Can You Catch the Breakaway?

This year’s Tour de France may already be decided. But the Tour de Fiddler remains in need of some serious solving!

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

Every year or so I try to run a cycling-related puzzle to coincide with the Tour de France. Previous puzzles have involved the team time trial, “King of the Mountain” points, the shape of the peloton, and being paced up a climb.

This time around, a lone rider in the Tour de Fiddler is being pursued by a group of four riders. The four riders have an advantage—they take equal turns being in the lead position, while the other three riders draft behind. At any given speed, being in the lead position (as well as riding solo) requires twice as much power as drafting.

Assume that every rider must maintain the exact same average power over time, whether they are the lone rider or in the pack of four. To be clear, their power can change over time, but the time-averaged value must be the same for every rider. Also, when leading a pursuing group or riding solo, one’s speed is directly proportional to one’s power. When drafting, one's speed matches that of the leader (again, at half the power output).

The pursuers just passed under a banner indicating there are 10 kilometers left in the stage. How far back of the lone rider can they afford to be, such that they still catch them at the finish line?

This Week’s Extra Credit

In today’s stage of the Tour de Fiddler, there are 176 total riders. Some riders are grouped together in a single breakaway, while the remainder are grouped together in the peloton.

The breakaway group is 10 kilometers from the finish, while the peloton is one kilometer behind (i.e., 11 kilometers from the finish).

As before, assume that every rider must maintain the exact same time-averaged power when leading a group, and that drafters can match the leader’s speed but with half the power.

What is the smallest number of riders that the breakaway needs to reach the finish line before the pursuing peloton?

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 The Big Internet Math-Off 2024, “a just-for-fun competition to find ‘The World’s Most Interesting Mathematician,’” organized by Christian Lawson-Perfect. Sixteen mathematicians were invited to compete according to the bracket below. Since I wasn’t invited, I think it’s fair to say that I must be the 17th most interesting mathematician in the world.

Anyway, in each round of this single-elimination tournament, mathematicians share write-ups of interesting mathematics. I highly recommend perusing the various rounds and write-ups!

One particular factoid was so interesting that it was actually shared twice (by Dave Richeson and by Matt Enlow). Consider the following progression: sin(1/5°), sin(1/55°), sin(1/555°), and so on. Without reading either write-up, can you figure out why this sequence is so interesting? And can you prove your result?

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.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Sam Miner 🎻 from Seattle, Washington. I received 52 timely submissions, of which 37 were correct—good for a 71 percent solve rate.

Last week, you were asked to consider a three-set tennis match. To win the match, you had to win at least two of the three sets. To win a set, you had to win a certain number of games. And to win a game, you had to win a certain number of points.

A game was won by the first player who scored four points. However, the winner had to “win by two.” So if both players had three points, they kept going until one player had two more points than their opponent.

A set was won by the first player to win six games within the set. However, if both players had exactly five games each, then they played exactly two more games. If one player won both of them, that player won the set. If each player won one of these games, so that they now had six games apiece, they proceeded to a tiebreak.

A tiebreak was essentially a game that went to seven points rather than four. The first to seven points won, and players again had to win by two.

Keeping all this in mind, what was the greatest percentage of points (including tiebreak points) you could lose in a three-set match, while still winning the match itself?

To win the match, you had to win two sets—at least that much was clear. Now, to tank your percentage of points won, it made sense to also play a third set and lose every single point of that set. With six games and four points in each game, that meant you won zero points and lost 24 in that losing set.

But what about the other two sets? To lose as many points as possible, it made sense that they both went to tiebreaks. But before reaching those tiebreaks, you had to win six games and lose six games. Among the six games you lost, you again won a total of zero points and lost a total of 24.

Among the six games you won, thanks to “win by two,” you could have won them four points to two, five points to three, six points to four, and so on. For now, let’s assume you won them four points to two—but we’ll circle back and see why that resulted in the lowest possible percentage. Among these six games, that meant you won a total of 24 points and lost a total of 12.

As for the tiebreaks, you could have won each of them seven points to five, eight points to six, nine points to seven, and so on. Again, let’s assume you won them seven points to five, and we’ll circle back.

Putting this all together, you lost all 24 points from one set. The other two sets each had 24 + 36 + 12, or 72 total points, of which you won 24 + 7, or 31. In total, you won 31 + 31, or 62 points, out of a total of 24 + 72 + 72, or 168 points. You lost the remaining 106 points, representing a whopping 63.1 percent of the total. Talk about not being deserving of the victory!

By the way, if you instead submitted an answer of 36.9 percent—the lowest percentage of points you could win while still winning the match—I still gave you full credit.

Okay, let’s return to the “win by two” situation I mentioned earlier. If you won a game four points to two, you lost about 33 percent of the points. But if you won the same game five points to three, you lost 37.5 percent. If you went crazy and won the game 11 points to nine, you lost 45 percent. The longer the game went on, and the higher the total score, the greater the percentage of points you lost. So then why, paradoxically, did we have games (and tiebreaks) end much earlier in the preceding computation?

Because these additional points always came in pairs—one point for you, one for your opponent—which mean they pulled the percentage of points lost toward 50 percent. In the case of a single game or tiebreak that you won, additional points pulled the percentage of points lost upward to 50 percent. However, since we already calculated that you could lose up to 63.1 percent of the points and still win the match, tacking on additional pairs of points would have pulled the percentage downward to 50 percent.

Finally, a neat extension for this problem was to look at five-set matches, or rather, matches with any odd number of sets. More specifically, suppose the number of sets in a match was 2N+1. Then you could afford to lose N of the sets, for a total of 24N points. The remaining N+1 sets you won, in which you won a total of 31·(N+1) points and lost a total of 41·(N+1) points. Overall, the greatest fraction of points you could lose was (65N+41)/(96N+72).

When N was 1 (meaning there were three sets), the fraction came to 106/168, just as we already calculated. When N was 2 (meaning there were five sets), the fraction came to 171/264, or about 64.8 percent. In the limit as the number of sets got very, very large, the fraction of points lost approached 65/96, or about 67.7 percent. So even if you played a match with a billion sets, there was no way for you to win if you lost 68 percent of the points.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Eric Widdison 🎻 from Kaysville, Utah. I received 15 timely submissions, of which 11 were correct—good for a 73 percent solve rate.

Last week, you were asked to analyze a three-set tennis match that featured two players who were evenly matched, so that each player had a 50 percent chance of winning any given point. Also, points were independent, so the outcome of one point didn’t affect the probability of who won subsequent points.

How likely was it that one of the players lost a majority of the points in a match while winning the match itself? (Here, I meant a strict majority, i.e., more than 50 percent of the points.)