By David Roher

These playoffs have been one for the road. In the Division Series, the home team went a combined 4-11, including a 1-7 showing in the ALDS. Why this happened isn’t that important — you can come up with whatever narrative floats your boat; it’s probably just random variation. But it did remind me of a curious finding I made last March, as I was calculating the value of an NBA playoff seed.

Let’s say the World Series always has two even opponents, identical in every way. The only exception is that one team gets home-field advantage, which is at the historical MLB average. The games in the best-of-seven series are independent; that is, the outcome of one game does not influence the outcome of another. Over a hundred years, how many series could the home teams expect to win?

51: Just one over a random 50.

Sounds weird, right? You wouldn’t believe me if I told you the original idea for the new All-Star Game slogan was “One Out of a Hundred Times It Counts.” Home-field advantage gets played up in the playoffs more than any other time: win a game on the road, and you break your opponent. But the math doesn’t back that up.

It all comes down to the length of the series. Consider the NFL, which only has a one-game playoff and a historical home-field advantage of around 60% for the home team. The chance that the home team will win the playoff game (assuming both teams are equal) is therefore also around 60%.

That’s a pretty big number, especially when you consider that the alternative isn’t really a neutral site, but rather the other team being home, and the original home team being away. This means that of ten evenly-matched NFL playoff games a year with home and away teams (i.e. excluding the Super Bowl), we could expect the outcome of around two games *per year* to change if the home-field advantage were flipped.

Playoff series, rather than do-or-die games, are a different story. The home-field advantage is that one team gets to play one more game at home than the other team. Look at it this way: in the NFL, one team gets 100% of the games in the Conference Championship at home. But in baseball’s LCS, one team gets only 57% of the games at home.

How do we calculate the value of home-field advantage in a series based on the value of home-field in a single game? It’s a bit tricky. First, we split up the series in 2 — games at the home of the team with the advantage, and home games at the team without. This way, each set of games has their own set probability of the advantaged team’s winning. Then, for a seven-game series, we determine four separate probabilities: that the advantaged team will win all four home games and none on the road, 3 at home and 1 on the road, 2 each, and 1 at home and 3 on the road. Then, we add up each of these numbers, and the result is the chance that the team with home-field will win the series.

What are the results? **Assuming a 54% single-game advantage, the team with home-field will win the series 51.3% of the time** (how I arrived at the answer to the initial question). What if you think that home-field is more important in the playoffs, and that the single-game figure is more like the NFL’s 60%? The advantaged team will still win only 53.0% of the time.

In 5-game series, where the percentage of home games for the advantaged team is higher, a 54% single-game advantage yields a 51.5% series advantage, and a 60% single-game yields a 53.9% series advantage.

What’s the point? Just that a few people might be overestimating the importance of home-field in a playoff series. There are extenuating circumstances I haven’t addressed here, such as the strategic advantage of being ahead in the series and the reality that not all teams are even. But I believe that the former is negligible and the latter actually decreases the effect of home-field even further (if the home team is more likely to win already, then the home-field odds bonus is worth a smaller percentage). The lack of a home-field advantage is yet another dynamic that makes MLB’s playoffs the most random of the major sports.

In the next couple weeks I plan to take a look at playoff home-field more empirically, and find the playoff advantage compared to the regular season’s (with the appropriate adjustments for the better team playing more games at home). I’ll also try to put up a graph to visualize the difference between single-game and series home-field advantage. Stay tuned.

Have you considered analyzing possible variation in the home field advantage for particular games as the series progresses (i.e. a stronger advantage when playing game 7 at home as compared to game 2 also played at home)?

Yeah, a result-based study is next. My initial reaction is that, if the effect exists, it might balance out: in a 2-3-2 series, the average game number played at each team’s stadium is game #4 in both cases. But games become progressively less likely to happen, so if you weight by the probability of a game occurring, it’s actually the disadvantaged team that gets to play more games at home later in the series.

In a 2-2-1 series, the road team’s average game is very slightly earlier (2 1/3 vs 2 2/3), but I think that belies the fact that games 1 and 2 are generally not strategic, whereas games 3-4 will be.

Thanks for the feedback, David. I’ll look forward to the results-based study. Great post – very interesting topic.

The 2-3-2 series format is much better than the travel heavy 2-2-1-1-1…. those were brutal.

Rohvember is coming. Nice work, David. Really interesting stuff.

I’d love to see a comparison with the NBA. They have a seven game series, yet the advantage is still quite large.

And that one out of 100 years was 1987, the only year the home team won every game. The Cardinals were a better team but the Twins had the Metrodome, which was too much to overcome. Obviously, this is unprovable but the Twins’ edge in home games was probably higher than just about anyone else has had.

I think the home team won every game in 2001 as well. Tough to tell, as I’ve repressed game 7 pretty deep.

Glad more people are pointing this out. I did something similar at the baseball blog “Beyond the Boxscore” a few years ago. Here is the link:

http://www.beyondtheboxscore.com/story/2006/10/17/15727/828

I had about the same probability for the team with the HFA to win

Hey Cyril, I’m a fan. I tried to search if something had been done previously, since it’s pretty intuitive, but nothing came up. Looks like you beat me to it by a few years. But I’m happy we came up with the same figures.

Thanks. When I first looked at this I tried to find the right combination or permutation equation to use but nothing ever made sense for me. Let me know if there is one

I actually made a mistake in one of my scenarios (I only had one loss in one 6 game series!). Changing that to a loss gave me a 51.3% chance of winning the series, just what you got.

I figured it out for that basketball post and it was trickier than I thought. Basically, it’s best to think of the series as going the full length no matter what, and measuring whether the team wins at least four games in it.

What I did was split up the home and away games and did a total of 7 binomial distributions: the prob that the home team wins 0 games at home, 1 game at home, etc. to 4, and then 0-3 games on the road. For the home games, p was .54, and for the away games it was .46.

Then you have four equations as I described in the post. Scenario one is that the team wins four home games, which means they can win either 0, 1, 2, or 3 road games. Given that covers all road probabilities, it’s P(win all 4 home games)*1.

In scenario 2, it’s 3 home games and either 1,2, or 3. And so on until a minimum of 1 home game won in the series. Then I added the four together.

Thanks!

I meant changing a win to a loss got the right number

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