Unavoidable Free Throw Percentages

This blog post is the closest I will ever get to playing basketball and it’s still pretty far away.

According to the internet, free throws are awarded to a player fouled in the act of shooting and when the defending team has committed 5 team fouls.

It then makes sense that there is a thing called a free throw percentage (FTP). Basically, for a given player, it’s the percentage of free throws they get that end up going in the ring. The best NBA players often have percentages above 90%.

Here’s something for you to think about:

Early in a season, a player has a FTP of below 90%. Later in the season, the same player has a FTP of above 90%. Is there a point in the season when their FTP is exactly 90%?

When I was asked this, my answer was: maybe, maybe not. The Intermediate Value Theorem sprung to mind:

A continuous function $f$ on $[a,b]$ must take on any given value between $f(a)$ and $f(b)$ at some point within the interval.

Of course, a player’s FTP is not continuous, so the IVT does not apply here.

For example, if they have scored 5 out of 8 free throws, then their FTP is equal to 5/8 or 62.5%. Say they score their next free throw; then their FTP becomes 6/9 or roughly 66.67%.

Therefore, surely it’s possible for a FTP to jump over 90% without equalling it.

It turns out, however, that 90% is unavoidable. That is, the player in question will have a FTP of exactly 90% at some point.

Proof. Let’s prove this with a little proof by contradiction. We start by assuming that they can jump over 90%. That is, one second they are below 90% and the next they are above it.

We will denote by $F$ the number of free throws they have been awarded when they are sitting just below 90% and we will denote by $f$ the number of these free throws that they have scored.

Thus, their current FTP is equal to $f/F$ .

For this to be above 90% on their next free throw, they must score on the next free throw. Their FTP now becomes $(f+1)/(F+1)$ . By our assumption, this is above 90%. This gives us the inequality

$\frac{f}{F} < \frac{9}{10} < \frac{f+1}{F+1}.$

We can rearrage the leftmost inequality to get $10f < 9F$ .

We can rearrange the rightmost inequality to get $9F < 10f + 1$ .

Then, putting these together gives us the inequality

$10f < 9F < 10f+1.$

However, this is a contradiction, for $10f$ , $9F$ and $10f+1$ are all integers. Furthermore, $10f$ and $10f+1$ are consecutive integers, and so there is no way you can squeeze another integer between them!

Therefore, the player’s FTP must be exactly equal to 90% at some point.

This is pretty wild stuff. It blew my noodle when I saw it.

Also, $9/10$ is not that special! The result works for any number of the form $(k-1)/k$ where $k$ is an integer. And being the avid blog reader that you are, I’ll end this blog here so that you can prove it.