*Oliver Roeder is a journalist, author and games player. He is a former senior writer for FiveThirtyEight, where he covered the World Chess Championship and other gaming pursuits. The following is adapted from his new book, “**Seven Games: A Human History**,” published by W.W. Norton. It’s in stores today.*

I still remember the first board on which I ever played chess. It was an irregular and heavy slab of walnut, maybe 14 inches on a side, onto which green squares of felt had been carefully glued by hand. It was a homemade Christmas gift from my mother’s siblings to their father, my grandfather Jack. The felt had curled up at its corners after hosting decades of battle in a small farmhouse in eastern Iowa. The pieces, purchased by my grandparents on their honeymoon to Mexico in 1949, were slender, made of ebony and ivory. By the time I arrived, one of the knights had lost its head. The other knights, I thought, had a look of terror in their eyes. I remember the sharp spikes on the rims of the queens’ crowns and the neat, crenelated battlements of the rooks. I remember the clunk of the dense pieces on wood.

As a child, I spent every summer on that farm, jumping on hay bales and swimming and playing game after game of chess in the late-afternoon light. Grandpa Jack was a strong player, the wielder of a conservative and positional style. And as a matter of strict principle, he never let kids win. Therefore, every chess player in my family has a cherished memory of their first victory. When I finally beat Grandpa Jack, around age nine, I remember running into the kitchen to tell Grandma Shirley, who hugged me and who was, to borrow her word, tickled.

At first, I saw the game like a machine, not unlike the Mechanical Turk, that impressive 18th-century hoax of a chess-playing automaton, its parts moving and interacting according to complex yet intuitive principles. I took chess apart like a child disassembling a motor — curious and eager but mostly hapless and messy. Occasionally, the reassembled machine ran smoothly, and I won. More often, it sputtered out, and I lost. I was, to use my favorite chess insult, a woodpusher, curious about how the pieces worked. Experimentation was enough for me.

As I grew older, I grew fascinated with chess theory and studied diagrams of the intricate machine picked apart by countless tinkerers over hundreds of years. I would fall asleep reading the heavy reference book “Modern Chess Openings,” comforted and delighted by the fine- grained taxonomy and analysis of just the first few possible moves in a game, and the names they’d acquired — the Halloween Gambit, the Maróczy Bind, the Accelerated Dragon, the Hedgehog Defense.

Today I enjoy chess on an aesthetic level. My competitive career never amounted to much; my fourth-grade Greenwood Elementary championship remains its only highlight. I lacked skill, and, just as important, I lacked the obsession to internalize — to actualize — all those volumes of theory. But even if I’ve never stretched a canvas, I can still appreciate Rothko and de Kooning — appreciate the beauty in the picture. And one can appreciate beauty in a game of chess; it is art. Lengthy tactical combinations, complex and previously unseen, can, like music, unfold as if they were ordained. Essences of gnarly, complex positions can, like painting, be distilled and altered and presented in pure form.

My progression mirrors how we taught our computers to play chess. The earliest programs, gawkish code running on ungainly mainframes, were woodpushers, capable of playing chess technically but not well. Their successors, running on sleeker supercomputers or speedier modern desktops, had mastered theory — openings and endgames, as well as the sophisticated tactics of the middle game — and now played better than any human. And their successors, the latest evolution, ungodly chess beings sprung from the secretive labs of trillion-dollar companies, play a hyper-advanced alien chess, exotic and beautiful, something no human is capable of fully understanding, let alone replicating, but so full of awesome style.

I invite you to visit tcec-chess.com, home to an online arena called the Top Chess Engine Championship. Twenty-four hours a day, seven days a week, elite computer chess programs — with names such as Ethereal, Fire, Fizbo, Komodo, Laser, Winter, and Xiphos — play against each other there, and you can watch them live. The engines each run on four high-end Intel Xeon processors with eighty-eight cores, analyzing tens of millions of positions per second. You can watch them consider lines dozens of moves into the future and evaluate positions to the hundredth of a pawn. They constantly produce some of the best chess ever played. You can even, if you are so inclined, watch an insolent human commentariat chat about the programs’ games in real time:

“ez draw.”

“it blundered away a good position.”

“pathetic play.”

But the machines don’t care, and they never stop playing.

One imagines a not-too-distant future, after the temperatures and oceans have risen and the coastal cities of the world have flooded and emptied, after the human population has migrated inland, after crops have died during droughts and species have gone extinct, after famine and economic collapse, where on a long-abandoned server, as long as the power holds out, the highest expression of our human culture, our last art, is created in chess games played in silence with no one watching.

*Reprinted from “**Seven Games: A Human History**” by Oliver Roeder. Copyright © 2022 by Oliver Roeder. With permission of the publisher, W. W. Norton & Company, Inc. All rights reserved.*

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, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p>
</p>">^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Chengyu Wu comes a puzzle that cuts right to the chase:

You have a square piece of paper. You fold it in half along an axis parallel to two of its sides. You then fold it in half along another axis that is perpendicular to the original one, so that you again have a square shape that is one-quarter the size of the original square.

At this point, you make three cuts along three straight lines, all the way through the folded paper. As you unfold the paper, what is the greatest number of separate pieces you could have?

*Extra credit:* Instead of three cuts, suppose you make *N* straight cuts. Now what is the greatest number of pieces you could have?

Rumor has it that readers of The Riddler Ellora Sarkar and Daniel Gomez are getting married this weekend. Congratulations to you both! In keeping with the hexagonal design of your wedding backdrop, I thought you might enjoy the following puzzle. (Riddler Nation, this one’s for you, too!)

The larger regular hexagon in the diagram below has a side length of 1.

What is the side length of the smaller regular hexagon?

*Extra credit: *If you look very closely, there are two more, even smaller hexagons on top. Can you see them? No? Maybe this animation will help:

What are the side lengths of these two even smaller hexagons?

Congratulations to Justin Ahmann of Bloomington, Indiana, and Ethan Levine of Chicago, Illinois, winners of last week’s Riddler Express.

Last week’s Express was a visual puzzle. Out of the 100 squares in the 10-by-10 grid below, 99 were colored gray, yellow, green, red or blue. One square, in the top left corner, had not yet been colored.

What color should this final square have been: gray, yellow, green, red or blue, or perhaps another color entirely?

First off, I want to say that — given the (intentional) ambiguity in the puzzle — there were multiple correct answers. In fact, *any* of the color choices could be justified. Among the hundreds of solvers, about half said the final square should be blue, while more than a quarter said red. (The remaining choices were less popular.) A red square in the top left, along with the three existing red squares, would have formed the four corners of yet another square. Similarly, the blue squares were all separated from their nearest blue neighbor by four squares in one direction and two squares in the perpendicular direction. If the top left square were blue, it would have followed this pattern as well.

The lone hint I offered was that the diagram had something to do with mathematicians Pingala, Hemachandra and Fibonacci. In retrospect, I should have also mentioned another name Fibonacci went by: Leonardo Pisano. The Pisano period is how long it takes the *last digits* of the Fibonacci numbers to repeat. In base 10, the Pisano period is 60. As noted by Justin, it was no coincidence that the most common color in the diagram (gray) occurred in precisely 60 squares. Perhaps the puzzle was a visualization of the Pisano period?!

But first, it’s worth recalling the first few Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 and 144. Their last digits, respectively, are 0, 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9 and 4. So, what does it mean for these last digits to “repeat”? After all, 1 has already occurred three times, but the sequence shows no signs of repeating.

That’s because each Fibonacci number (after the initial 0 and 1) is determined by adding the *two* numbers that precede it in the sequence. So, once you have *two consecutive* digits that repeat, the entire sequence will then repeat. We can treat consecutive last digits as ordered pairs: (0, 1), (1, 1), (1, 2), (2, 3), (3, 5) and so on. If we plot them in a grid, reading the first coordinate as the row and the second coordinate as the column, we can visualize the Pisano period as the 60 gray squares. For reference, here’s the grid again, now with rows and columns labeled with the digits from 0 to 9:

What about the remaining 40 squares? Well, if you started with the digits 0 and 2 — instead of the Fibonacci sequence’s 0 and 1 — you’d generate an entirely new sequence of ordered pairs: (0, 2), (2, 2), (2, 4), (4, 6), (6, 0), and so on. This time, the sequence repeated every 20 terms. These were the yellow squares in the grid.

The red, green and blue squares all represented other cycles of ordered pairs. The only outlier was the square in the top left. It had a cycle of length one, since (0, 0) repeated forever. In other words, if the Fibonacci sequence instead started with 0 and 0, it would just be a whole bunch of 0s (and not particularly interesting). That meant the top left square was a cycle unto itself, and therefore required **its own new color**. (I vote for #FF69B4.)

Finally, before you write to me to say how your interpretation of the puzzle was equally valid — I get it. There were *lots* of patterns at play here, which gave all sorts of interesting answers. I hope you at least found the connection to Pisano cycles interesting!

Congratulations to Jenny Mitchell of Nashville, Tennessee, winner of last week’s Riddler Classic.

Over the last few weeks, Wordle has taken the puzzling world by storm. In Wordle, you have six guesses to determine a five-letter mystery word. For each word that you guess, you are told which letters are correct and in the correct position (marked in green), which among the remaining letters are in the mystery word but are in the incorrect position (marked in yellow) and which letters are incorrect altogether.

And so, last week, your goal was to devise a strategy to maximize your probability of winning Wordle in *at most three guesses*. To do this, I provided you access to Wordle’s library of 2,315 mystery words as well as all 12,972 words you are allowed to guess. (For the record, I pulled both word lists from Wordle’s source code and listed them alphabetically.)

Before we get to the solution, it’s worth a reminder that Wordle’s rules get a little hairy when the mystery word or one of your guesses has a letter that appears more than once. To brush up on the rules, you may want to check out the following example in which the mystery word is MISOS, taken from last year’s Lingo-inspired Riddler Classic:

Okay, now let’s get to the solution. But before trying to win in three guesses, let’s try to win in two. (The math is very similar, but involves one less degree of complexity.)

With no information about the mystery word prior to your first guess, the optimal strategy (again, for winning in at most two guesses) always began with the same word. Upon guessing this word, you would see what I like to call a “constellation” of yellow, green and uncolored letters. In the example above — MAGIC — I had a green in the first position and a yellow in the fourth position. At this point, you knew that the mystery word must have been among the cluster of words that produced this constellation for your first guess.

For simplicity, suppose there were 100 total mystery words and three possible constellations. Let’s further suppose that 20 mystery words aligned with the first constellation, 30 words with the second and 50 with the third. So, with that first guess, the probability the first constellation shows up was 20/100, and then your probability of guessing correctly out of those 20 words was 1/20. Multiplying these probabilities together, the first constellation contributed 20/100·1/20, or 1/100, to your overall probability of guessing correctly. Meanwhile, the second constellation contributed 30/100·1/30, or another 1/100 to your overall probability. And the third constellation contributed 50/100·1/50, which was yet another 1/100. With 100 mystery words that corresponded to three constellations, your chances of guessing correctly were 3/100 — the number of constellations divided by the number of mystery words. It didn’t matter how many words were in each constellation, just the total number of constellations. (This was a pretty awesome moment for those who figured this out!)

Applying this to Wordle’s actual list of 2,315 mystery words, you can search for which word results in the greatest number of distinct constellations among the other 2,314 words (note that each word will always result in five greens when it is the mystery word itself). The optimal first word turned out to be TRACE, which resulted in a whopping 150 constellations. Keep in mind that the maximum number of constellations was 3^{5}, or 243, since each of the five letters was either yellow, green or uncolored — so 150 wasn’t too shabby. And that meant your probability of guessing the word in at most two guesses was 150/2,314, or about 6.5 percent, a result that *was* a little shabby.

Surely, with *three* guesses, your chances improved. Also, as I mentioned before, your strategy needed an additional layer of complexity. Whatever your first guess was, you would get some number of constellations (as many as 150, as we just learned). Then, for each constellation, you had to determine the second word you could guess that would fracture it into “sub-constellations.” Just as before, your probability of guessing correctly was equal to the total number of these sub-constellations divided by the total number of mystery words.

One wrinkle in all of this was that your second word didn’t have to look anything like the words in the constellation. The only requirement was that it helped you discriminate among all the words in the constellation as much as possible.

And so an ideal strategy was to search across different first guesses. For each first guess, identify the constellations. Then, for each constellation, identify the second word that resulted in the greatest number of sub-constellations. In the end, the best first word turned out to be, once again, **TRACE**. (It wasn’t a guarantee that the same word would be the best opener for winning in both two and three guesses, but it wasn’t exactly a surprise, either.)

If you chose your second word optimally, then there were a grand total of 1,388 sub-constellations, which meant your probability of winning in at most three guesses was very nearly **60 percent**. This optimal strategy is summarized in the chart below:

To play along with this #SolveInThree strategy:

- First, guess TRACE.
- Then, locate your resulting constellation on the ring. Note that the constellations should be read inside-out, meaning they’re not always oriented left-to-right.
- Next, guess the corresponding second word.
- Finally, guess any mystery word that is consistent with what you’ve seen so far.

The bars in the middle of the chart indicate how often you can expect to get each constellation, as well as how often you’ll win (in green). As I said, if you follow these steps, your chances of winning in three or fewer guesses is about 60 percent.

It shouldn’t be *too* surprising that your chances of guessing the mystery word skyrocketed when we moved from two turns to three. That’s because the theoretical maximum number of constellations increased by another factor of 243, bringing the total to a whopping 59,049 — well in excess of the number of five-letter words. If the English language were a little more uniform in terms of its letters and their placement, Wordle would be a heck of a lot easier!

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!

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

]]>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, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p>
</p>">^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

This week’s Express is a visual puzzle. Out of the 100 squares in the 10-by-10 grid below, 99 have been colored gray, yellow, green, red or blue. One square, in the top left corner, has not yet been colored.

What color should this final square be? Should it be gray, yellow, green, red, or blue, or perhaps another color entirely?

(If you’re not sure how to get started here, you might want to check in with a few mathematicians by the names of Pingala, Hemachandra and Fibonacci.)

Over the last few weeks, Wordle has taken the puzzling world by storm. Thousands of people (including yours truly) play daily, and the story of its creation has been well documented.

Wordle closely resembles the classic game show Lingo. In Wordle, you have six guesses to determine a five-letter mystery word. For each word that you guess, you are told which letters are correct and in the correct position (marked in green), which among the remaining letters are in the mystery word but are in the incorrect position (marked in yellow) and which letters are incorrect altogether.

This sounds straightforward enough. But things get a little hairier when the mystery word or one of your guesses has a letter that appears more than once. To brush up on the rules, you may want to check out the following example in which the mystery word is MISOS, taken from last year’s Lingo-inspired Riddler Classic:

In addition to the many people who play Wordle daily, some folks — including Friends-of-The-Riddler Laurent Lessard and Tyler Barron — have generated approaches for winning Wordle in relatively few guesses, no matter the mystery word. And this is closely related to the task for this week’s Riddler Classic.

Your goal is to devise a strategy to maximize your probability of winning Wordle in *at most three guesses*. After all, if Yvette Nicole Brown can do it, then so can your strategy! (No offense to Yvette Nicole Brown. She’s awesome.)

In particular, I want to know (1) your strategy, (2) the first word you would guess and (3) your probability of winning in three or fewer guesses.

To do this, you will need to access Wordle’s library of 2,315 mystery words as well as the additional 10,657 words you are allowed to guess. Both of these libraries can be accessed by viewing the source code on the Wordle site and then clicking on the link you’ll find at the very bottom (example shown below).

**Spoiler alert!** The mystery words are ordered sequentially by how they have appeared (and will appear) in the daily Wordle game. So if you enjoy playing daily and do not want to know the order of the upcoming mystery words, then don’t look too closely at the Wordle source code. You have been warned!

Congratulations to Brett of Urbandale, Iowa, winner of last week’s Riddler Express.

Last week, you were the Riddler Football League’s Arizona Ordinals against your opponent, the Detroit Lines, and your team was down by 14 points. You could assume that you had exactly two remaining possessions (i.e., opportunities to score), and that Detroit would score no more points.

For those unfamiliar with American football, a touchdown is worth 6 points. After each touchdown, you could decide whether to go for 1 extra point or 2 extra points. You happened to have a great kicker on your team, and your chances of scoring 1 extra point (should you have gone for it) were 100 percent. Meanwhile, scoring 2 extra points was no sure thing — suppose that your team’s probability of success was some value *p*.

If the teams were tied at the end of regulation, the game proceeded to overtime, which you had a 50 percent chance of winning.

What was the minimum value of *p* such that you’d go for 2 extra points after your team’s first touchdown (i.e., when you were down 8 points)?

At first, you might have thought that *p* should have been at least 50 percent for you to go for 2. That’s because when *p* was 50 percent, you could expect to earn 1 point on average after each touchdown, whether you were going for 1 or 2 extra points. But the fact that you had *two* scoring opportunities was important here. So let’s play out what might have happened.

If *p* was sufficiently small, then you were better off always going for 1 extra point. After two touchdowns, this would have tied the game and sent it into overtime, which you had a 50 percent chance of winning.

But if *p* was large enough, it was (presumably) to your advantage to go for 2 extra points after the first touchdown. If you were successful, then you could guarantee victory by going for 1 extra point after your second touchdown. But if you were unsuccessful, you had another opportunity to go for 2: after your second touchdown. This would tie the game and send it into overtime, where you still had a 50 percent chance of winning.

So then, what were your chances of winning in terms of *p*? If you made that first 2-point conversion (with probability *p*), you were guaranteed victory. But if you missed the conversion (with probability 1−*p*), you’d have to make the second 2-point conversation (again with probability *p*) — and even then, your probability of winning was still 1/2. Mathematically, this total probability of victory came to *p*+(1−*p*)·*p*/2, or 3*p*/2−*p*^{2}/2.

Now, when this expression was greater than 1/2, or 50 percent, that meant you were more likely to win than if you had simply gone for 1 extra point. And 3*p*/2−*p*^{2}/2 was greater than 1/2 when *p*^{2}−3*p*+1 was less than zero. Solving this quadratic inequality, *p* had to be greater than (3−√5)/2, or about 0.382. In other words, when *p* was at least **38.2 percent**, you should have gone for 2 after your first touchdown.

In the Riddler Football League, the coaches always make decisions based on logic and probability. Alas, the same cannot be said for the NFL, in which even the prisoner’s dilemma is clearly a foreign concept.

Congratulations to John Halmi of Annapolis, Maryland, winner of last week’s Riddler Classic.

Last week, Amare the ant was traveling within Triangle ABC, as shown below. Angle A measured 15 degrees, and sides AB and AC both had length 1.

Amare started at point B and wanted to ultimately arrive on side AC. However, the queen of his colony had asked him to make several stops along the way. Specifically, his path had to:

- Start at point B.
- Second, touch a point — any point — on side AC.
- Third, touch a point — any point — back on side AB.
- Finally, proceed to a point — any point — on side AC (not necessarily the same point he touched earlier).

What was the shortest distance Amare could travel to complete the queen’s desired path?

If Amare wanted to minimize the length of his trip’s first leg — from point B to side AC — he would have traveled along a line segment perpendicular to AC. With some trigonometry, that distance turned out to be sin(15°). Then, to minimize the length of the second leg, he would have traveled directly down to side AB, for an additional length of sin(15°)·cos(15°). Finally, his return to side AC would again have been perpendicular to that side, for a distance of sin(15°)·cos^{2}(15°). Adding these three lengths together gave a total distance of approximately 0.7503. But was there an *even more* efficient journey for Amare?

Indeed, there was. If Amare hadn’t initially taken the shortest route to side AC, instead veering a little closer to point A, then the last two legs of the trip would have been shorter. It turned out that this was a tradeoff worth making — walking a little farther on that first leg shortened the overall distance.

At this point, with this being a minimization problem and all, several solvers worked their way to a solution using calculus. For example, solver Jason Ash used three parameters to describe Amare’s journey — the distances from point A when he first touched side AC, when he next touched side AB and when he touched side AC again. With computer assistance, Jason found that the shortest such path had a length of approximately 0.7071. Not only was this an improvement over the “greedy” strategy of moving perpendicularly to each next line segment — it looked suspiciously like a rather well-known irrational number.

But first, I want to acknowledge that this puzzle was inspired by another problem I encountered several years ago from the American Mathematics Competitions. The original problem had a different triangle, with different side lengths and a different angle, but there was an elegant geometric solution, just like in this one.

As solvers Ed Parks, Laurent and nine-year-old (!) Mari Fujioka explain, you can imagine Amare’s second leg occurring in an identical triangle that was reflected across side AC. Meanwhile, you can imagine Amare’s third leg occurring in yet another identical triangle, this time reflected across leg AB’ (where point B’ is the reflection of point B in that second triangle). If you need a visualization, an animation by Colin Parker and this diagram from solver Emma Knight may help:

As we’ve been saying all along, the shortest distance to a line is perpendicular to that line. And so, the shortest distance from point B to line AC’ in Emma’s diagram is perpendicular to line AC. With three 15° angles stacked on top of each other, angle BAC’ is 45 degrees, and no advanced trigonometry (not to mention calculus!) is required. The shortest path had a total length of **1/√2**, which is indeed approximately 0.7071.

This wasn’t a brutal exercise in calculus after all. Or at least, it didn’t have to be.

That Amare is one clever ant, and I have it on good authority he traveled along the shortest path, much to the queen’s delight.

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!

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

]]>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, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p>
</p>">^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Leonard Cohen comes a timely matter of touchdowns:

In the Riddler Football League, you are coaching the Arizona Ordinals against your opponent, the Detroit Lines, and your team is down by 14 points. You can assume that you have exactly two remaining possessions (i.e., opportunities to score), and that Detroit will score no more points.

For those unfamiliar with American football, a touchdown is worth 6 points. After each touchdown, you can decide whether to go for 1 extra point or 2 extra points. You happen to have a great kicker on your team, and your chances of scoring 1 extra point (should you go for it) are 100 percent. Meanwhile, scoring 2 extra points is no sure thing — suppose that your team’s probability of success is some value *p*.

If the teams are tied at the end of regulation, the game proceeds to overtime, which you have a 50 percent chance of winning. (Assuming ties are not allowed.)

What is the minimum value of *p* such that you’d go for 2 extra points after your team’s first touchdown (i.e., when you’re down 8 points)?

Amare the ant is traveling within Triangle ABC, as shown below. Angle A measures 15 degrees, and sides AB and AC both have length 1.

Amare starts at point B and wants to ultimately arrive on side AC. However, the queen of his colony has asked him to make several stops along the way. Specifically, his path must:

- Start at point B.
- Second, touch a point — any point — on side AC.
- Third, touch a point — any point — back on side AB.
- Finally, proceed to a point — any point — on side AC (not necessarily the same point he touched earlier).

What is the shortest distance Amare can travel to complete the queen’s desired path?

Congratulations to Mayukha Vadari of Redmond, Washington, winner of the last Riddler Express.

Last week, you had to solve a cryptarithm based on a puzzle originally by Frank Mrazik:

As with any cryptarithm, each letter represented one of the digits from 0 to 9, and different letters represented different digits.

The catch? This puzzle had *two *possible solutions — that is, two distinct sets of letter-to-number assignments. Could you find both solutions?

To no one’s surprise, many solvers found the answers via brute force. There were nine letters in the cryptarithm, each of which had to be assigned to one of 10 digits without repetition. That meant there were 10·9·8·7·6·5·4·3·2, or 10!, cases to consider. That was well beyond one’s ability to manually check, but not a problem for a computer. Solvers Steven Goldburg and David Wanner wrote code that went through all of the approximately 3.6 million possible letter assignments, checking for any that followed the pattern in the cryptarithm.

However, it was (borderline) feasible to find the solutions deductively. Looking at the hundred-thousands place, where an L became an H, solver Rich Erikson immediately recognized that H was one more than L. (Technically, L could be 9 and H could be 0, but having numbers with a leading 0 would be bad form.) Meanwhile, in the ten-thousands place, H+I had be at least 10 (or 9 if there was carrying), and both H and I had to be greater than O.

At this point, you found that most values of L didn’t work. For example, if L was 0 (meaning H was 1), then I had to be 8 or 9, which then meant O had to be 0. But O couldn’t have been 0, since that meant it would have had the same value as L. In the end — after *a lot* of careful casework — you found that the only ordered pairs (L, H) with corresponding solutions were (5, 6) and (7, 8).

Here was the solution when (L, H) was (5, 6):

And here was the solution when (L, H) was (7, 8):

I hope this riddle kept you in puzzling shape over the holidays!

Congratulations to John Massman of Seattle, Washington, winner of the last Riddler Classic.

Last week’s Classic was an extension of the tax collector game, which was relayed to me by Fawn Nguyen.^{2} The game consisted of paychecks with different whole number dollar amounts, 1 to *N*. You chose different checks, one at a time. For any check you chose, the tax collector immediately took any remaining checks (i.e., not already taken by you or the tax collector) whose dollar values were factors of the one you chose. For example, if you chose the $10 check, then the tax collector would immediately take the $1, $2 and $5 checks — if any of them was available. Importantly, the tax collector *had to get something* for each paycheck you chose. So if the $10 check was available, but the $1, $2 and $5 checks were not, then you could not take the $10 check. When there were no more checks you could take, the game was over and all remaining checks went to the tax collector.

In the original version of the puzzle, your goal was to make more money than the tax collector when *N* was 12 (or 24 or 48). When *N* was 12, you could make $50 to the tax collector’s $28, meaning you won about 64 percent of the total. When *N* was 24, you could win about 61 percent of the total, and when *N* was 48, you could win about 62 percent.

For this puzzle, not only did you want to get more money than the tax collector, you also wanted to win the biggest possible fraction of the available money. Which value of *N* (greater than 1) would you have chosen, so that you could win the greatest fraction of available money?

Finding an optimal strategy for a given *N* (without the aid of a computer) was, well, *hard*. Most solvers realized your first move should always be to take the largest prime, because the tax collector would always take the $1 check away on your first move, and then you would never have the opportunity to get another prime.

But before we dive any further into the strategy of the game, let’s take a step back and try to find an upper bound. In other words, how close to 100 percent could you hope to get? Remember, the tax collector had to take at least one check for every check you took, meaning they got at least half the checks. So even if you took all the checks from the “top half” (i.e., from ⌈*N*/2⌉+1 to *N*), you could never exceed 75 percent of the total value. So then, how close to 75 percent could you get?

Solvers Nelson Daou and Jason Shaw next checked small values of *N*. When *N* was 2, you took $2 and left the tax collector with $1, meaning you got about 67 percent of the total. Not bad. But you could do even better when *N* was 6. Your first move was to take the largest prime, $5, while the tax collector took $1. Next, you had to decide between the $4 and $6 checks. If you took the $6 check, the game would end, and you’d have $11 out of a total of $21, or about 52.4 percent. But if you had taken the $4 check, the tax collector took the $2 check, leaving you to take the $6 check on your final turn. In the end, you earned $15, or about 71.4 percent. That was awfully close to 75 percent!

The next value of *N* where you could do even better was 10. Once again, you started by taking the largest prime, $7, while the tax collector took $1. Next, you could take $9, while the tax collector took $3. Then it was time to take $6, while the tax collector got $2. Finally, you could take the $8 and $10 checks in either order, leaving the tax collector with the $4 and $5 checks. Indeed, you got the “top half” of the checks, $6 to $10, for $40 out of a total of $55. That corresponded to 72.7 percent, even closer to the upper bound of 75 percent.

As I mentioned, finding the optimal strategy for a given *N* was hard. That’s why solvers like Mike Strong turned to their computers, using recursion to exhaustively search across all possible strategies. The chart below shows the greatest share you could win when *N* was between 2 and 701, courtesy of Brian Chess. The greatest amount of money you could win for each of these values of *N* also happens to be an OEIS sequence.

Sure enough, the best you could do was **72.7 percent, when ***N*** was 10**.

I am pleased to report that I received several good arguments as to why this was the optimal solution. As noted by solver Michael Haugh, for larger values of *N*, there was an increasing amount of prime numbers in the “top half” of checks that the tax collector would always win. Solver Emily Boyajian took a closer look, finding that the density of primes was sufficiently low for generating a result better than 72.7 percent when *N* was in the billions.

In the end, no one was able to submit a rigorous proof that 72.7 percent was the best you could do. To my knowledge, this remains an open problem. But as Emily concluded, “...if there is another value of *N* that allows for a greater proportion of winnings, it would take longer than a human lifetime to play the game, and the game would [probably] involve an amount of money greater than the world’s GDP.”

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!

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

]]>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, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday, Jan. 3. Have a great holiday!</p>
</p>">^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Due to the holidays, the next column will appear on Jan. 7, 2022. See you in the new year!

Reader Betts Slingluff enjoys holiday cryptarithms with the family and suggested that now was a good time for such a puzzle on The Riddler. This week’s Express is a spin on a cryptarithm originally by Frank Mrazik:

As with any cryptarithm, each letter represents one of the digits from 0 to 9, and different letters represent different digits.

The catch? This puzzle has *two *possible solutions — that is, two distinct sets of letter-to-number assignments. Can you find both solutions?

This week’s Classic is an extension of a puzzle originally by Dan Finkel (and relayed to me by Fawn Nguyen):

In the game of “tax collector,” there are paychecks with different whole number dollar amounts, 1 to *N*. You choose a paycheck, one at a time. For any check you choose, the tax collector immediately takes any remaining checks (i.e., not already taken by you or the tax collector) whose dollar values are factors of the one you chose. For example, if you choose the $10 check, then the tax collector will immediately take $1, $2 and $5 checks — if any of them is available. Importantly, the tax collector *must get something* for each paycheck you choose. So if the $10 check is available, but the $1, $2 and $5 checks are not, then you cannot take the $10 check. When there are no more checks you can take, the game is over and all remaining checks go to the tax collector.

In the original version of the puzzle, your goal was tsino make more money than the tax collector when *N* was 12 (or 24 or 48). When *N* was 12, you could make $50 to the tax collector’s $28, meaning you won about 64 percent of the total. When *N* was 24, you could win about 61 percent of the total, and when *N* was 48, you could win about 62 percent.

For this puzzle, not only do you want to get more money than the tax collector, you also want to win the biggest possible fraction of the available money. Which value of *N* (greater than 1) would you choose, so that you can win the greatest fraction of available money?

Congratulations to Rick Kneedler of Portland, Oregon, winner of last week’s Riddler Express.

Last week, you were tasked with painting a building that was shaped like a regular tetrahedron. When the building was viewed from above, the architect wanted it to appear as four congruent equilateral triangles — one central blue triangle surrounded by three white triangles.

That meant that three faces of the tetrahedron contained a blue kite, as shown in the animation below:

What was the measure of the smallest angle in this kite?

It helped to first realize that the fraction of the total area that was blue was the same for each face whether you were looking at it from above or straight on. (For a great explanation of why that is, check out 3Blue1Brown’s recent video.) In other words, one-quarter of each of the three faces was painted blue, and the kites met the lateral edges of the tetrahedron three-quarters of the way up.

The answer didn’t depend on the overall scale of the tetrahedron, so let’s say it had side length 4, as shown in the diagram below. That meant the two top edges of the kite had length 1, while the half-width of the kite was 0.5. The height of the kite was the same as the height of the equilateral triangle, or 2√3. Again, three-quarters of that height was below the kite’s half-width, for a total length of 1.5√3.

Finally, this was enough to determine *half* of the kite’s bottom angle. It formed a right triangle whose opposite side was 0.5 and whose adjacent side was 1.5√3. That meant the tangent of this angle was 1/(3√3), and so the angle was tan^{-1}(1/(3√3)), or about 10.89 degrees. Again, since this was *half* of the bottom angle, the bottom angle was approximately **21.79 degrees**.

Unlike their 2D cousins, regular tetrahedra have all sorts of angles that aren’t 30 degrees or 60 degrees, don’t they?

Congratulations to David Cohen of Silver Spring, Maryland, winner of last week’s Riddler Classic.

Last week, you were trying to whack a mole. Every second, the mole popped its head out from one of 100 holes arranged in a line. You didn’t know where the mole started, but you *did* know that it always moved from one hole to an adjacent hole each second. For example, if it popped out of the 47th hole, then one second later it popped out of either the 46th hole or the 48th hole. If it popped out of the first hole, then it was guaranteed to pop out of the second hole next; similarly, if it popped out of the 100th hole, then it was guaranteed to pop out of the 99th hole next.

Of course, there was a catch — the mole was camouflaged, and you had no idea where it was at any time until you actually whacked it. Each second, you could whack one hole of your choosing. Many sequences of whacks did not guarantee that you’d eventually get the mole (e.g., always whacking the first hole), but some did.

What was the shortest such sequence that guaranteed you could whack the mole, no matter where it started or how it moved around?

Before we deal with 100 holes, let’s consider a simpler case: four holes labeled 1, 2, 3 and 4. Suppose you first whacked Hole 1 and missed. Then the mole had to be in Holes 2, 3 or 4. On the next turn, after moving one hole left or right, the mole could have been in Holes 1, 2, 3 or 4. In other words, whacking the first hole did absolutely nothing to reduce the number of possible locations, and you were back to square one!

Suppose you instead started on Hole 2 and missed, in which case the mole could have been in Holes 1, 3 or 4. This time, on the next turn, the mole could have been in Holes 2, 3 or 4 — but not Hole 1, since it had to have moved from Hole 2, which you had just whacked. Next, if you whacked Hole 3 and missed, you could guarantee the mole was in either Hole 2 or 4. Whacking Hole 3 a second time and missing meant the mole now had to be in Hole 1. One last whack of Hole 2 ensured you got the mole.

At the end of the day, the best you could do with four holes was four whacks. The sequences 2-3-3-2 and 3-2-2-3 both guaranteed the job got done with minimal effort.

But what about 100 holes? Before we come back to that, let’s take one more look at the case of four holes, this time keeping in mind the initial parity of the mole’s position — as noted by solvers like Michael Coffey and Madeline Argent, and as suggested in the video that inspired this Classic. Every move, the mole either switched from an even-numbered hole to an odd-numbered hole, or vice versa. So every move, its parity flipped.

Suppose the mole had started in one of the even holes (i.e., 2 or 4). Checking Hole 2 and then Hole 3 guaranteed that you’d whack the mole. But had the mole started in an odd hole (i.e., 1 or 3), then after you checked Holes 2 and 3 it would *still* be in one of the odd holes. In this case, flipping your order from here on out — checking Hole 3 and then 2 — guaranteed that you’d whack the mole.

This exact line of reasoning extended to any number of holes, *N*. First, you could rule out all the even starting points by working your way up from 2 to *N*−1. Then, you could rule out all the odd starting points by returning from *N*−1 to 2. (This strategy even worked when *N* was odd.) So for *N* holes, you needed 2·(*N*−2) turns to guarantee that you whacked the mole. When *N* was 100, the answer was **196**.

If you’re still not convinced, here’s an animation showing 100 moles that initially occupy all 100 holes. Each mole “reproduces,” splitting into two moles in adjacent holes, unless it’s at one of the end points, in which case it simply moves left or right. Occupied holes are red, while empty holes are white. Meanwhile, the hole you are currently whacking is blue.

Sure enough, if you whacked your way up and down the row, you accounted for every possible initial position of the mole and every possible sequence of its moves.

Now if only we could handle tribbles with such efficiency.

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

]]>How many Americans use artificial Christmas trees? And when is the right time to put up holiday decorations? In this episode of The United Stats of America, host Galen Druke quizzes people in New York — including Santa Claus himself! — about what Americans think of the winter holidays.

]]>In 2021, FiveThirtyEight’s visual journalists told stories of the pandemic, political gridlock and the world of sports. We’ve fought to make our work more accessible and to sharpen our storytelling. Through it all, we kept it weird. Now we continue our tradition of celebrating our best — and wackiest — charts of the year. Here are some of our favorites, grouped by topic but in no particular order beyond that. If you want more context for these (weird and wonderful) charts, don’t be shy! You can click any of them to read the stories in which they originally ran.

And, last but not least, check out this calendar with the results from our debate about when each season begins.

Did you enjoy this long list of weird charts? Then boy, do we have content in the archives for you! Check out our lists from 2020, 2019, 2018, 2016, 2015 and 2014.

]]>Which magical bow should you pick to win the archery tournament? Quiz yourself with this brain teaser!

For more puzzles, check out our Riddler column every Friday morning.

]]>^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Steven Anisman and Giacomo “Mike” Peccini comes a troublesome, triangular tribulation:

You have been tasked with painting a modern building that is shaped like a regular tetrahedron. When the building is viewed from above, the architect wants it to appear as four congruent equilateral triangles — one central blue triangle surrounded by three white triangles.

That means that three faces of the tetrahedron contain a blue kite, as shown in the animation below:

What is the measure of the smallest angle in this kite?

This week’s Riddler Classic comes courtesy of George Mu, who is passing along a puzzle based on one discussed in a mock coding interview.^{2}

You are trying to whack a mole. Every second, it pops its head out from one of 100 holes arranged in a line. You don’t know where the mole starts, but you *do* know that it always moves from one hole to an adjacent hole each second. For example, if it pops out of the 47th hole, then one second later it pops out of either the 46th hole or the 48th hole. If it pops out of the first hole, then it’s guaranteed to pop out of the second hole next; similarly, if it pops out of the 100th hole, then it’s guaranteed to pop out of the 99th hole next.

Of course, there’s a catch — the mole is camouflaged, and you have no idea where it is at any time until you actually whack it. Each second, you can whack one hole of your choosing. Many sequences of whacks do not guarantee that you’ll eventually get the mole (e.g., always whacking the first hole), but some do.

What is the shortest such sequence that guarantees you will whack the mole, no matter where it starts or how it moves around? (In the aforementioned video, Ben and Dan discuss *a* sequence that is guaranteed to whack the mole, but it is not necessarily the shortest such sequence.)

Congratulations to Christopher Halverson of St. Louis, Missouri, winner of last week’s Riddler Express.

Last week, my condo complex had a single elevator that served four stories: the garage (G), the first floor (1), the second floor (2) and the third floor (3). Unfortunately, the elevator was malfunctioning and stopping at every single floor, no matter what. The elevator always went G, 1, 2, 3, 2, 1, G, 1, 2, etc.

I wanted to board the elevator on a random floor (with all four floors being equally likely). As I rounded the corner to approach the elevator, I heard that its doors had closed, but I had no further information about which floor it was on or whether the elevator was going up or down. The doors might have just closed on my floor, for all I knew.

On average, how many stops would the elevator have made until it opened on my floor (including the stop on my floor)? For example, if I was waiting on the second floor, and I heard the doors closing on the garage level, then the elevator would have opened on my floor in two stops.

While I was equally likely to be on any of the four floors, the *elevator* was not. The repeating unit of the elevator’s cyclical motion was G-1-2-3-2-1. That meant it had a 1 in 6 chance of being in the garage going up, a 1 in 6 chance of being on the first floor going up, a 1 in 6 chance of being on the second floor going up, a 1 in 6 chance of being on the third floor going down, a 1 in 6 chance of being on the second floor going down and a 1 in 6 chance of being on the first floor going down.

Suppose I was on the garage level. If I heard the doors closing in the garage going up, I would have waited six stops. If the doors closed on the first floor going up, I would have waited five more stops. Working through all six cases, my average wait time would have been 3.5 stops (the average of the whole numbers from 1 to 6). If I was on the third floor, the average wait time would again have been 3.5 stops.

Now suppose I was on the first floor. If I heard the doors closing in the garage going up, I would have waited one stop. Had the doors closed on the first floor going up, I would have waited four stops (2-3-2-1). Had they closed on the second floor going up, I would have waited three stops (3-2-1). Had they closed on the third floor going down, I would have waited two stops (2-1). Had they closed on the second floor going down, I would have waited one stop (1). And had they closed on the first floor going down, I would have waited two stops (G-1). The average of *these* six cases was 13/6, or about 2.17. By symmetry, this wait time was the same had I been on the second floor. (I thought it was kind of neat that my average wait time was shortest on the two middle floors.)

My average wait was 3.5 stops half the time, while the other half the time it was 13/6 stops. Averaging these together, my expected wait was 17/6, or about **2.83 stops**.

For extra credit, my condo now had *N* floors instead of four. Once again, how many stops would the elevator have made, on average, until it opened on my floor? This time, the solution was a little more involved, requiring a number of summations. The answer turned out to be **(4***N***+1)/6**, which indeed gave you the correct solution when *N* was four. To see a complete derivation, check out the write-ups of solvers Jake Gacuan and Awang B. Brantas.

Congratulations to Michael DeLyser of State College, Pennsylvania, winner of last week’s Riddler Classic.

Last week, you were the coach at Riddler Fencing Academy, where your three students were squaring off against a neighboring squad. Each of your students had a different probability of winning any given point in a match. The strongest fencer had a 75 percent chance of winning each point. The weakest had only a 25 percent chance of winning each point. The remaining fencer had a 50 percent probability of winning each point.

The match was a relay. First, one of your students faced off against an opponent. As soon as one of them reached a score of 15, they were both swapped out. Then, a different student of yours faced a different opponent, continuing from wherever the score left off. When one team reached 30, both fencers were swapped out. The remaining two fencers continued the relay until one team reached 45 points.

As the coach, you could choose the order in which your three students occupied the three positions in the relay: going first, second or third. How would you have ordered them? And then what were your team’s chances of winning the relay?

Solver Izumihara Ryoma realized that a good strategy had the strongest fencer going last. Remember, the player who went first could score at most 15 points before being subbed out, while the player who went last could win many more points. By this logic, if we labeled the three fencers “25,” “50” and “75” (based on their percent chance of scoring each point), you definitely wanted 75 to go last. So then was the optimal ordering 25-50-75, or was it 50-25-75?

One way to figure this out was to calculate the probability of victory for each ordering. The chart below shows all six orderings your team could have had. Each ordering corresponds to a grid of possible scores you could have seen during the relay, from 0-0 in the top left to 45-44 and 44-45 in the bottom right. To complete each grid, I started with a probability of 1 in the top left (the cell corresponding to a score of 0-0), and then split up this probability into two parts. One part, which represented *your* team scoring the next point, was added to the value of the cell directly below; the other part, which represented your opponent scoring the next point, was added to the cell directly to the right. Scores less likely to occur are darker, while likelier scores are lighter. You can see how much the ebb and flow of the match depended on the ordering:

For each grid, you could have computed your team’s precise chances of victory by summing the probabilities in the bottom row, which corresponded to winning scores from 45-0 through 45-44. Not surprisingly, having your weakest fencer go last was a terrible idea. With an ordering of 75-50-25, you won 6.8 percent of the time. Switching those first two fencers wasn’t much better, marginally increasing your chances to 7.5 percent.

Having your middle fencer go last led to diverging results. With an ordering of 75-25-50, your chances of winning were 17.4 percent. Meanwhile, 25-75-50 gave you an 82.6 percent chance of winning. In the latter case, your strongest fencer could dig your team out of the hole created by your weakest fencer, allowing your middle fencer to hopefully maintain that lead.

Sure enough, your best bet was to have your strongest fencer go last. And while an ordering of 50-25-75 gave you a 92.5 percent chance of winning, your best strategy was to order your team from weakest to strongest, **25-50-75**, which gave you a **93.2 percent** chance of victory. (You may have noticed that these six probabilities from the six orderings formed three pairs that each added up to 100 percent. Why? Symmetry!)

By the way, a few years — *cough, decades* — ago, I saw this riddle play out before my very eyes. I was in the stands at a Long Island fencing tournament, watching two teams face off in a relay to 45 points. Over the first two rounds, one team jumped out to a 30-15 lead. And so, to win the relay, the final fencer on the trailing team had to earn 30 points while keeping his opponent to fewer than 15. And that’s exactly what he did. Needless to say, the gymnasium went wild.

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

]]>What makes a house a home? How do you spread tidings of comfort and joy around your apartment? Will roasting chestnuts on an open fire really make you feel like your troubles are miles away?

We know, instinctually, that the way our homes look can affect how we feel — but it’s less clear how to produce the specific feelings we want. That’s the question Google and Johns Hopkins University researchers were trying to answer in 2019 when they set up an interactive art installation-slash-experiment at a high-end furniture trade fair in Milan, Italy. Scientists gave visitors wristbands that could monitor several different biomarkers, such as heart rate, and then guided them through a series of model living and dining rooms set up with similar furniture but different themes. Those themes were created by altering color selections, lighting, smells and fabric textures. Then the scientists asked participants which rooms made them feel the most sense of calm and ease.

And then something strange happened …

“Sometimes what people thought cognitively that they liked the best or where they thought they were most at ease [was] not what their biometric data said,” explained Tasha Golden, director of research for the International Arts + Mind Lab at Johns Hopkins School of Medicine. “There was a distinction between what people thought they liked the most and what their bodies seemed to respond to.”

At a time of the year when our thoughts naturally turn to hominess and coziness, it would be nice if science could tell us exactly how to create that hygge we’ve all heard so much about. Researchers are certainly trying, but in the process of quantifying comfort, they have also discovered that the objective study of subjective concepts can produce some really weird results. The good news is that disconnects like the one Johns Hopkins researchers discovered don’t mean the study of how and why we perceive aesthetics is pointless. Even if nobody can pin down what “hominess for the holidays” *should* look like, exploring our individual and cultural tastes can still help scientists better understand the brain, help you achieve better health outcomes, and maybe even help you avoid an expensive remodel on your house.

Turns out, it’s not uncommon for researchers to uncover differences between what people say they like and what their body actually responds to positively. Lindsay Graham, a research specialist at the Center for the Built Environment at UC Berkeley, said she has run across similar issues in her attempts to understand what kinds of architectural environments relax people. She attributed the disconnect to the simple fact that our aesthetic judgments don’t happen in a cultural vacuum.

“We see things in Instagram ads that are aesthetically pleasing, and maybe they do align with some part of ourselves or what we want,” Graham said. “But if it doesn’t actually fit who we are or what we need, then it’s not really going to accomplish what it is that we think it is.” In other words, there’s a difference between what you think you *should* like and what *actually *makes you feel at home.

And that’s not the only subjectivity stumbling block that researchers run into when studying coziness. Anjan Chatterjee is a professor of neurology at the University of Pennsylvania School of Medicine. What makes us feel comfortable and at home changes depending on where we are, he told me. People in his studies rate indoor spaces as being homier if they have elements from outside — plants, for instance, or natural materials and patterns. But outdoor environments rate as more homey if they have elements of order to them, like a tidy English garden.

Culture also seems to affect aesthetic judgments. So far, the scientists we spoke to said, most of the research into the science of aesthetic perception has been focused on subjects from the United States and European countries. But when it’s been conducted in other regions and nations, differences quickly arise. “We have a really large database of survey perceptions that look at what predicts satisfaction” with an architectural space, Graham said. “In our primarily North American population, we saw that it’s amount of space that matters. But in a recent study, we had found that it was cleanliness in Singaporean spaces that was most predictive of being satisfied.”

This kind of variability should make researchers wary of ascribing too much certainty to what they think they know about how humans make aesthetic judgments, said Bevil Conway, a senior investigator at the National Institutes of Health who studies the way our brains perceive color. It’s really easy to hook someone up to an fMRI machine or give them a bunch of surveys and assume you’re learning something universal about beauty — when what you’re really doing is testing how well people conform to preconceived ideas of what looks and feels nice.

But that also doesn’t mean it’s impossible to know anything about aesthetics. Scientists might not be able to tell you exactly how to decorate your house so that you and everyone who visits will feel cozy, but Golden said there’s evidence that a wide swath of humans find the sound of running water, and other natural noises, to be calming and peaceful.

That’s not necessarily a universal effect, but it’s pretty close, she told me — enough that you could think of it as being true for humans, generally, even if it isn’t true for every individual. Her team at Johns Hopkins is interested in this kind of research because they believe it can point towards techniques that improve health and wellness. For instance, those nature sounds might have a stress-relieving effect on some people. More important than hints like these are larger questions that Golden’s work is trying to answer about how the things we like impact our physiology. Even if the specific likes are distinct between individuals, the effects might be more universal. “When somebody just says, ‘I am cozy,’ what does that do for their well-being? Or how might the absence of that be affecting their health and well-being?” she said.

Those questions tie in with Chatterjee’s work, which has found that when people see a space that they describe as being beautiful, the experience triggers in the parts of the brain that scientists associate with our body’s internal reward systems — the same as what happens when we get pleasurable food or good sex.

Graham also thought there might be similarities between how people want spaces to make them feel, even if they have different ideas of what makes that space cozy. Her team has worked on studies that tried to figure out what emotional expectations people have for different rooms of the house. Their samples were from North America, but they found a lot of consistency here, she said, with respondents generally wanting one of two things from a bedroom: Restful comfort or romantic intimacy. This might seem obvious, she told me, but from an architect’s point of view, it can be beneficial to understand how clients want their rooms to feel — because that framework can guide discussions and ensure that clients are happier, for longer, in the spaces that have been designed for them.

Beyond the practical, though, there’s reason to be interested in the science of subjective aesthetics just because it’s really, really weird. “I can ask you, ‘Hey, Maggie, what’s the color of a banana?’ and you’re not even looking at one and you could visualize what it looks like,” Conway said. “So we know that the brain is doing this really interesting thing [with color], which is somehow integrating across time. It’s allowing you to sometimes predict the future and to remember the past.” His work has involved mapping the parts of the brain that help process color in both human and non-human primates and he’s found color processing circuits in places he didn’t expect, including parts of the frontal cortex usually associated with decision making.

The color of a banana may seem pretty removed from the comfort of your parent’s living room at Christmas, but both are aesthetic experiences that are as influenced by human culture as they are by science. From the perspective of physics, the yellow light of a candle is physically colder than the bluish light of a fluorescent bulb — but we call the yellow light warm, because we often associate it with the cozy feelings of a house in winter.

For people like Conway, understanding how our brains tie all these associations together, how those meanings can shift, and how a change in our society can affect how we think about a color … or a place … well, that’s just cool.

]]>^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Michael Fienberg comes a tower of not so much terror:

My condo complex has a single elevator that serves four stories: the garage (G), the first floor (1), the second floor (2) and the third floor (3). Unfortunately, the elevator is malfunctioning and stopping at every single floor, no matter what. The elevator always goes G, 1, 2, 3, 2, 1, G, 1, 2, etc.

I want to board the elevator on a random floor (with all four floors being equally likely). As I round the corner to approach the elevator, I hear that its doors have closed, but I have no further information about which floor it’s on or whether the elevator is going up or down. The doors might have just closed on my floor, for all I know.

On average, how many stops will the elevator make until it opens on my floor (including the stop on your floor)? For example, if I am waiting on the second floor, and I heard the doors closing on the garage level, then the elevator would open on my floor in two stops.

*Extra credit: *Instead of four floors, suppose my condo had *N* floors. On average, how many stops will the elevator make until it opens on my floor?

You are the coach at Riddler Fencing Academy, where your three students are squaring off against a neighboring squad. Each of your students has a different probability of winning any given point in a match. The strongest fencer has a 75 percent chance of winning each point. The weakest has only a 25 percent chance of winning each point. The remaining fencer has a 50 percent probability of winning each point.

The match will be a relay. First, one of your students will face off against an opponent. As soon as one of them reaches a score of 15, they are both swapped out. Then, a different student of yours faces a different opponent, continuing from wherever the score left off. When one team reaches 30 (not necessarily from the same team that first reached 15), both fencers are swapped out. The remaining two fencers continue the relay until one team reaches 45 points.

As the coach, you can choose the order in which your three students occupy the three positions in the relay: going first, second or third. How will you order them? And then what will be your team’s chances of winning the relay?

Congratulations to Steve Schaefer of Carlsbad, California, winner of last week’s Riddler Express.

Last week was Hanukkah, which meant it was time for some Menorah Math!

I had a most peculiar menorah. Like most menorahs, it had nine total candles — a central candle, called the shamash, four to the left of the shamash and another four to the right. But unlike most menorahs, the eight candles on either side of the shamash were numbered. The two candles adjacent to the shamash were both “1,” the next two candles out from the shamash were “2,” the next pair were “3,” and the outermost pair were “4.”

The shamash was always lit. How many ways were there to light the remaining eight candles so that sums on either side of the menorah were “balanced”? For example, one such way was to light candles 1 and 4 on one side and candles 2 and 3 on the other side. In this case, the sums on both sides were 5, so the menorah was balanced.

Solver Madeline Argent approached this puzzle by first noting that the sum on either side could be any integer from 0 (when no candles were lit) to 10 (when all four candles were lit). However, many sums were achievable in more than one way, such as the aforementioned 5, which was both 1+4 and 2+3.

Here were all the possible sums, as well as how many ways they could be achieved:

- There was one way to have a sum of 0 (no candles lit).
- There was one way to have a sum of 1 (just candle 1 was lit).
- There was one way to have a sum of 2 (just candle 2).
- There were two ways to have a sum of 3 (just candle 3, or candles 1 and 2).
- There were two ways to have a sum of 4 (just candle 4, or candles 1 and 3).
- There were two ways to have a sum of 5 (candles 1 and 4, or candles 2 and 3).
- There were two ways to have a sum of 6 (candles 2 and 4, or candles 1, 2 and 3).
- There were two ways to have a sum of 7 (candles 3 and 4, or candles 1, 2 and 4).
- There was one way to have a sum of 8 (candles 1, 3 and 4).
- There was one way to have a sum of 9 (candles 2, 3 and 4).
- There was one way to have a sum of 10 (candles 1, 2, 3 and 4).

When the sums on either side were 0, 1, 2, 8, 9, or 10, there was only one way to balance the menorah, since each of these sums were only possible one way. But when the sums on either side were 3, 4, 5, 6 or 7, there were *four* ways to balance the menorah. For example, if the sums were 5, then both sides of the menorah could have been 1+4, both sides could have been 2+3, the left side could have been 1+4 and the right side 2+3, or the left side could have been 2+3 and the right side 1+4.

In all, there were six sums with one balanced lighting, and another five sums with four balanced lightings. That meant there were **26** balanced ways to light the menorah.

Perhaps Hanukkah should be 26 nights, instead of eight. I, for one, would not complain.

By the way, if you didn’t count the case where both sums were 0, which meant your answer was 25, I still counted that as correct. Who wants an empty menorah? (Looking at you, students at The Hewitt School!)

Congratulations to Jake Gacuan of Manila, Philippines, winner of last week’s Riddler Classic.

Last week, official ice master and deliverer Kristoff and his trusty pal Sven were preparing ice cubes that would be sent to the hot springs, where Kristoff’s extended family lived.

They were studying a cube whose side lengths measured 1 meter, which meant its surface-area-to-volume ratio is 6. (Kristoff didn’t particularly care about the units here but knew they were inverse meters.) Sven was concerned that much of the ice would melt before they reached the hot springs, so he suggested that Kristoff cut the ice along a single plane to minimize the surface-area-to-volume ratio of the larger resulting piece.

Along which plane should Kristoff have cut the ice? And what was the resulting surface-area-to-volume ratio?

First off, several solvers noted that any cut, regardless of the resulting surface-area-to-volume ratio, would have made the ice melt faster. To that I say that Sven is just a reindeer, and his mathematical prowess exceeds his knowledge of thermodynamics.

Back to the ice — for any given surface area (or volume), the 3D solid with the minimal surface-area-to-volume ratio is the sphere. And so one reasonable approach was to make the cube of ice a little more “spherical” (somehow).

To convince yourself this was even possible, you could have cut off a small corner of the cube along a plane that intersected three edges a distance *x* from one vertex, as shown below.

In this case, you were slicing off a tetrahedron with volume *x*^{3}/6. You were also removing a surface composed of three right triangles with a combined area of 3*x*^{2}/2. At the same time, you were creating a *new* equilateral surface with side length *x*√2, and hence area (*x*^{2}√3)/2. Putting this all together, the surface-area-to-volume ratio was (6−(3−√3)*x*^{2}/2)/(1−*x*^{3}/6). For small values of *x*, the denominator was very close to 1, while the numerator was slightly less than 6. In other words, it was indeed possible for the ratio to be less than 6!

Minimizing the aforementioned rational function gave you an optimal ratio of roughly 5.962 when x was approximately 0.4254. And that’s the answer that several readers submitted. But was it possible for Kristoff to do even better?

Another way to make the cube slightly more “spherical” was to cut along a plane that was parallel to an edge:

If the distance from the lopped-off corners to where the plane intersected the edges was again *x*, the volume removed this time was *x*^{2}/2, while the surface area removed was *x*^{2}+2*x*. Meanwhile, the newly created surface area was *x*√2. This time around, the resulting ratio was (6−(2−√2)*x*−*x*^{2})/(1−*x*^{2}/2). Minimizing *this* rational function gave you a ratio of approximately **5.957** when *x* was about 0.1481. As it turned out, cutting parallel to an edge was slightly better than lopping off a corner.

This week’s winner, Jake, extended the puzzle by looking at what would happen if Kristoff made multiple cuts, slicing off multiple corners or edges. According to Jake, with eight cuts, Kristoff could achieve a surface-area-to-volume ratio as low as 5.66. And if you want to see how solver Tyler Barron got this week’s answer in real time, check out his work on Twitch!

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

]]>^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Tonight marks the sixth night of Hanukkah, which means it’s time for some more Menorah Math!

I have a most peculiar menorah. Like most menorahs, it has nine total candles — a central candle, called the shamash, four to the left of the shamash and another four to the right. But unlike most menorahs, the eight candles on either side of the shamash are numbered. The two candles adjacent to the shamash are both “1,” the next two candles out from the shamash are “2,” the next pair are “3,” and the outermost pair are “4.”

The shamash is always lit. How many ways are there to light the remaining eight candles so that sums on either side of the menorah are “balanced”? (For example, one such way is to light candles 1 and 4 on one side and candles 2 and 3 on the other side. In this case, the sums on both sides are 5, so the menorah is balanced.)

Winter is almost upon us, at least for those of us in the Northern Hemisphere. Official ice master and deliverer Kristoff and his trusty pal Sven are preparing ice cubes that will be sent to the hot springs, where Kristoff’s extended family lives.

At the moment, they are studying a cube whose side lengths measure 1 meter, which means its surface-area-to-volume ratio is 6. (Kristoff doesn’t particularly care about the units here, but knows they are inverse meters.) Sven is concerned that much of the ice will melt before they reach the hot springs, so he suggests that Kristoff cut the ice along a single plane to minimize the surface-area-to-volume ratio of the larger resulting piece.

Along which plane should Kristoff cut the ice? And what will the resulting surface-area-to-volume ratio be?

Congratulations to Derrick Butler of Lexington, Kentucky, winner of last week’s Riddler Express.

Last week, you were on the Food Network’s latest game show, Cranberries or Bust, where you had a choice between two doors: A and B. One door had a lifetime supply of cranberry sauce behind it, while the other door had absolutely nothing behind it. And boy, did you love cranberry sauce.

Of course, there was a twist. The host presented you with a coin with two sides, marked A and B, which corresponded to each door. The host told you that the coin was weighted in favor of the cranberry door — without telling you which door that was — and that door’s letter would turn up 60 percent of the time. For example, if the sauce had been behind door A, then the coin would have turned up A 60 percent of the time and B the remaining 40 percent of the time.

You could flip the coin twice, after which you had to make your selection. Assuming you optimized your strategy, what were your chances of choosing the door with the cranberry sauce?

Before working through this scenario with two coins, let’s take a step back and look at one coin. Since the coin was slightly weighted in favor of the cranberry door, your best strategy was to choose whichever door was indicated by the coin. Then you’d be right — and win your delectable cranberry sauce — 60 percent of the time.

Surely, you had better odds of winning with *two *flips instead of one. Right?

Wrong. To see why that was, solver Rebecca Harbison looked closer at the possible results of the two flips. Instead of A and B, let’s relabel the coin C (for “Cranberry sauce is behind this door”) and D (for “Dangit, no cranberry sauce behind this door”). As stated by the problem, the coin had a 60 percent chance of landing on C and a 40 percent chance of landing on D.

With two flips, there were four possible outcomes:

- The first flip was C and the second flip was also C, which occurred with probability (0.6)(0.6), or 0.36.
- The first flip was C and the second flip was D, which occurred with probability (0.6)(0.4), or 0.24.
- The first flip was D and the second flip was C, which occurred with probability (0.4)(0.6), or 0.24.
- The first flip was D and the second flip was D, which occurred with probability (0.4)(0.4), or 0.16.

Putting these together, there was a 48 percent chance that the two flips were different. When this happened, you had absolutely no information how the coin was weighted, since the coin came up either side an equal number of times. In other words, you had to guess, which meant you’d be right half the time.

The other 52 percent of the time the two flips were the same. The majority of the time (i.e., with probability 36/52, or 9/13) that meant both flips were C. So when the two flips were the same, your best move was to guess the door both flips corresponded to.

So then what were your chances of winning the cranberry sauce? Well, 48 percent of the time you had a one-half chance of winning, while the other 52 percent of the time you had a 9/13 chance of winning. Numerically, these combined to (12/25)(1/2) + (13/25)(9/13), which simplified to 6/25 + 9/25, or 15/25 — in other words, **60 percent**.

Shockingly (to me, at least), that second flip didn’t improve your chances one bit. You might as well have simply flipped the coin once and chosen the resulting door.

For extra credit, you looked at what happened when you were allowed three or four flips, instead of just two. For three flips, the optimal strategy was to choose whichever door was indicated by a majority of the flips. All three flips came up C with probability (0.6)^{3}, or 0.216. Meanwhile, the probability of two Cs and one D was 3(0.6)^{2}(0.4), or 0.432. Adding these together meant you’d win the cranberry sauce **64.8 percent** of the time — an improvement over your chances with just one or two flips.

But just as two flips were no better than one, four flips were no better than three. That fourth flip either confirmed your decision based on the first three flips, or changed things so that now there were two flips for one door and two flips for another (opening up twice as many possibilities, and forcing you to take a random guess).

Solvers Jake Gacuan and Emily Boyajian both found a general formula for your chances of winning the cranberry sauce as a function of the number of flips *N*. Sure enough, having an even number of flips was no better than having the preceding odd number of flips. And as *N* increased, your chances of choosing the correct door approached 100 percent.

Congratulations to Izumihara Ryoma of Toyooka, Japan, winner of last week’s Riddler Classic.

Last week, you were introduced to Trig the turkey, who could not fly. Instead of flying, he decided to jump as far as he could with a running start up one of the famed Sinusoidal Hills, all of which were precisely the same size.

Trig knew that he simply could not jump a horizontal distance of more than two hills. Also, he prefered a smooth takeoff and landing. That is, when he took off from the ground and landed on it again, the slope of his parabolic trajectory through the air had to perfectly match the instantaneous slope of the ground beneath him.

The animation below shows a jump where the takeoff and landing were precisely 1.2 hills apart.

What was the greatest horizontal distance Trig could have jumped, such that his takeoff and landing were smooth?

At first glance, it was tempting to think the answer was 1.5 hills. Trig would take off just as the concavity of one hill was flipping from positive to negative. Then, Trig would land at a similar inflection point 1.5 hills away, as the concavity was flipping from negative to positive. Indeed, several readers believed this was the answer.

But there was a slight problem with this trajectory. Zooming in to Trig’s launch (or landing) revealed that he would have needed to pass *through the ground* in order to match the instantaneous slope of the ground at takeoff and landing, as shown below. Clearly, there was more to this problem than first met the eye, and closer inspection was necessary.

This puzzle could be reframed as two mathematical conditions. Given a sine wave, you were looking for a downward-facing parabola such that (1) it was tangent to the sine wave at two distinct locations, and (2) the parabola was always greater than the sine wave between those two points of tangency.

Before going further, it was also worth noting that the parabola representing the longest possible jump had an axis of symmetry that passed through a trough of the sine wave (as shown in the animations above). As solver Arvind Hariharan explained, if that weren’t the case, then Trig’s jump could be further extended by making the parabola symmetric.

Now for the calculus. We can represent the hills as *h*(x) = −cos(*x*). Meanwhile, we can represent the parabola as *p*(*x*) = −*ax*^{2} + *c*. (There was no linear term in the parabola due to its aforementioned symmetry.) If we suppose these two functions were equal at *x*_{0} — where Trig launched (and symmetrically landed) — then *h*(*x*_{0}) = *p*(*x*_{0}) and *h’*(*x*_{0}) = *p’*(*x*_{0}).

Mathematically, that meant −cos(*x*_{0}) = −*ax*_{0}^{2} + *c*, and sin(*x*_{0}) = −2*ax*_{0}. You could solve for *a* in the second equation: *a* = −sin(*x*_{0})/(2*x*_{0}). Plugging that value into the first equation gave *c* = −cos(*x*_{0}) − *x*_{0}sin(*x*_{0})/2.

Having solved for the parabola’s coefficients, the final step was to ensure that the parabola never fell below the sine wave between the points of tangency. There was a transition between the parabola dipping below the sine wave and always staying above — a transition that occurred when the *second derivatives* were also equal, i.e., *h’’*(*x*_{0}) = *p’’*(*x*_{0}), which simplified to tan(*x*_{0}) = *x*_{0}. Solving this transcendental equation gave you the largest possible value of *x*_{0}.

The smallest nonzero solution was when *x*_{0} was approximately 4.493. The total distance Trig jumped was actually twice this, or about 8.987. Finally, if the Sinusoidal Hills were represented by the function *h*(x) = −cos(*x*), then the width of each hill was 2𝜋. So in terms of hill-widths, Trig’s longest jump was approximately **1.43 hills**.

Here’s an infinite loop of that longest jump, courtesy of Maxim Wang:

Trig may not have flown, but this past week he certainly soared into our hearts.

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

]]>^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Keith Wynrow comes a clever little puzzle that stumped a few folks on social media:

On the Food Network’s latest game show, Cranberries or Bust, you have a choice between two doors: A and B. One door has a lifetime supply of cranberry sauce behind it, while the other door has absolutely nothing behind it. And boy, do you love cranberry sauce.

Of course, there’s a twist. The host presents you with a coin with two sides, marked A and B, which correspond to each door. The host tells you that the coin is weighted in favor of the cranberry door — without telling you which door that is — and that door’s letter will turn up 60 percent of the time. For example, if the sauce is behind door A, then the coin will turn up A 60 percent of the time and B the remaining 40 percent of the time.

You can flip the coin twice, after which you must make your selection. Assuming you optimize your strategy, what are your chances of choosing the door with the cranberry sauce?

*Extra credit:* Instead of two flips, what if you are allowed three or four flips? Now what are your chances of choosing the door with the cranberry sauce?

Trig the turkey cannot fly. And so, instead of flying, he decides to jump as far as he can with a running start up one of the famed Sinusoidal Hills, all of which are precisely the same size.

Trig knows that he *cannot* jump a horizontal distance of more than two hills. Also, he prefers a smooth takeoff and landing. That is, when he takes off from the ground and lands on it again, the slope of his parabolic trajectory through the air must perfectly match the instantaneous slope of the ground beneath him.

The animation below shows a jump where the takeoff and landing are precisely 1.2 hills apart.

What is the greatest horizontal distance Trig can jump, such that his takeoff and landing are smooth? Again, keep in mind that Trig cannot possibly jump a horizontal distance greater than two hills.

Congratulations to Masa Miyazaki of Tokyo, Japan, winner of last week’s Riddler Express.

Last week, you decided to make some new friends at your local gym, which was open daily from 5 p.m. to 8 p.m.

Some gym members attended very often. Others barely showed up at all. As a matter of fact, there was a uniform distribution for how often the different members were in the gym — from 0 percent of the time that the gym was open to 100 percent of the time.

As a new member, you planned to be in the gym 50 percent of the time that it was open. While working out, you decided to make friends with the first person you saw. (Aw!)

What was the probability that this person visited the gym more often than you?

If you were to pick a random person from the gym’s membership list, then there was a 50 percent chance that person visited the gym more often than you. However, you weren’t picking a person from any list — you were comparing yourself to the first person *you saw at the gym*. And that made a difference!

For example, consider members who were at the gym close to 0 percent of the time. You would have been *very* unlikely to see them there at all. Meanwhile, members who spent close to 100 percent of the time at the gym were far more likely to be seen, since they were simply around more often.

In fact, the relative likelihood of seeing each person was proportional to how often they were at the gym. As noted (and sketched below) by solver Sion Verschraege, that meant the probability distribution was triangular:

The smaller triangle on the left represented the collective probability of seeing someone who was in the gym less often than you. Its dimensions were scaled down by a factor of 2, which meant it had 25 percent of the area of the larger triangle. The remaining area — the probability the person you saw was in the gym more often than you — was therefore **75 percent**.

So the next time you’re at the gym and you feel like everyone there is fitter than you, worry not. Those people are just overrepresented in the probability distribution! Solver David Ding found that if you want to feel like the “median” (or should I say, “golden”) gym member, you would actually need to be there about 61.8 percent of the time.

Before moving on, it’s worth noting that the puzzle never explicitly said how many members the gym had. If you assumed there were many members, then the above reasoning was perfectly fine, and the answer was 75 percent. But if there were only two people (one of whom was you), there was only a 50 percent chance you were at the gym more often than your counterpart. For three, four or five members, the math got trickier, but solver Emily Boyajian was up to the challenge, finding that the exact probabilities rapidly converged to 75 percent.

Congratulations to Joseph Wetherell of San Diego, California, winner of last week’s Riddler Classic.

Four people were trying to escape from a room. Guards placed a hat on each person’s head, and each hat was one of three colors: red, yellow or blue. The four people were arranged at the vertices of a square, with an obstacle in the middle. Each person could see the hats on the heads of those on adjacent vertices of the square, but they could not see the hat of the person diagonally across from them. They also did not know the color of the hat on their own head.

Each person had to guess the color of the hat on their own head. If at least one person guessed correctly, they could all escape the room together. No communication was allowed once the hats were placed on their heads, but they could coordinate on a strategy beforehand. They also knew how they would be arranged in the square.

How could they be guaranteed to escape the room?

Like last week’s Express, this puzzle was a little ambiguous. Several readers asked whether the four people could hear each other’s guesses. If they could, and if they guessed at different times, then there was a straightforward strategy for success: They would agree on who should guess first, that person would say the color of one of their neighbors in the square, and then everyone else would say that same color. The neighbor whose color was said first was guaranteed to be correct. Huzzah!

But was it possible for them to escape *without* this simplifying assumption, or any others for that matter? Indeed it was.

The puzzle’s submitter, Eric Dallal, offered one such strategy. First, Eric assigned numerical values to each hat color: 0 for red, 1 for yellow and 2 for blue. Next, he labeled the four players in order around the square: p1, p2, p3 and p4, such that p1’s neighbors were p2 and p4, p2’s neighbors were p3 and p1, etc. He also labeled the *actual *number values of the hats on their heads: c1, c2, c3 and c4.

With these variables in place, a winning strategy was:

- p1 guesses c2 + c4, modulo 3
- p2 guesses -(c1 + c3), modulo 3
- p3 guesses c2 − c4, modulo 3
- p4 guesses c3 − c1, modulo 3

To ensure this works, check out the animation below, which loops through all 81 possible hat arrangements and verifies that at least one person guesses correctly for each arrangement.

As Eric noted, it was interesting that either one person would guess correctly or all four would do so — but never two or three.

In any case, the room has officially been escaped from!

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

]]>While the emotional toll of the pandemic may have completely wrecked our perception of time, periodic glances out our windows plus our knowledge of orbital dynamics confirm that Earth is continuing its annual revolution around our closest star. So even if we lose track of days or weeks, we can at least observe nature’s cycles and find comfort in the changing seasons together.

Or we could *if* we were able to stop arguing long enough to come to an agreement on when each season actually starts.

Prompted by a heated Slack debate among FiveThirtyEight staffers, we conducted an informal poll (i.e., a Google form) of our readers to find out what they think marks the seasons. The response was impassioned: In the first two hours that the form was online, we received more than 750 responses. By the end of the four-day survey, 2,742 readers had shared their thoughts. The results were mixed. While no overwhelming consensus was reached concerning a canonical calendar, certain dates and cultural signifiers pulled ahead within each season.

We started by asking readers to select from a list of options to indicate when each season begins.^{1} Overall, they were more open to considering qualitative metrics for determining the start of the transitional seasons, fall and spring, than of winter or summer.

A plurality of respondents (37 percent) said spring begins when buds appear on the trees, while 33 percent said it starts on the spring or vernal equinox, which falls on or around Mar. 20 in the Northern Hemisphere.

For fall, 26 percent said the season begins on the autumnal equinox, which lands on or around Sept. 22 in the Northern Hemisphere. But 25 percent said fall starts when the leaves start to change color. Another 25 percent said fall begins the day after Labor Day, which is obviously incorrect, Galen.

For summer, 40 percent of respondents — the largest plurality we had for any season — said the season begins the day after Memorial Day (in 2021, that was June 1). Twenty-three percent said summer starts when school lets out, and 18 percent said it begins on the summer solstice, which is on or around June 20 in the Northern Hemisphere.

And in case you had any doubt that America is first and foremost a capitalist society, the most popular response for when winter begins was Black Friday, with 23 percent selecting that option. (Of course, it is also the day after Thanksgiving, which could be seen as the pinnacle of harvest festivities.) Twenty-one percent said winter starts on the winter solstice (on or around Dec. 21 in the Northern Hemisphere), while 12 percent said it was when the first snow falls — the same percentage who said winter starts when the first snow sticks to the ground.

We also asked readers whether they associated the start of a season with a specific date rather than with a phenomenon (like leaves changing colors) or a moving holiday (like Black Friday) and, if so, to mark it on a calendar. We got a wide range of dates, but a plurality (or, in the case of summer and winter, a majority) of respondents landed on the first day of a month near the start of that season. For spring, 25 percent said it begins on March 1, while 51 percent said summer starts on June 1. Sept. 1 got the most votes for fall, with 31 percent, and Dec. 1 was deemed the start of winter by 51 percent of our survey-takers. These dates correlate with the meteorological seasons, which are based on the annual temperature cycle, not the astronomical seasons (think the solstices and equinoxes), which are based on Earth’s tilt and the sun’s alignment over the equator.

According to David Macauley, a philosophy and environmental studies professor at Penn State University and editor of “The Seasons: Philosophical, Literary and Environmental Perspectives,” the fact that there were both a variety in responses and a fair amount of consensus around particular dates or markers shows how deeply imbued the traditional notions of the seasons are in the American psyche, even with all our regional and cultural differences.

Historically, said Macauley, the seasons were closely tied to our agricultural cycle: Spring was for planting, summer was for growing, fall was for harvesting, and in winter we lived off the stores. But, over time, as we moved away from a largely agrarian society, those practical associations with the changing seasons were overtaken by cultural signifiers. “The astronomical stuff got connected with the cultural stuff — Halloween, Easter, basically everything Norman Rockwell and others popularized in American culture. And that became part of our national identity,” he said.

In addition to our multiple-choice options for when each season begins, we allowed respondents to write their own answers. Some of these responses were very specific. For example, one reader wrote that spring begins when their “malamute/husky begins to shed her heavy winter undercoat.” Another wrote that they know it’s summer “when the urge to grill and barbecue becomes too much.”

These open-ended responses revealed how big an influence religious and cultural calendars have on many people’s views of the seasons. Though we included a few religious markers (like Easter or Passover for spring), readers shared other religious or cultural dates that help them mark the changing of the seasons. Eight readers said they considered it winter on the first day of Advent, the Christian liturgical season leading up to Christmas. Two readers marked the start of winter with Samhain, an ancient Celtic pagan festival signaling the end of harvest season. One reader said they measured the beginning of summer and winter by the Chinese solar terms of lì xià and lì dōng, respectively.

Another major cultural influence that measures the seasons: sports. Multiple respondents marked the start of spring with the first day of spring training for Major League Baseball or with opening day. Others thought opening day heralded the start of summer. Four respondents said March Madness signaled the start of spring for them. Some marked the beginning of fall by the start of the MLB playoffs or when football season (both college and professional or, in one instance, pee-wee) begins. One fan noted that winter begins “when the Steelers are out of the playoffs.”

Other responses reveal regional differences in seasonal perception. Considering that most of the U.S. doesn’t experience the prototypical New England-style four seasons, it’s understandable that gauging the seasons would look different depending on where you live. Using ZIP codes our readers provided, we found that in the area around Phoenix, summer starts when there are “multiple 100-plus [degrees Fahrenheit] days in a row,” while in Tampa, Florida, summer equals hurricane season; in the Pacific Northwest, it’s when it stops raining all the time. And we’ll give you one guess where the reader who gave these responses lives: Winter starts with “the first five-car pileup on the pike due to ice.”^{2} Spring comes when “the snow pile melts at the mall.” Summer is marked by “Red Sox opening day,” which was April 1 this year. And it’s only fall once “Sam Adams OctoberFest is on the shelves.”

Here’s how our reader-generated calendar of the seasons breaks down. Take a look and perhaps you’ll find something other than politics to argue about with your family this Thanksgiving — truly something to be thankful for.

]]>^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Michael Dell’Amico comes a riddle requiring some mental gymnastics:

After realizing that your friends on μετα (The Riddler Social Network) are more popular than you, you decide to make some new friends at your local gym, which is open daily from 5 p.m. to 8 p.m.

Some gym members attend very often. Others barely show up at all. As a matter of fact, there’s a uniform distribution for how often the different members are in the gym — from 0 percent of the time that the gym is open to 100 percent of the time.

As a new member, you plan to be in the gym 50 percent of the time that it’s open. While working out, you decide to make friends with the first person you see. (Aw!)

What is the probability that this person visits the gym more often than you?

From Eric Dallal comes a variant of an oldie but a goodie — deducing the color of the hat on your head!

Four people are trying to escape from a room. Guards have placed a hat on each person’s head, and each hat is one of three colors: red, yellow or blue. The four people are arranged at the vertices of a square, with an obstacle in the middle. Each person can see the hats on the heads of those on adjacent vertices of the square, but they cannot see the hat of the person diagonally across from them. They also do not know the color of the hat on their own head.

Each person must guess the color of the hat on their own head. If at least one person guesses correctly, they can all escape the room together. No communication is allowed once the hats are placed on their heads, but they can coordinate on a strategy beforehand. They also know how they will be arranged in the circle.

How can they be guaranteed to escape the room?

Congratulations to Donald Adamek of Canton, Michigan, winner of last week’s Riddler Express.

Last week, I had three dice (d4, d6, d8) on my desk that I fiddled with while working, much to the chagrin of my co-workers. For the uninitiated, the d4 is a tetrahedron that is equally likely to land on any of its four faces (numbered 1 through 4), the d6 is a cube that is equally likely to land on any of its six faces (numbered 1 through 6), and the d8 is an octahedron that is equally likely to land on any of its eight faces (numbered 1 through 8).

I was playing a game in which I rolled all three dice in “numerical” order: d4, then d6 and then d8. I won this game when the three rolls formed a strictly increasing sequence (such as 2-4-7, but *not* 2-4-4). What was my probability of winning?

There were four possible outcomes for the first roll, six outcomes for the second and eight for the third. All together, that meant there were 4·6·8, or 192, equally likely cases to consider. So then, for how many of these 192 cases did the rolls strictly increase?

Many solvers, like Angelos Tzelepis, figured this out by writing code to go through all the cases. Others, like Paige Kester (winner of last week’s Classic!), used a spreadsheet. In addition to these approaches, there were many ways to solve this by hand.

One such way was to first assume all three dice were d8s. Had that been the case, then each way of choosing three distinct values from 1 to 8 would have corresponded to a different possible sequence. And there were 8 choose 3 (or 56) ways to choose those three values.

But you didn’t really roll three d8s — you rolled a d4, a d6 and then a d8. So, out of those 56 sequences, four of them had an impossible first roll of 5 or 6: 5-6-7, 5-6-8, 5-7-8 and 6-7-8. And another four of them had an impossible second roll of 7: 1-7-8, 2-7-8, 3-7-8 and 4-7-8. Subtracting these eight impossible sequences left 48 that were indeed possible. The probability of such a sequence was therefore 48/192, or **1/4**.

For extra credit, instead of three dice, I now had six dice: d4, d6, d8, d10, d12 and d20. If I rolled all six dice in “numerical” order, what was the probability I’d get a strictly increasing sequence?

This time around, there were 4·6·8·10·12·20, or 460,800, equally likely cases, which meant using a computer was definitely a good idea. Solver Chris Sears tallied up the number of strictly increasing sequences on a calculator:

Of the 460,800 cases, 5,434 were strictly increasing. That meant the probability was 5,434/460,800, or **2,717/230,400** — just a shade over 1 percent.

This was a little surprising (to me at least). Since I was rolling dice with larger and larger possible values, it *felt* like the probability of an increasing sequence should have been more likely.

Congratulations to Paige Kester of Southlake, Texas, winner of last week’s Riddler Classic.

Last week, you played a game against a magic genie. You had a stick of length 1, while the genie had another stick behind his back, with a random length between 0 and 1 (chosen uniformly over that interval).

Next, you had to break your stick into two pieces and present one of those pieces to the genie. If that piece was longer than the genie’s hidden stick, then you won a prize of $1 million times the length of your remaining piece. For example, if you presented to the genie a length of 0.4, and that was longer than the genie’s stick, then you would win $1 million times 0.6, or $600,000. However, if the genie’s stick was longer, then you won nothing.

But then you had a thought. You asked the genie if you could have more than one turn. For example, if you presented the genie with a length of 0.4, but the genie’s stick was longer, could you break off a piece of the remaining length of 0.6 — say, a length of 0.5 — and then present *that* to the genie? To keep things fair, your winnings would still be proportional to the leftover length. So had the genie’s length indeed been between 0.4 and 0.5, your first and second guesses, then the remaining length would have been 0.1, and you would have won $100,000.

The genie didn’t think any of this really mattered and said you could have as many turns as you desired. If your goal was to maximize your expected winnings, what would have been your strategy? And how much money would you win on average?

First off, if you only had one chance to present a length to the genie, why would you have broken the stick in half? (Clearly, you knew something.) Suppose you presented a length *A* to the genie. The probability that it was greater than the genie’s random stick was also *A*, and your winnings were proportional to the remaining length, 1−*A*. That meant your average winnings were proportional to *A*·(1−*A*), a quadratic expression whose maximum occurred when *A* was 0.5. In other words, your best move was indeed to **split the stick in half**! Your average winnings were then **$250,000**.

Now, what if you had *two* opportunities to present lengths to the genie? Suppose your first length was *A* and your second length was *B*. For this to work, *B* had to be greater than *A* — otherwise, you were just being masochistic. Again, your expected winnings from the first length were *A*·(1−*A*). Meanwhile, the second length would win only when the genie’s length was between *A* and *B*, which occurred with probability *B*−*A*. And your winnings were proportional to 1−*A*−*B*, since that was all you had left after the two opportunities.

Putting all this together, your average winnings were proportional to the expression *A*·(1−*A*) + (*B*−*A*)·(1−*A*−*B*). Multiplying this out, you get *A* − *A*^{2} + *B* − *AB* − *B*^{2} − *A* + *A*^{2} + *AB*. Amazingly, almost all of the terms cancel, leaving you with just *B*−*B*^{2}, or *B*·(1−*B*). This was the very same quadratic expression from before, now with *B* instead of *A*. So with two opportunities, you could first present any length between 0 and 0.5 (not including those two extremes). It really didn’t matter what the first length was. Then, if you lost, the second length should have been 0.5. Again, your expected winnings were $250,000.

Many solvers, including Emily Boyajian and Arvind Hariharan, went on to demonstrate that with three or more presentations to the genie, your expected winnings always maxed out at $250,000. When all was said and done, you couldn’t really stick it to the genie, and you might as well have presented a length of 0.5 on your first try.

I will also note that Riddler Nation proposed several interesting extension ideas for this puzzle. What if you could glue the presented pieces back together? What if the genie had a new random stick length every time you broke off a piece of your own stick? Sounds like the makings of a future riddle in which you’d get another shot at sticking it to that genie.

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

]]>Gay people in America overwhelmingly vote for Democrats and have done so for decades. But they aren’t a monolithic group. In our second episode of “Political Outliers,” meet two gay Republicans — one in his 20s and one in his 60s — who have been lifelong conservatives. Hear how they developed their political ideologies and which issues motivate them to stay loyal to the GOP.

]]>^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Adam Milligan comes a puzzle about dice, dungeons and dragons (but mostly about dice):

I have three dice (d4, d6, d8) on my desk that I fiddle with while working, much to the chagrin of my co-workers. For the uninitiated, the d4 is a tetrahedron that is equally likely to land on any of its four faces (numbered 1 through 4), the d6 is a cube that is equally likely to land on any of its six faces (numbered 1 through 6), and the d8 is an octahedron that is equally likely to land on any of its eight faces (numbered 1 through 8).

I like to play a game in which I roll all three dice in “numerical” order: d4, then d6 and then d8. I win this game when the three rolls form a strictly increasing sequence (such as 2-4-7, but *not* 2-4-4). What is my probability of winning?

*Extra credit:* Instead of three dice, I now have six dice: d4, d6, d8, d10, d12 and d20. If I roll all six dice in “numerical” order, what is the probability I’ll get a strictly increasing sequence?

One day you stumble across a magic genie, who says that if you play a simple game with him, you could win fabulous riches. You take the genie up on his offer, and he hands you a stick of length 1. He says that behind his back is another stick, with a random length between 0 and 1 (chosen uniformly over that interval).

Next, you must break your stick into two pieces and present one of those pieces to the genie. If that piece is longer than the genie’s hidden stick, then you win a prize of $1 million times the length of your remaining piece. For example, if you present to the genie a length of 0.4, and that’s longer than the genie’s stick, then you win $1 million times 0.6, or $600,000. However, if the genie’s stick is longer, then you win nothing.

Being a regular reader of The Riddler, you crunch some numbers and prepare to break your stick in half. But then you have a thought. You ask the genie if you can have more than one turn. For example, if you present the genie with a length of 0.4, but the genie’s stick is longer, can you break off a piece of the remaining length of 0.6 — say, a length of 0.5 — and then present *that* to the genie? To keep things fair, your winnings will still be proportional to the leftover length. So had the genie’s length indeed been between 0.4 and 0.5, your first and second guesses, then the remaining length would have been 0.1, and you would have won $100,000.

The genie doesn’t think any of this really matters and says you can have as many turns as you desire. If your goal is to maximize your expected winnings, what will your strategy be? And how much money would you expect to win on average?

Congratulations to Peter Ji of Madison, Wisconsin, winner of last week’s Riddler Express.

Last week, The Riddler Social Network was being rebranded as μετα.

A group of 101 people joined μετα, and each person had a random, 50 percent chance of being friends with each of the other 100 people. Friendship was a symmetric relationship on μετα, so if you were friends with me, then I was also friends with you.

I picked a random person among the 101 by the name of Marcia. On average, how many friends would you have expected each of Marcia’s friends to have?

Before answering that question, let’s take a closer look at Marcia. How many friends would *she* have on average? Well, there were 100 other people on the network, and she had a 50 percent chance of being friends with each of them. So, on average, Marcia would have 50 friends.

But as most readers realized, that wasn’t the case for Marcia’s friends. Why? Because you were given one additional piece of information: By virtue of being friends with Marcia, each of these people had one *guaranteed* friend. If Randall was one of Marcia’s friends, then Randall still had a 50 percent chance of being friends with each of the other 99 people on the network. That meant his average number of friends was 1 + 99/2, or **50.5**.

Now there was a small chance (one in 2^{100}) that Marcia had *no* friends whatsoever, in which case it was meaningless to ask about her friends’ friends. A few sharp readers pointed out that the puzzle was therefore poorly defined. Based on the puzzle’s wording, I think it was fair to assume that Marcia had at least one friend. Nevertheless, I treated answers of “undefined” as correct as well.

A few solvers — including Matt Luedtke, Sanandan Swaminathan, Jeremy White and Emily Boyajian — connected this riddle to the well-known friendship paradox. This paradox says that in *any* social network, one’s friends will have more friends than you, on average. Apparently, this also applies to the probabilistic social network known as μετα.

Poke!

Congratulations to David Nash of Portland, Oregon, winner of last week’s Riddler Classic.

The sum of the factors of 36 — including 36 itself — is 91. Coincidentally, 36 inches rounded to the nearest centimeter is … 91 centimeters!

Last week, you had to find another whole number like 36, where you could “compute” the sum of its factors by converting it from inches to centimeters.

In 1959, an inch was defined as being exactly 2.54 cm. That meant if a number *N* had a sum of factors 𝝈(*N*), you wanted 2.54*N* to be very close to 𝝈(*N*). More specifically, to get the rounding right, you wanted the following inequality to be true: 𝝈(*N*) − 0.5 ≤ 2.54*N* < 𝝈(*N*) + 0.5.

Now you might have heard of a perfect number before — that’s a number whose sum of factors (other than the number itself) equals the number. In other words, the sum of all its factors (including the numbers itself) is *twice* the number. Meanwhile, numbers whose factors add up to *more* than twice the number are called abundant numbers. Because 2.54 is greater than 2, that meant the solution had to be an abundant number. Small abundant numbers tend to be the products of small primes, so they have many factors to add up despite their relatively small magnitude.

Nevertheless, many solvers, like Paulina Leperi and Razeen Hossain, wrote code to exhaustively search every number until a solution was found. This first such solution was **378**, which was equal to 2·3^{3}·7. Its factors were 1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 63, 126, 189 and, of course, 378. The sum of these factors was 960, whereas 2.54 times 378 was equal to 960.12, which indeed rounded down to 960.

For extra credit, you had to find a *third* whole number with this property, beyond 36 and 378. Many solvers continued running their code until they found another solution. Meanwhile, solver Austin Shapiro found such a solution analytically. First, Austin noted that for any relatively prime *X* and *Y*, it was always true that 𝝈(*X*)𝝈(*Y*) = 𝝈(*XY*), which further meant that 𝝈(*X*)/*X* · 𝝈(*Y*)/*Y* = 𝝈(*XY*)/(*XY*).

Here, we wanted 𝝈(*N*)/*N* to be approximately 2.54, or 127/50. That numerator of 127 was one less than a power of 2 (and also a Mersenne prime!), which meant it was also the sum of factors for a smaller power of 2 — in this case, 2^{6}, or 64. So Austin rewrote 127/50 as 127/64·32/25. Next, in a happy coincidence, Austin recognized that 32 = 𝝈(31), and that 31 = 𝝈(25). Putting this all together, 127/50 = 𝝈(64)/64 · 𝝈(31)/31 · 𝝈(25)/25. And because 64, 31 and 25 were all relatively prime, you could combine these using the aforementioned multiplicative property to get 127/50 = 𝝈(49,600)/49,600. Sure enough, **49,600** was the next smallest solution. Because of the exact nature of the math here, 49,600 inches was *precisely *125,984 cm, just as the sum of the factors of 49,600 was *precisely* 125,984.

Solvers Elliot Brown, John Adair, Brian Casaday, Eoin O’Brien and Neville Fogarty looked for solutions up through (and past) 1 million but did not find any others. Surely, more such numbers exist. And I bet that if we had a sufficiently long metric ruler, we’d find them.

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

]]>How many Americans really like pumpkin spice lattes? And how many celebrate Halloween? In our new game show, The United Stats Of America, host Galen Druke quizzes people in New York’s Central Park about what Americans think of fall.

]]>Can white voters who back a Black candidate still hold racist beliefs and views?

That question has come to the fore in the wake of Glenn Youngkin’s gubernatorial victory in the blueish state of Virginia. Conservatives were quick to counter claims that Youngkin’s win represented the effectiveness of stoking racial fears with results from Virginia’s down-ballot election for lieutenant governor — a contest where the Republican candidate, Winsome Sears, made history by becoming the first Black woman elected to statewide office in Virginia. The Wall Street Journal’s editorial board, for example, emphatically mocked the notion that “voters called white supremacists elected a Black Lt. Gov.” Conservative commentators on Fox News and Twitter, including Sears herself, also used the historic victory as an ostensible shield against accusations of Republican racism.

But supporting a Black candidate hardly precludes voters from harboring racist beliefs and motivations. Republicans are increasingly more likely than Democrats to hold prejudiced views of minorities, so Black Republicans like Sears often draw especially strong support from white Americans with otherwise anti-Black views simply because they draw most of their support from Republican voters.

A clear example of this was in the 2016 Republican presidential primary, when Ben Carson made a bid to become the GOP’s first African American presidential nominee. Support for Carson was positively correlated with the belief that Black Americans have too much influence on U.S. politics, according to data from Washington University in St. Louis’s American Panel Survey (TAPS) in late 2015:

Whites who thought African Americans had “far too little” influence disliked Carson and preferred Hillary Clinton by 60 percentage points in a hypothetical general election matchup. Meanwhile, Carson was very popular among whites who were most concerned about African Americans having “too much” influence in politics. So much so that whites who thought African Americans have “far too much” influence preferred Carson to Clinton by 45 points.

Again, much of that relationship is down to partisanship — Republicans are more likely to hold prejudiced views and also more likely to support a Republican candidate. But that’s the point: For many white GOP voters, anti-Black views don’t seem to get in the way of supporting a Black Republican.

You can see a similar pattern in the January 2016 American National Election Studies Pilot Study. Carson received more favorable evaluations among the sizable minority (40 percent) of overtly prejudiced whites who agreed with the racist stereotype that “most African Americans are more violent than most whites.” This group rated Carson significantly more favorably on a 0-100 scale than the white moderate Republican presidential candidate, Jeb Bush (52 to 39, respectively). Then-candidate Donald Trump was the only politician in the survey who was rated higher than Carson among overtly prejudiced whites.^{1}

The contrast between how prejudiced whites rated Carson and Obama is rather revealing, as well. The sharp negative relationship between support for Obama and the endorsement of anti-Black stereotypes is consistent with several studies showing that prejudice was an unusually strong predictor of opposition to Obama from the 2008 election through the end of his presidency. These patterns also fit well with other political science research showing that racially prejudiced whites tend to be more opposed to Black Democrats than to white Democrats.

To make sense of why racially prejudiced white Americans are willing to support some Black candidates, it is worth considering why they so strongly oppose Black Democrats in the first place. Given the racialized nature of the two-party system in the United States, most Black political candidates are Democrats who embrace liberal positions on issues of race and justice. When asked whether they would support such a candidate, research shows that racially prejudiced white voters worry that these candidates will represent the interests of Black Americans, both because of a shared African American identity and because Democrats are perceived as the party more supportive of Black interests. So, it makes sense that racially resentful white Americans oppose candidates like Obama, as his racial identity *and* partisanship signaled to voters that he was more supportive of Black interests than prior presidents.

Put another way: Racially prejudiced white voters are not opposed to Black candidates simply because they are Black, but because they believe that most Black candidates will fight for “those people” and not “people like us.”

Black Republicans, on the other hand, are perceived differently by racially prejudiced white Americans. Their embrace of the Republican Party and its conservative ideology help assure racially prejudiced whites that, unlike Black Democrats, they are not in the business of carrying water for their own racial group. Instead, they are viewed as distinct from other Black elites. If Blackness is viewed as intertwined with a kind of racial liberalism that is antagonistic to the interests of white Americans, Black Republicans’ partisan and ideological commitments allay concerns that they are for “them,” not “us.”

This argument is buttressed by more recent scholarship in political science, which has found that Black candidates who embrace a “bootstrap” ideology — an ideology that focuses on individual versus structural explanations of inequality — are more positively evaluated by racially prejudiced whites relative to their white competitors. Explaining this finding, the authors note that racially prejudiced white voters “might find black Republicans delivering an individualism message more favorable than they might find other candidates delivering a similar message precisely because the aesthetic character and the partisan affiliation of the messenger contradict racial and political expectations.” LaFleur Stephens-Dougan, a professor of political science at Princeton University, similarly shows in her book “Race to the Bottom” that racially resentful whites respond well to Black candidates who take stances against the expected positions of their racial group — a phenomenon she calls “racial distancing.”

Finally, voting for Black Republicans may also be especially appealing to racially prejudiced whites because it assuages concerns of being seen as racist by enabling them to say, in essence, “I can’t be racist! I voted for a Black candidate!” Psychologists call this “moral credentialing,” and there’s even some evidence that voters who expressed support for Obama shortly after the 2008 election felt more justified in favoring white Americans over Black Americans. Electing a Black Republican like Sears, who railed against critical race theory during the run-up to the election and supports voting restrictions that adversely affect racial minorities, is similarly used as a symbolic shield by the entire party from inevitable charges of championing racist policies. As we mentioned earlier, conservative media outlets and politicians are already weaponizing her victory against anyone who would dare suggest so.

But, of course, the role race and racism play in American politics is much more nuanced than those simplistic defenses suggest. When racially prejudiced whites oppose Black candidates, it’s not just because of the candidate’s skin color. It’s also because they perceive (sometimes wrongly) that Black candidates, especially Black Democrats, have ideological commitments that are at odds with the interests of white Americans. Likewise, when racially prejudiced whites support Black Republicans, it’s hardly the case that they’ve become progressive on race. Racially prejudiced whites did not vote for Sears because they appreciated her attachment and commitment to Black people. They were willing to support her because they discounted it.

Race, after all, is a social construct. It has meaning because we imbue it with meaning. Racially prejudiced whites are not hostile to Blackness, per se. They are hostile to a particular manifestation of Blackness — one that reflects a commitment to racial justice and the advancement of the group’s collective goals. Racially prejudiced whites are not bothered much by a manifestation of Blackness that is ideologically consistent with their own identities and attitudes. Why would they be? Sears’s conservative politics don’t threaten the racial hierarchy, and her candidacy provides cover for a party that’s often antagonistic to racial minorities. For racially prejudiced whites, the real question is what is there *not* to love about Black politicians like Sears?

After an excruciating wait for many families, children ages 5 and older can now be vaccinated against COVID-19. The decision marks a turning point in the pandemic for millions of Americans as they can worry a little less and live a little more.

For more than a year and a half we’ve been told that the worries of adults have changed the lives of kids in all sorts of deleterious ways. And so, on the eve of the Centers for Disease Control and Prevention’s decision to approve vaccinations for 5- to 11-year-olds, we wanted to know just how kids (and their parents) were doing. Are kids and teenagers stressed, depressed and scared? How do they see themselves and their families changed by the pandemic? And how has their behavior changed to help reduce the spread of COVID-19 in their daily life?

To help answer those questions, we partnered with our friends at Ipsos, the polling firm, to work on a poll that asked parents and kids how they’re doing. Between Oct. 25 and Nov. 2, FiveThirtyEight and Ipsos used Ipsos’ KnowledgePanel to survey 689 kids ages 5 to 11, 572 kids ages 12 to 17 (let’s call them “teens” for our purposes), and more than 1,500 of their parents.^{1} The answers were surprising! We found a population of kids who appear to be pretty resilient, even in the face of loneliness and isolation, and who are forming strong relationships with their parents and families. Any one kid who’s struggling because of the pandemic is a source of concern. Overall, though, America’s kids aren’t as downtrodden as they’re often made out to be.

You can see all the poll results here. These were our highlights:

The kids are all right — or, at least, they report being a lot more all right than we were expecting. The COVID-19 pandemic drastically changed the lives of children and teens, cutting them off from friends, activities and adults other than their parents. The combination of limited social contact, disrupted schooling and fear of the virus itself led many to anticipate a major mental health crisis looming in youth. A poll conducted this past spring by the Children’s Hospital of Chicago found 65 percent of parents believed the mental health consequences for kids will be worse than for adults. So we were surprised to find that the kids who participated in our survey almost universally don’t describe themselves as struggling.

For example, we asked teens, ages 12 to 17, about the current state of their mental health and a vast majority had good things to say.

Answers to questions about mental health and share of respondents from FiveThirtyEight/Ipsos poll on COVID-19, by cohort

How would you describe your mental health? | parents | teenagers |
---|---|---|

Very good or somewhat good | 88% | 89% |

Very poor or somewhat poor | 10 | 8 |

No response | 1 | 2 |

How concerned are you about your kid’s/your mental health? | parents | teenagers |

Very or somewhat concerned | 34% | 17% |

A little or not at all concerned | 65 | 80 |

No response | 1 | 2 |

In fact, their parents were a little more likely to report poor mental health — 11 percent of parents to 8 percent of the teens. FiveThirtyEight and Ipsos asked kids aged 5 to 11 a slightly different question — “How do you feel right now?” — and 96 percent of them said they felt very or somewhat good.

“That’s great. It’s really good news,” said Anna Gassman-Pines, a professor of psychology and public policy at Duke University, when we told her the results of the survey. “There’s a lot of good developmental science behind the idea that ultimately kids will be resilient.”

The good vibes extended to more specific parts of their lives, too. Overall, teens were less likely than their parents to be worried about their ability to perform in school, less worried about their ability to participate in activities, less worried about catching COVID-19, and much less worried about their mental health. While 34 percent of parents were concerned about their kids’ mental health, about half as many teens — 18 percent — found their own mental health concerning. More than half said their mental health hadn’t changed at all since the start of the pandemic — 24 percent said it had gotten worse and 20 percent said their mental health was now better.

Meanwhile, while parents were busy worrying, teens were reporting … having a good time with them. Twenty-seven percent of teens reported that the pandemic led to improvements in their home lives and 30 percent reported having better relationships with their parent(s). And more than 90 percent of kids of all ages felt that their parents had done a good job keeping them safe during the pandemic.

Gassman-Pines’s own research has focused on the impacts COVID-19 had on low-income families, and particularly the families of people who worked in the service industry during the pandemic. She was pleasantly surprised to find that the good news remained largely good even when we broke our respondents down by income and race. There were some differences, though. Black parents were more likely than white parents to report that they were concerned about their personal finances, and low-income parents reported worse finances and physical health than higher-income parents. And teens and children of color were more scared of COVID-19 and of being lonely. But the shares of teens who felt good about their home lives, relationship with parents, mental and physical health, and current loneliness were mostly consistent across race and income.^{2} Even when parents were having a harder time, and kids were more worried, the outcomes for those kids weren’t particularly worse. It’s possible, Gassman-Pines said, that this is an indicator of the positive, stabilizing impacts of stimulus and child tax credit payments — something that’s been observed in other research.

We were also surprised to see high rates of reported mask usage and other behaviors to keep people safe from COVID-19. Eighty percent of teens and younger kids reported that their teachers were consistently or always wearing masks at school, 70 percent of both groups reported that their friends were consistently doing the same, and more than 70 percent said other kids at school were. The kids and teens in our survey also reported that they, themselves, were mask wearing at rates higher than 70 percent.

Predictably, there were geographic differences in this data. Teens in the Northeast were most likely to say their teachers always wear masks — more than 75 percent of them. Those rates were less than 50 percent in the Midwest and South. This tracks with regulations. Schools in the Northeast have been more likely to implement and maintain masking mandates than schools in the South. Cases of COVID-19 in children have followed the same patterns, with higher rates in the Southeast and South than in the Northeast.

There were also differences in behavior by race. White kids were more likely to report having indoor playdates in the last month — a particularly interesting result given that white kids were also the ones least likely to be worried about contracting COVID-19 themselves.

While kids and teens have generally positive things to report about their overall mental health and their relationships at home, the pandemic does seem to have had an impact on how young people interact with each other and how they feel about those changes. Indoor activities are still not the norm for most kids. Fewer than 50 percent of both teens and younger children had had an indoor playdate in the last month; fewer than 40 percent had attended an indoor party; and only 41 percent of teens and 31 percent of younger children had attended an indoor club meeting or played sports inside. In contrast, 64 percent of teens and 77 percent of younger kids had played with their friends outside, an activity that is significantly less likely to lead to COVID-19 transmission.

And we see some big contrasts in how kids and teens think about their friends versus their family. While 30 percent of teens reported improved relationships with parents since COVID-19 began and only 7 percent reported those relationships worsening, the changes in relationships with friends were a lot more split, with 27 percent reporting that their friendships and social lives had gotten better and 29 percent reporting they had gotten worse.

Likewise, 26 percent of teens reported feeling less connected and more lonely since the start of the pandemic, while 25 percent reported feeling more connected and less lonely. While the raw numbers were small, we did see correlations that connected a strong sense of feeling lonely to a sense that one’s mental health had gotten worse.

Parents have spent a lot more time talking about the vaccine to other adults than teens have spent discussing it with their peers — 50 percent to 38 percent. But that doesn’t mean the teens haven’t formed opinions on the vaccine and how it should be rolled out. Those opinions tend to track with the opinions of their parents. For example, 60 percent of parents and 61 percent of teens agreed that schools should mandate vaccines for adults in schools, including teachers and administrators. And a smaller majority agreed that schools should mandate vaccines for students 12 and older: 54 percent of parents and 57 percent of teens.

About 50 percent of parents in the FiveThirtyEight/Ipsos poll were interested in vaccinating their under-12 children, depending on the age of the child, and between 50 and 60 percent of the teens we interviewed had already received at least one dose of a COVID-19 vaccine. (The survey was conducted before the CDC approved vaccinations for 5- to 11-year-olds.)

We found strong correlations between parents’ attitudes about personal vaccination and the attitudes of their children. If you focus just on teens whose parents had already gotten vaccinated or were very likely to, 79 percent of them said they were already vaccinated or were very likely to be soon — compared to just 9 percent of teens whose parents weren’t vaccinated and weren’t likely to be. We found similar patterns with younger kids: 46 percent of those whose parents were vaccinated said they were (or were going to be soon), compared to just 1 percent of kids whose parents weren’t vaccinated and weren’t likely to be.

Teens’ and kids’ attitudes about whether they should get the vaccine themselves were also highly correlated with parents’ attitudes. Among teens and kids who were unlikely to get vaccinated, parent opinions mattered significantly more than whether peers were getting vaccinated or not.

So, to summarize, kids are relatively happy and listening to their parents. Just as we always assumed.

*Additional reporting by Anna Rothschild.*