Can You Root for the Underdog?

It’s another week of March Madness! This time around, the weaker team in every match gets a boost from the fans. Is it possible for every team to win the tournament?

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

From Emilie Mitchell comes another March Madness mystery:

Once again, there are four teams remaining in a bracket: the 1-seed, the 2-seed, the 3-seed, and the 4-seed. In the first round, the 1-seed faces the 4-seed, while the 2-seed faces the 3-seed. The winners of these two matches then face each other in the regional final.

Also, each team possesses a “power index” equal to 5 minus that team’s seed. In other words:

• The 1-seed has a power index of 4.

• The 2-seed has a power index of 3.

• The 3-seed has a power index of 2.

• The 4-seed has a power index of 1.

In any given matchup, the team with the greater power index would emerge victorious. However, March Madness fans love to root for the underdog. As a result, the team with the lower power index gets an effective “boost” B, where B is some positive non-integer. For example, B could be 0.5, 133.7, or 2𝜋, but not 1 or 42. To be clear, B is a single constant throughout the tournament, for all matchups.

As an illustration, consider the matchup between the 2- and 3-seeds. The favored 2-seed has a power index of 3, while the underdog 3-seed has a power index of 2+B. When B is greater than 1, the 3-seed will defeat the 2-seed in an upset.

Depending on the value of B, different teams will win the tournament. Of the four teams, how many can never win, regardless of the value of B?

This Week’s Extra Credit

From Emilie Mitchell also comes some extra credit:

Instead of four teams, now there are 2⁶, or 64, seeded from 1 through 64. The power index of each team is equal to 65 minus that team’s seed.

The teams play in a traditional seeded tournament format. That is, in the first round, the sum of opponents’ seeds is 2⁶+1, or 65. If the stronger team always advances, then the sum of opponents’ seeds in the second round is 2⁵+1, or 33, and so on.

Once again, the underdog in every match gets a power index boost B, where B is some positive non-integer. Depending on the value of B, different teams will win the tournament. Of the 64 teams, how many can never win, regardless of the value of B?

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 another bracketology basketball problem that came my way courtesy of Goh Pi Han. This puzzle is originally from Jane Street Capital and resembles last week’s Fiddler—in particular, the probabilities of advancement from each matchup.

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. (As always, if you think you see a mistake in the standings, kindly let me know.)

This week’s puzzles are your last opportunity to earn points for Q1 2025. That means next week, I’ll be 👑 crowning 👑 the finest of Fiddlers! Also, now is a great time to get in on the action before I reset the leaderboard for Q2.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Robert Shore 🎻 from Los Angeles, California. I received 76 timely submissions, of which 73 were correct—good for a 96 percent solve rate.

In March Madness, a single-elimination tournament consists of 64 teams spread across four regions, each with teams seeded 1 through 16.

Last week, you supposed that in any matchup between teams with seeds M and N, the M-seed won with probability N/(M+N), while the N-seed won with probability M/(M+N). For example, if a 3-seed played a 5-seed, then the 3-seed won with probability 5/8, while the 5-seed won with probability 3/8.

In one of the brackets, the top four seeds remained (i.e., the 1-seed, the 2-seed, the 3-seed, and the 4-seed). At this point, the 1-seed and 4-seed faced off, as did the 2-seed and 3-seed. The winners then played each other.

What was the probability that the 1-seed emerged victorious from this region?

This was a fairly straightforward problem. (The same could not be said for the Extra Credit, but we’ll get to that later.) To win the region, the 1-seed first had to defeat the 4-seed. This occurred with probability 4/(1+4) = 4/5, or 80 percent.

From there, you had more than one case to consider, because either the 2-seed or the 3-seed could have advanced out of the region’s other semifinal game. The 2-seed won this semifinal with a probability of 3/5. Then, the final paired the 1-seed and the 2-seed, which the 1-seed won two-thirds of the time. So the probability the 2-seed advanced to the finals and the 1-seed defeated them was (3/5)·(2/3), which simplified to 2/5, or 40 percent.

Alternatively, the 3-seed could have won the other semifinal game, which occurred with a probability 2/5. The 1-seed then defeated the 3-seed in the finals with a probability of 3/4. So the probability the 3-seed advanced to the finals and the 1-seed defeated them was (2/5)·(3/4), which simplified to 3/10, or 30 percent.

Combining these two mutually exclusive cases meant the probability the 1-seed won in the finals was 2/5 + 3/10, or 7/10.

Finally, to emerge victorious from the region, the 1-seed had to win its semifinal (with probability 4/5) and win the final (with probability 7/10). The probability that both events occurred was (4/5)·(7/10), which came to 14/25, or 56 percent.

Solvers like Nick Hlavacek, Peter Exterkate, Eric Widdison, and John from Washington, DC further computed the probabilities that each of the remaining three teams won the region:

• As we already found, the 1-seed’s probability of winning was (4/5) · [(3/5)·(2/3) + (2/5)·(3/4)] = 14/25, or 56 percent.

• The 2-seed’s probability of winning was (3/5) · [(4/5)·(1/3) + (1/5)·(4/6)] = 6/25, or 24 percent.

• The 3-seed’s probability of winning was (2/5) · [(4/5)·(1/4) + (1/5)·(4/7)] = 22/175, or about 12.6 percent.

• The 4-seed’s probability of winning was (1/5) · [(3/5)·(2/6) + (2/5)·(3/7)] = 13/175, or about 7.4 percent.

It was kind of neat how each seed had approximately double the chances of winning as the next best seed.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Matt Farney 🎻 from Beavercreek, Ohio. I received 31 timely submissions, of which 22 were correct—good for a 71 percent solve rate.

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

The teams were seeded 1 through 2^k and played in a traditional seeded tournament format. That is, in the first round, the sum of opponents’ seeds was 2^k+1. If the stronger team always advanced, then the sum of opponents’ seeds in the second round was 2^k−1+1, and so on. Of course, stronger teams don’t always advance, but this convention told you which seeds could play which other seeds in each round.

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