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Can You Best The Mysterious Man In The Trench Coat?

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. EDT on Sunday — have a great weekend!

Before we get to the new puzzle, let’s return to last week’s. Congratulations to ÑÑâÐ Jakob Infuehr ÑÑâÐ of Austin, Texas, 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, a fun one which comes to us from Dan Oberste of Little Rock, Arkansas.

A man in a trench coat approaches you and pulls an envelope from his pocket. He tells you that it contains a sum of money in bills, anywhere from $1 up to$1,000. He says that if you can guess the exact amount, you can keep the money. After each of your guesses he will tell you if your guess is too high, or too low. But! You only get nine tries. What should your first guess be to maximize your expected winnings?

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, concerning the efficacy of the “granny shot” free throw. Thirty-four percent of you hit the mark with your submissions — just a tick worse than Andre Drummond. This solution is adapted from the clever answer submitted by our winner, Jakob.

The percentage of time you’ll make a “granny-style” free throw is about 90.4 percent. Instead of fixing the radius of the circle, I fixed the variance of the random variables to one. So I have to find the radius of the circle first. The circle equation, thinking back to high school geometry, is $$x^2+y^2\leq r^2$$. The distribution of $$x^2+y^2$$ is a Chi-squared distribution with 2 degrees of freedom. Thus, to find the radius, I solved the following equation for r: $$0.75=1-e^{-r^2/2}$$. That makes $$r\approx 1.665$$. Using this value, I can find $$p=F(1.665)-F(-1.665)$$, where $$F(x)$$ is the cumulative distribution function of a standard normal variable. This gives the solution of 0.904.

In other words, if you’re a 75-percent free-throw shooter using the “normal” method, you can expect to improve to above 90 percent shooting underhand!

Jon Dudley emailed me this great illustration of the solution:

The extension tweets came in hot and heavy this week, and you’re all going to have to share the ÑÑÐâ  Coolest Riddler Extension Award ÑÑÐâ , because I simply can’t make up my mind.

Nick Stanisha compared our Riddler model to some real-life pros:

Randi Goldman argued that Steph Curry should always go for the swish, rather than the backboard:

And, last but not least, Joel Walker, a physics professor, emailed a proposal for an Interstellar Basketball Association. He asked, “Given the scoring fraction p for a player in a D+1 dimensional space, aiming at a hypersphere of radius 1, what is the scoring fraction for a player in a corresponding compact D-dimensional space, keeping all else equal?” The results:

“It would seem that the advantage of the lower dimensional space is much more pronounced for poor shooters (as it is not possible to improve much on a player who does not miss often),” Joel wrote. I’ve forwarded Joel’s proposal to the International Olympic Committee. No word yet.

Have a great 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.