Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

The astronomers on planet Xiddler have made several remarkable discoveries. After inventing the telescope, they quickly discovered a new planet in their solar system!

Xiddler is very much like Earth. The planet orbits its star in a nearly circular path, with an average distance of 150 million kilometers, a period of one Earth year and a day that lasts 24 hours. But *unlike* Earth, there weren’t any other known planets in the solar system…until now.

Moments after the Xiddlerian sun set below the horizon, three astronomers happened to focus their telescopes at the zenith of the evening sky, all seeing the same new planet. In their excitement, the astronomers race to Xiddler’s Grand Minister to deliver the momentous news.

The first astronomer says that, by her calculations, the newly discovered planet orbits their sun with a radius of 50 million kilometers. The second astronomer says that, by *her* calculations, the planet in fact orbits their sun with a radius of 300 million kilometers. The third astronomer disagrees with the other two — by *her* calculations, the planet has a very similar orbit to Xiddler, with a radius of 150 million kilometers.

Which astronomer should the Grand Minister believe?

The solution to this Riddler Express can be found in the following week’s column.

## Riddler Classic

Some friends have invited you to a protest, and you’ll be making a sign with large lettering. You’re filling in the sign’s letters by drawing horizontal lines with a marker. The marker has a flat circular tip with a radius of 1 centimeter, and you’re holding the marker so that it’s upright, perpendicular to the sign.

Since the diameter of the marker’s tip is 2 centimeters, you decide to fill in the letters by drawing lines every 2 centimeters. However, this is the pattern you get:

The shading doesn’t look very uniform — each stroke is indeed 2 centimeters wide, but there appear to be gaps between the strokes. Of course, if you drew many, many lines all bunched together, you’d have a rather uniform shading.

But you don’t have all day to make this sign. If the lines can’t overlap by more than 1 centimeter — half the diameter of the marker tip — what should this overlap be, in order to achieve a shading that’s as uniform as possible? And how uniform will this shading be, say, as measured by the standard deviation in relative ink on the sign?

The solution to this Riddler Classic can be found in the following week’s column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Brian Leet 👏 of Burlington, Vermont, winner of last week’s Riddler Express.

Last week, I was playing one of my favorite video games, The Legend of Zelda: Breath of the Wild. Within the game, there were hundreds of hidden “Korok Seeds,” which I was having an increasingly difficult time finding.

Fortunately, I obtained a special mask that made a sound any time I was within a certain distance of a Korok Seed. While playing, I marked nine distinct locations on the game map, forming the 3-by-3 grid shown below:

Each leaf symbol was within range of a Korok Seed, while the point in the middle was *not* within range of a Korok Seed. Given this arrangement, what was the minimum possible number of Korok Seeds I could have detected?

There certainly could have been eight Korok Seeds, each near one of the eight leaves. But rather than start with larger numbers, many solvers jumped straight to the smaller ones. Was it possible to generate this pattern if there had been just one Korok Seed?

Solver Kiera Jones supposed for a moment that there *was* just one Korok Seed. It couldn’t have been at the center of the grid, since that would have resulted in a leaf icon in the middle. In fact, this one Korok Seed had to have been closer to all eight leaves than it was to the center. But the moment you were closer to one of the leaves, you’re farther away from its diametric opposite. For example, if you were closer to the leaf on the right than the middle, you *must* have been closer to the middle than the leaf on the left.

In short, Kiera proved there had to be at least two seeds. And as it turned out, the answer was indeed *exactly* **two**.

No information was given in the problem about what that “certain distance” was, within which the mask was able to detect a Korok Seed. You were left to assume it could be anything, and that meant the seeds could be arbitrarily far away from the nine points in the grid.

By placing the two seeds in opposite directions *outside* of the 3-by-3 grid, it was possible to have the eight points marked by leaves all be within some distance of a seed, while the middle point was outside that range. Here was one such placement of the two seeds, courtesy of Matt Jenny:

The large black circles show the regions where you would detect the Korok Seed at the center of the circle. Sure enough, the middle green point didn’t fall within one of the large black circles, while all eight surrounding green points fell within one of the two large circles. Countless other arrangements were also possible, but the key idea was to split the eight surrounding locations into two groups of four, each of which was in range of a single Korok Seed.

Raul Perera arrived at the same answer, but without the aid of any graphing technology. Instead, Raul dropped a disk onto some graph paper!

And even if you were to split the locations as Matt and Raul did, there were still many possible locations for the seeds. In the diagram below, the red and blue regions represent possible locations for the two seeds:

Those Koroks certainly are pesky. It may take me a few more years, but someday I will have found them all.

## Solution to last week’s Riddler Classic

Congratulations to 👏 John Bullock 👏 of Lafayette, Indiana, winner of last week’s Riddler Classic.

Last week, everyone in the U.S. (about 330 million people) joined a single Zoom meeting between 8 a.m. and 9 a.m. — to discuss the latest Riddler column, of course.

The attendees all followed the same steps in determining when to join and leave the meeting. Each person independently picked two random times between 8 a.m. and 9 a.m. — not rounded to the nearest minute, mind you, but *any* time within that range. They then joined the meeting at the earlier time and left the meeting at the later time.

What was the probability that at least one attendee was on the call with everyone else (i.e., the attendee’s time on the call overlapped with everyone else’s time on the call)?

You don’t typically see a problem jump straight to the case of 330 million people unless something funny is going on. To understand what was happening here, many solvers first tried the case in which only two people were joining the meeting. What were the chances they’d be on at the same time?

You could have solved this with calculus, but you also could have skipped all the integrals with a combinatorial approach. With two people, there were four total times being randomly picked: two starting times and two ending times. Suppose these four times, in order from earliest to latest, were A, B, C and D. Then, there were six ways to split these times among the two attendees — the first person could have picked times A and B, A and C, A and D, B and C, B and D or C and D, while the other person would have been left with the remaining two times. The only time the two people didn’t meet up was when the first person picked A and B (leaving the other person with C and D) or C and D (leaving the other person with A and B) — two out of the six cases. That meant the chances they *did* meet up stood at four out of six.

So when there were two people in the meeting, the probability at least one attendee saw all the others was 2/3. But what if there were three people, or four — or 330 million?

This general problem is more involved, as there were way more than six cases to consider. Jim Crimmins, the puzzle’s submitter, along with solver Allen Gu, both referenced a 1990 paper that addressed this very problem. But before revealing the answer, let’s check in with those who took a computational approach.

John Bullock simulated 330 million attendees in Python and ran a whopping 364,000 total simulations. Among these, 242,919 — very nearly two-thirds of the simulations — had at least one attendee who was on the call with everyone else. Peter Ji, Matt Lee and Benjamin Phillabaum all kindly shared their code as well, and all three generated results that were remarkably close to 2/3 as well. Could it be that the answer was always 2/3, whether there were two attendees, 330 million or anything in between?

Indeed, the answer was **2/3**. Solver Josh Silverman proved this result with some powerful combinatorial reasoning and by swapping meeting times among certain attendees so that one could be on the call with everyone else.

For extra credit, you were asked to find the probability that at least *two* attendees were on the call with everyone else. Once again, a beautiful result was found by both the analysts and those using a computational approach: **2/5**.

As Josh explains, the probability of having exactly *k* attendees on the call with everyone else turned out to be (*k*+1)!/(2*k*+1)!! (The two exclamation points denote a double factorial — multiplying up all the evens or odds up to that number. In this case, since 2*k*+1 is always odd, it means multiplying all the odds from 1 to 2*k*+1.) Here’s what that probability distribution looks like, courtesy of Allen Gu:

One very cool consequence of this result is that, in the limit when there were many, many attendees (e.g., when there are 330 million of them), the *average* number of attendees who were on the call with everyone else approached 𝜋/2, or about 1.57.

Sometimes riddles just can’t help but circle back to 𝜋.

## Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.