NYT wins the gold medal for medal visualization.

A few days into these Olympics, my friend Ryan lobbed me an alley-oop question via email:

Which brings to the next point, what is the ideal medal count ranking to your estimation?  I figure this is something you probably have the correct answer to.

Alas, I told him, I don’t. There are three basic options, all of them bad.

First, the standard solution is to rank by gold medals. But like many standard things, this is deeply problematic. Did Ireland really outperform Brazil, even though the latter won 7 more silvers and 7 more bronzes?

Second is an alternative practiced sometimes in the U.S. and never anywhere else: to rank by total number of medals, treating gold, silver, and bronze as equals. But this is no better. Which would you prefer: Great Britain’s 7 extra bronzes, or France’s 2 extra golds + 4 extra silvers?

The third solution is to strike a balance between these deficient extremes; that is, to weight the medals. A gold is worth X silvers, and a silver is worth Y bronzes. But this has its own problem: what weights do you use? Is a gold worth 2 silvers, or 10? Is a silver worth 1.5 bronzes, or 15? Who knows! It’s inescapably arbitrary.

So, in replying to Ryan, I just shrugged my shoulders:

No good solution on medal counts. Maybe each country should submit their proposal for how to weight medals (anything within reason, from Gold = Silver = Bronze to Gold = 1, Silver = Bronze = 0). Then you do the tables based on a simple average of those.

No idea what weights that would give, but at least they’d have the veneer of consensus.

Then, not an hour later, I came across the gorgeous and insane solution that the data visualization dreamers at the New York Times had concocted.

Rather than choosing a weight, they decided to show all possible weights.

How does this work? I suspect that only an admirably geeky fraction of NYT readers know, so let’s do a worked example. We need a specific country; Brazil will suit us nicely.

Now, each color-coded point in the country’s graph represents a different way of weighting medals. Thus, as you move around the graph, Brazil’s ranking will change, from as high as 12th to as low as 20th, depending on the weighting described at that point.

For example, in the bottom left corner, all medals count equally. A gold is worth 1 silver, and a silver is worth 1 bronze.

Brazil, with its bronze-heavy haul, benefits from this system, and winds up ranking 12th.

(In the NYT’s language, a bronze is always worth “1 point.”)

Now, as we climb upwards, the gold-to-silver ratio increases. At the top left corner of the graph, a single gold is worth 150 silvers, effectively meaning that only gold counts, and the sum of other medals (silver + bronze) is used merely as a tiebreaker.

Here, about 3/4 of the way up the graph, a gold is worth 4 silvers, and Brazil (with relatively few golds among its medals) drops to 16th. But it still benefits from the fact that bronzes and silvers are counted equally.

Meanwhile, by moving to the right, we increase the silver-to-bronze ratio.

For example, the bottom-right corner gives a peculiar system: a gold is still worth the same as a silver, but a silver is now worth (in effect) infinitely more than a bronze.

(Note that the axes are nonlinear. We’ll come back to this!)

The top-right corner sounds extravagant when you look at the numbers (23,000 points?!), but the approach is perfectly sensible. It simply means that you rank first by golds (which are vastly more valuable then silvers), then break ties based on silvers (which are vastly more valuable than bronzes), and finally break any remaining ties based on bronzes.

All this is a batty and brilliant solution to the question of medal weightings, and all Olympics long I’ve been delighting in checking these visualizations. I especially enjoyed the mid-Olympics moment when Canada sat in either 9th or 11th place, but definitely not in 10th place.

It reminds me of high school chemistry phase diagrams. Under those peculiar and brief conditions, Canada sublimated directly from a 9th-place solid to an 11th-place gas, never passing through 10th-place liquid form.

Alas, when it comes to the top of the rankings, it’s all moot. The U.S. tied China for golds, and handily won the most silvers and bronzes, so its triumph is boringly complete. Never have I chanted “USA! USA!” with a greater sense of letdown.

In all this, the NYT evidently punted on the value judgment. Instead of one canonical weighting, they gave us the two-dimensional sprawl of all possible weightings. Good on ’em: truly living up to their famous slogan of “all the news that’s fit to parametrize.”

But I can’t help asking. Can we somehow deduce a best weighting from all this?

One natural solution is to invoke an integral. That is, take a weighted average of weighted averages. For example, if half of your graph is 4th, a third is 3rd, and a sixth is 12th, then your weighted average ranking is 4/2 + 3/3 + 12/6 = 5th.

But here’s the problem: remember that weird scale on the axes?

It’s definitely not linear. But it’s not logarithmic, either.

Walking along the axis, the weighting grows by factors of 1.5 (long step), then 1.333 (short step), then 2.5 (long-ish step), then 4 (short step). Suffice to to say that the growth factors do not correspond in any obvious way to the length of the step. I can only presume that the axis was tuned by the NYT team to be visually pleasing; in particular, they must have given more area to the border regions where ranking is highly sensitive to small changes in weights.

This means that the NYT has not punted on all value judgment, and so just taking an integral of the graph is not an impartial calculation. Still, no matter how you weight the various virtues of data visualization, this one is worthy of gold.

UPDATE, 8/13/2024, 10:10am: My father, devoted blog reader and Operations Researcher extraordinaire, emailed me with a great suggestion for next Olympics.

Instead of a square, a triangle.

In the lower-left corner, we count only gold.

In the lower-right corner: gold + silver.

And at the top corner, all three medals: gold + silver + bronze.

“Any point in the triangle,” he explains, “would be a mixture of the three.” (Indeed, any point in a triangle can be seen as a convex combination of the three corners.)

For example, the center of the triangle weights all three corners equally. This is equivalent to 3 points per gold, 2 points per silver, and 1 point per bronze.

It is informationally equivalent to the NYT’s version, but with a few potential advantages. First, the corners are a bit easier to label and interpret. Second, “movement” between these corners is a bit easier to understand than the equivalent movements around the NYT’s square. And third, there is a natural way to “parametrize” the space, thereby avoiding (I believe) the fine-tuning that the NYT needed to do with their axes.

NYT, if you’re listening, I hope you give the triangle a shot for 2026!