An example of Braess’s Paradox in basketball
In one of this blog’s more popular posts, I made an analogy between a basketball offense and a traffic network. The rudimentary analysis involved suggested that taking the best shot each time down the court is not the same as running the most efficient possible offense. It’s an idea that caused a minor stir, I guess because people like basketball and counterintuitive phenomena.
The post did draw some criticism, though, largely for being too vague. And rightly so, I think; the post was a little bit of a tease. For example, I made this statement:
In the study of traffic, this was Braess’s Paradox: closing a road can improve traffic. In basketball, it gets called “The Ewing Theory”, and its implication is this: eliminating a scoring option can improve the efficiency of your offense.
But I never actually demonstrated the “Ewing Theory” in action. That is, I never gave a plausible example of an offense that gets better when a player is removed.
In this post I want to rectify that situation. I’ll try to diagram a particular play, and show how the effectiveness of the play might improve when one of the key players is removed. Doing so will further highlight the idea of the “price of anarchy”: that making the best choice at each step in the game is not the same as playing the most efficient possible game.
The drive-and-kick play
Consider the play diagrammed below, which is designed to produce a layup for either the point guard (which I label ), the shooting guard (), or the center ().
The play starts with the ball in the hands of either () or (), who drives from the top of the key toward the basket. When he arrives at the low post, the driver has two options: he can put up a shot at the basket or he can shovel the ball to (). If () receives the ball, he also has two options: he can take a shot or he can pass the ball to a cutting () or () — whichever guard it was that did not make the initial drive to the basket — for the layup. The diagram above shows this network of possible ball movements. Solid lines indicate a drive or shot attempt, and dotted lines represent passes.
Next to each pathway is a function that is meant to show the effectiveness of that step of the play, i.e. the probability that the given step will be completed successfully. The variable stands for the fraction of the time that the given step is used; if that option is never used and if the option is always used. For example, if () is always the one to drive the ball to the hoop, then for that step and the corresponding efficiency is . If () only drives the ball half the time, then and the efficiency of his drive is . In general, every step of the play has a law of diminishing returns: the more it is used, the more the defense learns to expect what’s coming, and the less effective it is.
My choice of efficiencies for each step is completely arbitrary, but I imagine the situation like this: the () guard is quick and a good ball handler, but is not so good at finishing around the rim. The () guard is worse at driving, but is much better at finishing around the basket. The center () is somewhere in the middle. For simplicity, I have assumed that all passes are successful 100% of the time.
The most “reasonable” solution
If you look at the play diagram above, and you start to reason about what is the best way to run the play, two things will become apparent:
- The () guard will never be more successful at driving the ball than the () guard. Even at his very worst, the () is successful on 50% of his drives, which is as good as the () guard can do. So apparently there is no reason to ever start the play with the () guard driving the ball.
- By the same logic, the () guard will never be more successful at finishing at the rim than the () guard or the center (). At their very worst, () and () are 50% effective at finishing at the rim. So apparently there is no reason to ever have the () guard try to shoot.
If you buy this reasoning, which seems pretty solid, then you can cross out two edges of the network above. The () guard will drive the ball 100% of the time and then pass to the center (), so that the only remaining question is how often () should pass the ball on to () for the layup, and how often he should try to finish himself.
The most straightforward method is to use the “Nash Equilibrium” solution, which goes like this: if () is shooting more efficiently than (), then () should keep passing the ball to () for the finish. The moment that () becomes less efficient than (), then () should take the shot himself. The result of this procedure is to reach an “equilibrium” defined by the point where both () and () are shooting the same percentage. You can check for yourself; this results in the center () taking the shot 33.3% of the time, and both () and () making the layup 66.7% of the time.
The final result of this strategy is that the “drive” portion works half of the time, and the layup is successful two thirds of the time, so that the overall success rate of the play is 1/3.
Remove the center
Before I proceed, I should make it clear that in our solution above the center () is absolutely key to the play. He touches the ball every single time the play is run, and he shoots 1/3 of the time with 67% accuracy. So it seems clear that the play should suffer if he is removed.
But does it? Imagine that the team loses its star center to injury, and a replacement player is used who doesn’t know how to run the play. This replacement center simply stands off to the side every time the play is run, and remains completely uninvolved.
As a result, the diagram of the play becomes something like this:
The pass part of the “drive and kick” is now completely eliminated, since the player who facilitated the pass is gone. Instead, there are now only two options for the play: a drive and layup by the () guard or a drive and layup by the () guard. () was good at the drive and bad at the shot, while () was bad at the drive and good at the shot, so on the whole they are equally effective at driving from the top of the key and scoring. As a result, they split their shot attempts 50/50, and each of them has a total efficiency . So the play will be successful 37.5% of the time.
This should be a shocking result. We removed the center from the play, thereby eliminating two potential passes and one scoring option. But the efficiency of the play went up, from 33% to 37.5%. This is Braess’s Paradox. Or, to my mind, it is a diagrammatic proof-of-concept for the infamous “Ewing Theory”.
It shouldn’t have come to this
If it seems like there’s something wrong with this result, that’s because there is. Removing elements of a network should never improve its real efficiency. So there must be something wrong with the “reasonable” solution above. Even though the logic seemed relatively unimpeachable, there is in fact a problem with the way we considered the offense one step at a time. For example, just because () is always better than () at driving the ball doesn’t mean that he should do it every single time.
The correct way to optimize this play is to consider all the options simultaneously. That is, start by making a list of all possibilities for the play (there are six). Then write the efficiency of each possibility as a function of the frequency that it (and the other possibilities) are used. If the efficiencies of the six options are , and the fraction of the time that each option is used are , then the overall efficiency of the play is . Optimize the total efficiency as a function of (by taking derivatives) and you’ll find the real optimum efficiency of the play. And this one can only get worse when possible paths are removed from the network.
For this particular play, the optimum efficiency is 42.3%, well above the 33% that we got from the “reasonable” solution. It involves the () guard driving 25% of the time while the () guard drives 75% of the time. As for the shot attempt, the center () should shoot 50% of the time, the () guard should shoot 45% of the time, and the () guard should shoot 5% of the time.
Just to belabor the point, I should make it clear that the true optimum solution has the (), (), and () shooting very different field goal percentages. In fact, at the true optimum () shoots 50%, () shoots 77.5%, and () shoots 62.5%. In light of this fact, it would seem obvious that we should take shot attempts away from the () and () and give them instead to the ().
But it isn’t true. The best solution is not the most obvious solution, nor is it the one that seems most fair to the players involved.