Why is there no home-court advantage in foul shooting?

There's a home-site advantage in every sport.

Why is that? Nobody knows. One hypothesis is that it's officials favoring the home team. One piece of data that appears to support that hypothesis is that when you look at situations that don't involve referee decisions, the home field advantage (HFA) tends to disappear. In "Scorecasting," for instance, the authors report that, in the NBA, the overall home and road free-throw percentages are an identical .759. Also, in the NHL, shootout results seem to be the same for home and road teams, and likewise for penalty kick results in soccer.

However, there's a good reason for the results to look close to identical even if HFA is caused by something completely unrelated to refereeing.

The reason is that free-throw shooting involves only one player. At the simplest level, you could argue that foul shooting is offense. All other points scored in basketball are a combination of offense and defense. Not only is the offense playing at home, but the defense is playing on the road, which, in a sense, "doubles" the advantage. Therefore, if the home free-throw shooting advantage is X, the home field-goal shooting advantage should be at least 2X.

That's an oversimplification. A better way to think about it is that a foul shot attempt is the work of one player. A field goal attempt, on the other hand, is the end result of the efforts of *ten* players. Not every player is directly involved in the eventual shot attempt, but every player has the potential to be. A missed defensive assignment could lead to an easy two points, and the offense will take advantage regardless of which of the five defensive players is at fault. The same for offense: if a player beats his man and gets open, he's much more likely to be the one who gets the shot away. The weakest or strongest link could be any one of the ten players on the court.

So it might be better to guess that the HFA for a possession is 10X, rather than just X. We can't say that for sure -- it could be that the things a player has to do on a normal possession are so much more complex than a free throw, that the correct number is 20X. Or it could be that a normal possession is less complex than a free throw, so perhaps 5X is better. I don't know the answer to this, but 10X seems like a reasonable first approximation.

------

What would the actual numbers look like?

The home court advantage in basketball is about three points. That means that instead of (say) 100-100, the average game winds up 101.5 to 98.5.

Three points per game, divided by 10 players, is 0.3 points per game per player. Over (say) 200 possessions, that's 0.0015 points per possession per player.

If home-court advantage were made up only of serious mistakes, mistakes that turn a normal 50 percent possession into a 100 percent or zero percent possession, then that works out to exactly one point per mistake. In that case, the average player would make one such extra mistake every 667 possessions. That's a little less than one every three games. If you assume that a mistake is worth only half a point, then it's one mistake per player for every 333 team possessions.

In reality, of course, it's probably not nearly as granular as "mistakes" or "good plays". It's probably something like this: a player plays his role with an overall average of 50 effectiveness units, random between possessions, plus or minus some variation. But that's an average of home, where he plays with an average of 51 effectiveness units, and road, with an average 49 effectiveness units.

Still, that doesn't matter to the argument: the important thing is HFA is one point per player for every 667 total team possessions, regardless of how it manifests itself.

------

Now, let's go back to free throws. I'm going to assume that a player's HFA on a single possession should be about the same as a player's HFA on a single free throw. Is that OK? It's a big assumption. I don't have any formal justification for it, but it doesn't seem unreasonable. I'd have to admit, though, that there are a lot of alternative assumptions that also wouldn't seem unreasonable.

But the point of this post is that it is NOT reasonable to assume that a player's HFA on a free throw should be the same as the overall HFA for an entire game. That wouldn't make any sense at all. That would be like seeing that the average team wins 50 percent of games, and therefore expecting that the average team should win 50 percent of championships. It would be like seeing that the Cavaliers are winning 17 percent of their games, and expecting that they score 17 percent of the total points.

In any case, the overall argument stays the same even if you argue that the HFA on a single possession should be twice that of a single free throw, or half, or three times. But I'll proceed anyway with the assumption that it's one time.

If the HFA on a free-throw is the same 0.0015 points per player as on a possession, then you'd expect the difference between home and road free throw percentages to be 0.15%. Instead of the observed .759 home and road, it should be something like .75975 home, and .75825 road.

Why don't we see this? Well, here's one possible explanation. Visiting teams are behind more often, so will commit more deliberate fouls late in the game. They will try to foul home players who are worse foul shooters. Therefore, the pool of home foul shooters is worse than the pool of road foul shooters, which is why it looks like there's no home field advantage in foul shooting.

Since we're talking about such a very small HFA in the first place, this doesn't seem like an unreasonable explanation. It would be interesting to run the numbers, but controlling for who the shooter is. I suspect if you have enough data, you'd spot a very small home-court advantage in foul shooting.