FiveThirtyEight - Life

Can You Stack The Deck With Suits?

Fri, 17 Mar 2023 14:00:00 +0200

Can You Stack The Deck With Suits?

Illustration by Guillaume Kurkdjian

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,¹ and you may get a shoutout in 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 or send me an email.

Riddler Express

From composer Grant Harville comes a musical mystery:

Grant is writing a musical composition. At one point in the piece, there’s an improvisational passage where musicians are instructed to repeatedly play a sequence of eight notes, which we can label as 1 through 8. The shortest such sequence is 12345678.

However, musicians can also revert to previous notes, replaying certain subsequences for additional flair. More specifically:

They must always play the next note (i.e., adding 1 to the previous note), unless they revert to a previous note.
At no point can they play the same note twice in a row.
Notes 1 and 8 — that is, the first and last notes — can be played only once.
They can only revert to a given note at most once.
Once they have reverted to a specific note, they cannot then revert to an earlier note in the sequence.

That’s a whole bunch of rules! To make this clearer, it may be helpful to see some examples. The following are examples of valid sequences:

12345678 (This is the shortest sequence.)
1234567-234567-34567-4567-567-678 (This is the longest sequence.)
1234-234567-678
1234567-345-4567-5678
123-234567-3456-45678

Meanwhile, here are examples of invalid sequences, for various reasons:

1245678 (This skips the 3.)
12437568 (Some notes are out of order.)
12345-34678 (This skips a note within a reversion, even though that note occurs earlier.)
1234-3456-345678 (This reverts to the same note twice.)
12345-456-2345678 (This reverts to an earlier note after reverting to a later one.)
12345-567-678 (This repeats a note twice in a row.)
123-1234567-5678 (This repeats note 1.)
1234-23456-5678-78 (This repeats note 8.)

How many different sequences of the eight notes are possible under these conditions?

Submit your answer

Riddler Classic

From Brett Humphreys comes a card-counting conundrum:

Brett plays poker with a large group of friends. With so many friends playing at the same time, Brett needs more than the 52 cards in a standard deck. This got Brett and his friends wondering about a deck with more than four suits.

Suppose you have a deck with more than four suits, but still 13 cards per suit. And further suppose that you’re playing a game of five-card stud — that is, each participant is dealt five cards that they can’t trade.

As the number of suits increases, the probability of each hand changes. With four suits, a straight is more likely than a full house (a three-of-a-kind and a different two-of-a-kind in the same hand). How many suits would the deck need so that a straight (not including a straight flush) is less likely than a full house?

Extra credit: Instead of five-card stud, suppose you’re playing seven-card stud. You are dealt seven cards, among which you pick the best five-card hand. How many suits would the deck need so that a straight (not including a straight flush) is less likely than a full house?

Submit your answer

Solution to the last Riddler Express

Congratulations to Kris Adams of Bartonville, Illinois, winner of last week’s Riddler Express.

Last week, Bill had four opaque bags, each of which had three marbles inside. Three of the bags contained two white marbles and one red marble, while the last bag contained three white marbles. The bags were otherwise indistinguishable.

Ted watched as Bill randomly selected a bag and reached in without looking to grab two marbles without replacement. It so happened that both marbles were white. Bill was about to reach in and grab the last marble in that bag.

What was the probability that this marble was red?

As with other famous riddles related to conditional probability (like Monty Hall and the two child problem), your intuition could lead you astray here.

Some readers observed that, after removing two white marbles from among the 12 total marbles, Bill was left with three red marbles out of a total of 10. Therefore, the probability the last marble was red should have been 3/10. However, this was not the right answer.

Other readers argued that because all three bags had two white marbles, drawing two white marbles offered no new information about which bag Bill had selected. Because three of the four bags had a red marble, the probability the last marble was red should have been 3/4. However, this too was not the right answer.

To see why, suppose Ted randomly selected two balls from a bag with a red marble. In this case, he had a two-thirds chance of picking one red and one white marble, as well as a one-third chance of picking two white marbles. But for the remaining bag with three white marbles, Bill was guaranteed to choose two white marbles.

That meant Bill was three times more likely to pick two white marbles from the bag without a red marble than he was from each bag with a red marble. At the same time, there were three times as many bags with a red marble as there were bags without a red marble. And so the final marble was equally likely to be red or white; the probability that was red was 50 percent.

If you’re still not convinced, you can simulate this for yourself at home. Set up the four bags, pick a random bag and then draw two marbles. But remember, if you happen to draw one red and one white marble, then you should discard that simulation. Only when you draw two white marbles is there a 50 percent chance that the last marble is white.

Solution to the last Riddler Classic

Congratulations to Tom Singer of Melbourne, Florida, winner of last week’s Riddler Classic.

Last week you decided to set up a marble race course. No Teflon was spared, resulting in a track that was effectively frictionless.

The start and end of the track were 1 meter apart, and both positions were 10 centimeters off the floor. It was up to you to design a speedy track. But the track always had to be at floor level or higher.

What was the fastest track you could design, and how long would it have taken the marble to complete the course?

From an introductory physics course, you know that the lower down the marble was, the less potential energy it had and the more kinetic energy it had, and thus the faster it moved. So one track design was to have the marble go straight down, at which point an infinitesimal lip redirected the marble horizontally along the floor. Once it was directly below the finish line, another infinitesimal lip redirected the marble straight up.

How long did it take for the marble to traverse such a course? If the initial descent (and, symmetrically, the final ascent) took t seconds, then you knew h = gt²/2, where h was the initial height of the marble (0.1 meters) and g was the acceleration due to gravity, approximately 9.8 m/s² at Earth’s surface. Solving this equation gave you t = 1/7 s. Meanwhile, the marble’s velocity along the floor was equal to the square root of 2gh, or 1.4 m/s. Traversing the floor at this speed took 5/7 s. Adding up all these times meant the marble reached the finish after precisely 1 second. But it was possible to get there even faster.

Last week I had said this puzzle was likely to “break your brachistochrone,” with that last word being the operative one. A brachistochrone is a path that takes an object from one place to another in minimal time using the force of gravity. But this exact path remained a mystery until it was solved by several big names in mathematics in the late 17th century. The brachistochrone turned out to be a segment of a cycloid (the path outlined by a single point on a rolling circle). Having the marble travel along a cycloid is faster than the aforementioned straight drop down. While a straight drop gets the marble to its maximum speed faster, the cycloid reduces the overall time by moving the marble closer to its destination as it accelerates.

Of course, this being The Riddler, the optimal path was not merely a cycloid. To travel a distance of 1 meter without any net change in elevation, a cycloid would have to dip 1/𝜋 m, or about 31.83 cm. This was impossible, as the puzzle stated the marble could not pass through the floorboards 10 cm below the starting point.

The solution was therefore to get the marble reasonably far along via a downward cycloid, then travel horizontally at high speed and finally return back up to the finish line along an upward cycloid. As noted by solvers Paige Kester and Laurent Lessard, having half a complete cycloid period (also known as a tautochrone) on either end, as shown below, did the trick. Solver Starvind even animated the marble along the track.

The time to traverse either tautochrone was 𝜋√(0.05/9.8), or about 0.2244 s. Once again, the marble again traveled at a speed of 1.4 m/s along the flat portion of the track, which was now 1−𝜋/10 m long. Adding up the times for the flat portion and the two tautochrones gave a total time of approximately 0.9387 s, which was indeed faster than the track with the initial straight drop. (Solvers who used different values of g got slightly different answers.)

In the end, you literally had to “break your brachistochrone” into two parts — separated by a flat track, of course.

Want more puzzles?

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

Want to submit a riddle?

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

There’s A Racial Bias on Police Facebook Pages

Mon, 13 Mar 2023 12:00:00 +0200

Police

There's A Racial Bias on Police Facebook Pages

PHOTO ILLUSTRATION BY EMILY SCHERER / GETTY IMAGES

There was nothing overtly biased about the way the Wilkes-Barre Township Police Department described a mugging on its Facebook page in February 2019. The first post simply described a Black suspect who was alleged to have threatened a victim with a gun and demanded cash in this small community in northeastern Pennsylvania. Two later Facebook posts about the case congratulated the police on catching the suspect.

But two years before, when a white man had robbed a gas station at gunpoint and fled the scene, the police department’s social media response was completely different. There was no mention of the case on social media at all, according to John Rappaport, a professor of law at the University of Chicago who is part of a team studying racial bias in law enforcement social media accounts. Not before the suspect was arrested, to warn the public and seek their help in an arrest. And not after, to reassure the community that the suspect had been caught. “The crimes are quite similar,” Rappaport said. “[It undermines] any notion that crime severity is straightforwardly driving the department’s posting decisions.”

This is just one example of a larger pattern of bias that Rappaport’s team found when they analyzed nearly 14,000 Facebook pages maintained by law enforcement agencies across the United States. They found that police Facebook pages consistently overreport crimes by Black suspects relative to local arrests rates: Between 2010 and 2019, Black suspects were described in 32 percent of posts but represented just 20 percent of arrestees. It mirrors statistics that show white Americans overestimate the percentage of crimes committed by Black Americans by as much 20 to 30 percent compared to the actual figures (numbers that, themselves, already reflect a bias in who gets arrested versus who actually commits crimes).

And scientists say it’s reasonable to suspect those two sets of statistics are connected to one another. “We really framed the paper as being less about ‘are police departments behaving well or badly,’ and more about the perspective of the reader,” Rappaport said. That’s because these biased accounts are likely part of feedback loops, reflecting bigger issues in society as police both respond to — and perpetuate — the myths white Americans already believe.

Wilkes-Barre isn’t uniquely problematic, Rappaport said. And not all of the law enforcement agencies his team looked at engaged in biased posting. But the totality of the data showed clear patterns that extended nationwide. Only a few areas didn’t overrepresent Black suspects, relative to actual arrests, including part of the Black Belt region of the South, where Black people make up the majority of the overall population.

The racial disparity in posts compared to arrests differed by type of crime, but was present across a variety of serious offenses. Car theft, for example, had the smallest disparity: There was less than a percentage point of difference between the percent of local auto thefts involving Black suspects and the percent of Facebook posts about auto thefts involving Black suspects. But the differences were much larger with other crimes. While Black suspects made up 22 percent of all theft arrests, 32 percent of Facebook posts about thefts involved Black suspects.¹

Overall, Black people’s involvement in violent crimes was being overreported by law enforcement Facebook pages by 11 percentage points and involvement in property crime was being overreported by 8 percentage points.

These differences may seem small, but Rappaport and outside researchers said the impacts of being exposed to these disparities can be wide reaching. I spoke with three other scientists, unaffiliated with Rappaport’s research, who also study American beliefs about race and crime. They all told me this paper is representative of larger, systemic issues with how race, crime and punishment are viewed in this country.

At a time when relationships between traditional media, like newspapers, and police have become strained, social media allows law enforcement to regain more control over narratives of crime, said Sarah Britto, a professor of criminal justice administration at California State University, Dominguez Hills. In decades past, researchers found evidence of traditional media overrepresenting Black people as perpetrators of crime and under-representing crimes committed by white suspects. That’s changed — newer research suggests Black Americans are now underrepresented as both suspects and victims of crime in cable and network news.

But back when researchers were finding clear bias in traditional media, research also suggested that portraying Black people as criminals had an impact on how viewers thought about criminals and criminality. For example, a study in the late 1990s exposed Los Angelenos to a news report in which an alleged perpetrator was identified as Black, white or without identifying information. It found that, when the perpetrator was Black, white viewers’ support for punitive laws increased by 6 percent — while that support only increased by 1 percent when the perpetrator in the news story was white.²

But what the public already believes about race and crime could also be shaping what police post. Rappaport’s study opens up a whole new direction in research, said Tony Cheng, a professor of criminology at the University of California, Irvine. One of the things Cheng said he’d like to see studied in the future is the motivations and practices within police departments that create racial disparities in social media posts. He suspects that the nature of social media incentivizes police to seek traffic and “likes” as much as any other group or individual who is trying to build an audience. If a post produces a lot of engagement, the posters are likely to try to repeat the success with similar content. But that becomes a problem if the most popular posts are all about Black people committing crimes, Cheng said.

The irony here is that public communication through social media channels is often lauded as part of the best practices that improve transparency in policing, he told me. “[This shows us that] those very practices can be exacerbating racial biases in public information and crime information in ways that we wouldn't even really think about,” Cheng said.

And those biases are powerful, said Nicholas Valentino, a professor of political science at the University of Michigan. There’s lots of research showing that, the more Americans perceive poor people to be overwhelmingly Black, the less support they have for social welfare policies aimed at helping the poor, he said. It makes sense that portraying Black people as more likely to be arrested for a crime than they actually are would have a similar impact on how Americans view crime policy.

“That's not a controversial thing to say,” Valentino said. “What's interesting here is that this is neither new, nor even unique to this domain of political communication. It's widespread, and we've known about it for maybe 30 or 40 years.”

How Red States Are Fighting Their Blue Cities

Mon, 13 Mar 2023 12:00:00 +0200

Partisanship

How Red States Are Fighting Their Blue Cities

PHOTO ILLUSTRATION BY EMILY SCHERER / GETTY IMAGES

In the second half of 2021, housing prices rose faster in Florida than in any other state. In some cities, rents soared by as much as 30 percent that summer. This put an enormous strain on families living in Florida and already struggling to pay rent, especially those who worked in the tourism industry for moderate wages, many of whom lost income during lockdowns of 2020. Evictions began to rise after a pandemic-related moratorium ended that summer.

Cynthia Laurent, a housing justice coordinator for the political advocacy group Florida Rising, said she heard from people struggling all over the state. In response, her organization worked with others to launch a campaign for rent-stabilization laws in the most affected cities. In Orange County, which includes Orlando, voters passed a referendum to establish rent stabilization for certain apartments for one year, keeping residents in place while the markets adjusted and families found stable footing. For many, this is exactly how local government is supposed to work: A need arises, and people put pressure on local officials or vote to change their local laws. “I believe it was the most popular item on the ballot,” Laurent said. “It wasn’t Democrat or Republican, folks from all walks of life, party, class voted yes for rent stabilization.”

But Orange County’s rent-stabilization ordinance will likely never go into effect, thanks to preemption — a type of law that lets states stop cities from setting their own agendas.

Preemption is an old, broadly used tool, and in the past decade, preemption bills have passed across the country, blocking local legislation on everything from culture-war issues to basic city governance. In Florida, a state Senate bill passed last week would prevent local governments from enacting rent control or rent stabilization. This year, other states are considering laws revoking local authority over school curriculum and punishing local district attorneys who don’t prioritize laws passed by the state legislature. Other states are threatening to take over whole chunks of city government. And there may not be much cities can do about it.

The tug of war between state and local power is an old one. Local governments, whose responsibilities are not outlined in the U.S. Constitution, have different levels of authority depending on the state, and it’s not always clear exactly what authorities localities have. “It is very much a gray zone,” said Christine Baker-Smith, a research director at the National League of Cities. “The only place where it’s clearly not a gray zone is when there is clear, clear guidance around a certain policy area.”

What has happened in the past decade is what many experts call a shift from “minimalist” preemption to “maximalist” preemption. An example of a minimalist preemption law is the minimum wage. No state can have a minimum wage that’s lower than the $7.25 set by the federal government,³ but they can go higher, and cities and counties can pass laws that set even higher minimums than their states … as long as their state hasn’t forbidden it through preemption laws.

The shift began during President Barack Obama’s presidency. He often struggled to advance progressive goals in Congress, and Republicans made electoral gains in statehouses around the country. Partisanship also became more clearly geographical: More urban populations became more solidly Democratic than ever, while rural areas became even more Republican. With progressive priorities blocked at the state and federal levels, more liberal-leaning cities began passing ordinances on issues like gun control, higher minimum wages, sick leave and LGBTQ rights. “Urban areas can’t go to the legislature to get their voice heard,” Jocelyn Johnston, a professor at American University’s School of Public Affairs, told Pew in 2015. “so they’re going to do something in-house. That’s why this is happening. Most state legislatures are not as liberal as urban interests are.”

What’s happening now is a pushback from conservative organizations and red-state legislatures. “[A]ctivists have begun targeting local governments to create big government policy that could not survive at the state capitol,” said a 2015 op-ed in RedState, arguing states should pass preemption laws to protect businesses from excessive regulation in these cities. And by that point, states were already doing just that. A 2020 Economic Policy Institute analysis found the use of preemption was more prevalent in southern states.

In the past few years, at least 25 states have prohibited local governments from raising the minimum wage. Eighteen states bar municipalities from banning plastic bags. At least 20 states have laws that prevent cities from banning gas stoves. Oklahoma is considering a bill that would prevent cities from banning combustion engines. Forty-two states preempt local legislators from passing gun regulations.⁴

Florida is one of 34 states that preempts many local housing laws, allowing rent stabilization only in an emergency; the bill that passed the state Senate last week would remove even that ability. The bill passed unanimously, but that was likely because the housing preemption was wrapped in a much larger bill, which includes measures to encourage mixed-use zoning and incentivize development of affordable housing. The bill’s proponents said it would help fix the housing shortage.

In many cases, these preemption laws were in direct response to cities’ actions. After widespread protests against police departments in the summer of 2020, states began preempting reforms or budget cuts to local police departments, with the governors of Florida and Georgia signing laws forbidding it.

Preemption laws keep expanding into new topic areas as well. This year, as of March 8, at least 493 preemption bills have been introduced into state legislatures around the country on a range of issues, according to the Local Solutions Support Center (LSSC), an organization that tracks certain preemption laws and advocates against them. Some of the biggest battles seem to be over LGBTQ rights and abortion, which fits a pattern, said Marissa Roy, head of the legal team at LSSC. She’s seen such bills originate with organizations like American Legislative Exchange Council and other think tanks. But preemption laws are also inspired by whatever culture wars are raging. “Pretty much any trends that you could note coming out in [the Conservative Political Action Conference] or on Fox News … you see them show up in preemption,” she said.

On abortion, the battle has turned to preempting local district attorneys from deciding how to use prosecutorial discretion. After the Supreme Court eliminated the constitutional right to abortion last summer, red states ramped up efforts to strictly limit the procedure, but some district attorneys in more urban, liberal areas pushed back, vowing not to prioritize enforcing those new laws. In Texas, lawmakers have introduced bills in the state House and Senate that would essentially require prosecutors to enforce all state laws or face penalties. Florida and Georgia are further enforcing preemption laws by penalizing local officials who don’t follow them. Florida Gov. Ron DeSantis suspended a Tampa-area state attorney after the attorney pledged not to enforce the state’s new abortion law, and DeSantis may suspend another one over a similar matter of enforcing state law. And the Georgia legislature is considering creating a commission with the power to remove prosecutors who “categorically” refuse to prosecute offenses that state law requires the prosecution of.

But local prosecutors have long had the discretion to set their own priorities, said Richard Briffault, a professor at Columbia University and an expert on preemption. “The state is saying, ‘No, you can’t do that for the hot-button issues that the state’s interested in,’” he said. “But at some point, they’re going to have to set priorities because they almost never are going to have the resources to prosecute everything, let alone the fact that some of these issues really do fly in the face of strong local preferences.”

It’s hard for cities to block these moves. Many state constitutions would come down on the side of the state, Roy said. According to her, reforms rooted in the 19th century gave local governments more authority than they’d had, but also allowed for state preemption. “The idea was that states would use preemption wisely to only ensure consistency where statewide consistency was needed,” Roy said. “Now, we’ve seen this abuse of preemption … and that is the exception that state legislatures have taken advantage of.”

Preemption of preemptions would need to come from changes to the state constitutions, or to fight for authority in new, specific policy areas, Baker-Smith said. And Roy added that maybe states could change some of these laws through ballot initiatives or the legislatures themselves, which she believed they’re unlikely to do. In Oklahoma, two Democrats in the state House of Representatives have introduced two separate pieces of legislation to repeal some of the state’s preemption laws, but they face an uphill battle in the Republican-dominated Legislature.

The reality is that in the U.S. today, cities are more likely to be Democratic and progressive, even when they’re in red states. When they pass more liberal laws, it can be too tempting of a target for Republicans in the state’s legislature. “They can strongly come out, whether it’s in their campaigns or whatever the case may be, and say, ‘Oh, I was against raising the minimum wage. And here’s how I stopped it,’ or, ‘Here’s how I stopped drag shows in our community,’” said Oklahoma Rep. Cyndi Munson, a Democrat who introduced one of the bills.

For those who oppose what they call its overuse, preemption undermines the basic idea behind local governance — that communities get to set priorities that reflect their own values. Laurent said that preemption laws have a longer-term, corrosive effect on local participation. State legislatures are often influenced by special interests, she said, and preempting local action removes a tool people have to fight against that. “The entire purpose of having representatives is for folks to go up there and reflect the needs that your community has,” she said. “But unfortunately, that’s being silenced.”

How Fast Can You Make The Track?

Fri, 10 Mar 2023 15:00:00 +0200

The Riddler

How Fast Can You Make The Track?

Illustration by Guillaume Kurkdjian

Riddler Express

From Bowen Kerins comes an opportunity to reclaim any marbles you may have lost:

Bill has four opaque bags, each of which has three marbles inside. Three of the bags contain two white marbles and one red marble, while the last bag contains three white marbles. The bags are otherwise indistinguishable.

Ted watches as Bill randomly selects a bag and reaches in without looking to grab two marbles without replacement. It so happens that both marbles are white. Bill is about to reach in and grab the last marble in that bag.

What is the probability that this marble is red?

Submit your answer

Riddler Classic

This week’s Classic is sure to break your brachistochrone:

While passing the time at home one evening, you decide to set up a marble race course. No Teflon is spared, resulting in a track that is effectively frictionless.

The start and end of the track are 1 meter apart, and both positions are 10 centimeters off the floor. It’s up to you to design a speedy track. But the track must always be at floor level or higher — please don’t dig a tunnel through your floorboards.

What’s the fastest track you can design, and how long will it take the marble to complete the course?

Submit your answer

Solution to the last Riddler Express

Congratulations to Michael Joyce of New Orleans, Louisiana, winner of last week’s Riddler Express.

Last week, you were adding up whole numbers in order. You started with 1, then added 2 to get 3. Then you added 3 to get 6. Then you added 4 to get 10, and so on.

If you kept going — adding larger and larger numbers in this fashion, and looking at the resulting sums — you might have noticed a few patterns among the last two digits. One such pattern was that no sum ever ended with the digits “23.” Meanwhile, some pairs of final digits were more common than others.

Which pairs of final digits were the most common? (Here, “23” was a distinct final pair of digits from “32.”)

Solver Nate Kinast was very practical and decided to compute the first million triangular numbers. Upon sorting them by their last two digits, a pattern emerged, as shown by Nate’s histogram:

We already said that no triangular number ended with the digits 23, but there were quite a few other pairs that never occurred as well. Of the 100 total two-digit pairs, only 44 of them appeared as the last two digits of triangular numbers. Among these 44, 40 of them each occurred for two percent of triangular numbers. But the last four pairs of digits occurred more frequently, each accounting for five percent of triangular numbers. These terminal digits were 03, 28, 53 and 78, the solution to last week’s puzzle.

How could you know that the pattern in the histogram continued, rather than some new behavior starting among very large triangular numbers? You knew this because — as observed by The Hewitt School Problem Solving And Problem Posing Class — the last two digits repeated themselves every 200 sums. To see why, consider the 200th triangular number, which was 200×201/2, or 20,100. This ended with two zeros, and you were effectively starting over again with the last two digits, adding 201, 202, etc. The 400th, 600th and 800th triangular numbers similarly ended with two zeros as the cycle started over again.

Solver Justin Ahmann approached the puzzle using modular arithmetic, analyzing the last two digits of triangular numbers modulo 4 and modulo 25. Modulo 4, the cycle was 1, 3, 2, 2, 3, 1, 0 and 0, which frankly wasn’t very interesting. Each of the four possible residues occurred with the same frequency. But modulo 25, the cycle was 1, 3, 6, 10, 15, 21, 3, 11, 20, 5, 16, 3, 16, 5, 20, 11, 3, 21, 15, 10, 6, 3, 1, 0 and 0. Here, each residue occurred twice, with the exception of 3, which occurred five times.² That meant final digits congruent to 3 (mod 25) were more likely to occur, again resulting in 03, 28, 53 and 78 and simultaneously explaining why it was that these values differed by 25.

The puzzle’s submitter, Oscar Lanzi, noted the connection between triangular numbers and p-adics, and how this particularly relates to triangular numbers ending with 3 or 8.

Finally, Tom Keith looked at triangular numbers written in different bases. Tom identified a number of interesting patterns, such as the fact that triangular numbers exhibit every possible final digital digit if and only if the base is a power of 2.

#ThisWeeksRiddler by @xaqwg asks about the last 2 digits of triangular numbers. This got me thinking about triangular numbers written in different bases. https://t.co/WzCbvqyeQW pic.twitter.com/iH27AC7ISp
— Tom Keith (@tom_keith1) March 6, 2023

Solution to the last Riddler Classic

Congratulations to Don Barkauskas of Carlsbad, California, winner of last week’s Riddler Classic.

Last week, you considered a single-elimination tournament with four teams that held a definitive ranking — that is, one team was the best, another team was second-best and so on. In each game, the better team won. When the tournament was complete, you knew for certain which team was the best. However, you couldn’t make similar claims about the remaining teams.

Suppose the winning team was A, and it played B in the first round. Meanwhile, the team that lost in the final was C, whose opponent in the first round was D. With this bracket, there were three possible rankings of teams from best to worst: A/B/C/D, A/C/B/D and A/C/D/B.

But instead of four teams, you had to consider a single-elimination tournament with eight teams. How many possible rankings of teams were possible for a given completed bracket?

If you tried listing these out by hand, you quickly became overwhelmed by all the possibilities. That’s why a number of solvers turned to their computers for assistance. With eight teams, there were 8! (or 40,320) rankings that were possible. Solver Jake Cuppels went through those 40,320 rankings and counted how many were compatible with a completed bracket with teams A through H. The answer turned out to be 315.

A number of solvers approached this puzzle analytically. In doing so, they effectively solved the extra credit, which had a tournament with 2^N teams, and then plugged in N = 3 to solve the original puzzle.

Andrea Andenna and Dan Baker both found a recursive pattern as the bracket doubled in size. Andrea let F(N) be the number of possible rankings in a bracket with 2^N teams. A bracket consisted of two halves with 2^N⁻¹ teams, which meant each half had F(N−1) possible rankings among its teams. The question then became how many ways you could merge the two halves into an overall ranking of 2^N teams. One team had to be the overall winner. From there, you had 2^N−1 choose 2^N⁻¹ ways to merge the remaining teams. This gave you the recursive formula F(N) = F(N−1)² · (2^N−1 choose 2^N⁻¹). Since we already knew F(2) = 3, the formula implied that F(3) = F(2)² · (7 choose 4), or 315 — this week’s answer. You could even use this formula to compute F(4) = F(3)² · (15 choose 8), which meant a bracket with 16 teams had a whopping 63,851,287 possible rankings!

A few solvers like Starvind used this recursive relationship to derive an explicit formula for F(N): (2^N)!/2^{2^}^N⁻¹, which had a corresponding sequence on OEIS. According to Starvind, when N was 6 — that is, when there were 64 teams — the number of possible rankings exceeded the total number of atoms in the observable universe.

So if you really want to know how all the teams compare at the conclusion of March Madness this year, don’t hold your breath.

Want more puzzles?

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

Want to submit a riddle?

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

How Many Brackets Can You Bust?

Fri, 03 Mar 2023 15:00:00 +0200

The Riddler

How Many Brackets Can You Bust?

Illustration by Guillaume Kurkdjian

Riddler Express

From Oscar Lanzi comes a matter of tricky triangular numbers:

Suppose you are adding up whole numbers in order. You start with 1, then add 2 to get 3. Then you add 3 to get 6. Then you add 4 to get 10, and so on.

If you keep going — adding larger and larger numbers in this fashion, and looking at the resulting sums — you might notice a few patterns among the last two digits. One such pattern is that no sum ever ends with the digits “23.” Meanwhile, some pairs of final digits are more common than others.

Which pairs of final digits are the most common? (Here, “23” is a distinct final pair of digits from “32.”)

Submit your answer

Riddler Classic

From Lucas Fagan comes a buster of brains and brackets that’s just in time for March Madness:

Consider a single-elimination tournament with four teams that hold a definitive ranking — that is, one team is the best, another team is second-best and so on. In each game, the better team wins. When the tournament is complete, you know for certain which team is the best. However, you can’t make similar claims about the remaining teams.

Suppose the winning team is A, and it plays B in the first round. Meanwhile, the team that loses in the final is C, whose opponent in the first round is D. With this bracket, there are three possible rankings of teams from best to worst: A/B/C/D, A/C/B/D and A/C/D/B.

For this week’s puzzle, instead of four teams, consider a single-elimination tournament with eight teams. How many possible rankings of teams are possible for a given completed bracket?

Extra credit: Instead of eight teams, suppose there are 2^N teams. In terms of N, how many possible rankings of teams are possible for a given completed bracket?

Submit your answer

Solution to the last Riddler Express

Congratulations to Carl Cepuran of Glen Ellyn, Illinois, winner of last week’s Riddler Express.

In long-track speed skating’s mass start competition, points awarded at three intermediate sprints (laps 4, 8 and 12) and one final sprint (lap 16). The first three skaters to finish each intermediate sprint earn 3 points, 2 points and 1 point, respectively. For the final sprint, the first six skaters to cross the line respectively earn 60, 40, 20, 10, 6 and 3 points.

In the end, competitors are ranked by how many points they accumulated during the race. When skaters with the same point total, whoever crossed the finish line first is ranked higher.

Last week, a particular skating competition consisted of two semifinal heats, with eight skaters from each heat advancing to the final.

What was the highest possible point total for a skater who fails to advance to the final?

In other words, you had to find the highest possible score for a semifinalist who came in ninth. And to do that, you wanted to spread out the points awarded among the top nine semifinalists as much as possible.

A good place to start was to look at the total number of points awarded in a race. There were 6 points awarded at each of the three intermediate sprints and then a whopping 139 points at the final sprint. In total, that amounted to 157 points. If all of these points were shared among nine individuals, then they averaged about 17.4 points each.

Now three racers had to come in first, second and third in the final sprint, each earning more than that average and altogether accounting for 120 points. That left 37 points for places four through nine, who therefore averaged about 6.2 points each. If we assume the fourth place finisher in the final sprint also earned no other points beyond their 10 at the finish line, that left 27 points for the five individuals who came in fifth through ninth overall. These five averaged 5.4 points each, which meant ninth place overall could have no more than 5 points.

But to complete the riddle, you had to show that there really was a way for ninth place to earn 5 points. Here was one such way:

The winners of the first intermediate sprint were A, B and C, in that order.
The winners of the second intermediate sprint were again A, B and C, in that order.
The winners of the third intermediate sprint were C, D and B, in that order.
The fifth-place skater in the final sprint was E, and the sixth-place skater in the final sprint was D.

Here, A and E finished with 6 points, while B, C and D all finished with 5 points. Since ties were broken by time across the finish line, B or C came in ninth place with 5 points, which was indeed the maximum score for a skater who failed to advance to the final.

Solution to the last Riddler Classic

Congratulations to Gary M. Gerken of Littleton, Colorado, winner of last week’s Riddler Classic.

Last week, I noted the recent hubbub around packing smaller squares inside larger squares. But squares were so cliche. Instead of squares, you had to pack three ellipses with a major axis of length 2 and a minor axis of length 1 into a larger ellipse with the same aspect ratio.

What was the smallest such larger ellipse you could find? Specifically, how long was its major axis?

Packing squares was hard enough. Changing the geometry to ellipses only made things harder. So let’s start with some more intuitive arrangements of the smaller ellipses and see where it gets us.

A reasonable place to start was by stacking the three ellipses together along their minor axes. Here was that arrangement:

This packing looked pretty efficient (to me at least). The major axis of the bounding ellipse was a shade longer than 4.5.

However, it was possible to do quite a bit better. Stepping away from ellipses for a moment and returning to circles, the best way to pack three unit circles was when the larger circle had a radius 1+2/√3, as shown below:

Compressing all four of these circles in the same direction by a factor of two resulted in three smaller ellipses packed into a larger ellipse with a major axis of 2+4/√3, or approximately 4.31, which was already an improvement over the stacked arrangement.

But the riddle didn’t end there. In the image above, it was possible to get an even tighter packing by symmetrically sliding the two bottom ellipses around the top one. To convince yourself this was true, you could rotate the two bottom ellipses by 15 degrees in opposite directions. This resulted in quite a bit of slack, as shown below. The three smaller ellipses could even be moved closer together, further shrinking the larger ellipse.

Solver Peter Ji specified the geometry using two parameters: the angle by which the two bottom ellipses were symmetrically rotated and the distance between the centers of these two ellipses. For any pair of these values, he deterministically nestled the third small ellipse between them. Peter found that the tightest bounding ellipse had a major axis of approximately 3.87, as shown below:

For extra credit, instead of three smaller ellipses, you considered other numbers of ellipses. Solver Laurent Lessard found excellent (but not necessarily optimal) packings when there were between two and 10 smaller ellipses:

Finally, I wanted to give a shoutout to reader Josh Silverman, who packed ellipses into a circle not with fancy optimization software but with a gravity simulator:

in #thisweeksriddler, I throw 200 ellipses into a gravitational field, and let them figure things out for a while… @xaqwg pic.twitter.com/2C8jUySJnT
— joshsilverman (@_joshsilverman) February 27, 2023

Given the apparent difficulty of packing curved shapes into other curved shapes in just two dimensions, I will forgo asking readers to pack Cheerios in a fishbowl. Instead, that might be a fun experiment to try at home.

Want more puzzles?

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

Want to submit a riddle?

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

Can You Pack The Ellipses?

Fri, 24 Feb 2023 15:00:00 +0200

The Riddler

Can You Pack The Ellipses?

Illustration by Guillaume Kurkdjian

Riddler Express

From Paul R. Pudaite comes speed-skating stumper:

In the end, competitors are ranked by how many points they accumulated during the race. When skaters with the same point total, whoever crossed the finish line first is ranked higher.

A particular skating competition consists of two semifinal heats, with eight skaters from each heat advancing to the final.

What is the highest possible point total for a skater who fails to advance to the final?

Submit your answer

Riddler Classic

For whatever reason, there is once again intrigue on social media concerning optimized packing. In particular, I have seen several tweets and even a web comic over the last month about packing unit squares into as small a larger square as possible. In particular, the case of 17 squares is known for its lack of symmetry and general aesthetic appeal:

I should add – this packing for 17 squares is not rigid – 3 of them can slide, and one even has room to wiggle, but it's not enough to allow any of the other squares to move inward, so it doesn't change the size of the containing square. pic.twitter.com/7pSz1HqQUU
— Daniel Piker (@KangarooPhysics) February 14, 2023

There have been similar packing puzzles on The Riddler before, but with this renewed interest, I figured it was time for another one. But circles and squares are so cliche.

This week, try packing three ellipses with a major axis of length 2 and a minor axis of length 1 into a larger ellipse with the same aspect ratio. What is the smallest such larger ellipse you can find? Specifically, how long is its major axis?

Extra credit: Instead of three smaller ellipses, what about other numbers of ellipses?

Submit your answer

Solution to the last Riddler Express

Congratulations to Jim Woest of Huntington Beach, California, winner of last week’s Riddler Express.

Last week, you had four squares to place on a large, flat table. You could place the squares so that their edges aligned, but their interiors could not overlap.

Your goal was to position the squares so that you could trace as many rectangles as possible using the edges of the squares. For example, if you had two squares instead of four, you could have placed the squares side by side, as shown below:

With this arrangement, it was possible to trace three rectangles: the square on the left, the square on the right and the larger rectangle around both squares.

How would you have arranged four squares to get as many rectangles as possible? And what was this number of rectangles?

Most solvers started by examining a few arrangements of squares. For example, you could have arranged the four squares in a two-by-two grid:

Here, there were four one-by-one squares, two rows, two columns and one larger two-by-two square. In total, there were nine distinct rectangles. But it was possible to do better.

Next, consider the four-by-one arrangement:

Again, there were four one-by-one squares, as well as three two-by-one rectangles, two three-by-one rectangles and one four-by-one rectangle. In total, there were 10 distinct rectangles. But it was possible to do better.

The two aforementioned arrangements had four contiguous squares. However, this wasn’t required by the puzzle. In fact, the mind-blowing optimal arrangement had four squares that only touched each other at the corners. Here was that arrangement, with the four squares colored for clarity:

While this may have looked like five squares, there was no square in the middle — that was just empty space. Nevertheless, the edges of the surrounding four rectangles could still be used to form rectangles. This time, there were five one-by-one squares, two horizontal two-by-one rectangles, two vertical two-by-one rectangles, one horizontal three-by-one rectangle and one vertical three-by-one rectangle. In total, there were 11 distinct rectangles.

Readers were evenly split between answers of 10 and 11. Solver Cass from Denver, Colorado, correctly answered 11 but still wasn’t convinced this was the solution, writing, “I think this is the answer, but I also think that, in the end, I have yet lost. Perhaps Riddler is defeated, perhaps he is not. But even if he is, I am not Batman. I am Joker now.”

Cass — this week, you are Batman!

Solution to the last Riddler Classic

Congratulations to Ryan Lafitte of Tucker, Georgia, winner of last week’s Riddler Classic.

Last week, you tried your hand at a version of solitaire played in southern Italy with a deck of 40 Neapolitan cards, with four suits numbered from 1 to 10. The deck was shuffled and then cards were turned over one at a time. Flipping over the first card you said “one,” the second card “two” and the third card “three.” You repeated this, saying “one” for the fourth card, “two” for the fifth card and “three” for the sixth card. You continued your way through the deck, until you at last said “one” for the 40th card.

If at any point the number you said matched the value of the card you flipped over, you lost.

What was your probability of winning the game?

This was a veritable combinatorial challenge. There were essentially four different card types to consider:

Four 1s, which could not occur in 14 places in the deck
Four 2s, which could not occur in 13 places in the deck
Four 3s, which could not occur in 13 places in the deck
28 other cards, which could occur anywhere in the deck

Several readers came up with a decent approximation for the correct solution. While flipping through the deck, you will at some point encounter the 12 cards numbered 1, 2 or 3. For each of these, you have a roughly two-in-three chance of saying the number not indicated on the card. That meant your probability of winning was approximately (2/3)¹², or 0.77 percent.

Austin Shapiro of Oakland, California put this approximation to the test by running 10 million solitaire simulations. It turned out your chances were slightly better than predicted: roughly 0.83 percent.

In my opinion, the exact calculation wasn’t particularly enlightening. It required you to consider all 40!/(4!·4!·4!·28!) — or 193,584,473,082,000 — cases, and figure out which among them had all the 1s, 2s and 3s in the right spots.

Solver Starvind worked through the detailed combinatorics here and even extended the puzzle so that victory allowed for a certain number of matched numbers greater than zero. According to Starvind, you should consider a variant of this solitaire that allows for up to three matches, which increases your chances of winning to almost 40 percent.

Want more puzzles?

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

Want to submit a riddle?

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

How Many Rectangles Can You Make?

Fri, 17 Feb 2023 15:00:00 +0200

The Riddler

How Many Rectangles Can You Make?

Illustration by Guillaume Kurkdjian

Riddler Express

You have four squares that you can place on a large, flat table. You can place the squares so that their edges align, but their interiors cannot overlap.

Your goal is to position the squares so that you can trace as many rectangles as possible using the edges of the squares. For example, if you had two squares instead of four, you could place the squares side by side, as shown below:

With this arrangement, it’s possible to trace three rectangles: the square on the left, the square on the right and the larger rectangle around both squares. How would you arrange four squares to get as many rectangles as possible? And what is this number of rectangles?

Submit your answer

Riddler Classic

From Nicola Paciolla comes a game that you can win (or more likely lose?) all by yourself:

There’s a version of solitaire played in southern Italy with a deck of 40 Neapolitan cards, with four suits numbered from 1 to 10. The deck is shuffled and then cards are turned over one at a time. Flipping over the first card you say “one,” the second card “two” and the third card “three.” You repeat this, saying “one” for the fourth card, “two” for the fifth card and “three” for the sixth card. You continue your way through the deck, until you at last say “one” for the 40th card.

If at any point the number you say matches the value of the card you flip over, you lose.

What is your probability of winning the game?

Submit your answer

Solution to the last Riddler Express

Congratulations to Kiera Jones of Cincinnati, Ohio, winner of last week’s Riddler Express.

Last week, you had to find all pairs of integers a and b that are solutions to the following equation: a·(a+1)·(a+2) = b²+4.

That was the entire riddle. It was short and sweet.

While it might have been tempting to go down a rabbit hole of algebra, expanding the cubic expression on the left and equating it to the quadratic on the right was no picnic. There was a simpler way, using ideas from number theory.

There were a few things you could immediately tell about the expression on the left, a·(a+1)·(a+2). For one, it had to be an even number. If a was even, then the product was also even, since an integer with an even factor must also be even. And if a was odd, then a+1 was even, again ensuring that the overall product was even.

You similarly knew that a·(a+1)·(a+2) was divisible by 3. That was because exactly one of the three factors had to be a multiple of 3. In number theory parlance, either a was a multiple of 3, a was congruent to 1 (mod 3),² in which case a+2 was a multiple of 3, or a was congruent to 2 (mod 3), in which case a+1 was a multiple of 3.

Finally, since a·(a+1)·(a+2) was a multiple of both 2 and 3, that meant it was also a multiple of 6. In terms of factors, that was everything you could say for sure about the left side of the equation.

Now what about the right side? At first, it wasn’t immediately clear how b²+4 might have related to factors of 2, 3 or 6. But let’s take a look. If b was odd, then b² was also odd, as was b²+4. Conversely, if b was even, then so were b² and b²+4. Since a·(a+1)·(a+2) was guaranteed to be even, that meant any solution had to have an even value for b.

If b was a multiple of 3, then so was b², which meant b²+4 was congruent to 1 (mod 3). If b was congruent to 1 (mod 3), then so was b², which meant b²+4 was congruent to 2 (mod 3). Finally, if b was congruent to 2 (mod 3), then b² was again congruent to 1 (mod 3), which meant b²+4 was congruent to 2 (mod 3). In other words, b²+4 was never congruent to 0 (mod 3). If you’re not convinced, check the values of b²+4 for different integer values of b. Sure enough, it’s never a multiple of 3!

At this point, you knew the left side of the equation had to be a multiple of 3. Meanwhile, the right side of the equation couldn’t be a multiple of 3. Therefore, this equation had no integer solutions. Sometimes the answer … is that there is no answer.

Solution to the last Riddler Classic

Congratulations to Michael Seifert of Quaker Hill, Connecticut, winner of last week’s Riddler Classic.

Last week, on a cloudless night, the sky above you contained an equal number of visible stars and planes. Suddenly, you spotted a point of light at a particular angle of elevation above the horizon. Based purely on this angle, you wanted to determine whether the point of light was more likely to be a star or a plane.

While figuring this out, you made the following simplifying assumptions:

Earth was a perfect sphere with a radius of 4,000 miles.
Stars were equally likely to be anywhere in the night sky.
Planes were equally likely to be flying anywhere over Earth (i.e., not just limited to established air routes) at an altitude of 6 miles. You could neglect takeoffs and landings.

After some quick thinking, you realized that at this particular angle, the point of light was equally likely to be a star or a plane. What was this angle of elevation?

This was a hard one!

Before getting into the math, let’s first try to understand what’s going on at a more qualitative level and convince ourselves that there is a solution. To do that, it’s helpful to reduce the problem from three dimensions to two. That is, suppose Earth is a circle (rather than a sphere) with a radius of 4,000 miles. Planes fly around a concentric circle with a radius of 4,006 miles. And the stars lie on a third concentric circle with a radius that is effectively infinite.

In this two-dimensional scenario, where in the sky were you more likely to see stars? Where were you more likely to see planes? As an illustration, the diagram below (not to scale) shows where planes appear at an angle of elevation above or below 45 degrees. While stars had an equal chance of being above or below this angle (again, in two dimensions), the same wasn’t true for planes. Of the portion of the 4,006-mile radius circle that was visible from the surface of the 4,000-mile radius circle, much of it was closer to the horizon. In fact, only about 3 percent of planes had an angle of elevation greater than 45 degrees, with the remaining 97 percent at lower angles.

The same phenomenon occurred in three dimensions. Once again, planes were more likely than stars to be closer to the horizon, while stars were more likely than planes to occur at higher angles. Solver Josh Silverman generated the following graph, which showed the probability distributions by angle of elevation for both stars (dashed black line) and planes (solid yellow line). Note that these distributions didn’t account for the fact that there was more spherical space at lower angles — but for comparative purposes, this omission didn’t matter.

Since the puzzle stated that there were an equal number of visible stars and planes in the sky, the intersection in the graph was precisely the angle at which the point was equally likely to be a star or a plane. Reading off the graph, this angle of elevation was approximately 0.1 radians, or about 6 degrees.

For those of you who went through the detailed calculus to determine these distributions as well as their intersection, I salute you! But I won’t be working through that derivation here. Instead, I would like to point readers to some excellent write-ups by Riddler Nation, including those of Josh, Peter Ji and Starvind. For the analytically inclined, the more precise answer turned out to be about 6.065 degrees.

Of course, there are other, more practical ways to distinguish stars and planes in the night sky. Planes visibly move and also have blinking lights. Orbiting satellites are interesting in that they also visibly move but generally don’t blink, as they are visible via reflected sunlight.

In any case, the next time you’re skygazing, keep an eye out for planes, stars and satellites. Unless you live right under a major air route, I bet you’ll see planes closer to the horizon, satellites further away from the horizon, and stars the most evenly distributed throughout the sky. Why? Because math!

Want more puzzles?

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

Want to submit a riddle?

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

More Americans Are Choosing To Stay Single

Tue, 14 Feb 2023 13:00:00 +0200

Marriage

More Americans Are Choosing To Stay Single

Happy Valentine’s Day.

Robert Voets / CBS via Getty Images

Looking for a last-minute Valentine’s Day reservation? It might be easier to get one this year than it was before the COVID-19 pandemic — and not just because some people are still avoiding dining indoors. A new survey from the American Enterprise Institute’s Survey Center on American Life shows that more people — particularly young women — are single than before the pandemic. And they might not be mad about it.

Marriage has been getting less common for a while. A Pew Research Center report published in 2021 found that the share of American adults ages 25 to 54 who are married fell by almost 15 percentage points between 1990 and 2019, from 67 percent to 53 percent. The rising share of unmarried people living with a partner accounted for a bit of that drop, but it was mostly driven by people living without a spouse or partner. In 1990, adults were more than twice as likely to be married than on their own, but by 2019, the gap between the two groups had narrowed considerably.

But how do Americans feel about the rise of single life? The new survey — and other research conducted within the past few years — shows that a decent chunk of people without a romantic partner are single by choice. According to the Survey Center on American Life, single people were about as likely to say that they’re not currently dating anyone and not looking to date (41 percent) as they were to say they were not dating someone but open to it (42 percent).¹ Older people had different justifications for not dating than younger people — but for the most part, it wasn’t because they felt undesirable. Instead, more people said that they like being single, can’t find someone who meets their expectations, or just have more important priorities right now.

Different generations have different reasons for not dating

Share of respondents from each generation who said that the following statements were a major or minor reason they were not currently dating or looking for a relationship

	Gen Z	Millennial	Gen X	Baby Boomer
Enjoy being single more than being in a relationship	53%	59	64	73
People aren’t interested in dating me	44	43	47	31
Cannot find someone who meets my expectations	56	53	65	62
Find it difficult to meet people	64	67	61	59
Have more important priorities right now	73	61	65	54

Daniel Cox, the primary researcher behind the new survey and a FiveThirtyEight contributor, told me that he was struck by the fact that Gen Z respondents² — particularly women — were more likely than older Americans to say that they were friends with their partner before they started dating. And bear in mind — these are the people who are most likely to have tried online dating. “There’s what people say they’re going to do on a survey, but then there’s also the choices and decisions they’re reporting, which is significant too,” Cox said. “This suggests a significant pushback [among young Americans] against online dating as a way to meet partners.”

It might not be especially romantic, but changing economic calculations could explain some of these shifts. In many ways, single people are more economically vulnerable than married people — they pay more in taxes, tend to have lower earnings and are slightly less likely to be employed. The challenges are even bigger for single mothers. But single women without kids may be in a better financial position than other singles. According to a 2022 analysis by the Federal Reserve Bank of St. Louis, single women without children had a median wealth that was more than nine times higher than single women with children — and slightly higher than single men with or without children.

The idea that women are better off financially without children isn’t new — the “motherhood penalty,” as it’s called by economists, is well studied — and it’s a reason why many women are delaying getting married and having kids. But it could help explain why college-educated women are more likely than women without college degrees to say they aren’t dating because they can’t find someone who meets their expectations. Being financially self-sufficient gives women a choice they didn’t have when men dominated the workforce. In the past, women might have married because of a need for financial stability, said Philip Cohen, a sociologist at the University of Maryland, College Park, who studies families. “It’s an advantage to not need to be married, in terms of economics or social pressure,” he said. “People can improve their career status and happiness on their own terms, and they can set the terms for potential mates in the future.”

Cox also thinks that women’s expectations for a partner may be shifting along with gender roles — and not in ways that benefit men who are interested in dating women. “Women who can provide for themselves might look for a partner who’s emotionally engaged, who’s attentive — instead of trying to find someone who can put food on the table,” he said. And if they can’t get those qualities from a romantic partner, they might look elsewhere. In another survey, Cox found that women are almost twice as likely as men to say they get emotional support from friends. Men looking to date women may be facing a double disadvantage, he said: They’re less likely to have friends who fill this emotional void and thus have fewer opportunities to develop the traits that potential partners want from them.

So are things worse for single men than single women? There do seem to be more single men who haven’t found what they’re looking for: The new study found that 52 percent of single young men (ages 18 to 29) were not dating but open to it, compared to 36 percent of single young women. Women were also more likely than men to say that they weren’t dating because they have other priorities right now.

But Cox said that it’s understandable why single women — particularly young single women without children — wouldn’t be in a hurry to find a partner. According to the survey, 38 percent of women under the age of 30 think marriage is old-fashioned and out of date, compared to 29 percent of young men. “If you’ve got friends and a job you like and the dating apps are full of people you’re skeptical about, what’s the rush?”

It’s A Star! It’s A Plane! It’s … The Riddler!

Fri, 10 Feb 2023 15:00:00 +0200

The Riddler

It’s A Star! It’s A Plane! It’s … The Riddler!

Illustration by Guillaume Kurkdjian

Riddler Express

Find all pairs of integers a and b that are solutions to the following equation: a·(a+1)·(a+2) = b²+4.

That’s it. That’s the riddle.

Submit your answer

Riddler Classic

One cloudless night, the sky above you contains an equal number of visible stars and planes. Suddenly, you spot a point of light at a particular angle of elevation above the horizon. Based purely on this angle, you want to determine whether the point of light is more likely to be a star or a plane.

While figuring this out, you make the following simplifying assumptions:

Earth is a perfect sphere with a radius of 4,000 miles.
Stars are equally likely to be anywhere in the night sky.
Planes are equally likely to be flying anywhere over Earth (i.e., not just limited to established air routes) at an altitude of 6 miles. You can neglect takeoffs and landings.

After some quick thinking, you realize that at this particular angle, the point of light is equally likely to be a star or a plane. What is this angle of elevation?

Submit your answer

Solution to the last Riddler Express

Congratulations to Christian Wolters of San Jose, California, winner of last week’s Riddler Express, which came just in time for Super Bowl LVII.

In football, a touchdown is worth 6 points, a point after touchdown (PAT) is worth 1 point, a 2-point conversion is worth 2 points, a field goal is worth 3 points, and a safety is worth 2 points.² A team may attempt a conversion only after it has scored a touchdown, and it must decide whether to attempt a PAT or a 2-point conversion.

Some methods of scoring points are more common than others. So when a team has scored 14 points, it’s safe to assume that they scored two touchdowns and two PATs. But that’s not necessarily how those 14 points were scored.

Using the aforementioned methods of scoring, how many distinct ways could a team have scored 14 points? Note that the sequence in which a team scores these points didn’t matter. So scoring a field goal and then a safety was the same as a safety and then a field goal (i.e., there was only one distinct way a team could score 5 points).

Suppose a team scored A touchdowns, B PATs, C 2-point conversions, D field goals and E safeties. Then the problem became finding the number of distinct ordered quintuples (A, B, C, D, E) such that 6A + B + 2C + 3D + 2E = 14, with the constraints that A, B, C, D, E ≥ 0 and B + C ≤ A (i.e., a team couldn’t have more conversions than touchdowns).

A few solvers wrote computer code to count up the number of such quintuples. But for those counting by hand, like organizing your thinking based on the number of touchdowns (A) was a good strategy, since touchdowns were worth the most points. There were three cases to consider: when A was 2, 1 or 0.

When A was 2, you somehow needed to account for 2 remaining points (with at most two conversions). There were three ways to do this: with two PATs, with one 2-point conversion and with one safety.

When A was one, you needed to account for 8 remaining points (with at most one conversion). Here, you could look at subcases when D (the number of field goals) was 2, 1 or 0. When D was 2, that again left 2 points remaining, either due to a 2-point conversion or a safety. When D was 1, that left 5 points remaining. The only way to account for this odd number of points was with two safeties and a PAT. And when D was 0, that left 8 points remaining. The two ways to generate 8 points were with four safeties or with three safeties and a 2-point conversion.

Finally, when A was 0, you needed to account for all 14 points without any conversions. At this point, you were left with the somewhat simpler equation 3D + 2E = 14, which had three non-negative integer solutions: D = 4 and E = 1, D = 2 and E = 4, and D = 0 and E = 7. (Ah yes, the seven-safety game!)

In the end, there were 11 ways to score 14 points. Again, here were all 11 ordered quintuples (A, B, C, D, E):

(2, 2, 0, 0, 0)
(2, 0, 1, 0, 0)
(2, 0, 0, 0, 1)
(1, 0, 1, 2, 0)
(1, 0, 0, 2, 1)
(1, 1, 0, 1, 2)
(1, 0, 0, 0, 4)
(1, 0, 1, 0, 3)
(0, 0, 0, 4, 1)
(0, 0, 0, 2, 4)
(0, 0, 0, 0, 7)

There were also other ways to arrive at the same solution. For example, to account for the constraint that B + C ≤ A, solver Andrea Andenna considered a touchdown and its corresponding conversion attempt as a single unit, worth either 6, 7 or 8 total points.

For extra credit, you had to similarly count the number of distinct ways a team could score 28 points. This was tougher to count by hand, and at this point most solvers, like Adnan Haque, resorted to computer assistance. It turned out there were 57 ways a team could score 28 points. (And yes, one of those ways was to score a whopping 14 safeties.)

Solution to the last Riddler Classic

Congratulations to Malachi Elkins of Huntsville, Alabama, winner of last week’s Riddler Classic.

Last week, you and your friends were singing the traditional song, “99 Bottles of Beer.” With each verse, you counted down the number of bottles. The first verse contained the lyrics “99 bottles of beer,” the second verse contained the lyrics “98 bottles of beer,” and so on. The last verse contained the lyrics “One bottle of beer.”

There was just one problem. When completing any given verse, your friends had a tendency to forget which verse they were on. When this happened, you finished the verse you were currently singing and then went back to the beginning of the song (with 99 bottles) on the next verse.

For each verse, suppose you had a 1 percent chance of forgetting which verse you were currently singing. On average, how many total verses did you expect to sing?

This sort of puzzle, with probabilities of moving from one state to another and a “final” absorbing state (i.e., the end of the song) was readily solved with a Markov chain. To construct a transition matrix, you knew that each verse (with n bottles) had a 99 percent chance of transitioning to the next verse (with n−1 bottles) and a 1 percent chance of transitioning back to the first verse (with 99 bottles).

You could equivalently have used this information to set up a rather large system of equations. Solver Mike Strong defined E(n) as the expected number of verses when starting with n bottles. Here, E(n) = 1+ 1/100·E(99) + 99/100·E(n−1). Plugging in 99 for n and rearranging gave the equation E(99) = E(98) + 100/99. Similarly, E(98) = E(97) + (100/99)², E(97) = E(96) + (100/99)³ and so on. Next, working backwards from the fact that E(0) = 0, Mike found that E(99) = 100/99 + (100/99)² + (100/99)³ + … + (100/99)⁹⁹.

This sum came to approximately 170.47, which was indeed the expected total number of verses. (Due to some ambiguity in the puzzle regarding whether you could forget while singing the last verse, I accepted any answer that was close to 170.) A number of solvers, including David Ding and Clement Lelievre were able to confirm this result via simulation.

For extra credit, you and your friends were singing “N Bottles of Beer” instead of “99 Bottles of Beer,” where N was some very, very large number. Your collective probability of forgetting where you were in the song is 1/N for each verse. If it took an average of K verses to finish the song, what value did the ratio of K/N approach?

Generalizing Mike’s equations, you now had the recursive relation E(n) = 1+ 1/N·E(N) + (1−1/N)·E(n−1). Working through the system of equations, you ultimately had E(N) = N/(N−1) + [N/(N−1)]² + [N/(N−1)]³ + … + [N/(N−1)]^N. This was a finite geometric series, and dividing this sum by N gave you the expression [N/(N−1)]^N−1. This might not have looked like much, but it could be rewritten as [1+(1/(N−1)]^N−1. In the limit as N went to infinity, subtracting 1 from N in the dominator made negligible difference, which meant the first term in the expression approached e. And so, as N increased, K/N approached e−1, or approximately 1.71828.

Solver Laurent Lessard plotted how the expected number of verses approaches this limiting ratio as N increased:

So forget “99 Bottles of Beer.” From now on, you and your friends will be singing “99 Bottles of B-e-r.” (Sorry, that was a terrible joke.)

Want more puzzles?

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

Want to submit a riddle?

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

Which Super Bowl Halftime Show Had The Most Star Power?

Fri, 10 Feb 2023 13:00:00 +0200

Super Bowl LVII

Which Super Bowl Halftime Show Had The Most Star Power?

Meet MARIAH, our newest metric.

Super Bowl LVII is two days away, can you believe it? While most of you will be watching the actual football going on, some of us will be keeping a closer eye on where the action really is — the halftime show! (Just kidding … mostly. One in 5 Americans who planned to watch last year’s Super Bowl said that halftime is their favorite part of the broadcast, so we’re not alone here!)

For our avid readers, you may remember FiveThirtyEight’s 2019 article on Super Bowl halftime shows, where we ranked the star power of performers according to their success on the Billboard Hot 100 chart. Well, we’re bringing back the Billboard Hot 100 this year, but we’re also introducing a new metric: MARIAH, or the Metric for Appraising Records, Indexed to Analyze Halftimes. It’s named after the queen of the Hot 100, Mariah Carey, who ranks No. 1 in our metric among all artists. (Remarkably, despite her high MARIAH score, she’s never performed at the Super Bowl halftime show. Maybe she’d make an appearance if the Super Bowl were set a couple months earlier, during the holidays.)

Our main goal with MARIAH is to give artists credit for songs that land on the Billboard Hot 100 chart, with extra kudos for a song’s staying power. Should a song with a 90-week run score bonus points for its endurance on the chart, compared with a song that charted for only a couple weeks? We think so. Then we used that metric to analyze Super Bowl halftime performers, performances and career arcs.

But we’ll get to all that in a bit. First, let’s break down how MARIAH works.

Methodology

To assess the star power for each Super Bowl halftime performance, we looked at each performer’s success on the Billboard Hot 100 chart since its inception in 1958. We use data compiled by the Whitburn Project as well as data we gathered ourselves. (The data in this article is as of Jan. 21, 2023.)

We narrowed our focus to analyzing halftime performances since 1993, when the King of Pop, Michael Jackson, changed the show forever, giving it a shot in the arm after a steep loss of viewers during the previous year’s halftime show. To retain viewership, halftime shows have been extremely star-studded ever since.

Here’s how we calculated our MARIAH score:

Artists gain points for each “run” of each of their songs that appears in the Billboard Hot 100.¹
In a big shift from the 2019 analysis, we considered only the artists listed on a track at equal billing — no featured artists.²
For the base point totals, we calculated the natural log of the number of weeks that a song (1) appeared on the Hot 100 chart, (2) appeared in the top 40, (3) appeared in the top 10 and (4) appeared at its peak position.³
We gave an extra boost in weight to songs according to how highly they charted: Songs in the top 40 got an extra weighting of 2, songs in the top 10 got an extra weighting of 5, songs in the top two to five slots got an extra weighting of 10 and songs in the No. 1 slot received an extra weighting of 15.⁴
Given how streaming has changed how listeners engage with music — and therefore, their listening trends as measured by the chart — we slightly penalized songs that first reached their peak in the charts in 2007 or later (digital streams were considered in the Billboard Hot 100 after July 2007). That means that charts since the year of the incorporation of streaming were weighted at 80 percent, while charts from before then were given their full value.
Similar to our methodology from 2019, we gave partial credit (one-third) to halftime performers who are associated with groups that also had hits. For example, Paul McCartney gets one-third credit for the Beatles songs that charted. The same also applies to performing groups whose members have hitmaking solo careers.⁵

Finally, to get an artist’s total MARIAH score, we added up each of their songs’ MARIAH scores.

So now that we have a somewhat objective way to rank performers, which Super Bowl was the most star-studded?⁶ Based on MARIAH, the winner is Super Bowl LVI, when Dr. Dre, Snoop Dogg, Mary J. Blige, Kendrick Lamar and Eminem took the stage, with special appearances by 50 Cent and Anderson .Paak.

Four of the seven performers at the 2022 show were major contributors to its MARIAH score, largely based on their work from over a decade ago. Relative newcomer Kendrick Lamar also put in a fair share of points, while Anderson .Paak is just getting his career started. And Eminem has put out a steady stream of charting songs over the past 20 years or so, though his newer material has been less likely to chart as high as his early hits.

The distribution in the chart above is not particularly surprising. In general, by the time artists play the Super Bowl, many have already made their biggest hits. Here’s a breakdown of performers since 1993, according to the share of their all-time artist MARIAH score that they’d produced by the time of their halftime-show appearance (or re-appearance, in the case of a few).

Not all halftime performers were well-established, however. For instance, during the early 2000s, performers like Nelly and Justin Timberlake were relative newcomers. But it’s probably no coincidence that the number of early -career performers stopped for a while after Timberlake and Janet Jackson’s controversial 2004 halftime show, which caused the Super Bowl to revert to tried-and-true (or, some might say, past-their-prime) artists unlikely to provoke an uproar.

But for many artists with less mainstream exposure, it seems that playing the halftime show would be a good career move (though our data shows only a very weak correlation with future success).⁷ Looking at performers who’d earned 80 percent or less of their career MARIAH score when they took the halftime stage, we can see that many went on to rack up hits (and points) after their Super Bowl appearances.

Which songs contributed the most to these performers’ MARIAH scores? Unsurprisingly, The Weeknd’s “Blinding Lights” took the top spot among songs that have been performed at the Super Bowl.⁸ That song held a spot on the Billboard Hot 100 for a record-breaking 90 weeks!

Where will Rihanna and this year’s halftime show rank after Sunday’s game? Although we don’t know whether she’ll have special guests, Rihanna holds her own. In Super Bowl halftime shows since 1993, she ranks ninth in artist MARIAH score at time of performance, and first among individual halftime performers. And if she sings “We Found Love,” we’ll have a new sixth-place entry on the list of most successful Super Bowl halftime songs. Now, all we need is for Mariah Carey herself to perform at a future halftime show, and our metric’s destiny will be complete.

Check out our latest NFL predictions.

Art direction by Emily Scherer. Additional research and copy editing by Andrew Mangan. Visual editing by Alex Newman. Story editing by Neil Paine.

Can You Take Down All The Bottles Of Beer?

Fri, 03 Feb 2023 15:00:00 +0200

The Riddler

Can You Take Down All The Bottles Of Beer?

Illustration by Guillaume Kurkdjian

Riddler Express

This week’s Express comes just in time for Super Bowl LVII:

In football, a touchdown is worth six points, a one-point conversion is worth one point, a two-point conversion is worth two points, a field goal is worth three points and a safety is worth two points.² A team may attempt a conversion only after it has scored a touchdown, and it must decide whether to attempt a one-point conversion or a two-point conversion.

Some methods of scoring points are more common than others. So when a team has scored 14 points, it’s safe to assume that they scored two touchdowns and two one-point conversions. But that’s not necessarily how those 14 points were scored.

Using the aforementioned methods of scoring, how many distinct ways can a team score 14 points? Note that the sequence in which a team scores these points doesn’t matter here. So scoring a field goal and then a safety is the same as a safety and then a field goal (i.e., there is only one distinct way a team can score 5 points).

Extra credit: Using the aforementioned methods of scoring, how many distinct ways can a team score 28 points?

Submit your answer

Riddler Classic

From Steven Brown comes a puzzle to take down and pass around:

You and your friends are singing the traditional song, “99 Bottles of Beer.” With each verse, you count down the number of bottles. The first verse contains the lyrics “99 bottles of beer,” the second verse contains the lyrics “98 bottles of beer,” and so on. The last verse contains the lyrics “1 bottle of beer.”

There’s just one problem. When completing any given verse, your group of friends has a tendency to forget which verse they’re on. When this happens, you finish the verse you are currently singing and then go back to the beginning of the song (with 99 bottles) on the next verse.

For each verse, suppose you have a 1 percent chance of forgetting which verse you are currently singing. On average, how many total verses will you sing in the song?

Extra credit: Instead of “99 Bottles of Beer,” suppose you and your friends are singing “N Bottles of Beer,” where N is some very, very large number. And suppose your collective probability of forgetting where you are in the song is 1/N for each verse. If it takes you an average of K verses to finish the song, what value does the ratio of K/N approach?

Submit your answer

Solution to the last Riddler Express

Congratulations to Ryan Calkin of Lathrup Village, Michigan, winner of last week’s Riddler Express.

Last week, you were playing darts and trying to maximize the number of points you earned with each throw. You were deciding which sector to aim for. Your dart had a 50 percent chance of landing in that sector and a 25 percent chance of landing in one of the two neighboring sectors. Reading clockwise, the sectors were worth 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12 and 5 points, as shown below. (For the purposes of this puzzle, you didn’t have to worry about the bullseye, the outer ring that was worth double or the inner ring that was worth triple.)

Which sector should you have aimed for to maximize your expected score?

This was a relatively straightforward computation. You could have targeted any of the 20 sectors, and for each sector you had to add 50 percent of that sector’s value plus 25 percent of the values for both neighboring sectors.

Now the average value of all the sectors was the sum of the numbers from 1 to 20 divided by 20, or 1/20*20(21)/2, or 10.5. Surely it was possible to do better than that.

Aiming for the 16-point sector did fairly well. Half of 16 plus a quarter of 7 plus a quarter of 8 resulted in an expected score of 11.75 points. Meanwhile, both the 19-point and 14-point sectors were better targets. Half of 19 plus a quarter of 3 plus a quarter of 7 resulted in an expected score of 12 points. And half of 14 plus a quarter of 9 plus a quarter of 11 similarly resulted in 12 points.

But the best sector to aim for was the 7-point sector. Half of 7 plus a quarter of 16 plus a quarter of 19 resulted in an expected score of 12.25 points. While the individual sectors ranged from 1 to 20 points, the expected scores formed a much tighter distribution around the mean of 10.5, ranging only from 8.75 to 12.25 points.

For extra credit, you had to “fairly” (by some definition of fair for you to define) assign the point values around a dartboard in some other way. Solver Michael Greenberg worked under the assumption that you still had a 50 percent chance of hitting your target sector and a 25 percent chance of hitting either neighbor. Michael then sought to have all 20 expected values be as close as possible to the average of 10.5.

He immediately realized they couldn’t all equal 10.5, since targeting the 20-point sector (with minimal neighbors worth 1 and 2 points) resulted in an expected score of 11.75 points. But Michael still came up with the following ordering of sectors: 20, 1, 19, 3, 17, 5, 15, 7, 13, 9, 11, 10, 12, 8, 14, 6, 16, 4, 18 and 2. The 20-point and 10-point sectors both had an expected score of 10.75 points, while the 11-point and 1-point sectors both had an expected score of 10.25 points. The remaining 16 sectors were right on target, with an expected score of 10.5 points. As noted by solver Emily Kelly, this arrangement decreased the standard deviation of the expected scores by an order of magnitude. Not bad!

Solution to the last Riddler Classic

Congratulations to Bill Neagle of Springfield, Missouri, winner of last week’s Riddler Classic.

Last week, you took TikTok’s #blindletterchallenge. You were presented with five letters, one at a time. Letters were picked randomly, but you could assume that no two letters were the same (i.e., letters were picked without replacement). As each letter was presented, you had to identify which of five slots you’d place it in. The goal was for the letters in all five slots to be in alphabetical order at the end.

For example, consider an attempt at the challenge by Michael DiCostanzo. The first letter is X. Since this occurs relatively late in the alphabet, he puts this in the fifth slot. The second letter is U. He puts that in the fourth slot, since it also comes relatively late (and the fifth slot is already occupied). Next, the third letter is E. He takes a gamble, and places E in the first slot. The fourth letter is D. Since D comes before E alphabetically, but no slots prior to E are now available, Michael loses this attempt.

If you played with an optimal strategy, always placing letters in slots to maximize your chances of victory, what was your probability of winning?

Solvers Izumihara Ryoma, Austin Shapiro and Mark Girard all started with a more general version of the problem. Instead of 26 letters and five slots, they considered L letters and S slots. Let’s write P(L, S) as the probability of winning this general version of the game using the optimal strategy. Suppose the first letter presented to you is x^th in the sequence of letters, so that x = 1 for A, x = 2 for B, and so on. If you were to put this letter in the i^th slot, what would be your probability of winning?

Well, you would need a few things to go right. First, among the remaining S-1 letters to be picked, you needed i-1 of those to come earlier in the alphabet than x^thto occupy the first i−1 slots, noting that there were x−1 such letters. You also needed S–i letters to come later in the alphabet than x^th to occupy the last S−i slots, noting that there were L–x such letters. The probability of both of these occurring was x−1 choose i−1 multiplied by L−x choose S−i, divided by L−1 choose S−1 — i.e., all the ways to choose among the L−1 letters for the remaining S−1 slots.

But that wasn’t all you needed to go right. Placing this first letter correctly did not guarantee victory. You still had to correctly place the remaining S−1 letters appropriately. In other words, you had to correctly place letters in the first i−1 slots (among x−1 potential letters) and you had to correctly place letters in the last S−i slots (among L−x potential letters). The probability of the former was P(x−1, i−1) and the probability of the latter was P(L−x, S−i). The probability of them both happening was the product of these.

Returning to the first letter you were presented with, we had said you placed it in the i^th slot. But what was the optimal value of i? It was the one that maximized your overall probability of winning from that point on, i.e., the value of i that maximized (x−1 choose i−1) × (L−x choose S−i) / (L−1 choose S−1) × P(x−1, i−1) × P(L−x, S−i).

But wait, there’s more! You didn’t know what letter you would be presented with, which meant x was equally likely to be anywhere from 1 to L. Summing that previous expression over all values of x and dividing by L gave your optimized probability of winning the challenge.

With this formula in hand, you could use recursion or dynamic programming to calculate P(L, S) for small values of L and S, making use of edge cases like P(n, n) = 1 and P(L, 1) = 1. The TikTok #blindletterchallenge had 26 letters and five slots. Therefore, the probability of winning with the optimal strategy was P(26, 5), or about 25.43 percent.

If you enjoyed this puzzle, check out some of the extra credit proposed by the puzzle’s submitter, Angela Zhou (a few of which have been answered by Mark):

Some extra credit for #thisweeksriddler classic @xaqwg pic.twitter.com/Y5a8wUTs22
— AZ (@a_____z___) January 30, 2023

Want more puzzles?

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

Want to submit a riddle?

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

Can You Defeat The TikTok Meme?

Fri, 27 Jan 2023 15:00:00 +0200

The Riddler

Can You Defeat The TikTok Meme?

Illustration by Guillaume Kurkdjian

Riddler Express

From Kyle Willstatter comes a puzzle that’s right on target:

You’re playing darts and trying to maximize the number of points you earn with each throw. You are deciding which sector to aim for. Your dart has a 50 percent chance of landing in that sector and a 25 percent chance of landing in one of the two neighboring sectors. Reading clockwise, the sectors are worth 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12 and 5 points, as shown below. (For the purposes of this puzzle, don’t worry about the bullseye, the outer ring that’s worth double or the inner ring that’s worth triple.)

Which sector should you aim for to maximize your expected score?

Extra credit: How would you “fairly” (by some definition of fair for you to define) assign the point values around a dartboard? Explain your thinking.

Submit your answer

Riddler Classic

From Angela Zhou comes a challenging meme analysis:

The #blindletterchallenge has recently taken TikTok by storm. In this challenge, you are presented with five letters, one at a time. Letters are picked randomly, but you can assume that no two letters are the same (i.e., letters are picked without replacement). As each letter is presented, you must identify which of five slots you will place it. The goal is for the letters in all five slots to be in alphabetical order at the end.

If you play with an optimal strategy, always placing letters in slots to maximize your chances of victory, what is your probability of winning?

Submit your answer

Solution to the last Riddler Express

Congratulations to Amy Leblang of Wayland, Massachusetts, winner of last week’s Riddler Express.

Last week, you were introduced to two friends who had birthdays on Feb. 9 and Nov. 18. When written numerically in MM/DD formatting, these dates were 02/09 and 11/18. Interestingly, the latter date included both the sum and the product of the values in the former date: 11 = 02 + 09 and 18 = 02 × 09.

How many pairs of dates were there such that one of the dates included both the product and the sum of the values in the other date (in either order)? Here, the order of the dates in the pair didn’t matter, so “02/09 and 11/18” was considered the same as “11/18 and 02/09.”

It turned out that there were quite a few such pairs. The key here was to carefully organize your counting so that you didn’t double-count any dates or miss any of them. Suppose the first date was A/B. Solver Jenny Mitchell distinctly considered when the second date was “sum-first” — A+B/A·B — or “product-first” — A·B/A+B. Sometimes, only one of these was possible. And occasionally, they were the same.

Here are the “sum-first” possibilities, organized by A:

When A was 1, the first date could be anywhere from 01/01 to 01/11.
When A was 2, the first date could be anywhere from 02/01 to 02/10.
When A was 3, the first date could be anywhere from 03/01 to 03/09.
When A was 4, the first date could be anywhere from 04/01 to 04/07.
When A was 5, the first date could be anywhere from 05/01 to 05/06.
When A was 6, the first date could be anywhere from 06/01 to 06/05.
When A was 7, the first date could be anywhere from 07/01 to 07/04.
When A was 8, the first date could be anywhere from 08/01 to 08/03.
When A was 9, the first date could be anywhere from 09/01 to 09/03.
When A was 10, the first date could be anywhere from 10/01 to 10/02.
When A was 11, the first date had to be 11/01.
A could not be 12.

In total, this accounted for 61 “sum-first” pairings. But what about when the second date was “product-first”?

When A was 1, the first date could be anywhere from 01/01 to 01/12.
When A was 2, the first date could be anywhere from 02/01 to 02/06.
When A was 3, the first date could be anywhere from 03/01 to 03/04.
When A was 4, the first date could be anywhere from 04/01 to 04/03.
When A was 5, the first date could be anywhere from 05/01 to 05/02.
When A was 6, the first date could be anywhere from 06/01 to 06/02.
When A was 7, the first date had to be 07/01.
When A was 8, the first date had to be 08/01.
When A was 9, the first date had to be 09/01.
When A was 10, the first date had to be 10/01.
When A was 11, the first date had to be 11/01.
When A was 12, the first date had to be 12/01.

This accounted for another 35 “product-first” pairings. In all, that meant you had 61 pairs plus another 35 pairs, which meant there were 96 possible pairs. Right?

Wrong. That’s because of the special case of 02/02. Its “sum-first” pair was 04/04, and its “product-first” pair was also 04/04. Since we counted this pair in both lists — rather than counting it once — the total tally was one too high. In the end, there were 95 pairs of dates.

Solution to the last Riddler Classic

Congratulations to Matt Frank of New York, New York, winner of last week’s Riddler Classic.

Last week, a restaurant at the center of Riddler City was testing an airborne drone delivery service against their existing fleet of scooters. The restaurant was at the center of a large Manhattan-like array of square city blocks, which the scooter had to follow.

Both vehicles traveled at the same speed, which meant drones could make more deliveries per unit of time. You could also assume that (1) Riddler City was circular in shape (2) deliveries were made to random locations throughout the city and (3) the city was much, much larger than its individual blocks.

In a (large) given amount of time, what was the expected ratio between the number of deliveries a drone could make to the number of deliveries a scooter could make?

This was equivalent to finding the ratio between the average distances — measured two different ways — from the center of Riddler City to a random location within Riddler City. For the drone, you simply needed the average Euclidean distance, or the straight-line distance.

For simplicity, let’s scale down Riddler City to a unit circle, centered at the origin and with radius 1. The distance between the origin and a point in the circle (x, y) was √(x²+y²). Since the city was circular, it actually made more sense to write this in polar notation: A point at (r, 𝜃) was a distance r from the origin. To find the average distance among all the points in the circle, you had to integrate r from 0 to 1, and 𝜃 from 0 to 2𝜋, using the area differential rdrd𝜃. Evaluating this integral gave you 2𝜋/3. Finally, you had to normalize by dividing by the total area of the circle, which was 𝜋. In the end, the average distance between a random point in a unit circle and the circle’s center was 2/3.

For our same point (x, y), the scooter traveled the “Manhattan distance,” or x+y. In polar form, this distance was |rcos𝜃| + |rsin𝜃|. After plugging this distance into the integral and separating variables, you were still integrating r² from 0 to 1, which was 1/3. As for 𝜃, you were now integrating |cos𝜃|+|sin𝜃| from 0 to 2𝜋, which was 8. The total integral was the product of these two separated integrals, or 8/3. Normalizing by the area of the circle meant the average Manhattan distance was 8/(3𝜋).

At this point, you had the average Euclidean and Manhattan distances. All that was left was to find the ratio of these two values, which was 2/3 ÷ 8/(3𝜋), or 𝜋/4. Again, this was the ratio of the average distances traveled by the drone and scooter. To find the ratio of their expected number of deliveries, you needed to take the reciprocal, since distance and delivery rate were inversely related. In the end, the ratio of drone deliveries to scooter deliveries was 4/𝜋, or about 1.273.

For extra credit, in addition to traveling parallel to the city blocks, scooters could also move diagonally from one corner of a block to the opposite corner of the block. With this additional motion in play, what was the expected ratio between the number of deliveries a drone could make and the number of deliveries a scooter could make?

Since the drone didn’t use these new paths, its average distance traveled in the unit circle remained 2/3. However, these new paths decreased the average distance for the scooter.

Consider points in the unit circle (x, y) with x > y > 0. (Note that these points make up one-eighth of the unit circle.) As shown below, the distance the scooter would travel to reach such a point was (x−y)+y√2.

In polar coordinates, this was rcos𝜃 + rsin𝜃(√2−1). Integrating and normalizing over the eighth of the unit circle resulted in an average distance of 16/(3𝜋)·(√2-1). And thanks to symmetry, this was the average for the other seven-eighths of the unit circle, meaning it was the average for the entire unit circle.

And so, with these diagonal routes now available, the ratio of drone deliveries to scooter deliveries was 8/𝜋(√2-1), or about 1.055. While the drone was still more efficient than the scooter, the additional diagonal paths impressively brought the scooter’s efficiency 80 percent closer to that of the drone.

Want more puzzles?

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

Want to submit a riddle?

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

Which Parents Are The Most Tired?

Fri, 27 Jan 2023 13:00:00 +0200

Pollapalooza

Which Parents Are The Most Tired?

PHOTO ILLUSTRATION BY FIVETHIRTYEIGHT / GETTY IMAGES

Welcome to Pollapalooza, our weekly polling roundup.

At some point last year — maybe around when my child woke up wailing at 4 a.m. for the thousandth time — I gave up wondering when I would stop being so tired. For parents of young children, “tired” isn’t a state of being that can be sloughed off with a few good nights’ sleep. It’s an innate condition — the thing I say reflexively when people ask me how I am, the excuse I use for days when everything I touch feels mediocre. Burnout, exhaustion — call it what you want, but I’m not the only one who can’t stop talking about how tired I am. Stories about parental exhaustion are ubiquitous.

Except that burden of fatigue isn’t evenly distributed, and parents are feeling a lot of other things, too. In a newly released survey of 3,757 parents of children under the age of 18 conducted last fall, the Pew Research Center dug into the drama of raising kids in the United States today, asking about parents’ worries and dreams for their children, how caring for kids is divvied up at home and — yes — how tired parents really are.

The survey found that the stress and worry of parenting are disproportionately affecting mothers and parents of color.¹ But that doesn’t mean the stress is getting to them — the groups that reported higher levels of stress, fatigue and worry were among the most likely to say that having children is rewarding and enjoyable all of the time. Perhaps it’s a kind of parental Stockholm syndrome, where the parents in the most arduous conditions grow to love their misery.

Fathers took on more caregiving responsibilities during the COVID-19 pandemic, but the Pew survey indicates that in most households, the emotional weight of parenting still falls on mothers. According to the survey, mothers are more likely than fathers to say that being a parent is tiring (47 percent vs. 34 percent) or stressful (33 percent vs. 24 percent) all or most of the time. Mothers are also more worried than fathers about whether their children will face hardships, like being bullied or struggling with anxiety and depression, and they’re more likely to say that they experience judgment about their parenting from friends, other parents in their community and other parents online.

Mothers in heterosexual relationships also reported that they do more child care tasks and their perceptions of the division of labor did not always line up with the way fathers saw things.² For example, a majority (58 percent) of mothers say they do more work providing comfort or emotional support to their children, while the same share (58 percent) of fathers said that this task was shared equally. The only area asked about in which mothers and fathers generally agreed that the work was shared equally was on disciplining their children — and even there, 31 percent of fathers said that they did more of the work, compared to 36 percent of mothers.

So who’s right? Data from the U.S. Bureau of Labor Statistics also supports the idea that women are spending more of their time on most forms of childcare. According to the latest American Time Use Survey, which measures the amount of time people spend on various activities throughout their day, mothers of children under the age of 18 report spending 1.76 hours per day with childcare as their main activity, while men only spent 1.02 hours. When the survey researchers broke it down, women reported spending more time than men on physical care for kids and activities related to their education — but men and women were spending about the same amount of time playing with their kids. (The BLS definition specifically excludes sports from “playing with children.”)

But mothers and fathers weren’t the only groups with different outlooks on parenting. There were also substantial divides by race and ethnicity. In the Pew survey, Black and Hispanic parents expressed more concern than white or Asian parents about their children facing challenges like being bullied, struggling with anxiety and depression, or being beaten up. Other groups suffered from different forms of anxiety: Asian parents were more likely than parents from other racial and ethnic groups to say they feel judged by their own parents at least sometimes, and white parents were more likely to say they feel judged by other parents in their community.

One of the biggest racial and ethnic divides wasn’t about the downsides of parenting, though — it was about the benefits. Black (39 percent) and Hispanic (39 percent) parents were more likely than white (18 percent) and Asian (13 percent) parents to say that they find being a parent to be enjoyable all the time. There’s a similar — although slightly less dramatic divide — when parents were asked whether they find parenting rewarding.

There’s a tension in those findings. Black and Hispanic parents were more likely to fear for their children’s safety — but they’re also the most likely to find consistent joy in being a parent. There was a similar pattern for lower-income parents, who were much more worried about a wide range of concerns — their children being bullied, kidnapped, beaten up, getting shot, or getting trouble with the police — than middle or higher-income parents, but also were substantially more likely to say they enjoy being a parent all the time. And all of the most worried groups — mothers, Black and Hispanic parents, and lower-income parents — were more likely than other parents to say that being a parent is the most important part of their identity.

Why are the most anxious parents in Pew’s survey also the most likely to find daily joy in raising children? Shouldn’t all that worry make parenting less fun? There could be a lot going on here, including differences in which respondents felt more comfortable reporting an emotion like worry (probably women), or more pressure to say they enjoy being a parent (again, probably women). But maybe it’s simply that the joys of parenting are inextricably linked with its frustrations and anxieties — and the more you have of one, the more you have of another. At least, that’s what I’ll tell myself the next time my daughter keeps me up all night.

Other polling bites

The American public has judged embattled Rep. George Santos, and the results are not pretty. A Data for Progress poll conducted from Jan. 20-23 found that only 11 percent of likely voters have a favorable view of Santos, who turns out to have lied about basically everything in his background. Half (50 percent) of respondents have an unfavorable view of Santos — including 38 percent who have a very unfavorable view — and 39 percent say they don’t know enough to say. For context: the poll found Santos well behind some of his Republican colleagues. Twenty-nine percent of Americans have a favorable view of House Speaker Kevin McCarthy, 20 percent have a favorable view of Senate Minority Leader Mitch McConnell and 17 percent have a favorable view of Rep. Marjorie Taylor Greene.
Speaking of the House GOP, a CNN poll conducted by SSRS from Jan. 19-22 found that nearly three-quarters (73 percent) of Americans think Republican leaders in the House haven’t paid enough attention to the nation’s most important problems, while 27 percent say they have paid enough attention. Of course, not all respondents likely agree on what the nation’s most important problems are.
Americans aren’t happy with the fact that classified documents were found in President Biden’s home in Delaware and a Washington, D.C. office, according to another CNN poll conducted by SSRS from Jan. 19-22 — and they think appointing a special counsel to investigate was the right call. About two-thirds (67 percent) of Americans think it’s a very or somewhat serious problem that documents were found in Biden’s home and office, and 84 percent approve of the Justice Department’s decision to appoint a special counsel to investigate.
Younger Americans are holding corporations to a high ethical standard, according to a newly released Gallup poll conducted in June 2022. The poll found that 77 percent of Americans ages 18-29 think it’s “extremely important” for businesses to operate in a way that is sustainable for the environment and a similar share (72 percent) say it’s extremely important for businesses to focus on long-term benefits to society instead of short-term profits. Both of those shares are substantially larger than any other age group.

Biden approval

According to FiveThirtyEight’s presidential approval tracker,³ 42 percent of Americans approve of the job Biden is doing as president, while 52.4 percent disapprove (a net approval rating of -10.4 points). At this time last week, 43.4 percent approved and 51.3 percent disapproved (a net approval rating of -7.9 points). One month ago, Biden had an approval rating of 43.4 percent and a disapproval rating of 51.5 percent, for a net approval rating of -8.1 points.

Who’s Favored To Win The Oscars This Year?

Tue, 24 Jan 2023 22:22:23 +0200

FiveThirtyEight’s senior elections analyst, Nathaniel Rakich, is a big movie buff. Each year, he keeps a spreadsheet of films to track nominations and predict which might win Academy Awards. He recently went on ABC News Live to talk about his 2023 Oscar predictions.

Can You Make A Speedy Delivery

Fri, 20 Jan 2023 15:00:00 +0200

The Riddler

Can You Make A Speedy Delivery

Illustration by Guillaume Kurkdjian

Riddler Express

From Jason Armstrong comes a curious conundrum of calendars:

Two friends of Jason, who happen to be married to each other, have birthdays on Feb. 9 and Nov. 18. When written numerically in MM/DD formatting, these dates are 02/09 and 11/18. Jason noted that the latter date includes both the sum and the product of the values in the former date. In other words, 11 = 02 + 09 and 18 = 02 × 09.

How many pairs of dates are there such that one of the dates includes both the product and the sum of the values in the other date (in either order)? Also, note that the order of the dates in the pair doesn’t matter, so “02/09 and 11/18” should be considered the same as “11/18 and 02/09.”

Submit your answer

Riddler Classic

From Graydon Snider comes a dilemma of delivery:

A restaurant at the center of Riddler City is testing an airborne drone delivery service against their existing fleet of scooters. The restaurant is at the center of a large Manhattan-like array of square city blocks, which the scooter must follow.

Both vehicles travel at the same speed, which means drones can make more deliveries per unit time. Assume that (1) Riddler City is circular in shape, as you may recall (2) deliveries are made to random locations throughout the city and (3) the city is much, much larger than its individual blocks.

In a given amount of time, what is the expected ratio between the number of deliveries a drone can make to the number of deliveries a scooter can make?

Extra credit: In addition to traveling parallel to the city blocks, suppose scooters can also move diagonally from one corner of a block to the opposite corner of the block. Now, what is the new expected ratio between the number of deliveries a drone can make and the number of deliveries a scooter can make?

Submit your answer

Solution to the last Riddler Express

Congratulations to Jim Lahey of Issaquah, Washington, winner of the last Riddler Express.

Last time, you and a friend were shooting some hoops at your local basketball court when she issued a challenge: She would name a number, which we’ll call N. Your goal was to score exactly N points in as many ways as possible using only 2-point and 3-point shots. The order of your shots did not matter.

For example, there were two ways you could score N = 8 points: four 2-pointers or two 3-pointers and one 2-pointer.

Your apparently sadistic friend chose 60 for the value of N. You tried to negotiate this number down, but to no avail. However, she said you were welcome to pick an even larger value of N. Did there exist an integer N greater than 60 such that there were fewer ways to score N points than there were ways to score 60 points?

At first, this all seemed rather counterintuitive. As N increased, the number of ways to score N, which we’ll call the function f(N), should also have increased. However, f(N) was not necessarily a monotonically increasing function.

Let’s see why that was, using smaller values of N. There were zero ways to reach N = 1. There was one way to reach N = 2 (one 2-pointer), 3 (one 3-pointer), 4 (two 2-pointers) and 5 (one 2-pointer and one 3-pointer, in either order). There were two ways to reach N = 6: three 2-pointers or two 3-pointers. Up to this point, as N increased, f(N) never decreased. But there was only one way to reach N = 7: two 2-pointers and one 3-pointer. In function notation, f(7) < f(6). Sure enough, f(N) was not monotonically increasing.

A similar dip occurred around when N was 60. To score 60 points, you could have made 30 2-pointers and no 3-pointers. At the other extreme, you could have made no 2-pointers and 20 3-pointers. Other combinations were possible, swapping three 2-pointers at a time for two 3-pointers. In the end, the number of 3-pointers had to be an even number between 0 and 20 (inclusive), which meant f(60) was 11.

When N was 61, you could have made 29 2-pointers and one 3-pointer. At the other extreme, you could have made two 2-pointers and 19 3-pointers. This time, the number of 3-pointers had to be an odd number between 1 and 19 (inclusive), which meant f(61) was 10. Because f(61) < f(60), 61 points was your desired target.

It turned out that f(N) always decreased when N went from being a multiple of 6 to one more than a multiple of 6. Solvers like Matthew Tanzy and Aidan Dunkelberg were able to prove this by finding an explicit formula for f(N). Matthew found that when N was even, f(N) was equal to floor(N/6)+1; when N was odd, f(N) was equal to floor(N/6+0.5). Sure enough, this confirmed the result that f(61) was less than f(60).

Solution to the last Riddler Classic

Congratulations to Jared Schmitthenner of Madison, Wisconsin, winner of the last Riddler Classic.

Last time, the astronomers of Planet Xiddler were back in action! They had used their telescopes to spot an armada of hostile alien warships on a direct course for Xiddler. The armada was scheduled to arrive in exactly 100 days. (Recall that, like Earth, there are 24 hours in a Xiddler day.)

Fortunately, Xiddler’s engineers had just completed construction of the planet’s first assembler, which was capable of producing any object. An assembler could be used to build a space fighter to defend the planet, which took one hour to produce. An assembler could also be used to build another assembler (which, in turn, could build other space fighters or assemblers). However, building an assembler was more time-consuming, requiring six whole days. Also, you could not use multiple assemblers to build one space fighter or assembler in a shorter period of time.

What was the greatest number of space fighters the Xiddlerian fleet could have when the alien armada arrived?

One strategy was to spend the entire 100 days building fighters. There are 2,400 hours in 100 days, so you would have had 2,400 fighters in the end. Alternatively, you could have spent the first six days building an assembler and then the last 94 days making fighters. There are 2,256 hours in 94 days, and with two assemblers that would have resulted in 4,512 fighters.

At this point, you might see how a good strategy was to invest in assemblers early on and then switch over to fighters at the end. After all, having assemblers build more assemblers resulted in exponential growth, whereas building fighters was just linear growth. And sooner or later, exponential growth always blows linear growth out of the water.

Taking this to the extreme, you could have gone through 16 cycles of building assemblers over the first 96 days. This would have resulted in 2¹⁶, or 65,536, assemblers. Over the final four days, or 96 hours, they would have collectively made 6,291,456 fighters. That’s a whole lot of fighters!

But it was possible to do even better. Working backward, solver Jen McTeague suggested starting with 15 (rather than 16) cycles of building assemblers over the first 90 days. With this strategy, you had half as many assemblers at the end. However, you also had more than twice as many days remaining in which to build fighters (i.e. 10 vs. four), meaning you’d net more fighters.

After those 15 cycles, you had 2¹⁵, or 32,768 assemblers. Over the final 10 days, or 240 hours, they would have collectively made 7,864,320 fighters. And that was the largest the Xiddlerian fleet could have possibly been.

But was it enough to fend off the alien armada? Find out next time on … The Xiddler!²

Want more puzzles?

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

Want to submit a riddle?

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

Can You Fend Off The Alien Armada?

Fri, 06 Jan 2023 15:00:00 +0200

The Riddler

Can You Fend Off The Alien Armada?

Illustration by Guillaume Kurkdjian

Note: There will be no column on Jan. 13. The next column will appear on Jan. 20.

Riddler Express

From Richard Jacobson comes a matter of bewildering basketball:

You and a friend are shooting some hoops at your local basketball court when she issues a challenge: She will name a number, which we’ll call N. Your goal is to score exactly N points in as many ways as possible using only 2-point and 3-point shots. The order of your shots does not matter.

For example, there are two ways you could score N = 8 points: four 2-pointers or two 3-pointers and one 2-pointer.

Your apparently sadistic friend chooses 60 for the value of N. You try to negotiate this number down, but to no avail. However, she says you are welcome to pick an even larger value of N. Does there exist an integer N greater than 60 such that there are fewer ways to score N points than there are ways to score 60 points?

Submit your answer

Riddler Classic

From James Anderson comes a puzzle to stave off galactic disaster:

The astronomers of Planet Xiddler are back in action! Unfortunately, this time they have used their telescopes to spot an armada of hostile alien warships on a direct course for Xiddler. The armada will be arriving in exactly 100 days. (Recall that, like Earth, there are 24 hours in a Xiddler day.)

Fortunately, Xiddler’s engineers have just completed construction of the planet’s first assembler, which is capable of producing any object. An assembler can be used to build a space fighter to defend the planet, which takes one hour to produce. An assembler can also be used to build another assembler (which, in turn, can build other space fighters or assemblers). However, building an assembler is more time-consuming, requiring six whole days. Also, you cannot use multiple assemblers to build one space fighter or assembler in a shorter period of time.

What is the greatest number of space fighters the Xiddlerian fleet can have when the alien armada arrives?

Submit your answer

Solution to the last Riddler Express

Congratulations to Paul Menchini of Hillsborough, North Carolina, winner of the last Riddler Express.

Last week, you had to make it to 2023. And if you’re reading this right now, that means you made it! More specifically, you had to make it to 2023 using a “tribonacci” sequence, which starts with three whole numbers, with each new term equal to the sum of the preceding three.

Of course, many tribonacci sequences included the number 2023. For example, if you had started with 23, 1000 and 1000, then the very next term would have been 2023. Your challenge was to find starting whole numbers a, b and c (with a ≤ b ≤ c) so that 2023 was somewhere in their tribonacci sequence and the sum a + b + c was as small as possible.

A few solvers tried out various values of a, b and c — some with computer assistance, like Ria Skies and Ryan McShane. But there was a way to solve this puzzle with a lot less guesswork, as with a similar puzzle from about a year ago that was about Fibonacci, rather than tribonacci, sequences.

No matter your two starting values, every Fibonacci sequence eventually converged to the same ratio between consecutive terms. Without proving this fact, a quick and dirty way to solve for this ratio was to assume that three consecutive numbers in the sequence had a common ratio, so we could call them x, ax and a²x. Since they were in a Fibonacci sequence, that meant x + ax = a²x. Dividing through by x and rearranging produced the quadratic equation a² − a − 1 = 0, which had a positive solution of (1+√5)/2, the golden ratio.

The same sort of pattern emerged for tribonacci numbers, no matter which three numbers you started with. This time, the common ratio was the solution to the cubic equation a³ − a² − a − 1 = 0, which had one real solution of approximately 1.8393.

At this point, let’s take a step back and return to the actual puzzle. You wanted a, b and c to be small, which meant you effectively wanted 2023 to show up as late as possible in the sequence. By that point, the ratios between consecutive terms should have converged somewhat. That meant the term prior to 2023 should have been close to 2023/1.8393, or about 1100. And the term before that should have been close to 1100/1.8393, or about 598.

From there, you could find each prior term in the sequence by taking a given term and subtracting the two preceding terms. Working backwards, the resulting sequence was 2023, 1100, 598, 325, 177, 96, 52, 29, 15, 8, 6, 1 and 1. It might have been tempting to go a step further (and generate another term, 4), but the requirement of a ≤ b ≤ c meant you had to stop there. In the end, this sequence had the minimum initial sum: a = 1, b = 1 and c = 6.

In case you’re looking ahead to next year, solver Darren L. of Grand Blanc, Michigan, found the smallest starting triplet for 2024 as well: a = 3, b = 4 and c = 20. But let’s not get ahead of ourselves.

Solution to the last Riddler Classic

Congratulations to Dan Potterton of Atlanta, winner of the last Riddler Classic.

This past Christmas, puzzle submitter Gary Yane and his family had a pairwise gift exchange. So if cousin Virginia gave uncle Justin a gift, then Justin gave Virginia a gift.

There were 20 people in the gift exchange. In the first round, everyone wrote down the name of a random person (other than themselves) and the names went in a hat. Then if two people randomly picked each other’s names out of that hat, they exchanged gifts and no longer participated in the drawing. The remaining family members went on to round two. Again, they wrote down the name of anyone left, and again, any two people who picked each other exchanged gifts.

This continued until everyone was paired up. And yes, if exactly two people remained, they still went through the process of selecting each other, even though they knew who their partner would ultimately be.

On average, what was the expected number of rounds until everyone was paired up?

First, as pointed out by several readers, this was admittedly a rather inefficient way to run a puzzle exchange. Instead, the 20 people should each have simply placed their own names into the hat, guaranteeing that every name was in there exactly once. With the convoluted way Gary’s family ran the exchange, it was quite likely that there were duplicate names as well as missing names in the hat. But I digress.

Second, the puzzle was ambiguous as to whether the family members picked names from the hat with replacement (i.e., once a name was selected it was placed back in the hat) or without replacement (i.e., once a name was selected it was not placed back in the hat). As is typical with gift exchanges, my own interpretation of the puzzle was to assume no replacement. But in the end, I accepted solutions that made either assumption.

Let’s return to the puzzle. Before taking on the case with 20 people, suppose there were just two (again, let’s call them Virginia and Justin). What was the probability that they paired up in a single round? That required whoever picked first — say, Virginia — to pick out Justin’s name rather than her own, which happened half the time. Then, since we assume no replacement, Justin picked Virginia’s name. The other half the time they each picked their own name. So with two people, there was a 50 percent chance they paired up in one round, a 25 percent chance they paired up in two rounds, a 12.5 percent chance they paired up in three rounds and so on. To find the expected number of rounds, you had to evaluate the arithmetico-geometric series 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + …, which had a sum of 2.

(With replacement, instead of picking each other’s names half the time, they only did this a quarter of time. This meant it took an average of four rounds to pair up. Based on this, it was apparent that replacement resulted in a greater expected number of rounds with 20 participants.)

Of course, that was just with two people. With four people in the exchange, things immediately got a lot more complicated. Each person now had three names to choose from when writing one down, resulting in 81 cases to consider. But in theory, it was possible to compute the transition probabilities that four people would result in no pairs (leaving all four people to play another round), one pair (leaving two people to play the next round) or two pairs (meaning the exchange was complete).

At this point, several readers got sidetracked by an article that referenced a game called “Look Up and Scream.” This game was similar to Gary’s gift exchange, with people arranged in a circle and pointing to each other at the same time. However, one key difference was that in the gift exchange, it was possible for you to pick your own name from the hat (i.e., if someone else wrote it down) — something that wasn’t possible in “Look Up and Scream.” So while “Look Up and Scream” lasted an average of about 17.16 rounds when starting with 20 people, Gary’s gift exchange lasted a little longer.

Through simulation, solver David Ding found that it took an average of about 22.1 rounds when there was no replacement and an average of about 26.7 rounds when there was no replacement. Here are the distributions for the number of rounds that David found in both cases:

By the way, if everyone had simply placed their own name in the hat (as several readers suggested) and then picked without replacement, the exchange would have been more efficient, lasting an average of about 20 rounds. And with replacement in this case, it would have lasted an average of about 24.2 rounds.

Want more puzzles?

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

Want to submit a riddle?

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

Watch: https://abcnews.go.com/fivethirtyeight/video/solve-pizza-puzzle-fivethirtyeights-riddler-84226315

How Americans Really Feel About Elon Musk

Fri, 23 Dec 2022 13:00:00 +0200

Twitter

How Americans Really Feel About Elon Musk

Hint: His unscientific polls don’t tell us much.

PHOTO ILLUSTRATION BY FIVETHIRTYEIGHT / GETTY IMAGES

Welcome to Pollapalooza, our weekly polling roundup.

Twitter’s new owner, Elon Musk, might not have any credibility as a pollster in FiveThirtyEight’s rating system, but he’s a pollster nonetheless. Kinda.

Soon after he took control of Twitter in October, the once-richest person in the world implemented a new management style that allowed users to make key decisions via polls. Should former President Donald Trump be allowed to rejoin the platform after supposedly being permanently banned last year? A slim majority of users said yes, so — “Vox Populi, Vox Dei,” as Musk wrote — he was back. Should Musk bring back Vine, the short-form video app which shut down in 2016? Sure! Maybe! The people have spoken!

The stakes of the polls escalated quickly. On Sunday, Musk put his own job security on the line, vowing to abide by the results of his own, unscientific poll. “Should I step down as head of Twitter?” he asked users. By Monday, he had an answer: By a 15-point margin — 57.5 percent to 42.5 percent — respondents said he should resign from his post atop the social media giant. Musk said on Tuesday he plans to honor the poll’s results as soon as he finds “someone foolish enough” to succeed him. It’s unclear when that will happen, or how much power he will actually be relinquishing.

It’s too bad for Musk that he didn’t take a more scientific approach, though, because according to a number of professionally conducted polls, Americans still have a somewhat favorable opinion of him — although they do hold negative views of social media companies generally.

Let’s kick things off with Musk’s own question of whether he should quit. Though a majority of respondents in his own survey said “yes,” an overnight poll conducted by HarrisX in mid-December found that a whopping 61 percent of U.S. Twitter users and 53 percent of U.S. adults actually want Musk to stay at the helm. Meanwhile, another December poll, this one from Quinnipiac University, found that Americans were almost evenly split on their feelings toward how Musk runs the social media giant: 37 percent said they approved of the way he’s operating Twitter, 37 percent disapproved and 25 percent offered no opinion.

And poll after poll shows that Musk isn’t overwhelmingly unpopular with the American public, either. According to that same Quinnipiac survey, 36 percent of Americans said they viewed Musk positively versus 33 percent who viewed him negatively. (Another 26 percent said they hadn’t heard enough about him to make an opinion either way.) A YouGov/The Economist poll, fielded in November, found that 41 percent of U.S. adults had a “very” or “somewhat” favorable view of Musk compared with 37 percent who viewed him “somewhat” or “very” unfavorably. These findings come despite evidence showing that, generally, Americans hold negative opinions about social media companies. Quinnipiac, for example, found that 70 percent think that social media giants like Twitter and Facebook “do more harm than good,” while 18 percent disagreed. Another spring 2022 survey from the Pew Research Center, which polled citizens in 19 advanced economies about their views on social media, technology and their influence on society, found that 79 percent of U.S. respondents believed that access to social media and the internet has made people more divided in their political opinions.

The fact that Musk isn’t overwhelmingly disliked might come as a surprise to people who have been closely following Twitter’s fate. In a matter of months, he gutted the company’s staff, drove away major advertisers and suspended (then unsuspended) the accounts of several prominent journalists — among many other things. And it’s worth underscoring that not everyone is over the moon with Twitter’s newest CEO. Per Quinnipiac, among U.S. adults, Republicans (63 percent) and white men (51 percent) were the most likely to view Musk favorably. Democrats (9 percent), Black respondents (17 percent) and women (25 percent) were the demographic groups least likely to harbor positive opinions toward Musk.

And, to be sure, it does look like Musk’s overall favorability numbers have ticked down since purchasing Twitter. Back in April, YouGov found that closer to half of U.S. adults (49 percent) had a “very” or “somewhat” favorable opinion of Musk compared with 31 percent who viewed him “somewhat” or “very” unfavorably.

Unfortunately, most polls that ask respondents their opinions toward Musk don’t ask why people feel the way they do. Is his wealth impacting people’s views of him? Is his high name ID giving him an added advantage? Did his suspension of journalists (which a majority of respondents in a December CivicScience poll viewed negatively) depress his favorability ratings? Topline survey findings don’t give us a lot of clues. What we do know, however, is that people view Musk as an influential and successful businessman — and maybe someone who’s a bit quirky, too. And that might be why, despite his many flubs at Twitter, Americans don’t have overwhelmingly negative views of him.

For example, YouGov’s April survey asked respondents how influential they felt Musk was in the tech world and the overwhelming majority of respondents (80 percent) said he was “very” or “somewhat” influential. Another question on the same polls asked the same sample to select terms that they felt described Musk. The winners were: rich (60 percent), an entrepreneur (49 percent), an innovator (39 percent) and eccentric (37 percent). Meanwhile, a December YouGov survey found that 58 percent of U.S. adults believe that Musk is a “successful business person” versus 22 percent who said he wasn’t.

So have Twitter users actually seen the last of Musk, then? It doesn’t seem like he’s planning to bow out entirely — or even partially. After announcing that he would resign as CEO once he could find a sufficiently foolish successor, he said that his next steps would be to “just run the software & servers teams.” So the main change to Twitter — at least in the short-term — might be the way Musk conducts his polls (he seemed to agree with a user’s comment which suggested that, from now on, only Twitter blue subscribers should be allowed to vote in “policy related” polls). Or maybe Musk actually will step back and open the door for someone else — hello, Snoop Dogg and Dionne Warwick — to take over at Twitter in 2023. Your guess is as good as mine.

Other polling bites

Would you consider yourself a poor gift wrapper? If so, your humility might put you in the minority, according to new December polling data from YouGov. Per their survey, 64 percent of U.S. adults said they considered themselves to be either “very” or “somewhat” good at gift wrapping. Nineteen percent of respondents said they were somewhat bad at gift-wrapping, while another 11 percent said they were very bad at it. You’re also in the minority, per YouGov, if you’ve never regifted a present to someone else. Only about one-third of adult respondents (31 percent) claimed that they’ve never done so, but most admitted to doing it: 29 percent said they’ve regifted once or twice, 18 percent said they’ve regifted “several times” and 13 percent of respondents have regifted “many times.”
The end of the calendar year has also led certain national pollsters — Marist, specifically — to find out which word or phrase U.S. adults find most irritating. This year’s winner/loser? “Woke.” According to their survey, about one-third of Americans (35 percent) agree that “woke” is the most annoying word used in conversation. Coming in second was the word “whatever” (22 percent), followed by “it is what it is” (15 percent). But while attacking “woke” and “wokeism” was initially the crusade-du-jour by the GOP, it appears that a number of survey respondents — regardless of party ID — found the term annoying. In fact, 31 percent of Democrats, 39 percent of Republicans and 38 percent of independent voters listed “woke” as the most bothersome word used in conversation. For context, 2021’s winners, according to Marist, were “Trump” and “coronavirus.”
With the 2022 midterm elections behind us, all eyes are now on the 2024 presidential race. And new polling data from Morning Consult seemingly suggests that voters might be looking at another head-to-head matchup between Trump and President Biden (assuming the latter runs for reelection, which looks increasingly likely). On the Republican side, Morning Consult’s tracking among potential GOP primary voters¹ gives Trump a wide lead (48 percent) over potential competitors like Florida Gov. Ron DeSantis, (33 percent) former Vice President Mike Pence (8 percent), Texas Sen. Ted Cruz (3 percent) and others. Pitted directly against DeSantis, however, Trump has a harder time breaking through. When potential GOP primary voters were asked who they’d vote for in a primary election or caucus if it were held in their state today, 45 percent of respondents said DeSantis while 44 percent said Trump. Eleven percent said they didn’t know or had no opinion.
According to newly released survey data from Gallup, Americans’ assessment of their own mental health is at an all-time low. Currently, just about 3-in-10 U.S. adults (31 percent) described their mental or emotional well-being as “excellent” — the lowest rating Gallup has recorded since it began asking respondents this question in 2001. The author notes, however, that part of the downward trend might be attributed to the COVID-19: Before then, Americans’ “excellent” ratings ranged in the 40s. Those numbers didn’t begin to tick down until late 2020; that year, the percentage of adults who felt that their mental health was in “excellent” condition dropped to the 30s for the first time at 34 percent. The demographic groups least likely to say that their mental health and emotional well-being were “excellent” were people between the ages of 18 and 34 (20 percent), people who made less than $40,000 annually (21 percent) and women (28 percent).
In a calendar year that featured a major election and the rippling effects of a global pandemic, which headlines stuck out most to Americans? According to Morning Consult, this year’s most salient news events, according to registered voters, were the Uvalde shooting (73 percent), the fall of Roe v. Wade (71 percent), Queen Elizabeth II’s death (71 percent) and Hurricane Ian (70 percent). But there was a partisan gap in news salience, too. While Democrats (78 percent) and independent voters (72 percent) both listed the Uvalde shooting — which left 21 people dead — as the top news event that they saw, heard or read “a lot” about, Republicans were more likely to put Hurricane Ian (70 percent) in the No. 1 slot. Among registered GOP voters, the Texas shooting ranked fourth (69 percent) after the hurricane, the fall of Roe (70 percent) and the queen’s death (70 percent).

Biden approval

According to FiveThirtyEight’s presidential approval tracker,² 43 percent of Americans approve of the job Biden is doing as president, while 51.6 percent disapprove (a net approval rating of -8.6 points). At this time last week, 43.0 percent approved and 51.3 percent disapproved (a net approval rating of -8.3 points). One month ago, Biden had an approval rating of 41.5 percent and a disapproval rating of 53.5 percent, for a net approval rating of -12.0 points.

33 Cool Charts We Made In 2022

Tue, 20 Dec 2022 13:00:00 +0200

2022 In Review

33 Cool Charts We Made In 2022

In 2022, FiveThirtyEight’s visual journalists covered the midterm elections, the end of Roe v. Wade and sports stories ranging from the World Cup to changes in Major League Baseball’s pitch timing rules. Here are some of the most interesting — and weird and colorful and complicated — charts we made in the last 12 months.

Charts are grouped by topic but are not in any particular order beyond that. Click any of them to read the story featuring that chart.

Politics

Sports

Science

Can You Make It To 2023?

Fri, 16 Dec 2022 15:00:00 +0200

The Riddler

Can You Make It To 2023?

Illustration by Guillaume Kurkdjian

Due to the holidays, the next column will appear on Jan. 6. See you in 2023!

Riddler Express

From Dean Ballard comes a puzzle to help us ring in the new year, 2023:

The Fibonacci sequence begins with the numbers 1 and 1,² with each new term in the sequence equal to the sum of the preceding two. The first few numbers of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 and so on.

One can also make variations of the Fibonacci sequence by starting with a different pair of

numbers. For example, the sequence that starts with 1 and 3 is 1, 3, 4, 7, 11, 18, 29, 47, 76 and so on. Generalizing further, a “tribonacci” sequence starts with three whole numbers, with each new term equal to the sum of the preceding three.

Many tribonacci sequences include the number 2023. For example, if you start with 23, 1000 and 1000, then the very next term will be 2023. Your challenge is to find starting whole numbers a, b and c so that 2023 is somewhere in their tribonacci sequence, a ≤ b ≤ c, and the sum a + b + c is as small as possible.

Submit your answer

Riddler Classic

From Gary Yane comes a a puzzle that’s just in time for Christmas:

Every Christmas, Gary’s family has a gift exchange. And every year, there is a big fight over how much folks should spend on the gifts. This year, they decided to pair up. So if Virginia gives Justin a gift, then Justin gives Virginia a gift. This way, while there will still be arguments, only two people will be involved in each argument.

There are 20 people in the gift exchange. In the first round, everyone writes down the name of a random person (other than themselves) and the names go in a hat. Then if two people randomly pick each other’s names out of that hat, they will exchange gifts, and they no longer participate in the drawing. The remaining family members go on to round two. Again, they write down the name of anyone left, and again, any two people who pick each other exchange gifts.

This continues until everyone is paired up. And yes, if exactly two people remain, they still go through the process of selecting each other, even though they know who their partner will be.

On average, what is the expected number of rounds until everyone is paired up?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to Nick Imholte of Charlotte, North Carolina, winner of last week’s Riddler Express.

World Cup group play consists of eight groups, each with four teams. The four teams in a group all play each other once (for a total of six matches), earning three points for a win, one point for a draw and zero points for a loss.

Last week, after group play in a particular group, all four teams had different numbers of points. The first-place team had A points, the second-place team B points, the third place team C points and the last-place team D points. Your task was to find all possible quadruples (A, B, C, D).

Several solvers brute-forced their way through this puzzle. With six matches, each resulting in one of three outcomes (one team wins, the other team wins or it’s a draw), there were a total of 3⁶, or 729, cases to consider. That wasn’t an unreasonable number of cases for those with computer assistance.

But for those who solved this by hand, I salute you!

One way to get started was to look for possible values of A. The best a team could have done was win all three matches, earning 9 points. But what was the minimum value of A? If A had been 4 — meaning the first-place team had a win, a draw and a loss — then there was no way for the four teams to have unique point totals and for the collective number of wins to equal the collective number of losses. That meant A had to be 5, 6, 7, 8 or 9.

Reader Kathy Estevez simultaneously narrowed down the possible values of A+B+C+D. If all the matches had been draws, this sum would have been 12; if none of them had been draws, the sum was 18. In other words, it had to be somewhere between 12 and 18. If it had been 12, then all four teams would have had 3 points each. If it had been 13, then (A, B, C, D) had to be (5, 3, 3, 2). That meant the sum had to be between 14 and 18.

All this forethought helped reduce the number of cases. For example, if A had been 5, then the only way for A+B+C+D to be at least 14 was if (A, B, C, D) was (5, 4, 3, 2). In the end, there were 13 possible quadruples:

(5, 4, 3, 2)
(6, 5, 4, 1)
(7, 4, 3, 1)
(7, 4, 3, 2)
(7, 5, 2, 1)
(7, 5, 3, 1)
(7, 5, 4, 0)
(7, 6, 2, 1)
(7, 6, 3, 1)
(7, 6, 4, 0)
(9, 4, 2, 1)
(9, 4, 3, 1)
(9, 6, 3, 0)

In this year’s World Cup, three of these unique scores occurred. Group A resulted in the quadruple (7, 6, 4, 0), Group B resulted in (7, 5, 3, 1) and Group F resulted in (7, 5, 4, 0).

Solution to last week’s Riddler Classic

Congratulations to Michael M. Amati of Geneseo, New York, winner of last week’s Riddler Classic.

Last week’s Riddler Football Playoff (RFP) consisted of four teams. Each team was assigned a random real number between 0 and 1, representing the “quality” of the team. If team A had quality a and team B had quality b, then the probability that team A defeated team B in a game was a/(a+b).³

In the semifinal games of the playoff, the team with the highest quality (the “1 seed”) played the team with the lowest quality (the “4 seed”), while the other two teams played each other as well. The two teams that won their respective semifinal games then played each other in the final.

On average, what was the quality of the RFP champion?

Before tackling (see what I did there?) the semifinal games, let’s look at what would have happened if two teams faced off. The values of a and b could have been anywhere from 0 to 1. Team A won with a probability of a/(a+b), in which case the quality of the winner was a. Meanwhile, team B won with a probability of b/(a+b), in which case the quality of the winner was b. That meant the average quality of the winner was a²/(a+b) + b²/(a+b), or (a²+b²)/(a+b). Integrating this expression over a and b gave you (4·ln(2)−1)/3 — not the prettiest expression! — or about 59.1 percent.

Well, if you thought the two-team case was complicated, the four-team playoff was a lot more so. And on top of that was another wrinkle: The teams were ranked and seeded for their semifinals.

Many solvers, like David Ding, simulated thousands or even millions of playoffs to find the expected value. After 10 million simulations, David found the average quality of the winning team was approximately 0.674.

As with the case of two teams, you could use integration to find an expression for the four-team case as well. Suppose the four teams by order of seed were A, B, C and D, with respective qualities a, b, c and d. The probability that team A was the champion was a/(a+d) · [a/(a+b) · b/(b+c) + a/(a+c) · c/(b+c)]. In other words, team A had to defeat team D and then go on to defeat whichever team won the other semifinal (B or C). After finding similar probabilities for teams B, C and D, to find the expected value of the winner you had to multiply each probability by each team’s respective quality, then add these values up.

But you weren’t done yet! You still had to integrate over all possible quadrupes (a, b, c, d) with the restriction a > b > c > d. One way to do this was to integrate over a from 0 to 1, b from 0 to a, c from 0 to b and d from 0 to c. Alternatively, you could have integrated over d from 0 to 1, c from d to 1, b from c to 1 and a from b to 1. Either way, this region of integration was a four-dimensional tetrahedron with a volume⁴ of 1/24. To find the average quality of the winner, you had to divide your quadruple integral by 1/24.

This integral was rather nasty, to put it mildly. Nevertheless, several solvers, including Izumihara Ryoma, Benjamin Phillabaum and Laurent Lessard, evaluated the integral, finding the champion’s average quality was 2/5 · (23 − (29+2𝜋²)ln(2) + 39ln²(2) − 8ln³(2) − 3𝜻(3)), or about 0.67354. Here, 𝜻 is the Riemann zeta function, which, to my knowledge, has never before appeared in this column. Solver Ryan McShane plotted the probability distribution for the winning team, which had a nice little skew:

Given the complexity of the double integral in the two-team scenario, and the jump in complexity of the quadruple integral in the four-team scenario, I dare not ask about an octuple integral in the eight-team scenario.

Want more puzzles?

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

Want to submit a riddle?

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

Can You Win The Riddler Football Playoff?

Fri, 09 Dec 2022 15:00:00 +0200

The Riddler

Can You Win The Riddler Football Playoff?

Illustration by Guillaume Kurkdjian

Riddler Express

From Jenny Mitchell comes another great puzzle about soccer and/or football:

After group play in a particular group, all four teams have different numbers of points. The first-place team has A points, the second-place team B points, the third place team C points and the last-place team D points. Find all possible quadruples (A, B, C, D).

Submit your answer

Riddler Classic

Speaking of “football,” the Riddler Football Playoff (RFP) consists of four teams. Each team is assigned a random real number between 0 and 1, representing the “quality” of the team. If team A has quality a and team B has quality b, then the probability that team A will defeat team B in a game is a/(a+b).²

In the semifinal games of the playoff, the team with the highest quality (the “1 seed”) plays the team with the lowest quality (the “4 seed”), while the other two teams play each other as well. The two teams that win their respective semifinal games then play each other in the final.

On average, what is the quality of the RFP champion?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to Pirmin Patel of London, winner of last week’s Riddler Express.

Last week, Maryam was playing billiards on a 1 meter by 1 meter square table. She placed the ball in one of the corners, aiming to strike the ball so that it traveled as far as possible before hitting a wall for the third time. Note that the ball didn’t necessarily have to hit three different walls of the table.

You could assume that the ball traveled in a straight path and that it bounced off a wall as you’d expect.³ You could also assume that it was impossible for Maryam to hit the ball precisely in one of the corners of the table. Instead, it would have hit both sides that were adjacent to the corner.

What was the farthest the ball could have traveled before hitting a wall for the third time?

Several solvers, like David Ding and Emily Boyajian, analyzed the different directions in which the ball could be struck, then split these into cases and found the optimal path with trigonometry and coordinate geometry.

But, as noted by solver Jenny Mitchell, this puzzle was reminiscent of another one from almost a year ago, in which Amare the ant had to find the shortest path around a triangle. Unlike billiards, Amare could change direction at any point. To find the answer, a good strategy was to reflect the triangle every time Amare reached an edge. So let’s try a similar approach here.

Solver Alex Livingston reflected the square table across any wall the ball hit, as shown below:

The blue border in the bottom right of the figure were all the places the ball could hit the wall for the third time. So the question became: Which point on the blue border was farthest from the starting location in the top left?

In the diagram above, there were two such points, one of which is shown. (The other was located symmetrically across the diagonal.) To maximize the distance traveled, Maryam had to aim one-third along one of the opposing sides. The ball then ricocheted and hit a point two-thirds along the side opposite that one, and then finally hit the corner opposite from where the ball started. Yes, the precise corner itself counted as hitting two walls, but you could instead assume the ball hit very, very close to the corner and the result would have effectively been the same.

With the Pythagorean theorem, you found that the total distance traveled was √(3²+1²), or √10 — roughly 3.16 meters.

For extra credit, you had to find the farthest the ball could travel before hitting a wall for the N^th time. Again, by reflecting the square table every time the ball hit a wall, you could create a similar triangular diagram. Instead of going down 3 meters and across 1 meter, this time the longest path called for going down N meters and across 1 meter. By the Pythagorean theorem, this distance was √(N²+1²).

By the way, a few solvers, like Ravi Fernando of Berkeley, California, made the connection between billiards and “Maryam” — in this case, Maryam Mirzakhani. Mirzakhani was Ravi’s pre-major advisor when he was a freshman at Stanford!

Solution to last week’s Riddler Classic

Congratulations to Andrew Love, Jr. of Columbia, Maryland, winner of last week’s Riddler Classic.

Last week, a certain hotel in Qatar was hosting 11 American fans and seven Dutch fans. Since no alcohol was available inside the stadiums, the fans spent the afternoon at the hotel bar before shuttle buses took them to a match. Then they haphazardly wrote their room numbers on a big board by the concierge desk.

To avoid any rowdiness between rival fans, shuttle bus drivers were instructed to ferry American and Dutch fans separately. To ensure this, a shuttle pulled up in front of the hotel, and its driver called out room numbers from the board, one by one at random. As long as they supported the same team, each fan climbed aboard the bus and their room number was erased. Once the driver called out the room number of a fan for the second team, the shuttle left with only the fans of the single team aboard. The next shuttle then pulled up and repeated the process.

What was the probability that the last shuttle ferried American fans?

Many readers thought the answer should have been proportional to the number of American fans. After all, if there were more American fans than Dutch fans, it made sense that it was more likely that the last fan to be picked up would have been American. If you looked at all 18 choose 7 ways the fans could have been ordered, the last fan was American in precisely 11/18 of them.

However, the answer was not 11/18. Why? Because every time a bus driver called a fan that was different from those they had previously called, that fan returned to the pool and wouldn’t necessarily have been called first by the next driver. If they had been, then the answer would indeed have been 11/18.

To find the correct answer, several solvers like Daniel Gershenson and Mike Donner used dynamic programming techniques, although these didn’t quite offer a satisfying explanation for what was going on.

Suppose there were a American fans and d Dutch fans, with both a and d greater than or equal to 1. Then one of three things could have happened when the next bus comes around:

The bus picked up all a American fans.
The bus picked up all d Dutch fans.
The bus picked up less than a American fans or less than d Dutch fans, in which case there were new numbers of American or Dutch fans remaining, still both greater than 1.

While it was tempting to dig into the third case and work out all the possibilities, this wasn’t necessary. That was because only this whole transportation scenario had to end with one of the first two cases, with the penultimate bus picking up all the fans on one side, and the last bus picking up all the remaining fans on the other side.

So, how likely were these first two cases? With a+d fans, and (a+d) choose a total orderings, there was only one way to order them so that all a American fans came before all d Dutch fans. And there was also only one way to order them so that all d Dutch fans came before all a American fans. In other words, the first two cases were equally likely. That meant the penultimate bus was equally likely to transport American or Dutch fans, and the same went for the last bus. The probability that the last bus shuttled American fans was 50 percent.

For extra credit, you had to find the expected number of shuttle buses needed to ferry all 18 fans. At this point, cleverly reasoning about symmetry was a lot less helpful than dynamic programming. The answer turned out to be about 9.545 buses, which was tantalizingly close to how many buses you’d expect (9.556) if each bus instead first boarded the passenger the previous bus had refused to take.

In any case, while either group of fans was likely to board the last bus, the Dutch fans had the last laugh.

Want more puzzles?

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

Want to submit a riddle?

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