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Should You Shoot Free Throws Underhand?

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world, including you, the readers. You’ll find this week’s puzzle below.

Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the form at the bottom. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible for the shoutout, I need to receive your correct answer before 11:59 p.m. EST on Sunday — have a great weekend!

Before we get to the new puzzle, let’s return to last week’s. Congratulations to рџ‘Џ Mark van Hoeij рџ‘Џ of Tallahassee, Florida, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

Now, here’s this week’s Riddler, which comes to us from Po-Shen Loh, a math professor at Carnegie Mellon University, the coach of U.S. International Math Olympiad team and the founder of

Hark! The NCAA Tournament starts next week, and the “granny shot” has reappeared as a free throw technique. Its proponents claim that it improves accuracy because there are fewer moving parts — the elbows and wrists are held more stable, for example, and the move is symmetric because one’s arms are, more or less, equal length. Let’s find out how effective the granny shot really is.


Illustration by Expii

Consider the following simplified model of free throws. Imagine the rim to be a circle (which we’ll call C) that has a radius of 1, and is centered at the origin (the point (0,0)). Let V be a random point in the plane, with coordinates X and Y, and where X and Y are independent normal random variables, with means equal to zero and each having equal variance — think of this as the point where your free throw winds up, in the rim’s plane. If V is in the circle, your shot goes in. Finally, suppose that the variance is chosen such that the probability that V is in C is exactly 75 percent (roughly the NBA free-throw average).

But suppose you switch it up, and go granny-style, which in this universe eliminates any possible left-right error in your free throws. What’s the probability you make your shot now? (Put another way, calculate the probability that |Y|

Extra credit: Let’s offer up a рџЏ† Coolest Riddler Extension Award рџЏ†. Surely you, dear Riddler readers, can improve on this simple model. Try out some alternate shooting techniques, or introduce some more realistic assumptions, or whatever! Have fun, and submit a description in the form below, or shoot me a link to your work on Twitter. And for some more basketball math, and problem sets of all kinds, check out

Need a hint? You can try asking me nicely. Want to submit a puzzle or problem? Email me.

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And here’s the solution to last week’s Riddler, which was about how to win a hot new game show. It comes to us from Stephen Mellendorf. About a quarter of you are ready for primetime — 26.7 percent of your submitted solutions were correct.

The “obvious” solution is to use a cutoff of 0.5, throwing away your number in exchange for another if it’s less than 0.5, and keeping it if it’s greater than that, since that strategy will yield the highest average final number. But that isn’t the optimal cutoff.

Let C be the optimal cutoff the players use. The key observation is that if the first number revealed is exactly C, then the probability of winning by keeping C equals the probability of winning by pressing the button again — you are indifferent. We can compute each of these probabilities, keeping in mind that the other player is also using C as their cutoff.

probability player 1 wins by keeping C = probability player 2 gets a number below C for both presses = \(C \cdot C\)

probability player 1 wins by pressing again = (probability player 2 presses again) * (probability player 1 wins by pressing again | player 2 presses again) + (probability player 2 keeps first number) * (probability player 1 wins by pressing again | player 2 keeps first number) = \((C) \cdot (1/2) + (1 – C) \cdot ( (1 – C) \cdot 1/2)\)

As noted above, the above two are equal, so

\(C \cdot C = (C) \cdot (1/2) + (1 – C) \cdot ( (1 – C) / 2)\)

which simplifies to

\(C^2 + C – 1 = 0\)

The quadratic formula gives the solution.

\(C = (\sqrt{5} – 1) / 2 = 0.618034\ldots \)

Note that this cutoff is the golden ratio minus one, known as the golden ratio conjugate. So using the golden ratio gives the best chance to win the gold bullion!

This problem, as usual, yielded some pretty pictures. Zach Wissner-Gross provided this (spoiler-free) illustration of his solution:

And Christopher Mullan — in the best email I’ve received in recent memory, thanks, Chris — sent along what he dubbed the Pringle of Probability:


He also sent a Frown of Probability, not pictured. But hey, smile, it’s Friday! I’m also proud to report that David Sklansky, poker god, correctly solved this problem, although he was not the randomly chosen winner. Sorry, David. Have a super weekend.

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.