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An example of Braess’s Paradox in basketball

November 30, 2009

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 1), the shooting guard (2), or the center (5).

The play starts with the ball in the hands of either (1) or (2), 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 (5).  If (5) receives the ball, he also has two options: he can take a shot or he can pass the ball to a cutting (1) or (2) — 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 x stands for the fraction of the time that the given step is used; x = 0 if that option is never used and x = 1 if the option is always used.  For example, if (1) is always the one to drive the ball to the hoop, then x = 1 for that step and the corresponding efficiency is 1 - 0.5 \times 1 = 0.5.  If (1) only drives the ball half the time, then x = 0.5 and the efficiency of his drive is 1 - 0.5 \times 0.5 = 0.75.  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 (1) guard is quick and a good ball handler, but is not so good at finishing around the rim.  The (2) guard is worse at driving, but is much better at finishing around the basket.  The center (5) 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:

  1. The (2) guard will never be more successful at driving the ball than the (1) guard.  Even at his very worst, the (1) is successful on 50% of his drives, which is as good as the (2) guard can do.   So apparently there is no reason to ever start the play with the (2) guard driving the ball.
  2. By the same logic, the (1) guard will never be more successful at finishing at the rim than the (2) guard or the center (5).  At their very worst, (2) and (5) are 50% effective at finishing at the rim.  So apparently there is no reason to ever have the (1) 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 (1) guard will drive the ball 100% of the time and then pass to the center (5), so that the only remaining question is how often (5) should pass the ball on to (2) 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 (2) is shooting more efficiently than (5), then (5) should keep passing the ball to (2) for the finish.  The moment that (2) becomes less efficient than (5), then (5) should take the shot himself.  The result of this procedure is to reach an “equilibrium” defined by the point where both (2) and (5) are shooting the same percentage.  You can check for yourself; this results in the center (5) taking the shot 33.3% of the time, and both (5) and (2) 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 (5) 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 (1) guard or a drive and layup by the (2) guard.  (1) was good at the drive and bad at the shot, while (2) 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 0.5 \times (1 - 0.5 \times 0.5) = 0.375.  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 (1) is always better than (2) 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 f_1, f_2, ... , f_6, and the fraction of the time that each option is used are x_1, x_2, ... , x_6, then the overall efficiency of the play is x_1 f_1 + x_2 f_2 + ... + x_6 f_6.  Optimize the total efficiency as a function of x_1, x_2, ... , x_6 (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 (2) guard driving 25% of the time while the (1) guard drives 75% of the time.  As for the shot attempt, the center (5) should shoot 50% of the time, the (2) guard should shoot 45% of the time, and the (1) guard should shoot 5% of the time.

Just to belabor the point, I should make it clear that the true optimum solution has the (1), (2), and (5) shooting very different field goal percentages.  In fact, at the true optimum (1) shoots 50%, (2) shoots 77.5%, and (5) shoots 62.5%.  In light of this fact, it would seem obvious that we should take shot attempts away from the (1) and (5) and give them instead to the (2).

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.

25 Comments leave one →
  1. November 30, 2009 10:35 pm

    What a great post, very clear!

    Does this sound right to you: In some setups (such as the traffic setting), people really are stuck with the Nash equilibrium due to self-interest. In this setup, the players are only stuck with it insofar as they fail to take into account the effect of their immediate actions on the overall efficiency of the various paths to the basket. The analog to self-interest here is “interest only in the current play.”

    • gravityandlevity permalink*
      December 1, 2009 9:06 am

      Yep, that’s exactly what I had in mind. In part that’s why I think this idea is worth discussing. The Nash equilibrium is a very natural state, but one that is completely avoidable.

  2. Noah permalink
    December 1, 2009 6:00 pm

    Unrelated, but blew my mind today. Write something on this!!

  3. December 1, 2009 7:40 pm

    Noah, there’s really no paradox. As it says, buried in that link, a rotating disk cannot be an inertial frame. When you recognize that there’s a centrifugal force, use the equivalence principle that says this force is locally indistinguishable from a gravitational pull, and work it all out, the effect is exactly due to the spacetime curvature you expect to give rise to that gravitational behavior.

  4. Lee permalink
    December 2, 2009 10:32 am

    Nice work! The phrase that comes to mind when reading your post is “keeping the defense honest.” Sometimes you need to run a lower percentage play so that your higher percentages works even more efficiently.

    Taking your suggestion from your previous post, I shall draw an analogy, this time from football. Every team in the league average 9+ yds/pass (topping out at 12.9), yet even the best team averages only 5.4 yds/rush, with some teams as low as 3.4. With statistics like this, why would you ever rush? It is all about keeping the defense honest, making them respect the run because it could result in much more yardage if all defenses did was focus on the pass. This is something you hear about ever Sat/Sun if you watch football, maybe less so if you watch basketball.

    Data from:

    • gravityandlevity permalink*
      December 2, 2009 1:34 pm

      An excellent point. This actually occurred to me a little while ago — that the example was much clearer in football, and probably the sort of approach I’m advocating is more straightforward there. It’s also much less surprising, since as you pointed out, people use this kind of reasoning all the time.

      Which isn’t to say that it isn’t a “puzzle” to some:

  5. peter permalink
    April 3, 2010 1:38 pm

    I had a quick question–my friend and I were talking about this, and were wondering: if we had a play that had two options: I always shoot (0%) or you shoot a non-0 percentage, under this optimal strategy solution would I still be shooting sometimes? Based on the logic above, the 0% shooter would have to shoot sometimes, right?

    • gravityandlevity permalink*
      April 3, 2010 7:55 pm

      That actually depends on the specifics of the non-zero option. In general, you should use the strategy that optimizes the total efficiency F = x * f(x) + (1 – x) *0, where x is the fraction of the time that I (the non-zero option) take the shot. The optimal strategy depends on the skill curve f(x).

      Take a couple examples:
      f(x) = 0.5, like the (1) guard from above: F = 0.5 x. The maximum possible efficiency is at x = 1. This is the obvious solution: there is no reason to ever use the zero option if you have a 50% option.

      f(x) = 1 – 0.5 x, like the (2) guard above: F = x (1 – x/2). This function has its maximum at x = 1. So again, the nonzero option should take every shot.

      f(x) = 1 – x. This is a more extreme example. Obviously in this case the nonzero option shouldn’t shoot every time; if he does, then he will shoot zero percent. In fact, the resulting total efficiency F = x (1 – x) has its maximum at x = 1/2.

      So the short answer to your question is that it depends. Passing the ball to a lower-percentage option is a tradeoff. You lose in the short term (by exchanging a high-percentage shot for a lower one), but you gain in the long term (by preserving the high efficiency of your better option). Whether or not it’s worthwhile to make that tradeoff depends on how quickly your good option declines with usage.

  6. thomasthethinkengine permalink
    May 10, 2010 2:49 am

    Hi. I followed a link from Love the argument you make here.

    I want to make a couple of points, one theoretical and one practical.

    1. Nash equilibria involve strategic interactions. This seems like an optimisation problem between cooperating agents. The involvement of a range of agents is not sufficient for game theory.

    The original Braess paradox differs in that drivers are not cooperating like you assume team-mates are. If the payoffs to the players were dependent on their own scoring percentages (and subsequent sponsorship deals…) then the interactions between the team-mates might be more strategic.

    Alternatively, the interaction between offense and defense might produce an interesting allocation of assets that constitutes a sub-optimal nash equilibrium.

    2. If the centre stands off to the side, the defence will focus on the other two players. Their efficiency is then likely to diminish faster than 0.5x.

    All that said, reading a new theory is always fun and gets the mind going. It’s much harder to make a theory than to poke (small) holes in one, so thanks to you for your work!

    • john permalink
      June 25, 2014 2:09 pm

      To reply to your first point, it’s not the players who are competing to create the Nash equilibrium, but the plays. The point that was made was that if players mistakenly think maximizing each play is best, the result is effectively having selfish plays competing.

      When players realize that the best strategy for the game is not to maximize each play individually, they won’t use the Nash equilibrium, and their overall performance will improve.

  7. October 28, 2011 7:33 am


    found your blog when looking for images for Braess’ paradox. I happen to be also in love with basketball. Great post.

    One observation for 2011: consider renaming this the “Rudy Gay theory”, reverse the example for Marc Gasol and Zach Randolph inside, and leave Rudy on the injured list.


  8. Bryce Sweeney permalink
    November 28, 2011 12:48 am

    I absolutely love this, a recent espn magazine had a short note on how NFL teams should be trying for a Nash Equilibrium on yards gained per play for runs & passes. I vehemently disagreed with that conclusion, but the only way to express it was by thinking that the reasons & goals of running vs passing were different. I think that this is about the same thing and explains it very well.

    • gravityandlevity permalink*
      November 28, 2011 9:23 am

      Thanks, Bryce. I tend to agree with you: I think it is in a team’s best interest to concertedly maintain a difference between their passing efficiency and running efficiency.

      This is a fairly controversial opinion, and I haven’t always had much success convincing people of it.

  9. john permalink
    June 25, 2014 1:54 pm

    I realize this is an old post, but I’ve just found a link to it from a blog trying (unsuccessfully in my opinion) to copy it in a soccer context. You straighten out all the issues I had with his post. He failed to point out the distinction between optimizing for each play, and optimizing for the game. He also gives the impression he thinks the original “solution” is optimal, period.

    Anyway, GOOD JOB on this post!

    Although I do have a couple small dissagreements.

    Since your model has passes being 100% effective, you should be able to seperate the steps of the play and maximize them independantly – you just can’t seperate the plays. Which guard brings the ball forward in no way affects who can take the shot.

    Solving this way, I get both guards being given equal number of opportunities to try to advance the ball with an overall success rate of 5/8. And I get the center and guard (2) equally sharing the shots (guard (1) not taking any) with an overall successfull shot ratio of 11/16. This would give an overall scoring efficiency of 55/128. I beleive that comes to 43%, just a little better than your 42.3%.

    The fact that I get guard (1) never taking shots makes me a little leary of my optimization, so I wouldn’t be too surprised if someone manages to improve on the 43%.

    And let me just repeat – GOOD JOB!

    • Brian permalink*
      June 25, 2014 4:35 pm

      Hi john. Thanks for the kind words, and for checking my math (it is indeed possible that I made an arithmetic mistake somewhere). It’s probably important for me to point out that this model is just a toy with no actual “reality” to it. All the play efficiency functions, the 0.5’s and 1 – 0.5x’s, etc. that I put in this diagram are things I just made up to try and concoct a fun example.

      That said, I still hold out a modicum of hope that this kind of network approach can be useful for basketball. I even have a little side project about the idea, and if I ever get that finished I’ll post about it some time here.

      • john permalink
        July 3, 2014 1:41 pm

        The words were earned, and the math was already done to “correct” the post that tried to copy yours (by the way, I didn’t try the overall 6 options approach as it appeared there would be way too many terms for me to reliably keep track of them all).

        I realize a model is not “reality”, but I would still say that your analysis is a proof of concept. Still, I appreciate you mentioning the distinction, as that is one of the 2 aspects of game theory use that most annoy me. Too often, an analysis of a simplistic model is applied to reality. William Spaniel’s analysis of soccer penalty kicks is a blatant example. He says kickers should kick more to their less accurate side. A simplistic model not too different from his gives the opposite result, and data he links to also suggests the opposite.

        Anyway, keep up the good work.

  10. Rufus T Barleysheath permalink
    October 10, 2014 9:54 am

    Forgive my ignorance, but I’m trying to learn here, and you seem like one to teach.

    I’ll frame my question with an example, if that’s fine.

    The Eagles run 333 plays and gain:

    6.5 Y/A on 203 pass attempts, 60.96% of plays
    3.8 Y/A on 130 rush attempts, 39.04% of plays

    How do I figure out their ideal pass/run ratio?

    Thanks for the great work. I’m learning a lot from you.

    • Brian permalink*
      October 10, 2014 11:10 am

      Hi Rufus (I love your name, by the way),

      The short answer is that the information you gave me isn’t enough to answer the question. A lot of people have looked at facts like the ones you just wrote and concluded that the Eagles should pass more and run less. But the true answer requires you to know how the success of passing/running depends on the frequency with which each play type is used. It might be that the pass is declining quickly with increased usage %, while the run gives about the same yardage regardless of how often it is run. If this is the case, then the Eagles’ play calling could be understood as something not far from optimal.

      • Rufus T Barleysheath permalink
        October 13, 2014 11:25 am

        Thanks, Brian. I figured if it were that easy, someone would have done it by now.

  11. David Zhang permalink
    January 25, 2016 12:45 pm

    I have a question about the 1-0.5X function. I understand that X is a percentage, so it’s between 0 and 1. Is it taken before or after an instance of a play? For example, if for the first play of the game, the 2 guard drives and shoots. Then on the second play the 1 guard drives and shoots. For this particular play, is his X equal to 0.5 or 0? It would be 0.5 if it takes into consideration of the current play, and 0 of it only considers previous plays (he hasn’t used his drive before).

    • Brian permalink*
      January 26, 2016 12:38 am

      Hi David,

      This is sort of a tricky level of description that I’m attempting. The quantity x is best thought of as a game-wide strategy for how often to use the play in question. The idea is that if the defense sees their opponent using a given play with frequency x, then they will defend the play in such a way that it is successful 1-0.5x percent of the time on average.

      This level of description shouldn’t be taken literally to the extent of insisting on the function 1- 0.5x after just one or two plays. It only makes sense when averaged over many play attempts.


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