FanPost

Red Sox chances of having home-field advantage about 7 out of 11

If reposting here what I posted on my WordPress baseball blog is a breach of etiquette, I apologize. I'm just assuming that nobody but my brother reads that other blog, so I was aiming to get a wider audience here.

The Red Sox will lose any tiebreaker against the Oakland A’s this season, should their records end up tied. As each team has only 4 games remaining, and the Red Sox have just a 1 game lead over the A’s, to end up ahead of the A’s, the Red Sox must at least match the A’s win for win.

There are 256 possible outcomes of the Red Sox and A’s remaining 8 games (16 outcomes for the Red Sox’s 4 games times 16 outcomes for the A’s 4 games). Outcomes here means, for example, win-loss-win-win, or win-win-win-loss (order matters).

When you examine all possible combinations of these outcomes, and even factor in their likelihoods of occurring, and keep in mind the statements made above, it turns out that the Red Sox have about a 7 out of 11 chance of securing home-field advantage over the A’s. When you factor in the remote possibility of the Red Sox beating out Oakland for home-field advantage, but not beating out Detroit, the odds are slightly lower, but still about 7 out of 11.

Now let’s break down some of the above statements to see what’s behind them.

WARNING: the rest of this post contains a bunch of logic and math. If that stuff bothers you, you may want to stop here.

First, why is it that the Red Sox will lose any tiebreaker against the Oakland A’s?

The first tiebreaker is record in head-to-head games between the teams. With each team winning 3 of the 6 games they’ve played against each other, that tiebreaker has no effect.

The next tiebreaker is intradivision record (record against the other teams within their own division). Each team will finish with 76 intradivision games, and currently each team has 30 intradivision losses. So this tiebreaker will go to the team that gets fewer intradivision losses the rest of the way.

This may seem hard to predict, but we can use two facts to our advantage here:

1) To end up tied, the Red Sox must lose exactly one more game the rest of the way than do the A’s.

2) There is only one non-intradivision game left for either team, and that is tonight’s Red Sox-Rockies game.

If the Red Sox win tonight’s game with the Rockies but end up tied with the A’s, it will be because they lost one more of their other games, all intradivision games, than did the A’s. Since both teams currently have the same number of intradivision losses, that will give the Red Sox one more intradivision loss than the A’s, and the worse intradivision record, and so they lose this tiebreaker to the A’s.

If on the other hand the Red Sox lose tonight’s game with the Rockies but end up tied with the A’s, it will be because they lost the same number of their other games, all intradivision games, than did the A’s. Both teams end up with the same intradivision record in this case, and so we move on to the next tiebreaker.

The next tiebreaker is higher winning percentage in the last half of intraleague games – games against other teams in the American League. Currently, the Red Sox are 40-28 in these games, and the A’s are 40-27. The Rockies game is again the only one of the remaining games that doesn’t contribute to this tiebreaker, which means, since we are considering the case in which the Red Sox and A’s lose the same number of the other games, all of which are intraleague games, the Red Sox would end up with one more loss in the same number of games for this tiebreaker, thus having the worse record, and losing the tiebreaker.

All of the above considered, the Red Sox will lose any tie with the A’s for best record, and therefore must maintain or grow their current one-game lead over the A’s to get home-field advantage.

Next we consider the question of why this means the Red Sox have a 7 out of 11 chance of getting home field advantage.

Of the 256 possible outcomes of the 8 remaining Red Sox and A’s games, in 163 of them the Red Sox at least match the A’s win for win, if not surpass them. This represents 63.7% of the 256 possible outcomes. If we assume all outcomes have an equal possibility of occurring, that means the Red Sox have a 63.7% chance of ending up with home field advantage over the A’s, or about 7 out of 11 (which is 63.64%). But saying all outcomes have an equal possibility of occurring is to assume that both teams have a 50-50 chance of winning each of their remaining games, and that’s probably not the case. If instead we assume a 60% chance for each team winning each remaining game (which essentially matches their winning percentages on the season), we can redo the calculation, weighting less likely outcomes (like loss-loss-loss-loss) lower than more likely outcomes (like win-win-win-win). When you do this, the Red Sox’s odds turn out to be just a little bit better, 64.0%, to end up with home field advantage over the A’s. Still pretty much 7 out of 11.

To get their chances of ending up with home field advantage, period, we have to subtract the likelihood of the Tigers tying or surpassing the Red Sox, while the A’s do not. These odds, right now, are very small. Assuming 50-50 chances in the games, the Red Sox have a 63.5% chance of ending up with home field advantage, and assuming 60-40 chances in the remaining games for each division leader, the Red Sox have a 63.9% chance.

No matter how you slice it, it’s pretty much 7 out of 11.

Of course you could argue that due to schedules, the odds are better now for one team or the other. But I think the schedules are not too slanted for one team or the other right now, so I’m sticking with 7 out of 11.

I hope somebody out there enjoys reading this even half as much as I enjoyed producing it.

My sources for the data and tiebreaker information in this post were:

http://www.overthemonster.com/2013/9/23/4761626/red-sox-athletics-playoff-tiebreaker-home-field-advantage

http://en.wikipedia.org/wiki/Major_League_Baseball_tie-breaking_procedures

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