How Home Field Advantage is Like Mike Trout by Jeff Sullivan September 26, 2013 Home-field advantage is a strange concept, or should I say, a strange reality. It doesn’t really matter, for our purposes, why it exists — it just matters that it exists. It’s there, all of the time, in every single baseball game, and while I wouldn’t say it’s an unspoken thing, it’s seldom thought of in depth. A team playing at home has an advantage it wouldn’t have in a neutral site. A team playing on the road is at a corresponding disadvantage. We accept that it is, and we don’t talk much about it, and when we talk about potential edges, it’s usually ignored in favor of pointing at match-ups. It’s almost too boring to point out Team X stands better odds because they’re playing in their ballpark. Someone’s always playing at their ballpark. But home-field advantage is exactly what the Reds and Pirates have to play for this weekend. Very fleeting home-field advantage — home-field advantage in the one-game wild-card playoff between the two rivals. The teams will play three before they play one, and the Pirates are 50-31 at home, while the Reds are 49-28. Each would prefer to play before its own partisan audience. It’s obvious that it matters who gets to play at home. But how much does it matter? What’s a way that we can think about this? Fortunately, it’s simple to figure out how home field will be decided. The Pirates, right now, are a game ahead of the Reds in the standings. In head-to-head match-ups, they’re even at 8-8. If the Pirates win this series, they’ll play at home. If the Reds sweep the series, they’ll play at home. If they win the series 2-1, they’ll play at home, still, because they’d have the head-to-head advantage, 10-9. The winner of this series gets home field, all uncomplicated-like. Let’s just ignore, for the sake of simplicity, that the Pirates are three back of the Cardinals. So what’s the history of home-field advantage? This, too, is pretty simple. All-time, in the playoffs, the home team has won about 54% of games. Just since 1995, the home team has won about 54% of games. In the regular season, the home team has won about 54% of games. We’re getting a pattern. We’re going to stay simple and use that 54%. The Reds and the Pirates are pretty even teams, in record and in talent. We’re going to say that the home team in the one-game playoff would win about 54% of the time. So what’s at stake this weekend is a 54% chance of advancing to a real playoff round. Lose, and it’s more like 46%. It’s simultaneously a big and small difference. The Reds and Pirates might argue that they have particular, unique home-field advantages, that are stronger. That they’re tailored to their ballparks, that they’re exceptionally familiar with their ballparks. Maybe, but probably not. Single-season home records have to be really strongly regressed, and it’s best to just use the average advantage. You’re free to disagree, but if you disagree, the following analysis probably won’t suit you. So what’s another way of looking at what’s at stake? For this part, we’re going to make use of the Log5 method, which I will link instead of explain. As noted earlier, the Reds and Pirates seem pretty even. Maybe you think they’re true-talent 90-win teams. For a one-game playoff, they could optimize, maybe briefly making them 95-win teams. These numbers don’t really matter. Let’s just accept that there isn’t a huge difference between the ballclubs. All things being equal, if you have even ballclubs, you’d expect each to win 50% of the time. So then: what does it take to turn 50% into 54%? What kind of boost do we have to give one of the teams to turn the winning odds into 54/46, to mirror home-field advantage? The answer: 6-7 wins. A 97-win team would beat a 90-win team 54.4% of the time. A 102-win team would beat a 95-win team 54.5% of the time. At least, that’s what Log5 says, and Log5 is smarter than I am. That’s one way to think of the strength of home-field advantage. It makes a team act like it’s several wins better. If you figure the Reds and Pirates are even, home-field advantage gives one a winning-percentage edge of about four percentage points. A way more fun way to imagine this is that home field would allow the Reds to exchange Jay Bruce for Mike Trout. It would allow the Pirates to exchange Pedro Alvarez for, say, healthy Miguel Cabrera. What’s at stake this weekend is home field in one game, and that’s the equivalent of one team being able to add a superstar before playing at a neutral site. So home-field advantage = added superstar + neither team at home. At the same time, that makes it seem big and small. There’s no more impactful player than Mike Trout. But, this isn’t basketball. One position player in baseball can mean only so much, and in one game, on average, you’d expect Trout to be worth 0.3 – 0.4 more runs than Bruce. You’re talking a fraction of a hit, a fraction of a total base. Incrementally better defense. There’s a lot that takes place that doesn’t involve one of many starting position players, and so a team at home can still lose quite often. In last year’s wild-card playoffs, the Rangers lost to the Orioles in Texas, and the Braves lost to the Cardinals in Georgia. Home-field advantage is real, even if we don’t quite know how and where it shows up, but like all advantages, it’s limited in strength. If you want to downplay the importance, you can focus on that 54%-46% separation. That’s a gap of just eight percentage points. Play at home, and you have a good chance of moving on. Play on the road, and your chance is just a little less good. Ehh. If you want to emphasize the importance, you can focus on swapping Jay Bruce for Mike Trout, or Pedro Alvarez for peak Miguel Cabrera. Ultimately, everything gets you to the same place, but our brains can see the same things so differently. Who wouldn’t want to rent a superstar for a day? The winner of this Reds/Pirates series, effectively, gets to do just that.