The Riddler

Can You Take Down All The Bottles Of Beer?

By Zach Wissner-Gross

Feb. 3, 2023, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">¹ and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.

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 10.75 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^th to 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):

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.

The Riddler

Can You Defeat The TikTok Meme?

By Zach Wissner-Gross

Jan. 27, 2023, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">³ 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 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.

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 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.

Pollapalooza

Which Parents Are The Most Tired?

By Amelia Thomson-DeVeaux

Jan. 27, 2023, at 6:00 AM

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.⁴ 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.

Watch: https://abcnews.go.com/fivethirtyeight/video/debt-ceiling-countdown-begins-fivethirtyeight-politics-podcast-96623875

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.⁵ 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.

Watch: https://abcnews.go.com/fivethirtyeight/video/george-santos-fivethirtyeight-politics-podcast-96623828

Biden approval

According to FiveThirtyEight’s presidential approval tracker,⁶ 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.

CORRECTION (Jan. 27, 2023, 10:45 a.m.): A previous version of this story included a chart with incorrect data on mothers’ and fathers’ thoughts about who provided more comfort or emotional support to their children. The shares of mothers who thought mothers did more, fathers did more and both did about equal were incorrect, as was the share of fathers who thought both did about equal. They have been updated.

Watch: https://abcnews.go.com/fivethirtyeight/video/debt-ceiling-countdown-begins-fivethirtyeight-politics-podcast-96623875

The Riddler

Can You Make A Speedy Delivery?

By Zach Wissner-Gross

Jan. 20, 2023, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">⁷ 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 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!⁸

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.

The Riddler

Can You Fend Off The Alien Armada?

By Zach Wissner-Gross

Jan. 6, 2023, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">⁹ 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.

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

Twitter

How Americans Really Feel About Elon Musk

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

By Alex Samuels

Dec. 23, 2022, at 6:00 AM

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 voters¹⁰ 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,¹¹ 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.

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

The Riddler

Can You Make It To 2023?

By Zach Wissner-Gross

Dec. 16, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">¹² 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.

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 1,¹³ 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).¹⁴

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 volume¹⁵ 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.

The Riddler

Can You Win The Riddler Football Playoff?

By Zach Wissner-Gross

Dec. 9, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">¹⁶ 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 Jenny Mitchell comes another great puzzle about soccer and/or football:

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.

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).¹⁷

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.¹⁸ 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.

Data Is Plural

The Datasets We're Looking At This Week

By Jeremy Singer-Vine

Dec. 7, 2022, at 12:00 PM

You’re reading Data Is Plural, a weekly newsletter of useful/curious datasets. Below you’ll find the Dec. 7, 2022, edition, reprinted with permission at FiveThirtyEight. After today’s edition, FiveThirtyEight will no longer be republishing Data Is Plural. However, you can still subscribe to the original newsletter at its website or by entering your email in the subscription box at the end of this article.

2022.12.07 edition

Work stoppages, avian flu detections, social media suppression, literature prizewinners and video games.

Work stoppages. The U.S. Bureau of Labor Statistics’ Work Stoppages program collects and publishes data on “major work stoppages involving 1,000 or more workers lasting one full shift or longer.” The program’s main dataset lists each major strike or lockout since 1993, the organizations involved, employer industry and ownership type, start and end dates, number of workers participating and total worker-days idle. Its annual dataset counts the number of major stoppages, workers involved and worker-days idle for each year since 1947. Related: The Federal Mediation and Conciliation Service used to publish data on all stoppages that its mediators entered into its case system (regardless of number of workers involved), but stopped doing so in late 2020. Forest Gregg has rescued and standardized the archived records. [h/t Chartr]

Avian flu detections. The USDA’s Animal and Plant Health Inspection Service has been tracking local cases of avian influenza detected this year in commercial and backyard flocks as well as in wild birds and in mammals. For each infected flock, the agency’s public data indicate the flock’s state, county, producer type, number of birds affected and confirmation date. The mammal and wild bird data indicate the state, county, date detected, flu strain and species. Read more: “More Than 52 Million Birds in the U.S. Are Dead Because of Avian Flu” (Smithsonian Magazine). [h/t Ed Vine]

Social media suppression. Researchers at Surfshark have been monitoring government-imposed social media shutdowns and restrictions, drawing on reports from NetBlocks, AccessNow (DIP 2021.11.03), news publications and other sources. The project’s spreadsheet highlights each case from 2015 to the present; it lists the country, start/end date, particular services affected (Facebook, Twitter, YouTube, etc.) and observed connections to political events (e.g., elections, protests, other turmoil). [h/t Agneska Sablovskaja]

Literature prizewinners. Claire Grossman et al. have compiled a dataset of “the winners and judges of prizes for prose, poetry, or unspecified genre between 1918 and 2020 with a purse of $10,000 and over.” The 7,100-plus entries, shared through the Post45 Data Collective, relate to 50 awards and fellowships, plus the Library of Congress’s poet laureateship. Each entry indicates the prize name, institution, type, genre, year and dollar amount, plus the winner/judge name, gender and educational affiliations. [h/t Melanie Walsh]

Video games. Developer Vladimir Belyaev has constructed a dataset representing “all games available” on Steam, the massively popular video game platform and store. It indicates each game’s name, price, developer, languages, genre categories, ratings from third-party websites and more. [h/t Saul Pwanson]

The Riddler

Can You Separate The World Cup Fans?

By Zach Wissner-Gross

Dec. 2, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">¹⁹ 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

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

Assume the ball travels in a straight path and that it bounces off a wall as you’d expect.²⁰ Also assume that it’s impossible for Maryam to hit the ball precisely in one of the corners of the table. Instead, it will hit both sides that are adjacent to the corner.

What is the farthest the ball can travel before hitting a wall for the third time?

Extra credit: What is the farthest the ball can travel before hitting a wall for the N^th time?

Submit your answer

Riddler Classic

From Starvind comes a timely puzzle about the FIFA World Cup:

A certain hotel in Qatar is hosting 11 American fans and seven Dutch fans. Since no alcohol is available inside the stadiums, the fans spend the afternoon at the hotel bar before shuttle buses will take them to a match. Then, they haphazardly write their room numbers on a big board by the concierge desk.

To avoid any rowdiness between rival fans, shuttle bus drivers have been instructed to ferry American and Dutch fans separately. To ensure this, a shuttle pulls up in front of the hotel, and its driver calls out room numbers from the board, one by one at random. As long as they support the same team, each fan climbs aboard the bus and their room number is erased. Once the driver calls out the room number of a fan for the second team, the shuttle leaves with only the fans of the single team aboard. The next shuttle then pulls up and repeats the process.

What is the probability that the last shuttle ferries American fans?

Extra credit: On average, what is the expected number of shuttle buses needed to ferry all 18 fans?

Submit your answer

Solution to the last Riddler Express

Congratulations to Michael Ringel of Jacksonville, Florida, winner of the last Riddler Express.

In the last Riddler Express, I was running a 5-kilometer “turkey trot” race. Before the race began, all the runners (including me) gathered behind the starting line in a random order. Once the race began, everyone started running at their own fixed pace.

I hadn’t run in several years, so I wasn’t sure how my pace would compare to that of the other racers. Nevertheless, once the race began, I found myself passing quite a few other runners — and being passed myself. It was safe to assume that if my pace was faster than that of any runner who started in front of me, I would pass them at some point during the race.

On average, what fraction of the other runners could I have expected to pass during the race? 

First off, this was equivalent to asking what fraction of the runners in front of me were also slower than I was.

While the puzzle gave you no information about the distribution of paces among the different runners, the shape of this distribution didn’t matter. Suppose you ordered all the runners from fastest to slowest. Since I didn’t know how my pace compared to that of the other runners, you could assume (for the purposes of this puzzle) that I was equally likely to be anywhere in that ordering. And so, on average, I was faster than half of the runners and slower than the other half.

Now suppose I had been the first person in the queue at the starting line. In this case, I couldn’t possibly have passed anyone. If instead I had been the last person in the queue, I would have passed half of the runners on average.

But what if I had started somewhere else in the queue? As noted by Lisa of Los Altos, California, since the queue was random, I could expect to be faster than half the people in front of me, no matter where I started. Meanwhile, the number of people in front of me ranged from no one to everyone (minus me, of course — but it was safe to assume there were many runners in the race), which meant that, on average, half of the runners were in front of me.

To find the expected number of slower runners in front of me, all I had to do was multiply these two halves together, and one-half times one-half equals one-quarter.

Unfortunately, thanks to symmetry, I also expected to be passed by a quarter of the racers.

Solution to the last Riddler Classic

Congratulations to Sanandan Swaminathan of San Jose, California, winner of the last Riddler Classic.

In the last Riddler Classic, I had five kinds of fair Platonic dice: tetrahedra (whose faces are numbered 1-4), cubes (numbered 1-6), octahedra (numbered 1-8), dodecahedra (numbered 1-12) and icosahedra (numbered 1-20).²¹

When I rolled two of the cubes, there was a single most likely sum: 7. But when I rolled one cube and two tetrahedra, there was no single most likely sum — 8 and 9 were both equally likely.

Which whole numbers were never the single most likely sum, no matter which combinations of dice I picked?

As we already said, rolling two cubes had a most likely sum of 7. Similarly, rolling two tetrahedra had a most likely sum of 5, rolling two octahedra had a most likely sum of 9, rolling two dodecahedra had a most likely sum of 13 and rolling two icosahedra had a most likely sum of 21. (In general, adding two numbers chosen randomly, uniformly and independently from 1 to N had a most likely sum of N+1.)

From there, solver Michael Bradley of London, England, looked at what happened when you combined pairs of identical dice. For example, Michael found that if you rolled a pair of cubes and another pair of cubes, the most likely sum was 7 plus another 7, or 14. And if you rolled a pair of tetrahedra and a pair of icosahedra, the most likely sum was 5 plus 21, or 26.

Why did this work? Finding the sum of random variables from two different distributions is mathematically equivalent to convolving those two distributions. Convolving the symmetric triangular distributions you got from rolling two identical dice gave you more complicated distributions, but they always featured a unique central peak.

This meant you could pick pairs of dice to get any most likely number that was a linear combination of 5, 7, 9, 13 and 21.²² The only whole numbers that weren’t linear combinations of those five numbers were 1, 2, 3, 4, 6, 8 and 11.

Several readers stopped there, but 11 turned out to be a rather interesting case. If you had three tetrahedra and one cube, then the average sum was 2.5 + 2.5 + 2.5 + 3.5, or 11. But 11 also turned out to be the single most likely sum in this case, with a probability of 11/48. Meanwhile, the probabilities of 10 or 12 were both slightly lower, at 53/384.

As for 8, it might have been tempting to roll a cube and an octahedron. Here, the average sum was 3.5 + 4.5, which was indeed 8. However, 8 was not the single most likely sum, as 7, 8 and 9 were equally likely with a probability of 1/8.

In the end, the only single most likely sums that were not obtainable were 1, 2, 3, 4, 6 and 8.

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.

Data Is Plural

The Datasets We're Looking At This Week

By Jeremy Singer-Vine

Nov. 30, 2022, at 12:04 PM

You’re reading Data Is Plural, a weekly newsletter of useful/curious datasets. Below you’ll find the Nov. 30, 2022, edition, reprinted with permission at FiveThirtyEight.

2022.11.30 edition

Pills, per-pupil spending, travelers’ coronavirus variants, Indonesia earthquake intensities and more roadkill.

Pills. From its launch in 2009 until its retirement last year, the National Library of Medicine’s Pillbox project collected and created 8,600-plus photographs of medical pills. The images, which are still available to download, are accompanied by a dataset that provides information on 83,000-plus pills’ shape, size, color, markings, dosage and other characteristics derived from drug labels. Related: The library’s DailyMed service provides frequently updated images and data from 140,000-plus labels submitted to the FDA for drugs and other regulated products. As seen in: Jon Keegan’s Pillbox overview in Beautiful Public Data. [h/t Giuseppe Sollazzo]

Per-pupil spending. The National Education Resource Database on Schools (“NERD$”) describes itself as the “first-ever national dataset of public K-12 spending by school.” Its researchers, based at Georgetown University, aggregate and standardize the expenditure disclosures that the Every Student Succeeds Act requires states to publish. You can explore and download the data they’ve processed for fiscal year 2019, including spending totals, enrollment counts and normalized figures that facilitate cross-state comparisons. For 2020 to 2022, you can access “the raw files we obtain from states while our team conducts validation checks and norms the data.” As seen in: “How much money do states spend on education?” (USAFacts). [h/t Douglas Hummel-Price]

Travelers’ coronavirus variants. In the past year, the CDC’s Traveler-Based Genomic Surveillance program has collected over 60,000 voluntary nasal swabs from people disembarking international flights at four major U.S. airports. The agency uses the samples as an “early warning system” to detect emerging SARS-CoV-2 variants and publishes weekly metrics that include participation counts, positivity rates (per pooled sample) and variant distributions. Read more: An interview with two private industry experts working on the program, by the COVID-19 Data Dispatch’s Betsy Ladyzhets.

Indonesia earthquake intensities. Gempa Nusantara, a database compiled by Stacey S. Martin et al., uses historical documents to catalog over 7,300 “macroseismic effects” of 1,200 earthquakes near Indonesia during a four-century span, from 1546 to 1950. It provides summaries of the local reports and categorizes the effects according to the European Macroseismic Scale, which focuses on the intensity of ground-shaking and potential impacts on buildings and terrain.

More roadkill. Florian Heigl et al. have compiled a pair of datasets containing 15,000-plus reports of vertebrate roadkill from 2014 to 2020, submitted by 900-plus people through a phone app. The datasets differ in identification confidence, but both provide locations, dates and taxonomic classifications. Although the records span 40-plus countries, the majority come from Austria, where the project is now focused. Previously: Andean roadkill (DIP 2021.07.07).

The Riddler

How Many Turkey Trotters Can You Pass?

By Zach Wissner-Gross

Nov. 18, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">²³ 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.

Due to the holidays, the next column will appear on Dec. 2. See you then!

Riddler Express

I recently competed in a 5-kilometer “turkey trot” race. Before the race began, all the runners (including me) gathered behind the starting line in a random order. Once the race began, everyone started running at their own fixed pace.

I hadn’t run in several years, so I wasn’t sure how my pace would compare to that of the other racers. Nevertheless, once the race began, I found myself passing quite a few other runners — and being passed myself.

On average, what fraction of the other runners could I expect to pass during the race? (Assume that if my pace is faster than that of another runner’s who starts in front of me, I will pass them at some point during the race.)

Submit your answer

Riddler Classic

From Michael Branicky comes a challenge involve many, many dice:

I have five kinds of fair Platonic dice: tetrahedra (whose faces are numbered 1-4), cubes (numbered 1-6), octahedra (numbered 1-8), dodecahedra (numbered 1-12) and icosahedra (numbered 1-20).²⁴

When I roll two of the cubes, there is a single most likely sum: seven. But when I roll one cube and two tetrahedra, there is no single most likely sum — eight and nine are both equally likely.

Which whole numbers are never the single most likely sum, no matter which combinations of dice I pick?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to Andrew Busch of London, winner of last week’s Riddler Express.

Last week, in an effort to break open the gates of the city Tinas Mirith, an army of orcs first tried using a battering ram, but to no avail. They next erected a 100-foot pole with a very massive weight at the top (i.e., the weight was much, much heavier than the rest of the pole). The pole was also anchored at the bottom, so that as the weight fell the entire pole rotated around its bottom without slipping.

How far away should the orcs have positioned the vertical pole from the gates so that when the weight came crashing down, its horizontal speed was as great as possible?

First, let’s examine an animation of the falling weight:

As the weight fell, it moved faster. At the same time, due to the rotation, its horizontal motion turns into vertical motion. Now at some point, its horizontal speed was at a maximum. One way to find this maximum was to do what many students do time and again in their first physics course: Write an equation describing conservation of energy.

As the weight came down, its gravitational potential energy became kinetic energy. If we call the pole length L (100 feet in the puzzle), then by the time the pole made an angle 𝜃 with the ground, the weight was at a height Lsin𝜃. That meant the change in the height was L−Lsin𝜃, or L(1−sin𝜃). Therefore, the change in gravitational potential energy was mgL(1−sin𝜃), where m was the mass of the weight and g was the acceleration due to gravity — roughly 32 feet/sec².

This difference in potential energy was what became kinetic energy, which could be expressed as 1/2mv², where v was the speed of the weight. Setting these equal gave you v = √(2gL(1−sin𝜃)). But you didn’t want the velocity, you wanted the horizontal velocity, which was equal to vsin(𝜃). And so the function you were trying to maximize was sin𝜃√(2gL(1−sin𝜃)). Pulling out a few constants that didn’t matter, this became sin𝜃√(1−sin𝜃). To maximize this function, you had to take the derivative with respect to 𝜃 and set that equal to zero. After some friendly cancelation, this left you with sin𝜃 = 2/3.

That was a nice result, but this wasn’t what the riddle was asking for. What was the right distance between the pole and the gates? That was L times the cosine of this optimal angle. If sin𝜃 was 2/3, then cos𝜃 was (√5)/3, which meant the distance was 100/3·√5, or about 74.5 feet.

A few folks, like Ricky Reusser, opted for a more challenging, “bad approach.” As Ricky stated: “This problem is very easily solved with a simple energy argument, but let’s brute force it from the equations of motion instead!” Ricky proceeded to solve second-order differential equations with Jacobi elliptic functions. In the end, the answer was the same.

After placing their pole in just the right spot, the orcs successfully knocked down the gates of Tinas Mirith, securing a victory for Sord Lauron. And that was the end of all things.

Solution to last week’s Riddler Classic

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

Last week, I was slicing a square peanut butter and jelly sandwich. But rather than making a standard horizontal or diagonal cut, I instead picked two random points along the perimeter of the sandwich and made a straight cut from one point to the other. (These points could have been on the same side.)

My slice was “reasonable” if I cut the square into two pieces and the smaller resulting piece had an area that was at least one-quarter of the whole area. What was the probability that my slice was reasonable?

To see what’s happening here, we can start by looking at special cases, like when the first point I picked was one of the square’s corners or at the midpoint of one of its sides, as shown below.

When the first point was in a corner, the cut was reasonable whenever the other point was closer to the opposite corner than any other corner, a region that made up a quarter of the square’s perimeter. And when the first point was a midpoint, the cut was reasonable whenever the other point was somewhere on the opposite side, another region that made up a quarter of the square’s perimeter. And so, for both of these special cases, my probability of making a reasonable cut was 1/4. Could the answer then simply have been 1/4?

This being a Classic, that was not the answer. As it turned out, whenever the first point was anywhere else, the probability of making a reasonable cut was greater than 1/4. Solvers Trey Goesh and Starvind made animated graphs suggesting this was the case.

To determine the exact probability, suppose the first point was a distance x from the nearest corner of the square, and let’s assume that the square had side length 1, as shown below. Note here that x was equally likely to be anywhere from 0 to 0.5.

If the second point was on the opposite side, the cut was reasonable if it was within 1/2+x of the opposite corner, creating a trapezoid whose area was at least 1/4. But if the second point was on the other side touching that opposite corner, then the cut was reasonable when that second point was within (1−2x)/(2−2x) of that corner, creating a triangle whose area was at least 1/4.

To find the probability that the cut was reasonable, you had to add these two distances together, divide by 4 (the total perimeter of the square), integrate over x from 0 to 0.5, and finally divide by 0.5 (i.e., multiply by 2), since we wanted the average probability over this range of x. The result of this integral was relatively concise: (7−ln(16))/16, or about 26.4 percent.

In the end, 25 percent was not a bad guess. It seems I am only slightly more reasonable than that.

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.

Data Is Plural

The Datasets We're Looking At This Week

By Jeremy Singer-Vine

Nov. 16, 2022, at 12:00 PM

You’re reading Data Is Plural, a weekly newsletter of useful/curious datasets. Below you’ll find the Nov. 16, 2022, edition, reprinted with permission at FiveThirtyEight.

2022.11.16 edition

Big emitters, disease outbreaks, permissively licensed code, impact craters and tinned fish.

Big emitters. Climate Trace, a nonprofit coalition launched in 2020, uses satellite imagery, sector-specific datasets and other sources to estimate greenhouse gas emissions in detail. Their most recent inventory, released last week, highlights 70,000-plus individual sites that “represent the top known sources of emissions in the power sector, oil and gas production and refining, shipping, aviation, mining, waste, agriculture, road transportation, and the production of steel, cement and aluminum.” You can download the data, explore sector- and country-level estimates and browse a map of the sites. Read more: Coverage in The New York Times. [h/t Ian Johnson]

Disease outbreaks. Juan Armando Torres Munguía et al. have built a dataset of infectious disease outbreaks based on information extracted from the World Health Organization’s Disease Outbreak News alerts (DIP 2022.03.30) and its coronavirus dashboard. The authors have clustered the outbreaks by disease (classified by ICD-10 and ICD-11 codes), country and year. Excluding the COVID-19 pandemic, this leads to 1,500-plus total combinations between January 1996 and March 2022, spanning 60-plus diseases and 200-plus countries/territories. [h/t Konstantin M. Wacker]

Permissively licensed code. The Stack, a new dataset from the BigCode project, “contains over 3TB of permissively licensed source code files covering 30 programming languages crawled from GitHub.” Those terabytes hold more than 300 million files extracted from repositories whose licenses place “minimal restrictions on how the software can be copied, modified and redistributed.” The dataset provides the contents of each file along with its repository name, path, size, programming language, detected licenses and several high-level metrics. Read more: An introductory Twitter thread and preprint paper. [h/t Karsten Johansson]

Impact craters. The Earth Impact Database, maintained by the University of New Brunswick’s Planetary and Space Science Centre, catalogs nearly 200 impact craters caused by meteorites that have crashed into the planet. It presents the name, location, diameter, estimated age, geology and other features of the craters, as well as photographs and bibliographies. Related: Cody Winchester has scraped the crater characteristics into CSV and GeoJSON files.

Tinned fish. Rainbow Tomatoes Garden is a farm in East Greenville, Pennsylvania, that also happens to run an online store selling “the largest selection of tinned seafood in the world.” Curator-owner Dan Waber publishes a spreadsheet of the store’s 630-plus offerings, listing each product’s name, type of seafood, brand, country of origin, tin size and price; whether it’s organic, certified kosher, smoked, boneless and/or skinless; and more. [h/t George Ho]

The Riddler

Can You Knock Down The Gates?

By Zach Wissner-Gross

Nov. 11, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">²⁵ 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 the fantastical land of Central Earth comes a physics riddle that will break down your doors:

In an effort to break open the gates of the city Tinas Mirith, an army of orcs first tried using a battering ram, but to no avail. They next erected a 100-foot pole with a very massive weight at the top (i.e., the weight is much, much heavier than the rest of the pole). The pole is also anchored at the bottom, so that as the weight falls the entire pole rotates around its bottom without slipping.

How far away should the orcs position the vertical pole from the gates so that when the weight comes crashing down on the gates, its horizontal speed is as great as possible?

Submit your answer

Riddler Classic

I have made a square peanut butter and jelly sandwich, and now it’s time to slice it. But rather than making a standard horizontal or diagonal cut, I instead pick two random points along the perimeter of the sandwich and make a straight cut from one point to the other. (These points can be on the same side.)

My slice is “reasonable” if I cut the square into two pieces and the smaller resulting piece has an area that is at least one-quarter of the whole area. What is the probability that my slice is reasonable?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to Ben Gundry of San Jose, California, winner of last week’s Riddler Express.

Last week, you were challenged to improve upon the Gregorian calendar. Now, each solar year consists of approximately 365.24217 mean solar days. That’s pretty close to 365.25, which is why it makes sense to have an extra day every four years. However, the Gregorian calendar is a little more precise: There are 97 leap years every 400 years, averaging out to 365.2425 days per year.

But could you make a better approximation than the Gregorian calendar? More specifically, you were asked to find numbers L and N (where N was less than 400) such that if every cycle of N years included L leap years, the average number of days per year was as close as possible to 365.24217.

Many solvers used a “brute force” approach, checking all the values of N from 1 to 399. For each value of N, you had to find the whole number L that resulted in L/N being as close as possible to 0.24217. Solver Tiago Batalhao knew this meant L was either floor(0.24217·N) or ceiling(0.24217·N) — the whole numbers on either side of 0.24217N. In fact, you could find the best L for a given N by rounding 0.24217·N to the nearest whole number. Finally, the best pair of N and L minimized the absolute difference between L/N and 0.24217, which could be written as a function of N: abs(round(0.24217·N)/N − 0.24217).

But brute force wasn’t the only way to solve this puzzle. A particularly elegant approach I’d like to highlight involved mediants and Farey sequences. Without getting into the details, we could start with the fractions 0/1 and 1/1. The mediant of these two fractions was calculated by adding across the numerators and denominators and denominators, which was 1/2. Since 0.24217 was between 0/1 and 1/2, we next calculated the mediant of these two fractions: 1/3. It was also between 0/1 and 1/3, and so the next mediant of interest was 1/4. And after that, it was 1/5 — the first mediant that was less than 0.24217. From there, you wanted the mediant of 1/5 and 1/4, which was 2/9.

Continuing in this fashion, the last mediant that was less than 0.24217 was 7/29, after which came 8/33. Their mediant was 15/62, after which came 23/95, 31/128 and 54/223. These last two fractions were on either side of 0.24217, and their mediant was 85/351, or approximately 0.242165. This turned out to be the best approximation where the denominator was less than 400, meaning N was 351 and L was 85.

In the end, having 85 leap years every 351 years was about 70 times more accurate (in terms of averaging out to the right number of solar days per solar year) than the Gregorian calendar’s 97 leap years out of 400. That means we might have to skip a scheduled leap day … in a few thousand years or so.

Solution to last week’s Riddler Classic

Congratulations to Adam Richardson of Old Hickory, Tennessee, winner of last week’s Riddler Classic.

Last week, it was peak fall foliage season in Riddler Nation, where the trees changed color in a rather particular way. Each tree independently began changing color at a random time between the autumnal equinox and the winter solstice. Then, at a random later time for each tree — between when that tree’s leaves had begun changing color and the winter solstice — the leaves of that tree would all fall off at once.

At a certain time of year, the fraction of trees with changing leaves was expected to peak. What was this maximal fraction?

Solver Tom Keith simulated thousands of trees — quite beautifully, I might add — finding that the peak appeared to occur 63 percent of the way through the fall. And at this peak, almost 37 percent of the trees had changing leaves.

Meanwhile, many solvers like Andrea Andenna were able to calculate the exact answer. To do this, you wanted to first determine the probability p(t) of any given tree having changing leaves as a function of time t, which you could conveniently rescale between 0 (representing the autumnal equinox) and 1 (representing the winter solstice). Then, you could use calculus to maximize p(t). 

So what was this probability distribution? A tree had changing leaves at time t if the leaves started changing before t and — assuming that was true — the leaves fell after t. The probability of the former was simply t. Now, assuming the leaves started changing at a time x prior to t, what was the probability that the leaves fell at a time y after t? It was equal to the ratio of the amount of time after t to the amount of time after x, or (1−t)/(1−x). Meanwhile, x was equally likely to be anywhere between 0 and t, which meant you had to integrate this ratio over this range and then normalize by dividing by t. In the end, this integral came out to (t−1)/t·ln(1−t). Multiplying this by t  — again, that was the probability that the leaves started changing before t — gave you p(t) = (t−1)·ln(1−t).

To maximize this function, you took its derivative, which was ln(1−t)+1, and set it equal to zero. That meant the peak occurred a fraction 1−1/e, or about 63.21 percent, of the way through the fall. That places the peak around Nov. 18, so there’s still time to see them if you happen to live in Riddler Nation!

However, the puzzle asked for the maximum fraction of trees with changing leaves, not when this maximum occurred. You could find that fraction by evaluating p(1−1/e), which was an even nicer-looking expression: 1/e, or about 36.79 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.

The Riddler

When Will The Fall Colors Peak?

By Zach Wissner-Gross

Nov. 4, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">²⁶ 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

The end of daylight saving time here on the East Coast of the U.S. got me thinking more generally about the calendar year. Each solar year consists of approximately 365.24217 mean solar days. That’s pretty close to 365.25, which is why it makes sense to have an extra day every four years. However, the Gregorian calendar is a little more precise: There are 97 leap years every 400 years, averaging out to 365.2425 days per year.

Can you make a better approximation than the Gregorian calendar? Find numbers L and N (where N is less than 400) such that if every cycle of N years includes L leap years, the average number of days per year is as close as possible to 365.24217.

Submit your answer

Riddler Classic

It’s peak fall foliage season in Riddler Nation, where the trees change color in a rather particular way. Each tree independently begins changing color at a random time between the autumnal equinox and the winter solstice. Then, at a random later time for each tree — between when that tree’s leaves began changing color and the winter solstice — the leaves of that tree will all fall off at once.

At a certain time of year, the fraction of trees with changing leaves will peak. What is this maximal fraction?

Submit your answer

Solution to last week’s Riddler Express

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

Last week, the winner of a particular baseball game was determined by the next pitch. The pitcher either threw a fastball or an offspeed pitch, while the batter was similarly anticipating a fastball or an offspeed pitch. If the batter correctly guessed the pitch would be a fastball, they had a 1-in-5 chance of hitting a home run. If the batter correctly guessed the pitch would be offspeed, they had a 1-in-2 chance of hitting a home run. But if the batter guessed incorrectly, they struck out and lost the game. (The batter was guaranteed to swing either way.)

To spice things up, the pitcher truthfully announced the probability with which they’d throw a fastball. Then the batter truthfully announced the probability with which they’d anticipate a fastball.

Assuming both pitcher and batter were excellent logicians, what was the probability that the batter hit a home run?

A good place to start was the mindset of the pitcher. You didn’t yet know what probability the batter would announce, but let’s call it b. Meanwhile, suppose you were considering announcing a fastball probability of p. What were the batter’s chances of hitting a home run, in terms of p and b?

The batter would correctly anticipate a fastball with probability pb, so the probability of hitting a home run off a fastball was pb/5. The batter would correctly anticipate an offspeed pitch with probability (1−p)(1−b), so the probability of hitting a home run off an offspeed pitch was (1−p)(1−b)/2. Adding these probabilities together gave you the total probability of a home run, which was 7/10pb − p/2 − b/2 + 1/2, which we’ll call P.

At this point, the Problem Solving & Posing class at The Hewitt School decided to examine the partial derivative of P with respect to b, which was 7/10p − 1/2. This was equal to zero when p was 5/7. When p was less than 5/7 (meaning the partial derivative was negative), the batter could improve their chances of hitting a home run by lowering the value of b. For example, if the batter announced b was 0, then P was 1/2 − p/2, which was always greater than 1/7. And when p was greater than 5/7 (meaning the partial derivative was positive), the batter could improve their home run odds by similarly increasing the value of b, again resulting in P being always greater than 1/7.

But when p was exactly 5/7, the batter was effectively pinned. The value of P was 1/7 and didn’t change no matter what probability the batter announced. And so the probability of a home run was 1/7. For MLB, that would be a rather high probability.

Solution to last week’s Riddler Classic

Congratulations to Christian Wolters of San Jose, California, winner of last week’s Riddler Classic.

Last week, you purchased 150 pieces of candy to distribute for Halloween. However, you weren’t sure how many trick-or-treaters would visit you. Based on previous years, it could have been anywhere from 50 to 150 (inclusive), with each number being equally likely.

As the trick-or-treaters arrived, you could have decided to give each of them one, two or three candies. You wanted to avoid running out of candy, but you also wanted to avoid having any candy left over. Let X represent the number of trick-or-treaters who didn’t get candy (if you did run out) or the number of leftover pieces (if you didn’t run out).

The day before Halloween, you came up with a strategy to minimize the expected value of X. What was this minimum expected value?

The “MassMutual Crew” of Springfield, Massachusetts, came up with a strategy that resulted in a fairly low value of X. First, they gave out 100 candies to the first 50 trick-or-treaters. (Whether it was exactly two candies per kid or alternated between one and three candies made no difference.) From there, they gave out one candy per kid for however many kids showed up. For this strategy, the expected value of X was precisely the expected value of the difference between the number of trick-or-treaters and 100, which turned out to be 50·51/101, or about 25.2475.

Solver Rohan Lewis proved this was indeed the minimum expected value of X by recognizing that X could be zero for at most one value, and then increased by at least 1 for each trick-or-treater more or less than that value. By symmetrically placing that “zero case” in the middle between 50 and 150 (i.e., when there were 100 trick-or-treaters) and making sure everyone beyond the 50th trick-or-treater got at most one candy, Rohan arrived at the same strategy as the MassMutual Crew.

Finally, solver Starvind extended the puzzle, analyzing cases where your regret for each leftover candy was not necessarily the same as your regret for each empty-handed trick-or-treater. Starvid found that If you regretted each empty-handed trick-or-treater k times more than each leftover candy, then instead of planning for 100 trick-or-treaters you should plan for (99+301k)/(2+2k) trick-or-treaters.

I for one had way too much leftover candy. So much regret.

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.

Data Is Plural

The Datasets We're Looking At This Week

By Jeremy Singer-Vine

Nov. 2, 2022, at 12:00 PM

You’re reading Data Is Plural, a weekly newsletter of useful/curious datasets. Below you’ll find the Nov. 2, 2022, edition, reprinted with permission at FiveThirtyEight.

2022.11.02 edition

Nuclear stockpiles, decades of river widths, flood insurance changes, the weight of the web and Swiss apartment layouts.

Nuclear stockpiles. As of early 2022, a total of nine countries possessed approximately 12,700 nuclear warheads, according to estimates from the Federation of American Scientists. Although “the exact number of nuclear weapons in each country’s possession is a closely held national secret,” the researchers say that “publicly available information, careful analysis of historical records and occasional leaks” make the estimates possible, albeit “with significant uncertainty.” The report includes each country’s current warhead count and subtotals by status, as well as annual totals for each country since 1945. As seen in: Our World In Data. Previously: Nuclear capabilities (DIP 2016.02.24) and explosions (DIP 2016.03.23). [h/t u/jcceagle]

Decades of river widths. Dongmei Feng et al. have applied an algorithmic approach to calculating the widths of the world’s largest rivers over time. Their dataset contains more than 1 billion measurements of 2.7 million fluvial cross-sections (focusing on those wider than 90 meters), based on roughly 1.2 million satellite images captured between 1984 and 2020. Previously: Free-flowing rivers (DIP 2019.07.24) and U.S. hydrography (DIP 2022.10.12). [h/t Colin Gleason]

Flood insurance changes. The Federal Emergency Management Agency recently revamped its method of pricing U.S. flood insurance, aiming for “rates that are actuarily sound, equitable, easier to understand and better reflect a property’s flood risk.” A series of datasets and dashboards from the agency summarizes the expected changes in premiums, which began taking effect last year. They count the number of policies for which monthly payments were projected to increase/decrease by a given amount, bucketed into ten-dollar increments, for each state, county and ZIP code. As seen in: “How have flood insurance premiums changed?” (USAFacts).

The weight of the web. Researchers at the HTTP Archive, a project of the Internet Archive, “periodically crawl the top sites on the web and record detailed information about fetched resources, used web platform APIs and features and execution traces of each page.” They make the raw data available via Google BigQuery, and also publish aggregate data tracking metrics such as loading speed and page weight (measured in kilobytes transferred). As seen in: “Why web pages can have a size problem” (Datawrapper).

Swiss apartment layouts. Swiss Dwellings “contains detailed data on over 42,500 apartments (250,000 rooms) in ~3,100 buildings including their geometries, room typology as well as their visual, acoustical, topological and daylight characteristics,” sourced from Archilyse AG, a company that analyzes building plans. The details include the placement of rooms, features (e.g., sinks and bathtubs), walls, windows, doors and more. [h/t Matthias Standfest + India in Pixels]

Dataset suggestions? Criticism? Praise? Send feedback to jsvine@gmail.com. Looking for past datasets? This spreadsheet contains them all. Visit data-is-plural.com to subscribe and browse past editions.

The Riddler

Can You Hand Out All The Candy?

By Zach Wissner-Gross

Oct. 28, 2022, at 8:00 AM

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, win , I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!</p> </p>">²⁷ 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 Irwin Altrows comes a “high-speed” express:

The winner of a particular baseball game will be determined by the next pitch. The pitcher will either throw a fastball or an offspeed pitch, while the batter will similarly be anticipating a fastball or an offspeed pitch. If the batter correctly guesses the pitch will be a fastball, they have a 1-in-5 chance of hitting a home run. If the batter correctly guesses the pitch will be offspeed, they have a 1-in-2 chance of hitting a home run. But if the batter guesses incorrectly, they will strike out and lose the game. (The batter is guaranteed to swing either way.)

To spice things up, the pitcher truthfully announces the probability with which they will throw a fastball. Then the batter truthfully announces the probability with which they will anticipate a fastball.

Assuming both pitcher and batter are excellent logicians, what is the probability that the batter will hit a home run?

Submit your answer

Riddler Classic

For Halloween this year, you have purchased 150 pieces of candy. However, you’re not sure how many trick-or-treaters will visit you. Based on previous years, it could be anywhere from 50 to 150 (inclusive), with each number being equally likely.

As the trick-or-treaters arrive, you can decide to give each of them one, two or three candies. You want to avoid running out of candy, but you also want to avoid having any candy left over. Let X represent the number of trick-or-treaters who won’t get candy (if you do run out) or the number of leftover pieces (if you don’t run out).

This year, the day before Halloween, you come up with a strategy to minimize the expected value of X. What is this minimum expected value?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to Christian Wolters of San Jose, California, winner of last week’s Riddler Express.

Last week, my son noticed that when he held a fidget spinner in front of a television (with a 60 hertz refresh rate) and gave it a whirl, it appeared to suddenly spin backward a few times before coming to a halt. While many fidget spinners have three lobes, this particular spinner had five lobes, as shown below.

After giving it a spin, we clearly saw it spin backward three times before it stopped. How fast could it have been spinning at the beginning?

As stated in the problem, the television’s refresh rate was 60 hertz, so you could think of the spinner as having 60 snapshots taken per second. Now, what happened when the spinner was turning at exactly 12 revolutions per second (or 720 revolutions per minute)? Since 12 was one-fifth of 60, that meant the fidget spinner made one-fifth of a complete rotation for every snapshot. In other words, each lobe rotated from its current position to the position of the next lobe. In front of the television, the spinner appeared to be stationary!

As the spinner slowed to a halt (due to friction) and passed 12 revolutions per second, each lobe no longer made it all the way to the next lobe’s position with each snapshot, which made the spinner appear to spin backward. By the way, this is closely related to the wagon-wheel effect, where wheels caught on camera appear to spin backward when they’re actually spinning forward.

Not only did this backward motion appear to occur at speeds just below 12 revolutions per second, but it also occurred for speeds that were slightly less than integer multiples of 12 revolutions per second. For my son to have seen the effect occur three times, that meant the initial speed of the fidget spinner had to have been between 36 and 48 revolutions per second.

If you’re still not convinced, here’s an animation of a spinner that starts at 45 clockwise revolutions per second and slows down to zero, with 60 snapshots taken per second — all slowed down for the purposes of visualization, of course. As you can see, at speeds just below 36, 24 and 12 revolutions per second, the lobes appear to spin backward for a brief period of time.

For extra credit, you were presented with some additional information. Upon closer examination, my son saw the spinner go backward another three times, but in these cases, it appeared to have twice as many lobes (i.e., 10). Now, how fast could it have been spinning at the beginning?

We already said that the spinner appeared to go backward when the speed was slightly less than integer multiples of 12 revolutions per second. But what happened at half-integer multiples? For example, when the speed was 6 revolutions per second, every other snapshot appeared to be in the same position, while the remaining snapshots were halfway between. This made the spinner appear to have 10 lobes rather than five. And as the spinner slowed down past these half-integer multiples, the 10 lobes again appeared to spin backward. If you look closely, you can see these backward spins around the half-integer multiples of 12 revolutions per second in the above animation.

And so the answer to the extra credit was that the spinner’s initial speed was between 30 and 42 revolutions per second. The speeds at which there appeared to be 10 backward-moving lobes were just below 30, 18 and 6 revolutions per second.

By the way, if you want to see a real recording of this phenomenon, albeit for a three-lobed fidget spinner, check out this video from reader Danny Sleator.

Solution to last week’s Riddler Classic

Congratulations to Sanandan Swaminathan of San Jose, California, winner of last week’s Riddler Classic.

Last week, a thousand people were playing Lotería, also known as Mexican bingo. The game consisted of a deck of 54 cards, each with a unique picture. Each player had a board with 16 of the 54 pictures, arranged in a 4-by-4 grid. The boards were randomly generated, such that each board had 16 distinct pictures that were equally likely to be any of the 54.

During the game, one card from the deck was drawn at a time, and anyone whose board included that card’s picture marked it on their board. A player won by marking four pictures that formed one of four patterns, as exemplified below: any entire row, any entire column, the four corners of the grid and any 2-by-2 square.

After the fourth card had been drawn, there were no winners. What was the probability that there would be exactly one winner when the fifth card was drawn?

Before getting into the specifics of Lotería, suppose the probability that one particular person won on the fifth draw given that they didn’t win in the first four draws was p. Then the probability that this particular person would have been the only winner among the thousand was p∙(1−p)⁹⁹⁹. That was the same probability for another player being the only winner, and another — and for all thousand players, as a matter of fact. So the probability that anyone was that lone winner was 1,000∙p∙(1−p)⁹⁹⁹.

At this point, we still had to determine the value of p. Solver Max Candocia did this by first recognizing that there were 54 choose 5 ways to select the first five cards. Of these, 18∙50, or 900, resulted in a victory for a particular board. The 18 came from the fact that there were 18 winning patterns in total (four rows, four columns, one set of corners, and nine 2-by-2 squares), while the 50 came from the fact that the one card not involved in the winning pattern could have been any of the remaining 54−4, or 50, cards.

However, 20 percent of the time, these five-card sets resulted in victory after only four cards had been drawn — that is, when that unhelpful card happened to be drawn fifth from the deck. That meant p was equal to 0.8∙18∙50/(54 choose 5), or about 0.0002277. Plugging this back into the previous expression, the probability of having exactly one winner on the fifth draw (given no winners after four draws) was approximately 18.1 percent.

Solver Laurent Lessard went a step further, solving the general case for when there were N distinct cards in the deck (i.e., not necessarily 54) and M people playing (i.e., not necessarily 1,000). Laurent found a quartic relation (because four cards, of course!) between M and N that maximized the likelihood of having a single winner when the fifth card was drawn. This quartic relation is plotted below:

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.

Data Is Plural

The Datasets We're Looking At This Week

By Jeremy Singer-Vine

Oct. 26, 2022, at 12:00 PM

You’re reading Data Is Plural, a weekly newsletter of useful/curious datasets. Below you’ll find the Oct. 26, 2022, edition, reprinted with permission at FiveThirtyEight.

2022.10.26 edition

Strategic petroleum, internet service offers, Boston’s first women voters, Euro-area securities and gargantuan gourds.

Strategic petroleum. The U.S. Energy Information Administration maintains a dataset tracking the monthly volume of the country’s Strategic Petroleum Reserve, measured in the thousands of barrels. The figures go back to 1977, the year the first crude oil was delivered to the reserve, but lag by a couple of months; the end-of-August volume is scheduled for publication on October 31. Read more: The Department of Energy’s history of reserve releases. Previously: Petroleum Supply Monthly reports (DIP 2017.08.16) and weekly gas prices (DIP 2021.06.09), both also published by the EIA. [h/t u/CountBayesie]

Internet service offers. For an investigation into speed disparities in internet service offers, published last week at The Markup, reporters Leon Yin and Aaron Sankin examined more than 1 million address-specific offers across dozens of U.S. cities. To support the findings, they’ve shared the raw data gathered from ISPs’ websites, as well as tabular files that summarize each offer and attach the contextual variables used for the analysis. (Disclosure: I served, and am credited, as a “Data Coach” for this project.)

Boston’s first women voters. The City of Boston’s Mary Eliza Project has been compiling a dataset of women who registered to vote in 1920, the year the 19th Amendment granted them that right. The dataset, transcribed from the original registration books, “is updated periodically as additional voter registers are transcribed.” It contains 6,000-plus entries so far, each listing a voter’s name, registration date, ward, precinct, address, age, country of birth, occupation, husband’s information and more. [h/t Julie Rosier]

Euro-area securities. The European Central Bank collects detailed records concerning the financial instruments issued and held by organizations and individuals under its jurisdiction. Its quarterly-updated Securities Holdings Statistics dataset, available through the ECB’s data warehouse, aggregates the latter by investor type (bank, non-bank company, pension fund, household, etc.), investor country of residence, issuer country, type of financial instrument and more. [h/t Martijn Boermans et al.]

Gargantuan gourds. At BigPumpkins.com, you can find annual “weigh-off” results from 100-plus local competitions affiliated with the Great Pumpkin Commonwealth, an international standards-setting organization. Although pumpkins represent the titular attraction, the site also publishes results for the squash, long gourd, watermelon, tomato, field pumpkin, bushel gourd and marrow competition classes. HTML tables list each specimen’s weight, grower, location, weigh-off site and lineage. [h/t Julia Silge + Tidy Tuesday]

Dataset suggestions? Criticism? Praise? Send feedback to jsvine@gmail.com. Looking for past datasets? This spreadsheet contains them all. Visit data-is-plural.com to subscribe and browse past editions.