Basketball Shooting Drills Difficulty

The best way to become a better basketball shooter is to put up a lot of shots. However, shooting shot after shot can get monotonous. To make shooting interesting, drills can add a challenge, such as making a certain number of shots within a time limit. Often, these drills have a psychological challenge to them as well, as you must stay focused the entire time and not let misses or makes change your approach. Of course, any drill is easier for a higher accuracy shooter, but what success can we expect for a shooter given his typical accuracy? We’ll investigate this for three shooting drills.

Shoot 50% over 100 shots

This is a simple drill but it’s useful to think about the chances of success because they impact other drills. Let’s say you are a 40% 3 point shooter. If you shoot 100 shots, the average and the most likely percentage of shots you would make is 40%. What is the probability that you make more than 50% of the 100 shots?

I simulated ten thousand drills with each having one hundred shots. Here’s what the distribution looks like.

This is a binomial distribution but it is close to a normal distribution with a mean of 40.0 and a standard deviation of 4.9. We can use the properties of normal distributions to say that the chance of making 50% or more is 2.0%.

The more shots you take, the harder it is to get away from your long term accuracy. If you were to shoot one thousand shots each drill, the distribution would be much tighter.

The mean is the same, but the standard deviation is now only 1.6, meaning there is a 95% chance you will make between 36.9% and 43.1% of the shots. There is a 1 in 10 billion chance you would make 50% or more.

How can this be generalized to any shooting accuracy and shots per drill? The useful principle here is the standard error of the mean. Let’s say you are measuring the diameter of a basketball. Measuring once will give you an ok estimate of the true diameter. Measuring it again and averaging the two measurements would give you a better estimate. You can imagine that the more measurements you take, the closer your estimate will be and the more confidence you will have in your estimate. That’s basically what is happening here. By shooting 100 shots, you are “measuring” your true shooting ability. After one shot, that estimate will be bad: either 0% or 100%. But as you shoot more shots, the average make percentage will converge to your true shooting percentage, as we saw above. The equation that describes this is

$\sigma \approx \frac{\sigma_m}{\sqrt{n}}$

The $\sigma$ is the standard deviation of your estimate. For the 100 shot drill, this was 4.9. The $n$ is the number of measurements. The $\sigma_m$ is the standard deviation of each individual measurement. If you’re measuring something, that would be similar to the precision of each measurement.

Here, the standard deviation of each measurement can be calculated by taking a bunch of measurements and calculating the standard deviation of that group. This will be a function of the shooting accuracy. For a shooting accuracy of 50%, the group of measurements could look like

[100%, 100%, 0%, 100%, 100%, 0%, 0%, 100%, 0%, 0%]

The standard deviation of this group is 50. This will be the maximum standard deviation because the values are as spaced apart as possible. A shooting accuracy of 100% would mean each measurement is the same: 100%. The standard deviation of this is 0.

I did this calculation of the standard deviation for each shooting accuracy:

We now have a numerical way to provide the likelihood of making a certain number of shots given the shooter’s accuracy and number of shots in the drill. If we can come up with an equation for this measurement standard deviation curve, we would have a fully analytical way. The curve looks like a semicircle. Why is that?

Let’s start with the definition of standard deviation:

$\sigma=\sqrt{\frac{\sum{(x-\bar{x})^2}}{n}}$

$x$ is each one of the measurements. $a$ will represent the accuracy of the shooter. It is equal to the mean of the measurements. For instance, with a 40% shooting accuracy, the measurements could be

[100%, 100%, 0%, 0%, 0%, 100%, 0%, 100%, 0%, 0%]

The mean of that group is 40%.

$a=\bar{x}=\frac{\sum{x}}{n}$

We can substitute in $a$ and multiply out the squared term.

$\sigma=\sqrt{\frac{\sum{(x^2-2ax+a^2)}}{n}}$

$\sigma=\sqrt{\frac{\sum{x^2}}{n}-\frac{\sum{2ax}}{n}+\frac{\sum{a^2}}{n}}$

The third term is the average of a constant so it is just the constant. For the second term, if you pull out the $2a$ it is the mean of the measurements, which is $a$ .

The first term is a little tricky. If you think about what the measurements consist of, it becomes clearer. In the group above for the 40% accuracy, there are 6 measurements that are 0% and 4 that are 100%. Applying the sum to that group would be

$\sum{x^2}=6(0^2)+4(100^2)$

More generally, the number of 100%s is

$\frac{a}{100}n$

So, the first term is

$\frac{\frac{a}{100}n100^2}{n}=100a$

Putting the three terms together

$\sigma=\sqrt{100a-2a^2+a^2}=\sqrt{100a-a^2}$

If you pull out the $a$ , it is actually the standard deviation of the Bernoulli distribution.

$\sigma=\sqrt{a(100-a)}$

So this is a derivation of that property, which is pretty cool.

At this point, we have two terms of a quadratic. I can complete the square to factor it and end up with this equation.

$\sigma=\sqrt{50^2-(a-50)^2}$

This describes the shape of the top half of a circle with radius 50 and centered at x=50. Let’s plot that shape with the simulated curve from before.

Boom.

Putting the measurement standard deviation together with the standard error of the mean equation, we get the same result as here.

$\sigma=\sqrt{\frac{a(100-a)}{n}}$

Two comparisons to check if the equations are correct: First, I simulated a few drills ten thousand times with increasing numbers of shots per drill. I calculated the standard deviation of the number of makes across the ten thousand attempts and plotted them on the same figure as the expected standard deviation from the equations. They lined up perfectly.

Second, I created a normal distribution PDF with the theoretical mean and standard deviation. I plotted this on the histogram we saw above.

Make Two In A Row

This drill challenges your 3 point shooting and ability to hit big shots. There are five stations around the three point line and you must make two shots in a row to move to the next station. The goal is to go all the way around in the fewest shots possible.

I simulated one million attempts of the drill for a shooting accuracy of 35%. This is the distribution of how many shots it would take to complete the drill.

This is not a normal distribution because it is skewed to the left. However, we can still calculate a confidence interval numerically. 90% of the time, it would take between 22 and 88 shots to complete the drill and the mean number of shots is 55.

Higher shooting accuracies yield lower number of shots required, and also less of a spread. If you’re shooting 55%, the mean is 26 shots and the 90% confidence interval is 13 to 38 shots. A bad shooter is more likely to have a bad day and need to take a ton of shots to finish the drill.

Clearly the mean number of shots gets smaller as the shooting accuracy improves. Here’s what that looks like for having to hit two shots in a row.

This is the experimental curve from the simulations. Like before, let’s try to come up with an analytical way to predict the mean number of shots as a function of the shooting accuracy and how many shots in a row are required.

$s=\mbox{ shots required}$

$n=\mbox{ number of shots in a row}$

$p=\mbox{ chance of making a shot}$

$q=\mbox{ chance of missing a shot }=1-p$

The key goal of the drill is to hit $n$ shots in a row. If you shoot two shots, the chance of that happening for that set of two shots is

$p^2$

The expected number of attempts required is the inverse of that. For instance, if you have a 50% chance of making one in a row, you have a 25% chance of making two in a row, and you would expect that to happen once every four attempts to make two in a row. If you expect to be successful after four attempts on average, would it take eight shots on average to move to the next station? It’s actually fewer. Each attempt to make two in a row resets after a miss. The maximum length of an attempt is two but that is not the average length. For $n=2$ , there are three possibilities.

Missing the first shot. This means you have no chance of making two in a row on the second shot, so that attempt has ended and a new attempt starts on shot #2. The length of that attempt was only 1.
Making the first and then missing the second shot. Now you reset and start your second attempt on shot #3, giving the attempt a length of 2.
Making and first and making the second shot. This attempt is successful and also has a length of 2.

The average length of an attempt is the sum of all the possibility probabilities weighted by their length. In order, this is

$1q+2pq+2p^2$

For $p=0.5$ , this equals 1.5, which is less than 2. In general, the length of attempts will be short for low $p$ because a miss is likely to happen early on and end that attempt. To generalize this for other $n$ , let’s do the same thing for $n=3$ and look for a pattern.

There are now four possibilities for an attempt.

Miss
Make, Miss
Make, Make, Miss
Make, Make, Make

Average length is

$1q+2pq+3p^2q+3p^3$

Looking just at the unsuccessful possibilities, they each have a $q$ in them. The order of $p$ starts at $n-1$ and goes down to 0. The coefficient in front starts at $n$ and goes down to 1. So, the general form of the average attempt length is

$np^n+q \sum_{m=1}^{n}{mp^{m-1}}$

The final estimate for the shots required in the drill is the expected number of attempts to be successful once, times the average shots per attempt, times the 5 stations.

$s=5p^{-n}(np^n+q \sum_{m=1}^{n}{mp^{m-1}})$

Plotting this equation on the same curve as before shows a spot on agreement!

The equation indicates that the more makes in a row are required, the harder the drill gets.

If you shot 40% on threes and only had to hit two in a row, on average it would take 43 shots. But if you had to hit four in a row, it would take 316 shots! Increasing the shots in a row by a factor of 2 increased the difficulty by about a factor of 7 in this case. This is an easy way to make the drill more challenging.

Don’t Miss Two In A Row

This drill is about the opposite of the “Make Two in a Row” drill. The goal is to make as many shots as you can before you miss two shots in a row.

Similar to before, I simulated one million drills for several shooting percentages. This is how many shots you would expect to make for 35% accuracy.

The most likely outcome is that you miss the first two shots and end up with a total of zero. Probabilities decline sharply from there. The mean is only 1.4 shots. 90% of the shots made are between 0 and 4.

For 55%, results in the drill are better: the mean is 3.9 shots and the 90% confidence interval is 0 to 10.

These distributions are not normal or half of a normal distribution so calculating standard deviation isn’t that useful.

The shots made as a function of shooting percentage before missing two in a row has an increasing exponential behavior.

In the “Make Two in a Row” drill, there were large improvements in performance for shooting accuracy improvements when under about 50% accuracy. For this drill, the big improvements in performance don’t come until accuracy is well above 50%.

As you might imagine, ending the drill at three misses in a row instead of two enables higher scores. However, the big effect doesn’t kick in until the shooting accuracy is quite high. At 40% the increase in shots made is only 1.9, but at 70% it is 25.3.

We can analytically calculate the shots made as a function of shots in a row and shooting percentage using the result from the “Make Two in a Row” drill as a starting point.

$s=5p^{-n}(np^n+q \sum_{m=1}^{n}{mp^{m-1}})$

The drill ends when you miss two in a row, not when you make two in a row, so all the $p$ s and $q$ s are simply switched. There is only one station, not five. The previous equation measures the amount of shots taken, while this drill is about the number of shots made. To calculate that, we multiply the shots taken by the shooting accuracy to get the expected number of shots made.

$s=pq^{-n}(nq^n+p \sum_{m=1}^{n}{mq^{m-1}})$

Plotting the equation with the simulation matches very well like before.

We looked at the distribution of shots made for a given shooting percentage. It’s also interesting to look at the inverse distribution: shooting percentage for a given number of shots made.

If you make 5 shots in this drill, it’s most likely that you are a 58% shooter. However, it’s also possible to make 5 shots while being a 20% or 90% shooter.

Higher performance is more exclusive. If you’re going to make 50 shots, you need to shoot at 74% to 95% accuracy, with 90% confidence.

Using math and simulations, we evaluated the expected results from three shooting drills based on the shooter’s accuracy. Having these statistics in mind may give you a benchmark to compete against as you try these drills. You could measure your long run shooting percentage with the “Shoot 50% over 100 Shots” drill, and then see if you are underperforming or overperforming on the other drills.

Shoot 50% over 100 shots

Make Two In A Row

Don’t Miss Two In A Row

Share this:

Leave a Reply Cancel reply