Can You Defend Your Trivia Knowledge?

In “LearnedLeague,” you and your opponent answer the same six questions, and then assign point values for each other. How easy (or hard) is it to defend your knowledge of trivia?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “Extra Credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

A new season of LearnedLeague recently kicked off! Many folks may not be familiar with this daily trivia platform. I learned about it a few years ago and joined after my turn as a game show contestant.

Anyway, here’s how it works: Every day, you and your opponent for the day are presented with the same six trivia questions. You each do your best to answer these, and you assign point values for your opponent, without knowing (until the following day) which questions your opponent answered correctly. You must assign point values of 0, 1, 1, 2, 2, and 3 to the six questions.

For example, suppose I assigned values of 1, 2, 0, 3, 2, and 1 to the six questions, in order. Unbeknownst to me, my opponent answers the first, third, and fourth questions correctly. That means they get 1 + 0 + 3, or 4 points for the match. My own score depends on which questions I got right, and how these were scored by my opponent. If I get more than 4 points, I win the match.

Now, when someone answers three questions correctly, like my opponent just hypothetically did, the fewest points they can earn is 0 + 1 + 1 = 2, while the most points they can earn is 2 + 2 + 3 = 7. The fact that they got 4 points wasn’t great (from my perspective), but wasn’t terrible. In LearnedLeague, my defensive efficiency is defined as the maximum possible points allowed minus actual points allowed, divided by the maximum possible points allowed minus the minimum possible points allowed. Here, that was (7−4)/(7−2), which simplified to 3/5, or 60 percent.

By this definition, defensive efficiency ranges somewhere between 0 and 100 percent. (That is, assuming it’s even defined—which it’s not when your opponent gets zero questions right or all six questions right.)

Suppose you know for a fact that your opponent will get two questions right. However, you have absolutely no idea which two questions these are, and so you randomly apply the six point values to the six questions.

What is the probability that your defensive efficiency for the day will be greater than 50 percent?

This Week’s Extra Credit

Now suppose your opponent is equally likely to get one, two, three, four, or five questions correct.

As before, you randomly apply the six point values (0, 1, 1, 2, 2, 3) to the six questions.

What is the probability that your defensive efficiency will be greater than 50 percent?

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing that the problems from the 2025 American Invitational Math Exam (AIME) have been posted.

These are always good fun. If I can’t find three hours to sit down and try them out in a simulated exam session, I usually save them for a long car ride and see how many I can solve in my head. Solving while driving is a lot slower, but it definitely forces you to brainstorm approaches that require the less actual computation.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. (If you think you see a mistake in the standings, kindly let me know.)

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Dave Polidori 🎻 from Rancho Santa Fe, California. I received 49 timely submissions, of which 40 were correct—good for an 82 percent solve rate.

Last week, you analyzed a heart shape that included a unit square (i.e., a square with side length 1), with semicircles (each with radius 1/2) attached to adjacent edges, as shown in the diagram below:

What was the radius of the smallest circle that contained this heart shape?

The enclosing circle clearly passed through the bottommost point (or “apex”) of the heart. The question was then where along the two semicircles were the points that were tangent to the larger circle. You might have assumed that these tangent points were halfway along each of the semicircular arcs, but this was not correct. The illustration above indicated that the tangent points were both a little closer to the apex than that.

To be fair, you didn’t actually have to locate these points of tangency. But you did need to set up equations that assumed the enclosing circle was tangent to each of the semicircles, and then use these equations to solve for the radius of the larger circle. There were several ways to go about doing this.

One method, courtesy of solver Steve Curry, relied on the fact that when two circles are tangent to each other, their centers are collinear with the point of tangency, as illustrated in the following diagram:

Steve defined R as the radius of the larger circle and x as the vertical distance between the center of the larger circle and the centers of the two semicircles.

Starting from the apex of the heart and moving up to the midpoint of the semicircle’s centers gave you one equation: R + x = 3/4·√2. You could rearrange this to express x in terms of R: x = 3/4·√2 − R. Meanwhile, applying the Pythagorean theorem to the smallest right triangle in the diagram gave you a second equation: x² + 1/8 = (R − 1/2)².

Plugging the first equation into the second gave you a single equation in terms of R: (3/4·√2 − R)² + 1/8 = (R − 1/2)². Expanding the quadratic terms then led to cancellation of the R² terms, leaving you with a linear equation to solve. With a little rearrangement, you found that R = 2/(3√2 − 2). If you were so inclined, you could “rationalize the denominator” by multiplying the numerator and denominator by 3√2 + 2, which gave you R = (3√2 + 2)/7, or approximately 0.8918.

I’ll note that there were quite a few other ways to solve this, which generally ended up in linear or quadratic equations—and the correct answer, of course. How lovely.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Ed Foley 🎻 from Redmond, Washington. I received 19 timely submissions, of which 18 were correct—good for a 95 percent solve rate.

Instead of containing one heart shape, now your circle had to contain two heart shapes. Again, each heart consisted of a unit square and two semicircular lobes. The two hearts were not allowed to overlap.

What was the radius of the smallest circle that contained these two hearts?