Can You Shape the Peloton?

It’s a triangle! It’s a rhombus! It’s—something in between!

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. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

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

Earlier this month, I watched La Vuelta, one of the three grand tours of cycling (a trio that includes the Tour de France). I noticed the peloton—the main group of riders—took on an aerodynamic profile that sometimes looked like a triangle, sometimes looked like a rhombus, and sometimes looked somewhere in between.

For example, the figure below shows the four possible formations between a triangle and a rhombus when the peloton’s maximum width is four riders:

For certain numbers of riders, multiple formations like these are possible. In particular, there are two formations with 15 riders: a triangle that’s five riders wide at the base and an almost-rhombus that’s four riders wide in the middle, but missing the bottommost rider, as shown below.

After 15, what is the next smallest number of riders that similarly has two distinct formations between a triangle and a rhombus?

(Note: As I’m writing this, the sequence of numbers for which there are at least two arrangements does not appear to be on OEIS. I smell an opportunity here…)

Extra Credit

In this week’s Fiddler, you’re looking for numbers of riders where two distinct formations between a triangle and a rhombus are possible.

For Extra Credit, find the fewest number of riders needed so that there are three distinct formations between triangle and a rhombus.

(Note: The sequence of numbers for which there are at least three arrangements also appears to be missing from OEIS. The same goes for the sequence that has the smallest number needed for two formations, followed by the smallest number needed for three formations, followed by the smallest number needed for four formations, and so on. Who will be first to submit all these sequences?)

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 a (very) oldie but a goodie, and one that’s closely related to a discovery made earlier this year.

Suppose you have three elements that we’ll call x, y, and z. There are quite a few ways to form sets of these elements (without necessarily including all of them) such that no set is a subset of another.

For example, one way to do this is {x, y}, {y, z} and {x, z}. Note that none of these three sets is a subset of the other two! Another way to do this is just via an empty set: { }. And yet another way is {x, y} and {z}.

So—how many ways are there to make sets of these three elements such that no set is a subset of another?

This problem was recently solved in the case of nine elements (as opposed to the paltry three elements mentioned above). To learn more about this puzzle and the recent discovery, check out this article over at Quanta Magazine.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Mikolaj Franaszczuk 🎻, from Philadelphia, Pennsylvania. I received 91 timely submissions, of which 69 were correct—good for a 76 percent solve rate.

About halfway through the current Major League Baseball (MLB) season, all five teams in the American League East division had better records (i.e., winning percentages, or percent of games won) than all five teams in the American League Central region.

Inspired by this surprising fact, you were asked about Fiddler League Baseball, which, like the MLB, had six divisions, with five teams in each division. For simplicity, you could suppose each team had a winning percentage chosen randomly, uniformly, and independently between zero percent and 100 percent.

You specifically looked at two divisions: The Enigma League East division and the Enigma League Central division. What was the probability that every team in the Enigma League East division had a higher winning percentage than every team in the Enigma League Central division?

Now wait just a minute! There was no way that the winning percentages of teams could have been independent. Overall, the total number of wins among all 30 teams had to equal the number of losses, since every game had a winner and a loser. That meant if you knew the winning percentages of 29 teams, you could deduce the record of the final team, which meant it absolutely wasn’t independent.

So why did I word the puzzle this way? To keep things simple. I could have (and probably should have) said something like this:

Each team had a 50 percent chance of winning each game. Each team played an equal number of games against all the other teams. No two teams finished with exactly the same winning percentage.

This would have made the puzzle realistic. But it also would have resulted in the same answer, thanks to symmetry. In any case, let’s see what that answer was.

Solver Emily Kelly recognized that this wasn’t a matter of probability so much as it was a matter of combinatorics—the branch of mathematics concerned with counting and ordering. Suppose we label the five East division teams with an “E” and the five Central division teams with a “C.” Each possible collective ranking of the 10 teams from best to worst can be represented with a string of 10 letters (five E’s and five C’s), such as EECCECCEEEC.

Every arrangement of these 10 letters was equally likely, and there were 10 choose 5, or 252, of them. But only one of these ways resulted in an East division that was entirely superior to the Central division: EEEEECCCCC. The probability of this occurring was therefore 1/252, or about 0.4 percent. In other words, it wasn’t very likely.

A few readers gave an answer that was twice this value. However, this corresponded to the probability that either division was superior to the other, whereas the puzzle asked for the probability that the East division was superior to the Central division.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Sanandan Swaminathan 🎻, from San Jose, California. For the Extra Credit, I received 33 timely submissions, of which 20 were correct—good for a 61 percent solve rate. (That’s okay, this was a tough one!)

For last week’s Extra Credit, you had to consider all six divisions in Fiddler League Baseball. What was the probability that there existed two divisions such that every team in one division had a higher winning percentage than every team in another division? Note that this included cases where multiple divisions were better or worse than others, such as having two divisions that both had higher winning percentages than some third division.

Quite a few readers were tempted to multiply the result from last week’s Fiddler by the number of ways you could choose pairs of divisions, with one being superior and another being inferior. There were six divisions to choose from as superior, and then five remaining divisions as inferior, for a total of 30 ways to choose. According to this line of reasoning, the probability of having a pair of such divisions was therefore 30/252, or about 11.9 percent.

However, this was not the answer. That’s because when one division was superior to another, it was more likely to be superior to other divisions as well. In other words, occurrences of superior and inferior divisions were clumpy—they occurred less than 11.9 percent of the time, but when they did occur, you often had multiple pairs of superior/inferior divisions at the same time.

Finding the exact probability proved quite the challenge. This time around, there were far more than 252 cases to consider. If you assigned the same letter to the five teams in each of the six divisions (with each division getting a different letter), there were 30!/(5!)⁶ different 30-letter strings you could generate. Those 88 quintillion strings were nothing to sneeze at. For each string, you had to determine whether the fifth occurrence of any of the six letters preceded the first occurrence of any of the other five letters.

Solver Bradon Zhang found a decent approximation by running 10 million simulations of Fiddler League Baseball’s 30 teams. About 8.6 percent of the time, there was at least one pair of superior/inferior divisions. Some solvers, like David Kravitz, sought exact solutions. With dynamic programming, David was able to determine the probability was 2,565,091/29,745,716, or about 8.6234 percent. Colman Humphrey somehow arrived at the same solution via calculus. Wowzers!

So in this fantastical world where the records of all 30 teams were magically independent, you’d have one division superior to another every decade or so. In reality, where teams play within their own division with greater frequency, this hasn’t happened since the MLB expanded to 30 teams in 1993. In fact, no division over the last three decades has been superior to any other division within the last three decades. The “best worst” team was the 2005 Washington Nationals, who finished last in the National League East with an 81-81 record, while the “worst best” team was the 2005 San Diego Padres, who finished atop the National League West with an 82-80 record.

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.