If Brandon Jennings Could Shoot Like Stephen Curry

The NBA season is coming to an end soon. It is around this time of the year that “Awards Watch” talks begin to heat up. Typically the two most relevant and interesting races are the MVP and Rookie-Of-The-Year awards. This year LeBron James has had the MVP award locked up for some time now.

That leaves us to ponder over the top rookie title. Unlike the MVP race, this award has plenty of qualified candidates. The media, who vote for the award, have generally focused on three players as the most likely to win – Tyreke Evans, Stephen Curry, and Brandon Jennings.

What I want to focus on today is the play of Milwaukee Bucks guard  Brandon Jennings. More specifically, I want to talk about his ability (more like inability) to put the ball through the basket. In order to win a basketball game it is very important to make the shots you take. This idea is really simple and I think every basketball fan generally knows this to some degree. However when “experts” discuss the ability and productivity of a basketball player, especially in comparison to other players, this idea is generally ignored. The most cited stat of choice, sometimes the only reference of comparison, is Points Per Game. If you score a lot of points in the NBA you are generally considered a good player, no matter how many shot attempts it costs you to achieve that score. To most NBA observers a player scoring 25 PPG is very good, most of the time they’re right. However, what is often ignored is how they amass those points. Does it take them 20 shots or 30 to 40?. There is a big difference and a team has only a limited number of possessions to use. Empty (scoreless) possessions put a team in a hole. All of this brings me to note that making shots is something Brandon Jennings struggles to do. Because of this I don’t think Jennings should be considered the top choice for Rookie Of The Year.

When I say Jennings is struggling, I think I need to put it in perspective. Brandon Jennings field goal percentage (37%) is the lowest of any player, who shoots as often as he does, in the last 30 years (Evidence here). Now Jennings is a good 3 point and free throw shooter, so his low FG% is a bit misleading. However, when you look at his True Shooting Percentage (Which takes into account 3 pointers and free throws), he still comes up short. His TS% is 47.4%, he ranks 32nd of 38 among qualified rookies according to ESPN.

I have established that Jennings struggles making the shots he takes but what does this mean to his team? I’ve taken a long time to get to the point in the title but the question I ask is “What if Brandon Jennings could shoot like Stephen Curry?” I chose Curry as a point of comparison for a few reasons. First, they are both rookie point guards. They both play generally the same role on their team. And they have very similar non-shooting related statistics. Most important, Curry is a very good shooter.

If Jennings could shoot like Curry, what would it mean to the Bucks? This is where statistics come in. Curry scores .96 points per possession while Jennings scores .85. The Milwaukee Bucks as a team have an offensive efficiency of 101.7(Basically, their teams Points Per Possession times 100). Milwaukee as a team uses 91.2 possessions a game. Of these possessions Brandon Jennings uses 17.9 (on average). Using these possessions Brandon Jennings manages to average 15.4 points. If Jennings shot the ball like Curry he would average 17.18 PPG. That is a difference of a whole 1.78 points. That would give the Bucks an Offensive Efficiency of 103.5. If you subtract a teams defensive efficiency from its offensive efficiency you get its efficiency differential (Basically, how badly a team tends to beat their opponent). The Bucks currently sport a differential of 1.2. If BJ shot like Curry it would be 3.0.

Blah, blah, blah…what does this all mean? Well, efficiency differential is the best tool to use in determining which basketball team is the best. As such, it is the best tool for predicting future performance as well. As an example, Duke University led the nation this year in efficiency differential. A team with a zero point differential should finish 41-41 in an NBA season. A typical NBA team with an efficiency differential of 1.2 should win between 44-45 games. A team with an efficiency differential of 3.0 should win about 49 games. (I’m saying this because NBAStuffer.com tells me one point of efficiency differential = 2.7 wins, if this is incorrect please let me know). So if Brandon Jennings could shoot like Stephen Curry the Milwaukee Bucks would be 4 to 5 wins better. That is a significant amount considering it is just one aspect of one players production. So that answers my original question.

As I said Curry is an efficient scorer but what if Brandon Jennings scored like Chauncey Billups, who is one of the most efficient point guards in the NBA (1.08 PPP). If Jennings scored as efficiently as Billups then the Bucks would be looking at a projected record of 54 to 55 wins (a ten win difference!). No wonder a Chauncey Billups led team has reached the conference finals for six consecutive seasons (Now that we know how important making your shots is to winning, who would ever trade Billups for Allen Iverson?).


3 Comments on “If Brandon Jennings Could Shoot Like Stephen Curry”

  1. Palamida says:

    Robbie, completely off-topic. In Berri’s blog u asked how does one derive exp. wins from eff. diff.
    I wanted to answer u but I figured since I can’t reply there directly, u might miss it, so… i’ll do it here :p
    Without getting into the “Why” (you can look that up if u want) this is how u do it:
    let’s say a team has a diff. of 3, for example.
    a.) take the differential and multiply by 2.4 – 3*2.4=7.2 (mind u the 2.4 is a constant u should use for every given eff. diff.)
    b.) u combine the result of step a. with 41 – 7.2+41=48.2.
    there’s ur answer :p
    A team with an efficiency differential of 3 has a pythagorean expectation of 48.2 Wins.
    Hope this helps, enjoy.

  2. Thank you, that is really helpful! I have seen other sites claim each efficiency point is worth 2.7 wins. But the difference between 2.4 and 2.7 is not all that much.

  3. […] As you can see, Ty Lawson was the most productive player (per minute) of these four players. To add some context Darren Collison was a rookie and TJ Ford struggled to stay healthy, with either of these factors you can expect a player’s WP48 to be lower. However, Ty Lawson was also a rookie and struggled with injuries (missing 17 games). Despite this, overall Lawson was still more productive than the other three point guards (He was even better in the 8 games in which he was the starter). The difference in production is mainly attributable to Lawson being above average with respect to shooting efficiency and avoiding turnovers. Collison was above average in regards to shooting efficiency but struggled mightily in protecting the basketball. Watson and Ford were below average in both areas. It doesn’t take a genius to note that making the shots you take and avoiding turnovers are two of the most important aspects involved in winning on the court (Actually, one genius made such an observation – HERE). […]

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