Can You Hack the Olympics?

In 2002, a figure skater jumped from silver medal position to gold, long after she had skated. How likely was this sort of quirk of the ranking system?

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 the 2002 Winter Olympics, figure skaters first completed a “short program,” for which they received a set of numerical scores (out of 6.0). Based on these numerical scores, skaters were ranked from 1 (first) to N (last), where N was 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, here were the rankings of the others’ free skates:

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 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. (To no one’s surprise, the figure skating governing body overhauled its scoring shortly thereafter.)

In retrospect, was this phenomenon truly an anomaly, or was it something we might have expected to occur with some frequency? Let’s find out!

Suppose there are four figure skaters, all of equal merit. In other words, each has 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 is independent of their short program result. To keep things (relatively) simple, assume the order in which the skaters perform is the same for both the short program and the free skate.

What is 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)?

This Week’s Extra Credit

Instead of four skaters, now suppose there are N skaters, all of equal merit. Again, each has an equal chance of ranking anywhere from first to N^th for both the short program and the free skate, and these rankings are independent. Skate order is once again consistent between the short program and the free skate.

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 grows very, very large, what value does p_N approach?

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, what I’m sharing isn’t exactly a puzzle, but definitely some mathematical food for thought.

In keeping with this week’s Olympic theme, I’m sharing some interesting approaches to tracking countries’ total medal count.

First off, I’ve seen plenty of criticism of The New York Times’ medal tracker, which defaults to prioritizing the total number of medals (rather than the number of gold medals), conveniently placing the USA on top:

Perhaps in response to the criticism, the Times posted an interactive tracker showing phase diagrams like the one below. One axis shows the exchange rate between gold and silver (i.e., how many silver medals is one gold medal worth?), while the other shows the corresponding rate between silver and bronze.

Cool stuff. But even cooler is the article, “Population-adjusted national rankings in the Olympics” (also featured in the Times), which essentially asks: How unlikely is it for a country with a given population to have won as many medals as it did? These rankings are being tracked for the ongoing 2024 games here.

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: 🎻 Ivor Traber 🎻 from Waterloo, Ontario, Canada. I received 61 timely submissions, of which 56 were correct—good for a 92 percent solve rate.

Last week, you (player A) and a friend (player B) were playing a game in which you alternated rolling a die. The order of play was AB|AB|AB, and so on. (The vertical bars here were just for organizational purposes, and did not signify anything special that happened.) The first player to roll a five won the game. As it turned out, whoever went first had a distinct advantage!

That said, there were other ways you and your friend could take turns, ways that resulted in a fairer game. You were specifically asked consider the “snake” method, in which the order was reversed after each time you both rolled: AB|BA|AB|BA, and so on.

Assuming you were the first to roll, what was the probability you would win the game?

Before we get to the snake method, let’s confirm that A had an advantage when the two players alternated rolls. Player A won on the first roll with probability 1/6. A’s next opportunity to win was on the third roll, which required A to not win on the first roll (probability 5/6), B to not win on the second roll (probability 5/6), and then A to win on the third roll (probability 1/6). Since these were independent events, the probability A won on the third roll was the product of the three probabilities: 1/6·(5/6)². After the third roll, A’s next opportunity to win was on the fifth roll, which A did with probability 1/6·(5/6)⁴. After that, A could win on the seventh roll, the ninth roll, and so on.

All together, A’s probability of winning was 1/6 · [1 + (5/6)² + (5/6)⁴ + (5/6)⁶ + …], a geometric series. For any (convergent) geometric series, the sum is equal to the first term dividing by one minus the common ratio. For this particular series, the sum was 1/6/[1−(5/6)²], or 6/11—about 54.5 percent. Sure enough, A had an edge when players alternated rolls.

Now, let’s return to the snake method: AB|BA|AB|BA, and so on. It was notable that this sequence repeated every four rolls, meaning it could be reorganized as ABBA|ABBA, and so on.

This time, player A had opportunities to win on the first roll, the fourth roll, the fifth roll, the eighth roll, and so on. Let’s start by looking at the first and fourth rolls.

As before, the probability A won on the first roll was 1/6. Meanwhile, winning on the fourth roll required A to not win on the first roll (probability 5/6), B to not win on the second roll (probability 5/6), B to not win on the third roll (probability 5/6), and A to win on the fourth roll (probability 1/6). The probability of all this occurring was 1/6·(5/6)³.

Since the sequence repeated every four rolls, A’s next opportunities to win (the fifth and eighth rolls) came four rolls after the first and fourth rolls, respectively. And then A’s next opportunities to win—the ninth and 12th rolls—came another four rolls after those. Overall, A’s probability of winning was 1/6·[1 + (5/6)³ + (5/6)⁴ + (5/6)⁷ + …]. Pairing up consecutive terms inside the brackets and factoring out [1 + (5/6)³] gave you another geometric series: 1/6·[1 + (5/6)³]·[1 + (5/6)⁴ + (5/6)⁸ + (5/6)¹² + …].

This time, the sum turned out to be 1/6·[1 + (5/6)³]/[1−(5/6)⁴], which could be rewritten rather elegantly as (6³ + 5³)/(6⁴ − 5⁴). This simplified to 31/61, or about 50.8 percent. This was a lot closer to 50 percent than A’s chances of winning were when the players alternated, representing a more than fivefold improvement in approaching even odds.

But, as we’ll see in last week’s Extra Credit, it was possible to get even closer to 50-50.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Adam DeBruler 🎻 from Atlanta, Georgia. I received 36 timely submissions, of which 31 were correct—good for an 86 percent solve rate.

Another way to take turns in the game from last week’s Fiddler was to use the Thue-Morse sequence, where the entire history of the order was flipped to its complement (from A to B and from B to A) after each round. As an illustration, consider the first few rounds:

• Round 1: Player A went first.

• Round 2: Only A went in the first round. So now player B went.

• Round 3: Up to this point, the order was AB. Flipping this to its complement, round 3’s order was BA.

• Round 4: Up to this point, the order was ABBA. Flipping this to its complement, round 4’s order was BAAB.

• Round 5: Up to this point, the order was ABBABAAB. Flipping this to its complement, round 5’s order was BAABABBA.

Writing this out as a single sequence of turns, the order was A|B|BA|BAAB|BAABABBA, and so on.

Assuming you were the first to roll, what was the probability you would win the game?