Can You Hack Gymnastics?

Last week you looked at bizarre scoring rules for Olympic figure skating. This week, let's look at bizarre scoring rules for Olympic gymnastics!

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

In artistic gymnastics, competitors earn two subscores: one for difficulty and one for execution. Difficulty, as its name implies, is a measure of how challenging a routine is. While there’s technically no upper bound on the difficulty score, the toughest routines will earn around 6 or 7 points, depending on the apparatus. Meanwhile, execution scores have a maximum value of 10, essentially measuring how cleanly a competitor performed their routine.

What’s weird (to me at least) about gymnastics scoring is that the total score is difficulty plus execution, rather than difficulty times execution.

Why is that weird? Suppose two gymnasts have the same execution score, (say, 8.0 out of 10), but one’s difficulty is 20 percent greater (say, 6.0 to 5.0). You’d expect that gymnast’s overall score to be 20 percent greater, right? But instead, because we’re adding the two subscores, there’s only a 7.7 percent difference in overall score (14.0 to 13.0).

I guess no one is claiming that the total score should scale linearly with both difficulty and execution. But, you know, it really should.

This past week, Brazil’s Rebeca Andrade won the gold medal for her floor routine while American Simone Biles earned the silver. Andrade's total score was 5.9 + 8.266, or 14.166, and Biles’s was 6.9 + 7.233, or 14.133. (Don’t get me started on the rounding.) But if you had multiplied their subscores (like a rational human being), Biles would instead have come out on top, 49.9 to 48.8. 

So yeah, the decision to add versus multiply the subscores really matters.

Suppose gymnast A has a difficulty score of 6.0, while gymnast B has a difficulty score of 5.0. If both gymnasts receive independent, random execution scores between 0 and 10 (quite the range, I know), what is the probability that their relative ranking would be the same, regardless of whether the subscores were added or multiplied? (Here, you should assume the execution scores are real numbers that can go to any number of decimal places.)

This Week’s Extra Credit

In addition to their execution scores, now assume that both difficulty scores are now also independent, random values between 0 and 10 (again, quite the range).

What is the probability that their relative ranking would be the same, regardless of whether the subscores were added or multiplied? (Again, assume all subscores are real numbers that can go to any number of decimal places.)

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 fact I was recently (warning: link contains spoilers!) reminded of, or perhaps learned for the very first time—I honestly can’t recall.

Consider the volume of an n-dimensional ball with a radius r. In two dimensions, that’s a circle whose volume (i.e., area) is 𝜋·r². In three dimensions it’s a sphere with a volume of 4/3𝜋·r³. As it turns out, a four-dimensional hypersphere has a hypervolume of 𝜋²/2·r⁴.

Now, just consider the coefficients in these formulas (or, alternatively, the volumes when r = 1): 𝜋, 4/3𝜋, and 𝜋²/2. Numerically, that’s roughly 3.14, 4.19, and 4.93—an increasing sequence. Will these coefficients increase forever, or does the “unit ball” have a maximum volume in some finite number of dimensions?

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: 🎻 Paul Juster 🎻 from Walnut Creek, California. I received 19 timely submissions, of which 10 were correct—good for a 53 percent solve rate. This was a tough one!

In the 2002 Winter Olympics, figure skaters first completed a “short program,” for which they received a set of numerical scores (out of 6.0). These scores were averaged, and then skaters were ranked from 1 to N (the total number of skaters). The next day, skaters similarly completed a longer, “free skate” that received numerical and ordinal scores in the same way.

If a skater’s short program ranking was S and their free skate ranking was F, they received an overall score of S/2 + F, which was used for the final ranking of the competitors. Lower scores were better, and the lowest score earned a gold medal. In the event two or more skaters had the same overall score, whoever had a lower value of F (i.e., a better ranking in the free skate) received the better overall ranking.

In the ladies’ event, there were four medal contenders. Here were their rankings after the short program:

1. Michelle Kwan

2. Irina Slutskaya

3. Sasha Cohen

4. Sarah Hughes

The next day, Slutskaya was scheduled to perform her free skate last. Immediately prior to her taking the ice, the rankings of the others’ free skates were:

1. Sarah Hughes

2. Michelle Kwan

3. Sasha Cohen

That meant the overall standings at this point were:

1. Michelle Kwan (1/2 + 2 = 2.5 points)

2. Sarah Hughes (4/2 + 1 = 3 points)

3. Sasha Cohen (3/2 + 3 = 4.5 points)

To recap, Kwan was in the lead, and Slutskaya was the final competitor. Surely, one of these two would win the gold medal, right?

Wrong! Slutskaya finished between Hughes and Kwan in the free skate, so that the final standings were:

1. Sarah Hughes (4/2 + 1 = 3 points)

2. Irina Slutskaya (2/2 + 2 = 3 points)

3. Michelle Kwan (1/2 + 3 = 3.5 points)

4. Sasha Cohen (3/2 + 4 = 5.5 points)

While Hughes and Slutskaya both finished with 3 points, Hughes won the tiebreaker because she had the better-ranked free skate.

The result was a bizarre situation (which we’ll henceforth refer to as “madness”) in which the ultimate gold medalist was not in first place after completing her free skate, but rather after subsequent skaters had completed their free skates.

Last week, you explored whether this skating madness was truly an anomaly, or if it was something we might have expected.

To do this, you considered four figure skaters of equal merit. In other words, each had an equal chance of ranking anywhere from first to fourth for both the short program and the free skate. Each skater’s free skate result was independent of their short program result. To keep things (relatively) simple, the order in which the skaters performed was the same for both the short program and the free skate. (This turned out not to matter, as long as the free skate order was random and didn’t depend on short program performance in some way.)

What was the probability that the ultimate gold medalist was not in first place at some point after her free skate (among skaters who had completed their free skate)?

Oof. That was perhaps the longest setup I’ve ever given for a puzzle, but I think that context was helpful. And also, as hard as it is to believe, all this actually happened.

I remember watching the 2002 Olympics as a teenager, working out the numbers in advance of Slutskaya’s free skate. I realized that if she did well—but not too well—she would split Hughes’s and Kwan’s scores and allow Hughes to make the jump to gold. When Slutskaya landed too far forward on a triple flip late in her free skate, I knew if she executed cleanly thereafter that Hughes would win—and that’s exactly what happened. I grew up in the same town as Hughes, so we got to have a “Sarah Hughes Day,” a parade, and everything. It was pretty great!

Okay, back to the actual puzzle. A solid first step was to identify all the possible winning scores. There were a total of 15 points awarded, which meant skaters received an average of 3.75 points, and the gold medalist had to receive a total that was less than 3.75. With a little more work, you could show that the winning score had to be 1.5, 2, 2.5, or 3.

Now, the madness occurred when a skater did poorly in the short program and well in the free skate (i.e., Hughes), while a skater who didn’t do quite as well in the free skate (Kwan) was pushed down the rankings by a later skater (Slutskaya). Paul, this week’s winner, analyzed which gold medal scores could make this scenario possible:

• A score of 1.5 meant a skater ultimately ranked first in both the short program in free skate, and would always be on top.

• A score of 2 meant they ultimately ranked second in the short program and first in the free skate, and would again always be on top.

• There were two ways to score 2.5: first in the short program and second in the free skate (0.5+2), or third in the short program and first in the free skate (1.5+1). In either case, there was no way for someone who did worse in the free skate to ever be ahead of them.

• There were two ways to score 3: second in the short program and second in the free skate (1+2), or fourth in the short program and first in the free skate (2+1). In the first case (1+2), there was no way for someone who did worse in the free skate to ever be ahead of them.

So the only way for madness to occur was when the overall winner came in fourth during the short program and first in the free skate, just as Hughes did. And the only way for another skater to be ahead, at least temporarily, was for that skater to be first in the short program and (temporarily) second in the free skate.

Let’s call the four skaters A, B, C, and D, based on their ranking in the short program (i.e., A came in first, B came in second, etc.). There were 4! (i.e., 4·3·2·1, or 24 total) ways they could have been ranked in the free skate, and also 4! orders in which they could have skated. That meant there were a total of (4!)·(4!), or 576 rank/order combinations to consider.

We already said that madness occurred when D finished first in the free skate. Skater A was the only one who could temporarily rank ahead of D. But for A not to win gold, A had to ultimately finish third or fourth in the free skate. That left four possible free skate rankings to consider, each of which came with several orderings that resulted in madness. These scenarios are listed in the diagram below:

In total, there were 24 rank/order combinations that resulted in madness, out of the total 576. That meant the probability of madness was 24/576, which simplified to 1/24, or about 4.2 percent.

So the scoring anomaly that occurred in the 2002 games wasn’t particularly likely, but it was certainly within the realm of possibility. Whatever commission came up with this scoring method should have said, “You know, look how crazy this can get. Maybe we should avoid adding ordinal scores.” Then again, this was smack in the middle of college football doing the exact same thing. Live and learn, I guess.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Peter Ji 🎻 from Madison, Wisconsin. I only received 5 timely submissions, of which 2 were correct. This tied the record for the fewest number of solvers for a puzzle, and served as a testament to this one’s difficulty.

Instead of four skaters, now there were N skaters, all of equal merit. Again, each had an equal chance of ranking anywhere from first to N^th for both the short program and the free skate, and these rankings were independent. Skate order remained consistent between the short program and the free skate. (Again, this didn’t matter so long as the order in the free skate was random.)

Let p_N denote the probability that the ultimate gold medalist was not in first place at some point after her free skate (among skaters who completed their free skate).

As N grew very, very large, what value did p_N approach?