Can You Play the Favorite?

Get ready for a March Madness-themed puzzle! This time, you’re tasked with figuring out how likely a 1-seed is to emerge from their region.

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

March Madness—the NCAA basketball tournament—is here!

The single-elimination tournament consists of 64 teams spread across four regions, each with teams seeded 1 through 16. (In recent years, additional teams beyond the 64 have been added, but you needn’t worry about these teams for this week’s puzzle.)

Suppose in any matchup between teams with seeds M and N, the M-seed wins with probability N/(M+N), while the N-seed wins with probability M/(M+N). For example, if a 3-seed plays a 5-seed, then the 3-seed wins with probability 5/8, while the 5-seed wins with probability 3/8.

In one of the brackets, the top four seeds remain (i.e., the 1-seed, the 2-seed, the 3-seed, and the 4-seed). If case you’re not familiar with how such brackets work, at this point the 1-seed and 4-seed face off, as do the 2-seed and 3-seed. The winners then play each other.

What is the probability that the 1-seed will emerge victorious from this region?

This Week’s Extra Credit

As before, the probability that an M-seed defeats an N-seed is N/(M+N). But instead of 16 teams in a region, now suppose there are 2^k teams, where k is a very large whole number.

The teams are seeded 1 through 2^k, and play in a traditional seeded tournament format. That is, in the first round, the sum of opponents’ seeds is 2^k+1. If the stronger team always advances, then the sum of opponents’ seeds in the second round is 2^k−1+1, and so on. Of course, stronger teams may not always advance, but this convention tells you which seeds can play which other seeds in each round.

For any such region with 2^k teams, what is the probability that the 1-seed emerges victorious from the region?

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 something a little different—at the intersection of statistics, analysis, and basketball.

Earlier this week, I published the results of a regression that ranked college basketball teams, in preparation for “March Madness.” As you may already know, I’m far from the only person who created a model to help fill out brackets. If you enjoyed my column, be sure to check out the writings and predictions of my fellow FiveThirtyEight alumni like Neil Paine and, of course, Nate Silver.

There are some striking similarities between these more advanced models (which plug aggregations of various ratings systems into logistic regression models and simulate outcomes) and my own regression. I’m still not sure if that’s utterly fascinating or utterly predictable.

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. Next week (Mar. 28, 2025) will be the final set of puzzles for the first quarter. (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: 🎻 Jeffrey Ling 🎻 from Redwood City, California. I received 96 timely submissions, of which 79 were correct—good for an 82 percent solve rate.

Last week, you were planning a picnic on the remote tropical island of 𝜋-land. The island’s shape was a perfect semi-disk with two beaches, as illustrated below: Semicircular Beach (along the northern semicircular edge of the disk) and Diametric Beach (along the southern diameter of the disk).

If you picked a random spot on 𝜋-land for your picnic, what was the probability that it would be closer to Diametric Beach than to Semicircular Beach? (Unlike the illustrative diagram above, you assumed the beaches had zero width.)

First off, kudos to those who commented that the diagram above, which I painstakingly produced over the course of many seconds, was reminiscent of the street signs on Sesame Street. I mean … they’re not wrong. The island of 𝜋-land is right there with a “123” inside of it:

I admit that graphic design is not my forte. Anyway, back to the puzzle!

When I first heard this problem from its submitter, I immediately thought of Voronoi diagrams. To make a Voronoi diagram, you start with a set of points, or markers. Then, you partition the plane into “cells,” or regions that are each closer to one respective marker than any of the others. When producing such diagrams, the key is to draw the boundaries between cells. Each boundary is a locus, or set of points, equidistant between two or more of the markers. On either side of the boundary, you are now closer to one marker than the others, but along the boundary you maintain a balance.

Finding that boundary—where you were the same distance from Diametric Beach and Semicircular Beach—was a great strategy for this problem as well. Once you had that boundary, then all the points below it were closer to Diametric Beach, while all the points above it were closer to Semicircular Beach. The probability your picnic spot was closer to Diametric Beach was then the ratio of the area below the boundary to the area of the entire semicircle.

That said, determining the boundary itself was no simple feat.

To get started, let’s rescale the island to a unit disk. That is, we set the midpoint of Diametric Beach to the point (0, 0), while Semicircular Beach lay on the top half of the unit circle, defined by the equation x² + y² = 1. To define our boundary, let’s find all the points on the island that are the same distance d from both beaches.

To be a distance d from Diametric Beach, our point had to be on the line y = d. And to be a distance d from Semicircular Beach, our point had to be on a circle that was d inland from the beach itself, meaning it was on the circle x² + y² = (1−d)². The points where the line and the circle intersected, for various values of d, represented the boundary. The following animation sketches out this boundary by letting d vary from 0 to 0.5:

The path traced out by the intersections was some kind of arc—perhaps circular, parabolic, or hyperbolic—but certainly a conic section, given the quadratic nature of the equations we’ve looked at so far.

At the intersection of the line and circle, we needed both y = d and x² + y² = (1−d)² to be true. Plugging in d for y in the second equation, you got x² + d² = (1−d)² = 1 − 2d + d². With some cancellation, this became x² = 1 − 2d. Recalling that y = d, we can now plug y back in to get x² = 1 − 2y. Finally, solving for y gave you y = (1−x²)/2. This was a parabola, and was precisely the equation of the boundary mapped out in the animation above.

So any point in the unit semi-disk below the parabola y = (1−x²)/2 was closer to Diametric Beach, while any point above the parabola was closer to Semicircular Beach.

Next, you needed to find the area of the region between the parabola and the x-axis. To compute this area, calculus was your friend. In particular, the area was the integral of (1−x²)/2 from x = -1 to x = 1, which came to 2/3.

The area on the island below the curve was 2/3, while the area of the entire island (i.e., half the area of a unit disk) was 𝜋/2. As we already said, the probability your picnic spot was closer to Diametric Beach was the ratio of these two areas, (2/3)/(𝜋/2), which simplified to 4/(3𝜋), or approximately 42.44 percent.

As observed by solver David Ding, this puzzle had a little bit of everything: probability, geometry, algebra, and calculus. What a fitting way to have celebrated Pi Day!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Q P Liu 🎻 from Santa Cruz, California. I received 50 timely submissions, of which 40 were correct—good for an 80 percent solve rate.

Suppose the island of 𝜋-land, as described above, had a radius of 1 mile. That is, Diametric Beach had a length of 2 miles.

Again, you were picking a random point on the island for a picnic. On average, what was the expected shortest distance to shore?