the event of a tie:

1. Head-to-head winning percentage during the 2013 regular season.
2. Higher winning percentage in intradivision games.
3. Higher winning percentage in the last half of intraleague games.
4. Higher winning percentage in the last half plus one intraleague game, provided that such additional game was not between the two tied clubs. Continue to go back one intraleague game at a time until the tie has been broken.

And God help them if both teams have the same intra-and-interleague records.

With the Red Sox and Athletics having played to a 3-3 tie in their season series, the first criteria is no good. So we move onto intradivision games. There, the Red Sox have a 43-30 to 41-29 lead over the Athletics, which as you may-or-may-not have noticed, is ridiculously fragile. The Athletics play their last six games against intradivision opponents in Seattle and Los Angeles, and a 6-0 record would put the tiebreaker out of reach for Boston, forcing them to go 5-0 to stay ahead in the standings.

For what it's worth, though, Seattle has been a bit of a bugbear for Oakland this year, leaving their record against those two teams right at .500.

Unfortunately, the only competition the Red Sox will have to boost their intradivision record is the Orioles, who have had their number this year much like last year and the year before. If the Red Sox can't pull off a 2-1 series win in Baltimore to end the year, they're going to end up very vulnerable. Just a .500 record against the Angels and Mariners would allow them to match Boston at 44-32, leaving it up to the next tier of tiebreakers.

There the Red Sox and Athletics are in almost the exact same position, but unfortunately for the Sox, that almost makes a big difference. 40-28 for the Red Sox over their last 68 intraleague games, 39-26 for the Athletics over their last 65 means that 1-2 against the Orioles is not good enough to hold up against 3-3 for the Athletics.

To make a long story short, then, the Red Sox really can't afford to slip up very much, particularly against the Orioles. Going 0-2 in Colorado and then 3-0 in Baltimore would likely be enough for Boston to stay ahead, as 1-1 in Colorado and 2-1 in Baltimore, assuming Oakland drops at least two of their last six games. But 2-0 in Colorado isn't guaranteed to make up for a 1-2 series loss in Baltimore, since a 5-1 run for Oakland is not all that hard to imagine given their competition.

And even if by some miracle the Athletics don't overtake the Sox in the event of a 1-4 or 0-5 finish, that would still leave the door open to a strong push from the Tigers, who own the head-to-head tiebreaker over Boston.

In other words: just win.

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