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

Riddler Nation is competing against Conundrum Country at an Olympic archery event. Each team fires three arrows toward a circular target 70 meters away. Hitting the bull’s-eye earns a team 10 points, while regions successively farther away from the bull’s-eye are worth fewer and fewer points.

Whichever team has more points after three rounds wins. However, if the teams are tied after each team has taken three shots, both sides will fire another three arrows. (If they remain tied, they will continue firing three arrows each until the tie is broken.)

For every shot, each archer of Riddler Nation has a one-third chance of hitting the bull’s-eye (i.e., earning 10 points), a one-third chance of earning 9 points and a one-third chance of earning 5 points.

Meanwhile, each archer of Conundrum Country earns 8 points with every arrow.

Which team is favored to win?

*Extra credit:* What is the probability that the team you identified as the favorite *will* win?

From Eli Luberoff (courtesy of Iñaki Huarte) comes a geometric curiosity:

Suppose you have a chain with infinitely many flat (i.e., one-dimensional) links. The first link has length 1, and the length of each successive link is a fraction *f* of the previous link’s length. As you might expect, *f* is less than 1. You place the chain flat on a table and some ink at the very end of the chain (i.e., the end with the infinitesimal links).

Initially, the chain forms a straight line segment, and the longest link is fixed in place. From there, the links are constrained to move in a very specific way: The angle between each chain and the next, smaller link is always the same throughout the chain. For example, if the *N*^{th} link and the *N*+1^{st} link form a 40 degree clockwise angle, then so do the *N*+1^{st} link and the *N*+2^{nd} link.

After you move the chain around as much as you can, what shape is drawn by the ink that was at the tail end of the chain?

Congratulations to 👏 Jackson 👏 of Salt Lake City, Utah, winner of last week’s Riddler Express.

Last week, you navigated an 8-by-8 chessboard, each of whose squares had one piece. You began at the white bishop on the green square. You could then move from white piece to white piece via the following rules:

- If you were on a pawn, you could move up one space diagonally (left or right).
- If you were on a knight, you could move in an “L” shape — two spaces up, down, left or right, and then one space in a perpendicular direction.
- If you were on a bishop, you could move one space in any diagonal direction.
- If you were on a rook, you could move one space up, down, left or right.
- If you were on a queen, you could move one space in any direction (horizontal, vertical or diagonal).

For example, suppose your first move from the bishop was diagonally down and to the right. You would then be at a white rook, so your possible moves were left or up to a pawn or right to a knight.

Your objective was to reach one of the four black kings on the grid. However, at no point could you land on any of the other black pieces. (Knights were allowed to hop over the black pieces.)

What sequence of moves allowed you to reach a king?

It so happened that there were multiple solutions here. Solver Tо̄saka Rin of Kobe, Japan, started by working backwards from the four kings. The king in the top right could only be reached by the knight two squares down and one square to its left. The king in the top middle could be reached both by the rook one square to its right and the knight one square down and two squares to its left. The king in the top left could only be reached by the knight one square down and two squares to its right. Meanwhile, there was no way to reach the king in the bottom left corner.

Solver Andrew Heairet was kind enough to animate a few of the resulting solutions:

While most solvers worked out a viable pathway by hand, some, like David Wanner of St. Louis, wrote code to do it for them. I bet it took a little longer, but was just as glorious.

Congratulations to 👏 Austin Shapiro 👏 of Oakland, California, winner of last week’s Riddler Classic.

Last week marked the beginning of the Summer Olympics! One of the brand-new events this year is sport climbing, in which competitors try their hands (and feet) at lead climbing, speed climbing and bouldering.

You assumed the event’s organizers accidentally forgot to place all the climbing holds and had to do it last-minute for their 10-meter wall, which was the regulation height for the purposes of this riddle. Climbers didn’t have any trouble moving horizontally along the wall. However, climbers couldn’t move between holds that were more than 1 meter apart vertically.

In a rush, the organizers placed climbing holds randomly until there were no vertical gaps in excess of 1 meter between climbing holds (including the bottom and top of the wall). Once they were done placing the holds, how many were there on average (not including the bottom and top of the wall)?

Unlike last week’s Express, this puzzle was perhaps better solved (or at least approximated) via simulation. Most solvers wrote some code that added climbing holds one at a time, checking for the largest gap between neighboring holds after each addition. As long as this largest gap was greater than 1 meter, then you should have added another hold! Otherwise, you then recorded how many were placed and simulate the whole thing over again.

This is precisely what solver Peter Ji did. After 10,000 simulations, Peter found that the organizers placed about 43 climbing holds on average. Meanwhile, Rohan Lewis was able to be a little more precise by running a total of 10 *million* simulations, finding the average number of holds was closer to 43.05.

Meanwhile, a few solvers were able to find the exact solution analytically, albeit with the help of a computer. Finding the probability distribution for the largest spacing between values chosen randomly from a uniform distribution is not a new problem.

As discussed by solver Peter Norvig (who in turn gives a tip of the hat to George Hauser), if you were to pick any *k* gaps (among the *N*+1 gaps created by *N* holds), the probability that they were *all* greater than one-tenth of the wall’s height was given by the expression (1−0.1*k*)* ^{N}*, as long as

Even with this result in hand, there was still quite a bit of math to do. Peter used the probability that any *k* gaps were all greater than 1 meter to compute the probability that exactly one — the largest gap — was greater than 1 meter. He started by adding up all the probabilities where at least one specific gap exceeded 1 meter. However, this overcounted the cases where at least two gaps both exceeded 1 meter, so these probabilities had to be subtracted off. This result then *undercounted* cases where at least three gaps all exceeded 1 meter, and so on.

Using the probability that exactly one gap was greater than 1 meter, you could compute the probabilities for each number of holds. For example, the probability that the organizers had to add exactly 30 holds was equal to the probability that there was a gap greater than 1 meter at 29 holds and that there was no such gap at 30 holds. After some further simplification, Peter hits the nail on the head: The average number of holds is approximately **43.04683**.

For extra credit, the climbers found it just as difficult to move horizontally as vertically, meaning they couldn’t move between any two holds that were more than 1 meter apart in *any* direction. You were also told that the climbing wall was a 10-by-10 meter square. If the organizers again placed the holds randomly, how many had to be placed on average until it was possible to climb the wall?

As if the one-dimensional version of this puzzle wasn’t hard enough, now there were two dimensions to consider! This was especially tricky, since viable paths up the climbing wall were allowed to go down as well as up, as demonstrated by some of Peter’s simulations:

Over the course of 1 million simulations, solver Emily Boyajian added one climbing hold at a time, keeping track of whether it was accessible via the bottom or made any other holds accessible. Once a hold was accessible from the top of the wall, the simulation ended. With this approach, Emily found that the answer was very close to **143**.

Those were a lot of holds. Hopefully the Olympic organizers don’t plan on placing them randomly.

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

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{2} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Tom Hanrahan comes a maze for grandmasters:

The following 8-by-8 grid is covered with a total of 64 chess pieces, with one piece on each square. You should begin this puzzle at the white bishop on the green square. You can then move from white piece to white piece via the following rules:

- If you are on a pawn, move up one space diagonally (left or right).
- If you are on a knight, move in an “L” shape — two spaces up, down, left or right, and then one space in a perpendicular direction.
- If you are on a bishop, move one space in any diagonal direction.
- If you are on a rook, move one space up, down, left or right.
- If you are on a queen, move one space in any direction (horizontal, vertical or diagonal).

For example, suppose your first move from the bishop is diagonally down and to the right. Now you’re at a white rook, so your possible moves are left or up to a pawn or right to a knight.

Your objective is to reach one of the four black kings on the grid. However, at no point can you land on any of the other black pieces. (Knights are allowed to hop over the black pieces.)

What sequence of moves will allow you to reach a king?

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

Today marks the beginning of the Summer Olympics! One of the brand-new events this year is sport climbing, in which competitors try their hands (and feet) at lead climbing, speed climbing and bouldering.

Suppose the event’s organizers accidentally forgot to place all the climbing holds on and had to do it last-minute for their 10-meter wall (the regulation height for the purposes of this riddle). Climbers won’t have any trouble moving horizontally along the wall. However, climbers can’t move between holds that are more than 1 meter apart vertically.

In a rush, the organizers place climbing holds randomly until there are no vertical gaps between climbing holds (including the bottom and top of the wall). Once they are done placing the holds, how many will there be on average (not including the bottom and top of the wall)?

*Extra credit:* Now suppose climbers find it just as difficult to move horizontally as vertically, meaning they can’t move between any two holds that are more than 1 meter apart in any direction. Suppose also that the climbing wall is a 10-by-10 meter square. If the organizers again place the holds randomly, how many have to be placed on average until it’s possible to climb the wall?

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

Congratulations to 👏 Brian Mercurio 👏 of Binghamton, New York, winner of last week’s Riddler Express.

Last week, I found a brown 12-inch stick for my three dogs (Fatch, Fetch and Fitch) to play with. I marked the top and bottom of the stick and then threw it for Fatch. Fatch, a Dalmatian, bit it in a random spot — leaving a mark — and returned it to me. In her honor, I painted the stick black from the top to the bite and white from the bottom to the bite.

I subsequently threw the stick for Fetch and then for Fitch, each of whom retrieved the stick by biting a random spot. What was the probability that Fetch and Fitch both bit the same color (i.e., both black or both white)?

By and large, solvers took three different approaches here. First, you could have simulated thousands or even millions of random dog bites, and then counted up how often Fetch and Fitch both bit the same color. Angelos Tzelepis happened to simulate 5 million trios of bites, finding that 66.8 percent of them (a result suspiciously close to two-thirds) were the same color.

Meanwhile, Elaine H. of Tampa, Florida, arrived at an exact solution using calculus. Elaine set the length of the stick to 1 and supposed Fatch’s bite was a distance *x* from the top. The probability that Fetch and Fitch both bit *above* Fatch was then *x*^{2}, while the probability they both bit below Fatch was (1−*x*)^{2}. Adding these together gave the probability that Fetch and Fitch bit the same color: *x*^{2} + (1−*x*)^{2}.

Since *x* was equally likely to take on any value between 0 and 1, you could find the total probability by integrating *x*^{2} + (1−*x*)^{2} from 0 to 1. Not surprisingly, this integral came to **two-thirds**, matching Angelos’ computational results.

Two approaches down, one to go. Solver Winston Luo leveraged the inherent symmetry in the puzzle. All three bites followed an identical uniform probability distribution, which meant all six relative orderings of the bites from top to bottom (Fatch-Fetch-Fitch, Fatch-Fitch-Fetch, Fetch-Fatch-Fitch, Fetch-Fitch-Fatch, Fitch-Fatch-Fetch and Fitch-Fetch-Fatch) were equally likely. Of these six permutations, Fetch and Fitch bit the same color in four of them — when Fatch was either the topmost or bottommost bite. And so the answer was four-sixths, which simplified to two-thirds.

Woof!

Congratulations to 👏 Betts Slingluff 👏 of Rockport, Massachusetts, winner of last week’s Riddler Classic.

Two weeks ago, Italy defeated England in a heartbreaking (for England) European Championship that came down to a penalty shootout. In a shootout, teams alternate taking shots over the course of five rounds. If, at any point, a team is guaranteed to have outscored its opponent after five rounds, the shootout ends prematurely, even if each side has not yet taken five shots. If the teams are tied after five rounds, they continue one round at a time until one team scores and another misses.

If each player had a 70 percent chance of making any given penalty shot, then how many total shots would have been taken on average?

As with last week’s Express, some solvers simulated thousands of shootouts to compute the average. Getting the stopping conditions was the trickiest part. After all, it was possible for the shootout to end after six shots, seven, eight, nine or 10 shots, or even in sudden death.

As tricky as it was to write code to simulate the shootout, solving it analytically was even trickier. Sanandan Swaminathan of San Jose, California, accomplished this by finding the probability the shootout would end for each number of shots.

As I mentioned, the fewest possible number of shots was six, when either team made its first three shots and the other team missed its first three (leaving it with two remaining shots — a guaranteed loss). The probability of this occurring was 0.7^{3}×0.3^{3}, then doubled, since either England or Italy could have been the team that went up 3-0. This came to a nice, even 0.018522.

Sanandan continued working through the increasingly complicated combinatorics to calculate the chances for each shootout duration: seven (with probability 0.076734), eight (0.14518602), nine (0.22835064) and 10 (0.2596431852).

But after 10 shots, you had to work out the math behind sudden death, which ended when one team made a shot and the other missed. The probability of this happening was 0.7×0.3, again doubled because either team could be the victor. This product was 0.42, which meant the average number of sudden death rounds played (if sudden death was reached) was 0.42 + 2×(0.58)^{1}×0.42 + 3×(0.58)^{2}×0.42 + 4×(0.58)^{3}×0.42 + …. The sum of this infinite arithmetic-geometric series was 1/0.42, or about 2.381.

But wait! That was the average number of additional *rounds*. The average number of shots was twice this, since there were two shots per round. And so the additional number of shots was 2/0.42, or about 4.762.

Multiplying each of the respective number of shots by their probabilities, and then tacking on the 4.762 sudden death shots (again, in the event the shootout went to sudden death) resulted in an average total of **approximately 10.47 shots**, just edging into sudden death territory.

While the average number of shots was greater than 10, the median number was exactly 10, as shown by the frequency distribution Angelos generated below. This meant that, more often than not, the shootout did *not* go to sudden death.

Solver Grant Larsen worked out the more general case when the probability of any given goal being made was *p*. Defining *y* as the quantity *p*(1−*p*), Grant found that the average number of shots was 1/*y* + *y*(56 − *y*(203 − 4*y*(93 − 70*y*))). This was a fascinating result for two reasons. First, this function was symmetric about *p* = 0.5. So if each player had a 30 percent chance of making a goal, you’d get the same average number of kicks as when each player had a 70 percent chance.

Second, this meant the minimum occurred when players had a 50 percent chance of making any given shot, and this minimum was 10.03125 shots. In other words, when all the players have the same chance of making a goal, the average number of kicks will *always *exceed 10.

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

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

]]>FiveThirtyEight is seeking an intern to join our interactives and graphics team during the fall of 2021. We’re looking for a current student journalist or recent graduate who is excited about using data visualization to create powerful journalism. This is a highly collaborative role, so we’re also looking for candidates who are excited to work across our newsroom.

The ideal candidate will have demonstrated promise in at least two of the following areas: journalism, data visualization and programming. Roughly two-thirds of the internship will be spent working on daily charts, with a heavy emphasis on mentorship and constructive editing throughout. The rest of the internship will be designed — in collaboration with the Senior Editor for Data Visualization — around the intern’s own interests. That can take numerous forms of reader-facing and/or internal projects. Previous intern projects included watching hours of baseball in order to report out their own story, building a Twitter bot highlighting random Americans via census data, and working with a reporter to build out a standalone interactive project.

If you don’t check every box below, that’s OK. Apply with a cover letter and resume at the Disney jobs portal by Aug. 2. Applicants should be able to commit to a full-time schedule from mid-September through December and must be a current student or have graduated in the past 18 months. This is a U.S.-based remote position.

**Responsibilities**

- Crafting daily charts to FiveThirtyEight publication standards under the supervision of editors or more senior Visual Journalists.
- Regularly attending newsroom meetings and participating in our editorial workflow.
- Pitching in on larger team initiatives with the help and guidance of senior team members.
- Contributing original ideas for chart forms, stories and interactive features.
- Designing, alongside their manager, a growth-oriented plan for their non-charts time.

**Basic Qualifications**

- Experience working in Adobe Illustrator.
- Demonstrated ability in at least two of the following: journalism, data visualization and programming. School work or self-published pieces count!
- Knowledge of FiveThirtyEight and the kinds of journalism we create.

**Preferred Qualifications**

- An interest in sports and/or politics.
- A background in statistics.

FiveThirtyEight is seeking an organized, process-savvy and detail-oriented Data Editor to empower some of our newsroom’s most ambitious work. This full-time role is an opportunity to apply your analytical and creative powers to all stages of a story’s evolution, from conception to publication. In particular, the Data Editor will be responsible for reviewing the methodological and statistical choices made in our stories, checking the accuracy of statistical and empirical claims made across all areas of coverage (including politics, sports and science), publishing datasets we’ve created, maintaining internal reference databases (including our one-of-a-kind database of public-opinion polling) and supervising the work of two research assistants.

The Data Editor will report to the Copy Chief and work closely with the Interactives Editor to support our polling and data infrastructure. This full-time role with benefits is a U.S.-based position and remote work may be considered. To apply, please send a cover letter and résumé through the Disney jobs portal. Due to the number of applications we receive, individual emails to the hiring manager cannot be answered.

- Function as the quantitative editor on daily and feature stories — i.e., vet the methodological choices underlying the data analysis by FiveThirtyEight reporters and freelancers, audit the code in that analysis, and double-check the statistical and empirical claims being made.
- Collaborate with the Copy Desk to manage daily and weekly editorial workflow, ensuring that quantitative edits are completed in a timely manner.
- Maintain our proprietary polls database with the help of two part-time research assistants, whom you will supervise, plus support from the Interactives Team.
- Publish datasets to our public-data repository and work with the Interactives Team to maintain and improve those pages.
- Collaborate with reporters and the Interactives Team on data-intensive projects — especially those involving polls — from conception to publication.

Basic Qualifications:

- Experience conducting original data analysis and/or critically evaluating others’ work.
- Demonstrated success collaborating with kindness and generosity on large, multi-stakeholder projects and completing complex work under deadline pressure.
- A deep understanding of statistics (particularly causal inference), survey methodologies, and the uses of public-opinion polling.
- Fluency in spreadsheets, conceptual understanding of relational databases, and facility with at least one programming language (ideally R) plus an enthusiasm for learning new technologies as needed.
- The ability to derive satisfaction from patiently and incrementally improving and documenting evergreen data resources.

- Journalism experience or a robust understanding of the editorial process. If your experience was as a copy editor or fact-checker, that’s even better, as having meticulous attention to detail is key.
- Experience reviewing and evaluating academic papers to identify and correct methodological flaws.
- Experience supervising the work of others.
- A solid understanding of U.S. politics, particularly electoral politics.
- An enthusiasm for U.S. sports and sports statistics.
- Fluency in Ruby and experience with Ruby on Rails.
- A robust understanding of statistical models, such as those used to forecast sports or election results.

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,^{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.

From Irwin Altrows comes a rather sticky situation:

I have three dogs: Fatch, Fetch and Fitch. Yesterday, I found a brown 12-inch stick for them to play with. I marked the top and bottom of the stick and then threw it for Fatch. Fatch, a Dalmatian, bit it in a random spot — leaving a mark — and returned it to me. In her honor, I painted the stick black from the top to the bite and white from the bottom to the bite.

I subsequently threw the stick for Fetch and then for Fitch, each of whom retrieved the stick by biting a random spot. What is the probability that Fetch and Fitch both bit the same color (i.e., both black or both white)?

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

From Steven Pratt comes a probabilistic puzzle that you’ll get a kick out of:

Italy defeated England in a heartbreaking (for England) European Championship that came down to a penalty shootout. In a shootout, teams alternate taking shots over the course of five rounds. If, at any point, a team is guaranteed to have outscored its opponent after five rounds, the shootout ends prematurely, even if each side has not yet taken five shots. If teams are tied after five rounds, they continue one round at a time until one team scores and another misses.

If each player has a 70 percent chance of making any given penalty shot, then how many total shots will be taken on average?

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

Congratulations to 👏 James Anderson 👏 of Charlotte, North Carolina, winner of last week’s Riddler Express.

Earlier this year, a new generation of Brood X cicadas had emerged in many parts of the U.S. This particular brood emerges every 17 years, while some other cicada broods emerge every 13 years. Both 13 and 17 are prime numbers — and relatively prime with one another — which means these broods are rarely in phase with other predators or each other.

Last week, you analyzed two broods of cicadas, with periods of *A* and *B *years, that had just emerged in the same season. However, these two broods could have also interfered with each other one year *after* they emerged due to a resulting lack of available food. For example, if *A* was 5 and *B* was 7, then *B*’s emergence in year 14 (i.e., 2 times 7) meant that when *A* emerged in year 15 (i.e., 3 times 5) there wasn’t enough food to go around.

If *A* and *B* were relatively prime and both less than or equal to 20, what was the longest stretch these two broods could have gone without interfering with each other’s cycle? (Remember, both broods just emerged *this* *year*.)

Had the broods only interfered during their years of emergence, then you would have had to find a pair of whole numbers (less than or equal to 20) with the greatest least common multiple (LCM). In that case, the answer would have been 19 and 20, whose least common multiple was 380.

However, that wrinkle of interference occurring one year post-emergence meant that if *A* were 19 and *B* were 20, then there wouldn’t have been enough food to go around in year 20. Thus, this puzzle was more than simply finding an LCM.

Since both *A* and *B* were between 1 and 20, that meant there were 20^{2}, or 400, total cases to check. (There were actually far fewer than 400 scenarios to check, since the puzzle stated that *A* and *B* had to be relatively prime.) Solvers like Rohan Lewis and Benjamin Dickman checked them all:

As you can see from Benjamin’s table, the latest possible interference occurred when *A* (the values along the horizontal axis) and *B* (the values along the vertical axis) were 17 and 19. The first few multiples of 17 were 17, 34, 51, 68, 85, 102, 119, 136 and 153. Meanwhile, the first few multiples of 19 were 19, 38, 57, 76, 95, 114, 133 and 152. The first time their multiples were within 1 of each other was when they were 153 and 152, meaning the answer was **153**. (I also accepted answers of 152.)

In general, consecutive odd values are always relatively prime and go a pretty long time without interfering. Given an odd value of *N*, when do *N* and *N*+2 first interfere? Since *N* is odd, that means *N*−1 and *N*+1 are both even, which in turn means that (*N*−1)/2 and (*N*+1)/2 are both integers. Multiplying *N*+2 by (*N*−1)/2 gives the expression (*N*^{2}+*N*)/2−1, while multiplying *N* by (*N*+1)/2 gives the expression (*N*^{2}+*N*)/2. Indeed, these two expressions differ by one, and the interference occurs after (*N*^{2}+*N*)/2 years. You can convince yourself it won’t happen sooner using modular arithmetic.

Should cicadas ever start interfering with each other like this, it wouldn’t surprise me if the brood cycles adapt over the course of the next million years or so.

Congratulations to 👏 Jordan Rader 👏 of Knoxville, Tennessee, winner of last week’s Riddler Classic.

Last week, the astronomers of Planet Xiddler were back!

This time, they had identified three planets that circularly orbited a neighboring star. Planet A was three astronomical units away from its star and completed its orbit in three years. Planet B was four astronomical units away from the star and completed its orbit in four years. Finally, Planet C was five astronomical units away from the star and completed its orbit in five years.^{4}

They reported their findings to Xiddler’s Grand Minister, along with the auspicious news that all three planets were currently lined up (i.e., they were collinear) with their star. However, the Grand Minister was far more interested in the three planets than the star and wanted to know how long it would be until the planets were next aligned.

How many years would it be until the three planets were again collinear (not necessarily including the star)?

First off, to all those who pointed out the non-physicality of this planetary system: Thanks, but seriously, read the footnotes.

Now, back to the puzzle. Like last week’s Express, it looked like this puzzle would involve LCMs. To find the next time all three planets returned to their original positions, meaning they were all collinear with their star, you had to compute the least common multiple of 3, 4 and 5, which was 60.

But wait! In *half* that time — or 30 years, the planets with periods of three years and five years would have returned to their original positions, while the planet with a period of four years would have been on the exact opposite side of the star from where it started. This meant they were all collinear after just 30 years.

But wait (again)! The puzzle said that only the *three planets* had to be on the same line. There was no need for the star to be collinear with them. That meant this was no longer a number theory puzzle involving LCMs but rather a dynamics puzzle involving trigonometry!

To see this in action, check out solver Tyler Barron’s animation of the three planets:

It appears that they lined up quite a few times before even hitting the 30-year mark.

There were a few different strategies to precisely solve for collinearity. The first step was always to write the coordinates of the three planets as a function of the number of years *t*: (3cos(2𝜋*t*/3), 3sin(2𝜋*t*/3)), (4cos(2𝜋*t*/4), 4sin(2𝜋*t*/4)) and (5cos(2𝜋*t*/5), 5sin(2𝜋*t*/5)).

From there, some solvers, like Frank van Biesen, wrote expressions for the distances between these three points over time, and then used Heron’s formula to calculate the area of the triangle whose side lengths were these three distances. Here was the resulting graph:

When the area was 0, that meant the triangle was degenerate and the three points were collinear. Sure enough, the graph first hits zero after approximately **7.77 years** — the correct answer.

Meanwhile, other solvers used dot products, cross products and even complex analysis, always arriving at the same result.

But wait (one more time)! The puzzle never specified that all three planets were orbiting in the same direction around the star (i.e., clockwise or counterclockwise). Solver Jason Weisman found that if only Planet A was moving in the opposite direction, the planets would be collinear after about 5.31 years. But if Planet B was the one moving the opposite way, collinearity occurred after just 1.6 years. Therefore, I accepted 1.6 years as a correct solution as well.

These Xiddlerian astronomers have been on a roll. I’m sure we’ll see them again soon — or perhaps they’ll see us!

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

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{5} 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.

Earlier this year, a new generation of Brood X cicadas had emerged in many parts of the U.S. This particular brood emerges every 17 years, while some other cicada broods emerge every 13 years. Both 13 and 17 are prime numbers — and relatively prime with one another — which means these broods are rarely in phase with other predators or each other. In fact, cicadas following a 13-year cycle and cicadas following a 17-year cycle will only emerge in the same season once every 221 (i.e., 13 times 17) years!

Now, suppose there are two broods of cicadas, with periods of *A* and *B *years, that have just emerged in the same season. However, these two broods can also interfere with each other one year *after* they emerge due to a resulting lack of available food. For example, if *A* is 5 and *B* is 7, then *B*’s emergence in year 14 (i.e., 2 times 7) means that when *A* emerges in year 15 (i.e., 3 times 5) there won’t be enough food to go around.

If both *A* and *B* are relatively prime and are both less than or equal to 20, what is the longest stretch these two broods can go without interfering with one another’s cycle? (Remember, both broods just emerged *this* *year*.) For example, if *A* is 5 and *B* is 7, then the interference would occur in year 15.

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

The astronomers of Planet Xiddler are back!

This time, they have identified three planets that circularly orbit a neighboring star. Planet A is three astronomical units away from its star and completes its orbit in three years. Planet B is four astronomical units away from the star and completes its orbit in four years. Finally, Planet C is five astronomical units away from the star and completes its orbit in five years.^{6}

They report their findings to Xiddler’s Grand Minister, along with the auspicious news that all three planets are currently lined up (i.e., they are collinear) with their star. However, the Grand Minister is far more interested in the three planets than the star and wants to know how long it will be until the planets are next aligned.

How many years will it be until the three planets are again collinear (not necessarily including the star)?

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

Congratulations to 👏 Paul Juster 👏 of Walnut Creek, California, winner of last week’s Riddler Express.

Last week, you studied a full, 200-episode season of *Riddler Jeopardy!*. The first episode of the season featured three brand-new contestants. Each subsequent episode included a returning champion (the winner of the previous episode) as well as two new challengers.

Throughout the season, it so happened that the returning champions were particularly strong, with each one winning five consecutive episodes before being dethroned on the sixth.

If you picked a contestant at random from the season, what was the probability that they were a *Riddler Jeopardy!* champion (meaning they won at least one episode)?

To figure this out, you had to determine both the total number of champions in a season as well as the total number of contestants. The solution was then the former divided by the latter.

The number of champions wasn’t too tough. There were 200 episodes, with each champion winning five episodes. That meant there were 200 divided by 5 — or 40 — champions.

But what about the total number of contestants? Counting them up in games with new champions as opposed to returning champions was, for many readers, rather tricky.

Solver Jesus Petry of Porto Alegre, Brazil, worked this out in a straightforward way. Jesus knew that regardless of who was the winner, every episode introduced two new contestants. Every episode, that is, except the first, which introduced *three* new contestants. That meant the total was 199·2+3, or 401.

With 401 total contestants and just 40 champions, your chances of picking a winner were **40/401, or about 9.975 percent**.

What does this mean? Well, you might intuitively think that a third of *Jeopardy!* contestants are winners, since each episode has three contestants and one of them wins. However, because winners come back to play in the next episode, that effectively reduces how many winners there are.

But as unlikely as it might be to become a *Jeopardy!* champion, it’s still better than your chances of being the next host.

Congratulations to 👏 Peter Exterkate 👏 of Sydney, Australia, winner of last week’s Riddler Classic.

On June 24, the following challenge was posted on Twitter:

In this challenge, the numbers from 1 to 11 were arranged in a circle in a particular order: 1, 4, 8, 7, 11, 2, 5, 9, 3, 6, 10. You then had to connect pairs of numbers with straight line segments that didn’t intersect, and your score was the sum of the products of the joined numbers. For example, with the connections {1, 4}, {8, 10}, {3, 7}, {5, 9}, and {2, 11} (and the 6 left by itself), you got a score of 1·4 + 8·10 + 3·7 + 5·9 + 2·11, or 172.

The *best* score you could achieve with this ordering of 1 through 11 around the circle was 237, which you could get with the following connections: {6, 10}, {3, 4}, {7, 8}, {9, 11} and {2, 5} (and the 1 left by itself).

But what if you wanted the *lowest* possible maximum score? That is, how could you have ordered the numbers from 1 to 11 around the circle so that the maximum possible score was as low as possible? And what was the resulting score?

This puzzle was doubly challenging. For any given permutation of numbers around the circle, finding the maximum score was hard enough. But on top of that, you *also* had to find the arrangement that resulted in the smallest maximum score. I can report that everyone who found the correct answer did so with the assistance of a computer and by checking all possibilities.

Without further ado, here was one such optimal arrangement:

Before reading on, if you haven’t completely solved the puzzle yet, you can still take a crack at it! Yes, this is an optimal arrangement, but what’s the highest score you can get with it? Read on for the complete solution …

The most points you could squeeze out of the above arrangement was **224**. As it turned out, there were *two* different ways to do it. Here was one way:

And here was another:

As I said, this was one optimal arrangement. Peter, as well as Emily Boyajian of Minneapolis, Minnesota, found another 21 (for a total of 22) that were all distinct, meaning no two were rotations or reflections of each other.

Emily went one step further, studying how likely different maximum point totals were. The lowest such score was 224, while the highest was 250. With random arrangements of the 11 numbers around the circle, it turned out that higher-scoring totals were more likely:

Personally, I found it rather surprising that when all was said and done, the range of maximum scores only varied by about 10 percent. Now if only my bowling scores were this consistent.

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{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.

There are 200 episodes in a season of *Riddler Jeopardy!*. The first episode of the season features three brand-new contestants. Each subsequent episode includes a returning champion (the winner of the previous episode) as well as two new challengers.

Throughout the season, it so happens that the returning champions are particularly strong, with each one winning five consecutive episodes before being dethroned on the sixth.

If you pick a contestant at random from the season, what is the probability that they are a *Riddler Jeopardy!* champion (meaning they won at least one episode)?

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

On June 24, the following challenge was posted on Twitter:

In this challenge, the numbers from 1 to 11 are arranged in a circle in a particular order: 1, 4, 8, 7, 11, 2, 5, 9, 3, 6, 10. You then have to connect pairs of numbers with straight line segments that don’t intersect, and your score is the sum of the products of the joined numbers. For example, with the connections {1, 4}, {8, 10}, {3, 7}, {5, 9}, and {2, 11} (and the 6 left by itself), you get a score of 1·4 + 8·10 + 3·7 + 5·9 + 2·11, or 172.

The *best* score you can achieve with this ordering of 1 through 11 around the circle is 237, which you can get with the following connections: {6, 10}, {3, 4}, {7, 8}, {9, 11} and {2, 5} (and the 1 left by itself).

This got Friend-of-The-Riddler Tyler Barron and me thinking about possible extensions of this challenge. If you want the highest possible maximum score, then you can rearrange the numbers from 1 to 11 so that they are in numerical order around the circle. (With this arrangement, the maximum score is 250.)

But what if you want the *lowest* possible maximum score? That is, how can you order the numbers from 1 to 11 around the circle so that the maximum possible score is as low as possible? And what is the resulting score?

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

Congratulations to 👏 Joshua Korn 👏 of Cambridge, Massachusetts, winner of last week’s Riddler Express.

Last week, you had a bag containing 100 marbles. Each marble was one of three different colors. If you drew three marbles at random, the probability that you got one of each color was *exactly* 20 percent.

How many marbles of each color were in the bag?

Suppose the colors are red, green and blue, and that *r*, *g* and *b* represented the number of marbles of each color. You immediately knew that their sum, *r*+*g*+*b*, equaled 100 — the total number of marbles in the bag.

Next, there were six different ways to draw three marbles of different colors:

- red, green, blue
- red, blue, green
- green, red, blue
- green, blue, red
- blue, red, green
- blue, green, red

Moreover, each of these six permutations was equally likely. For example, the probability of first drawing a red was *r*/100. The probability of *then* drawing a green was *g*/99, since there were now only 99 marbles remaining in the bag. Finally, the probability of drawing a blue was then *b*/98. That meant the probability of each permutation was *rgb*/(100·99·98). To find the probability of drawing three different colors, you simply had to add these six probabilities together, giving you 6·*rgb*/(100·99·98), or *rgb*/161,700.

Finally, you were told that this probability equaled 20 percent, which told you that *rgb*/161,700 = 1/5. Multiplying both sides by 161,700 gave you *rgb* = 32,340.

At this point, you were looking for three whole numbers whose sum was 100 and whose product was 32,340. For some solvers, It was helpful to write 32,340 as a product of primes: 2^{2}·3·5·7^{2}·11. From here, you could try out different ways of separating these primes into three larger factors. For example, suppose these factors were 2^{2}·5, 3·7 and 7·11. Their product was indeed 32,340 (no surprise), but their sum was 20+21+77, or 118. Because the sum was *not* 100, these factors were *not* the correct values of *r*, *g* and *b*.

Many solvers found a trio of factors that worked by trial and error. Pramod Mathai applied Maclaurin’s inequality to show that *r*, *g* and *b* were all at least 20 and at most 50, which significantly reduced the number of possibilities to check.

Matt Enlow, the puzzle’s author, restricted this range a little further. Suppose the largest of the three factors was 49. To maximize their product, the other two factors (which added up to 51) would have been 25 and 26. However, 25·26·49 = 31,850, a few hundred shy of the required product — 32,340. That meant all three factors were at most 48, and that the primes 7, 7 and 11 all had to occur in different factors. In the end, the unique trio of factors was **21, 35 and 44**.

By the way, another way to interpret the problem was to place each marble back in the bag after is was selected (known as picking *with replacement*). Again, you needed *r*+*g*+*b *to equal 100. However, this time, the product *rgb* had to equal approximately 33,333.33. Since there was no way to multiply integers and get a non-integer product, picking with replacement was not an option.

Congratulations to 👏 Chris Bedell 👏 of Ashtead, England, winner of last week’s Riddler Classic.

Last week, you were playing Riddler Solitaire, a game with 11 cards: an ace, a two, a three, a four, a five, a six, a seven, an eight, a nine, a 10 and a joker. Each card was worth its face value in points, while the ace counted for 1 point. To play a game, you shuffled the cards so they were randomly ordered, and then turned them over one by one. You started with 0 points, and as you flipped over each card your score increased by that card’s points — as long as the joker hadn’t shown up. The moment the joker appeared, the game was over and your score was 0. The key was that you could stop at any moment and walk away with a nonzero score.

What strategy maximized your expected number of points?

To solve this, Guy D. Moore and Ming Chan both asked what would happen with the very next card you picked. Suppose the cards you had already turned over had a sum of *S*. Since the sum of all 10 cards (excluding the joker) was 55, that meant there were 55−*S* points remaining in your hand. Furthermore, suppose you had *N* cards remaining in your hand. On average, how many points would the *next* card have been worth?

If the next card was the joker, it would have zeroed out your score. In this case, it would have technically been worth −*S* points. That meant the total number of points remaining in your hand was 55−*S* (from the non-joker cards) *plus* another −*S* (from the joker). In other words, there were 55−2*S* points among *N* cards, meaning the next card was worth an average of (55−2*S*)/*N* points. The expected value was positive as long as 2*S* was less than 55, or *S* was less than 27.5. This result translated into the following optimal strategy:

**Whenever your current score was 27 or less, flip over another card.****Whenever your current score was 28 or greater, stop.**

For extra credit, you had to determine *how many* points this optimal strategy would earn you on average. There were 11! — or 39,916,800 — equally likely orderings for the 11 cards in the deck, which meant a lot of processing power was required.

Or was it? Solver Goh Pi Han of Malaysia reduced this figure by nearly an order of magnitude, since you would always flip the first three cards (whose sum was at most 27). In the end, the average score was **10,709/693, or about 15.453 points**.

Several solvers, like Emma Knight, observed that if you played optimally, you flipped over the joker and scored 0 half the time. Allen Gu explained this phenomenon by pairing each ordering of cards with the reverse ordering. For example, in a certain ordering you might have 26 points *before* the joker and 29 points *after* it; with the reverse ordering, you would have 29 points *before* the joker and 26 points *after* it. Within each pair, the ordering with 28 or more points before the joker would net you those points! Meanwhile, the other ordering would always net you 0 points. In which case, the joker’s on *you*!

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

]]>During the pandemic, many sex workers found themselves without support. Even sex workers — like strippers — whose jobs are legal were denied unemployment benefits and stimulus money. As a result, a number of in-person sex workers turned to online work. But online performers have struggled to get paid for years, long before the pandemic. As the industry changes and more performers move online, will the very platforms that sex workers made famous turn on them?

Watch: https://abcnews.go.com/fivethirtyeight/video/year-karaoke-fivethirtyeight-77653577

Watch: https://abcnews.go.com/fivethirtyeight/video/republicans-starting-make-climate-agenda-fivethirtyeight-politics-podcast-78546775

]]>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,^{8} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

From Matt Enlow comes an original puzzle of his that previously appeared in Math Horizons. (At first it may seem like there’s information missing, but I assure you that is not the case.)

A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is *exactly* 20 percent.

How many marbles of each color are in the bag?

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

This week’s Classic comes from Henk Tijms, author of *Basic Probability, What Every Math Student Should Know*.

Riddler solitaire is played with 11 cards: an ace, a two, a three, a four, a five, a six, a seven, an eight, a nine, a 10 and a joker. Each card is worth its face value in points, while the ace counts for 1 point. To play a game, you shuffle the cards so they are randomly ordered, and then turn them over one by one. You start with 0 points, and as you flip over each card your score increases by that card’s points — as long as the joker hasn’t shown up. The moment the joker appears, the game is over and your score is 0. The key is that you can stop any moment and walk away with a nonzero score.

What strategy maximizes your expected number of points?

*Extra credit: *With an optimal strategy, how many points would you earn on average in a game of Riddler solitaire?

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

Congratulations to 👏 Brian Gregorich 👏 of Las Vegas, Nevada, winner of last week’s Riddler Express.

Last week, Magritte went bowling. Now, scoring in bowling can be a tricky matter. There are 10 frames, and in each frame you get two chances to knock down as many of the 10 pins as you can. Each pin knocked down is worth 1 point, and the pins are reset after each frame. Your score is then the sum of the scores across all 10 frames.

If only it were that simple. There are special rules for strikes (when you knock down all 10 pins with your first ball). Whenever you bowl a strike, that frame is scored as 10 *plus* the scores of your next two rolls. For example, if you bowled three strikes and missed every subsequent shot (i.e., they were gutter balls), your third frame would be worth 10 points, your second frame would be worth 20 and your first frame 30. Your final score would be 30 + 20 + 10, or 60 points.

Anyway, the bowling gods had decided that Magritte would bowl *exactly* three strikes in three randomly selected frames. All the other frames would consist of nothing but gutter balls. Magritte also lacked patience for bowling’s particular rules. If one of his three strikes came on the 10th and final frame and he was prompted to bowl further to complete the game, he would bowl gutter balls out of frustration.

What score could Magritte expect to achieve? (That is, with three randomly placed strikes, what was his average score?)

First off, there were 10 choose 3 — that is, (10·9·8)/(3·2·1), or 120 — cases to consider. Now if no two strikes were consecutive, Magritte would simply score 30 points. If *all three* occurred in a row, Magritte would score 60 points. And if exactly *two* were consecutive, then the second of them was worth 20 points, while the other two were worth 10 points — good for 40 points in total.

From here, all that was left was to figure out how many of the 120 cases were worth 30, 40 or 60 points.

There were only eight ways to score 60 points (i.e., three consecutive strikes), each corresponding to a first strike between the first and eighth frame. But counting up the 40-point and 60-point cases was trickier, and there were several ways to do this.

One clever approach for counting up the 40-point cases, suggested by solver Denward Chung of Louisville, Kentucky, was to group cases by where their consecutive strikes occurred. So if Magritte rolled strikes on the *first and second* frames, then to score 40 the third strike had to come between the fourth and 10th frames — seven cases in total. Had Magritte rolled strikes on the *second and third* frames, the third strike had to come between the fifth and 10th frame — six cases in total. As it turned out, there were seven ways to score 40 points when the strikes came on the first and second frames or the ninth and 10th frames; otherwise, there were six ways. Adding this up, there were 56 ways for Magritte to earn 40 points.

But what about three non-consecutive strikes? We already said among the 120 total cases, eight resulted in 60 points and 56 resulted in 40 points. That meant that the remaining number — 56 again! — resulted in 30 points. With an equal number of cases scoring 30 and 40 points, it was fair to say that the probabilities here were “split.”^{9}

Finally, to compute Magritte’s expected score, you had to average out all the cases, which gave you (8·60 + 56·40 + 56·30)/120, or **about 36.7 points**.

Disappointed with this rather low score, Magritte decided to give up on the sport of bowling and instead stick to bowler hats.

Congratulations to 👏 David Ding 👏 of Natick, Massachusetts, winner of last week’s Riddler Classic.

Last week, eight two-way roads all converged at a single intersection, as shown in the diagram below.

Two cars were heading toward the single intersection from different directions chosen at random. Upon reaching the intersection, they both turned in a random direction (where proceeding straight is a possible “turn”) — however, neither car pulled a U-turn (i.e., heading back the way it came).

In some cases, the paths of the cars could be drawn so that they did not cross. In this case, all was well.

However, in other cases, the paths *had to* cross. In this event, the cars crashed.

What was the probability the cars crashed? (If both cars headed off in the same direction, that also counted as a crash.)

As with last week’s Express, this was a matter of counting up cases. To simplify the problem, suppose the “red” car came in from the east. Then there were seven possible directions (i.e., *not* east) that it could have traveled upon leaving the intersection. Meanwhile, there were seven directions that the “blue” car could have come from (again, *not* east) and another seven (i.e., *not* the direction it came from) that it could have traveled upon leaving the intersection. In total, that was a whopping 7·7·7, or 343, cases to consider. Yikes!

Due to the ambiguity of the problem, there were some “edge cases” for which it wasn’t entirely clear if the cars would crash. For example, what if the cars exchanged directions? In theory, as long as each stayed in the right line, they would never crash.

Because of this ambiguity, I accepted multiple “correct” answers. David had a conservative definition of which cases constituted a crash, arriving at an answer of **16/49**. Meanwhile, Emma Knight assumed more edge cases resulted in a crash, giving an answer of **29/49**. And Rohan Lewis was in between the other two solvers, justifying an answer of **22/49**.

Suffice it to say that everyone was right last week.

Fortunately, all these edge cases disappeared in the grand scheme of things once you got to last week’s extra credit — the best part of the puzzle. As the number of two-way roads converging at the intersection approached infinity, what value did the probability of crashing approach?

Many solvers had already studied the general case of the riddle with *N* streets (rather than eight streets). This allowed them to explore the limit of their expressions as *N* approached infinity.

That was all well and good. But a few solvers, like Tim Curwick of Maple Grove, Minnesota, realized an elegant, equivalent way to restate this problem: Randomly pick four points on a circle. Then randomly pick two of them — these two will be the start and end locations for the red car. The remaining two points will be the start and end locations for the blue car. Now draw two chords that connect each pair of points. If the chords cross, that means the cars will crash!

For *any* four distinct points on a circle (whether chosen randomly or otherwise), there are exactly three ways to split them into two pairs of points. And the resulting chords will only intersect in one of these three ways. Therefore, the probability the chords will intersect — that is, the probability the cars will crash — is exactly **1/3**.^{10}

If you’re still not convinced, Paulina Leperi simulated 100,000 drive-bys, finding that very nearly a third of them resulted in crashes.

Until next time, drive safely out there. And be sure to avoid intersections with infinitely many roads!

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{11} 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.

Scoring in bowling can be a tricky matter. There are 10 frames, and in each frame you get two chances to knock down as many of the 10 pins as you can. Each pin knocked down is worth 1 point, and the pins are reset after each frame. Your score is then the sum of the scores across all 10 frames.

If only it were that simple. There are special rules for spares (when you’ve knocked down all 10 pins with the second ball of a frame) and strikes (when you knock down all 10 pins with your first ball). Whenever you bowl a strike, that frame is scored as 10 *plus* the scores of your next two rolls. This can lead to some dependency issues at the end of the game, which means the final frame has its own set of rules that I won’t go into here.

For example, if you bowled three strikes and missed every subsequent shot (i.e., they were gutter balls), your third frame would be worth 10 points, your second frame would be worth 20 and your first frame 30. Your final score would be 30 + 20 + 10, or 60 points.

This week, Magritte is going bowling, and the bowling gods have decided that he will bowl *exactly* three strikes in three randomly selected frames. All the other frames will consist of nothing but gutter balls. Magritte also lacks patience for bowling’s particular rules. If one of his three strikes comes on the 10th and final frame and he is prompted to bowl further to complete the game, he will bowl gutter balls out of frustration.

What score can Magritte expect to achieve? (That is, with three randomly placed strikes, what is his average score?)

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

From Ben Edelstein comes a crash course in probability:

Eight two-way roads all converge at a single intersection, as shown in the diagram below.

Two cars are heading toward the single intersection from different directions chosen at random. Upon reaching the intersection, they both turn in a random direction (where proceeding straight is a possible “turn”) — however, neither car pulls a U-turn (i.e., heads back the way it came).

In some cases, the paths of the cars can be drawn so that they do not cross. In this case, all is well.

However, in other cases, the paths *must* cross. In this event, the cars will crash.

What is the probability the cars will crash? (If both cars head off in the same direction, that also counts as a crash.)

*Extra credit:* As the number of two-way roads converging at the intersection approaches infinity, what value does the probability of crashing approach?

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

Congratulations to 👏 Al Shaheen 👏 of New Haven, Indiana, winner of last week’s Riddler Express.

Last week, you were hired to design a new logo for Riddler City. The mayor was a little eccentric and had requested that the logo have at least two lines of symmetry that intersected at an angle of precisely 1 radian, or 180/𝜋 (approximately 57.3) degrees.

What sorts of logos met the mayor’s requirements? (You were specifically asked to give an example or describe what all possible logos had in common.)

One way to solve this was to start with a few points and see what happened. Because the logo was symmetric across two lines, that meant that for any point you drew, there had to be a corresponding point on the other side of the line. (Otherwise the figure would not be symmetric!)

If you happened to start with three points and alternated reflecting them across either of the two lines, here was the resulting pattern:

Holy smoke (rings) — the logo consisted of **concentric circles**,^{12} centered at the lines’ intersection.

Several readers recognized the inevitability of this result. As noted by solver Pramod Mathai, with each reflection, the distance between a point and the lines’ intersection was preserved. Meanwhile, because of the irrationality of the angle measure at the intersection (i.e., it was 1/(2𝜋) of a complete rotation), each subsequent reflection resulted in a brand new point. The points in the resulting locus were all the same distance away from the intersection. In other words, they formed a circle!

By the way, you might have noticed that each ring in the animation was a little “clumpy” at first. In fact, before the animation speeds up, each ring appears to have 22 clumps. Meanwhile, one well-known approximation for 𝜋 is 22/7 — again, the number 22! Could it be a coincidence?

On a separate note, before anyone shoots off an angry email to me (or to Riddler City’s mayor, for that matter), the logo didn’t technically *have* to be a complete circle. For those ready to venture into a little set theory, read on.

A circle on the real coordinate plane consists of ℵ_{1} points, since the points can be put in a one-to-one correspondence with the real numbers. Meanwhile, the countable process in the animation above, if allowed to run forever, will contain ℵ_{0} points. This set of points will be dense on the actual circle, but would make up an infinitesimal amount of the circle, similar to the set of rational numbers on a real number line.

Since the mayor likes continuous logos, concentric circles will do!

Congratulations to 👏 Peter Exterkate 👏 of Sydney, Australia, winner of last week’s Riddler Classic.

Earlier this month, Adam Kotsko asked the Twitterverse to choose four contiguous U.S. states for a new breakaway nation:

This got Philip Bump wondering about states you could pick such that the remaining (previously) contiguous states were no longer contiguous, but rather broken up into two near-halves by area. Treating the Upper Peninsula of Michigan as distinct from the Lower Peninsula, Philip hypothesized that removing Illinois, Missouri, Oklahoma and New Mexico would create two near-equal halves.

Last week, you were asked to remove a set of states (not necessarily four) so that you had two distinct contiguous regions among the lower 48 states, where the larger region had area *A* and the smaller region had area *B*. What states should you have removed to maximize area *B*?

It was ambiguous whether you should have used states’ land area or total area. It turned out these gave different answers!

But before we get to that, let’s see how Philip’s hypothesis did. By removing Illinois, Missouri, Oklahoma and New Mexico, Philip created eastern and western halves of the country. The eastern half was smaller, consisting of 35.39 percent of the land area and 37.19 percent of the total area. Not bad, considering only four states were selected.

When it came to *land* area, solver Jenny Mitchell was among those with the largest smaller region (i.e., the greatest value of *B*):

By removing **Idaho, Wyoming, Nebraska, Missouri, Arkansas and Louisiana**, the western half made up 42.65 percent of the land area — remarkably close to the 43.12 percent of the eastern half.

When it came to *tota*l area, solver Shawn Feils removed New Mexico, Oklahoma, Missouri, Iowa and Minnesota. This was very similar to Philip’s original idea, but resulted in a larger eastern half that produced a better balance. Here, the eastern half was again the smaller of the two, but it now made up 42.73 percent of the total area.

Again, due to the ambiguity of the problem, I received a number of solutions between 40 and 45 percent that appeared to make slightly different assumptions (e.g., land area vs. total area, whether or not to include Washington, D.C., etc.). So if you are convinced that you beat the results above, that’s totally cool. Share your map like Jenny did!

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{13} 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.

You’ve been hired to design a new logo for Riddler City. The mayor is a little eccentric and has requested that the logo have at least two lines of symmetry that intersect at an angle of precisely 1 radian, or 180/𝜋 (approximately 57.3) degrees.

What sorts of logos meet the mayor’s requirements? (You can give one example or describe what all possible logos have in common.)

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

Earlier this month, Adam Kotsko asked the Twitterverse to choose four contiguous U.S. states for a new breakaway nation:

This got Philip Bump wondering about states you could pick such that the remaining (previously) contiguous states were no longer contiguous, but rather broken up into two near-halves by area. Treating the Upper Peninsula of Michigan as distinct from the Lower Peninsula, Philip hypothesized that removing Illinois, Missouri, Oklahoma and New Mexico would create two near-equal halves.

Suppose you remove a set of states (not necessarily four) so that you have two distinct contiguous regions among the lower 48 states, where the larger region has area *A* and the smaller region has area *B*. What states should you remove to maximize area *B*? (Was Philip’s intuition correct?) And what percentage of the lower 48 states’ combined area does B represent?

For reference, here are the areas of the states according to the U.S. Census Bureau. Like Philip, you can treat Michigan’s peninsulas as distinct “states” here. Michigan’s Upper Peninsula has an area of 16,377 square miles. Also, don’t worry about islands, Minnesota’s Northwest Angle, etc. — you can assume they are contiguous with the rest of their respective states.

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

Congratulations to 👏 Max G. 👏 of Chicago, winner of last week’s Riddler Express.

Last week, Max the Mathemagician (no relation to this week’s winner, as far as I know!) was calling for volunteers. He had a magic wand of length 10 that could be broken anywhere along its length (fractional and decimal lengths were allowed). After the volunteer chose these breakpoints, Max would multiply the lengths of the resulting pieces. For example, if they broke the wand near its midpoint and nowhere else, the resulting product would have been 5×5, or 25. If the product was the largest possible, they’d win a free backstage pass to his next show. (Amazing, right?)

You raised your hand to volunteer, and you and Max briefly made eye contact. As he called you up to the stage, you knew you had this in the bag. What was the maximum product you could have achieved?

First off, for any given number of pieces, the product was always maximized when they were all the same length. For example, had you broken the wand into two pieces, if the length of the first piece was x, then the length of the other was (10−*x*). You could experiment with different values, but the maximum of *x*·(10−*x*) occurred when *x* was 5, giving you the optimal product of 25. To see why this was true, you could graph the function *f*(*x*) = *x*·(10−*x*) and spot the peak of its parabolic arch, or you could take the derivative of the function and set it equal to zero.

When you had three pieces, again the product was maximized when all three were the same. Dividing a length of 10 among three pieces meant each piece had a length of 10/3. Multiplying these values together gave you (10/3)^{3}, or about 37.04. And with four pieces, each segment had a length of 10/4, or 2.5. Their product was 2.5^{4}, or 39.0625.

So far, as the number of pieces increased, the product increased as well. Would this pattern continue?

No, it would not. With five pieces, the product was 2^{5}, or 32. Sure enough, the maximum product occurred when there were exactly four pieces, which meant the optimal product was indeed **39.0625**.

But what was so special about the number 2.5? Why did *that* length piece result in a larger product than the other values mentioned so far?

If you generalized to the case of *N* pieces, the product became (10/*N*)* ^{N}*. Let’s rewrite this product as a function of the length

However, it didn’t make sense to have lengths that were *precisely e*, since you needed a whole number of pieces that were all equal in length. And that’s why the optimal length was 2.5 — because 2.5 was fairly close to *e*.

For extra credit, Zax the Mathemagician (no relation to either Max) had the same routine in his show, only his wand has a length of 100. What was the maximum product now?

Again, your best bet was to look for piece lengths close to *e*. Splitting Zax’s wand into 37 equal pieces meant that each piece was about 2.703 units long, which was as close to *e* as you could get. And so the answer to the extra credit was (100/37)^{37}, or about **9.47 quadrillion**.

Looks like everyone is winning free backstage passes to some mathemagic shows!

Congratulations to 👏 Brian Schoep 👏 of Highlands Ranch, Colorado, winner of last week’s Riddler Classic.

Last week’s Classic came courtesy of Alexander Zhang of Lynbrook High School, California. Alexander won first place in the mathematics category at this year’s International Science and Engineering Fair, and has long had an interest in topology — which *just might *have been related to the puzzle.

You were asked to consider the following image, which showed a particular uppercase sans serif font:

Alexander thought many of these letters were equivalent, but he left it to you to figure out how and why. He also had a message for you:

It might not have looked like much, but Alexander assured me that it was equivalent to exactly one word in the English language.

What was Alexander’s message?

There wasn’t a lot to go on here, and the message provided — “YIRTHA” — was woefully insufficient for reverse engineering a cipher.

The clues were Alexander’s interest in topology, and the fact that he thought many of the letters were somehow equivalent. From there, many solvers recognized that some letters in the alphabet are homeomorphic, which is how a topologist might classify objects as being equivalent. More specifically, two letters are homeomorphic if one can be deformed into the other without cutting or gluing (and vice versa). So while certain features like holes and three-way intersections must be preserved, curves and lines make no difference.

Almost half of the letters in the typeface shown above can be made with a single open (i.e., there are no holes) stroke: {C, G, I, J, L, M, N, S, U, V, W, Z}. These 12 letters are all homeomorphic and make up the largest equivalence class.

Meanwhile, four letters have three “tails” that all meet up at a single point, with no holes: {E, F, T, Y}. Two letters have a single hole: {D, O}. Two letters have one hole and one tail: {P, Q}. (Q’s topology is particularly dependent on the choice of font, which is why an image of a specific font had to be used in this puzzle.) Two more letters have one hole and two tails: {A, R}. One letter has two holes: {B}. Another letter has four tails: {X}. Finally, two letters have a “bar” with two tails on either end: {H, K}. (Like Q, K’s topology depends on the font. In the image above, K’s bar is rather short.)

So what did these equivalence classes have to do with Alexander’s message? If you took each letter in “YIRTHA” and replaced it with another letter in its equivalence class (or left the letter untouched), there were 1,536 possible six-letter combinations. Many solvers worked out the solution by hand, letter by letter. Others, like Elaine H. of Tampa, Florida and Stefano Perfetti of Zurich, Switzerland, wrote code to search through all the letter combinations and check them against a dictionary. Either way, only *one* equivalent combination was an English word:

Yes, Alexander’s message was “EUREKA,” the famous interjection uttered by Archimedes and the motto of Alexander’s home state of California.

Solver Laurent Lessard got in on the fun with his own encoded message describing topological equivalence.

I will say, I found it rather surprising how many words in the English language have unique topologies. Of the 15,000 or so six-letter words (courtesy of Peter Norvig’s word list), almost 25 percent of them are unique. That was definitely my YIRTHA! moment last week.

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{14} 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.

Max the Mathemagician is calling for volunteers. He has a magic wand of length 10 that can be broken anywhere along its length (fractional and decimal lengths are allowed). After the volunteer chooses these breakpoints, Max will multiply the lengths of the resulting pieces. For example, if they break the wand near its midpoint and nowhere else, the resulting product is 5×5, or 25. If the product is the largest possible, they will win a free backstage pass to his next show. (Amazing, right?)

You raise your hand to volunteer, and you and Max briefly make eye contact. As he calls you up to the stage, you know you have this in the bag. What is the maximum product you can achieve?

*Extra credit: *Zax the Mathemagician (no relation to Max) has the same routine in his show, only the wand has a length of 100. What is the maximum product now?

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

This week’s Classic comes courtesy of Alexander Zhang of Lynbrook High School, California. Alexander won first place in the mathematics category at this year’s International Science and Engineering Fair for his work at the intersection of topology and medicine. He developed his own highly efficient algorithms to detect and remove defects (like “handles” or “tunnels”) from three-dimensional scans (e.g., MRI). Alexander has long had an interest in topology, which *just might *be related to his submitted puzzle.

Consider the following image showing a particular uppercase sans serif font:

Alexander thinks many of these letters are equivalent, but he leaves it to you to figure out how and why. He also has a message for you:

It may not look like much, but Alexander assures me that it is equivalent to exactly one word in the English language.

What is Alexander’s message?

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

Congratulations to 👏 Chris Smith 👏 of Buford, Georgia, winner of last week’s Riddler Express.

Last week, due to all the no-hitters that had been pitched this MLB season, you looked at perfect games. To achieve a perfect game, a pitcher must record 27 consecutive outs without allowing anyone from the opposing team on base. There have been only 23 perfect games in MLB history. Two were thrown in 2010, and three were thrown in 2012.

How low did a batter’s chances of reaching base have to be for you to expect one perfect game per season? (You were asked to make the following simplifying assumptions: All batters had the same chances of reaching base; at-bats were independent from each other; there were 30 MLB teams, and each club played 162 games; and no games went into extra innings.)

First, suppose the probability a batter reached base was *p*. The probability of pitching a perfect game was then the probability that none of the first 27 batters reached base, or (1−*p*)^{27}.

You also needed to figure out how many games there were in a season. As the number of games increased, the expected number of perfect games also increased. With 30 teams each playing 162 games, and with two teams in each game, there were 30·162/2, or 2,430 games in a season. However, each game had *two* pitchers who could each potentially pitch a perfect game, meaning there could be as many as 4,860 perfect games in a season.

The expected number of perfect games was then the probability of a perfect game, (1−*p*)^{27}, times the number of pitched games, 4,860. Setting this equal to 1 gave the equation 4,860·(1−*p*)^{27} = 1. Solving for *p* gave you a value of 1 − ^{27}√(1/4,860), which corresponded to a probability of about 26.977 percent. In baseball parlance, you’d call that an average of about **.270**.^{15}

To average two perfect games per season, *p* had to be .250. And when *p* was .204, you could expect about 10 perfect games each season!

So if the probability of getting on base ever drops from one-in-three to one-in-four, it won’t just be no-hitters that become relatively pedestrian, but perfect games as well.

Congratulations to 👏 Daniel Ko 👏 of Atlanta, Georgia, winner of last week’s Riddler Classic.

Last week, Dakota Jones, The Riddler’s greatest action hero/archaeologist, was back at work! To gain access to a hidden temple deep in the Riddlerian Jungle, she needed a crystal key.

She already knew the crystal was a polyhedron. And according to an ancient text, it had exactly six edges, five of which were 1 inch long. Cryptically, the text did not specify the length of the sixth edge. Instead, it said that the key was the largest such polyhedron (i.e., with six edges, five of which have length 1) by volume.

Once again, Dakota Jones needed your help. What was the volume of the crystal key?

There was only one kind of polyhedron with just six edges: a tetrahedron. Tetrahedrons have four faces, and each edge straddles two faces. So if five of the edges had the same length — meaning the sixth edge was the only one with a potentially different length — that meant the two faces *not* straddled by this sixth edge were equilateral triangles.

At this point, you could think of the crystal as two equilateral triangles connected by a common edge. But the two triangles were not coplanar; there was some dihedral angle between them. The vertices of these two triangles were connected, forming the sixth edge of the tetrahedron.

So which of these tetrahedrons had the greatest volume? Solver Vamshi Jandhyala tackled this problem head on, using the Cayley-Menger determinant, which can be used to calculate the volume of any tetrahedron (or higher dimensional simplex) using its edge lengths.

Meanwhile, Emily Boyajian solved this without the use of linear algebra by treating one of the equilateral triangles as the base. This base had an area of √3/4, while the height of the other equilateral triangle was √3/2. The volume of a tetrahedron (or any pyramid) is one-third times the area of the base times the perpendicular height. Therefore, the volume was maximized when the dihedral angle was 90 degrees, so that the perpendicular height was exactly the height of the triangle, or √3/2. This same idea was animated by Laurent Lessard:

Plugging these values into the volume formula, the tetrahedron’s volume turned out to be 1/3 · √3/4 · √3/2, which simplified to **1/8**. By the way, for this largest crystal shape, that sixth side had a length of √6/2 — just in case you were curious.

Thanks to the help of Riddler Nation, Dakota Jones was able to precisely fashion herself a crystal key and gain entrance to the hidden temple. Stay tuned to learn what was inside …

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{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.

From Nilay Shroff comes a puzzle that’s perfect for baseball season:

As of this week, there have already been six no-hitters this MLB season, well on pace to break the record for no-hitters for a season in the modern era, which stands at nine in 1990.

To achieve a no-hitter, a pitcher must pitch a complete game (recording 27 outs over nine innings) without allowing a hit (i.e., walks and other means of reaching base are allowed). However, to achieve a *perfect game*, a pitcher must record 27 consecutive outs without allowing anyone from the opposing team on base. There have been only 23 perfect games in MLB history. Two were thrown in 2010, and three were thrown in 2012.

In 2009, the leaguewide on-base percentage was 0.333. That figure has fallen over the past decade, and this year, it’s all the way down to 0.313, which helps explain the surge in no-hitters.

How low would a batter’s chances of reaching base have to be for you to expect one perfect game per season? (You can make the following simplifying assumptions: All batters have the same chances of reaching base; at-bats are independent from each other; there are 30 MLB teams, and each club plays 162 games; and no games go into extra innings.)

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

Dakota Jones is back in action! To gain access to a hidden temple deep in the Riddlerian Jungle, she needs a crystal key.

She already knows the crystal is a polyhedron. And according to an ancient text, it has exactly six edges, five of which are 1 inch long. Cryptically, the text does not specify the length of the sixth edge. Instead, it says that the key is the largest such polyhedron (i.e., with six edges, five of which have length 1) by volume.

Once again, Dakota Jones needs your help. What is the volume of the crystal key?

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

Congratulations to 👏 Seth Hollar 👏 of Raleigh, North Carolina, winner of last week’s Riddler Express.

Last week, the Riddler Cheese Company was producing “craft triples” — triangular slices of cheese whose side lengths were Pythagorean triples when measured in inches.

However, the company’s slicing machine had recently malfunctioned, producing a stock of square slices with side lengths of 5 inches. To salvage this situation, what was the greatest number of whole Pythagorean slices that could be made from each 5-inch square? (Note: You could only cut pieces out of the square. No melting or gluing pieces together was allowed!)

With a little work, you could immediately find an upper bound for the solution. The 5-inch square had an area of 25 square inches. Meanwhile, the smallest right triangle with whole number side lengths was 3-4-5, which had an area of 6 square inches. That meant you couldn’t have more than 25/6 (i.e., 4, because partial triangles were not allowed).

But was it possible to fit **four** craft tiples in a 5-inch square? Indeed it was! John from Washington, D.C., accomplished this *using actual cheddar* by lining up the four hypotenuses along the four sides of the square:

Because the non-right angles in each right triangle were complementary, you could prove that none of the four triangles overlapped. And, as expected, there was 1 square inch of cheese left over in the middle.

For extra credit, you had to find the smallest square of cheese such that 100 percent of it could be partitioned into craft triples.

If you were using just 3-4-5 triangles — again, with an area of 6 square inches — then the square’s area had to be a multiple of 6. The smallest such square had side lengths of 6 inches and an area of 36 square inches. However, there was no way to squeeze six 3-4-5 triangles into this space.

The next-smallest square had side lengths of 12 inches and an area of 144 square inches. Sure enough, it was possible to carve 24 3-4-5 triangles out of this square. You could also use two 5-12-13 triangles and another 14 3-4-5 triangles, as demonstrated by solver Pierre Bierre below. It turned out that these were the smallest such squares.

With this puzzle, I sincerely hope you had fun cutting the cheese!

Congratulations to 👏 Lowell Vaughn 👏 of Bellevue, Washington, winner of last week’s Riddler Classic.

Last week, Jordan Ellenberg (author of the new book, “Shape”) was asking about what he called *anti*-isosceles sets.

If you considered an *N*×*N* grid of points, what was the greatest subset of these *N*^{2} points such that *no three of them* formed an isosceles triangle? (Note: Degenerate triangles, or triangles with zero area, counted as triangles here, as long as their three vertices are distinct.)

As shown below, the largest anti-isosceles set in a 2×2 grid had two points; for a 3×3 grid, the largest anti-isosceles set had four points. (Note: For both grids, there were multiple sets with these maximum numbers of points.)

How many points were in the largest anti-isosceles set for a 4×4 grid?

As is often the case, there was a way to computationally brute force your way through this problem. With 16 total points in the grid, that meant there were 2^{16} — or 65,536 — ways to group these points into sets, since each point was either in or out of a given set. Several solvers wrote programs to check all of these sets by choosing each grouping of three points within the set. If the distances between any two of these were the same, the entire set could not be anti-isosceles.

It’s worth noting that many of these sets were rotations or reflections of each other, meaning if one of them was anti-isosceles then others were as well. But since the total number of sets was not prohibitively large (for a computer) large for a computer, you didn’t have to worry about being extra efficient with your code.

If your code worked correctly, it revealed that there were exactly four anti-isosceles sets, each with the maximal **six** points. They were all rotations of each other, and here’s one of them:

You can check for yourself: With six total points, there were 20 ways to choose three of them. And *none* of these 20 triangles was isosceles.

For extra credit, you had to find the largest anti-isosceles sets in a 5×5 grid and a 6×6 grid (and even larger grids if you dared). A 5×5 grid had 2^{25} — or 33,554,432 — total subsets, just at the edge of what was possible for a desktop computer to sift through in a few minutes. For larger grids, solvers like Eric Dallal of Boston, Massachusetts, had to use “every optimization I could think of.”

Meanwhile, Goh Pi Han of Petaling Jaya, Malaysia, went a different route, running simulations that added one point at a time in positions that didn’t result in any isosceles triangles. This was a clever way to identify lower bounds on the sizes of anti-isosceles grids with too many points to exhaustively handle.

Here are examples of the largest non-isosceles sets for the 5×5 grid and the 6×6 grid, which had **seven** and **nine** points, respectively.

Friend-of-The-Riddler Dean Ballard explored non-isosceles sets in three-dimensional grids. In a 5×5×5 grid, the largest such set he found had 16 points!

Finally, returning to two-dimensional grids, Jordan wanted to know if anyone could show that the size of the largest anti-isosceles set for an *N*×*N* grid was of the order *N** ^{k}*, with 1 <

However, to my knowledge, no one has (yet) proven upper or lower bounds on *k* between 1 and 2. Jordan’s challenge remains an open one!

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{17} 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.

The Riddler Cheese Company is producing what are called “craft triples” — triangular slices of cheese whose side lengths are Pythagorean triples, when measured in inches.

However, the company’s slicing machine recently malfunctioned and produced a stock of square slices with side lengths of 5 inches. To salvage this situation, what is the greatest number of whole Pythagorean slices that can be made from each 5-inch square? (Note: You can only cut pieces out of the square. No melting or gluing pieces together!)

*Extra credit: *What is the smallest square of cheese such that 100 percent of the square can be partitioned into craft triples?

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

This week’s Classic is being served up by Jordan Ellenberg, whose new book, “Shape,” goes on sale on May 25.

One of the many geometers who appears in “Shape” is the great Paul Erdős. In 1946, Erdős himself posed a riddle about what are called “isosceles sets”: How many points can you place in *d*-dimensional space such that any three of them form an isosceles triangle?

In two dimensions, the largest isosceles set has exactly six points, with five forming the vertices of a regular pentagon and the sixth point in the middle. In three dimensions, the largest isosceles set has eight points; in four dimensions, the largest set has 11 points. And in dimensions higher than eight, no one knows what the largest isosceles sets might be!

But this week, Jordan is asking about what he calls *anti*-isosceles sets. Consider a *N*×*N* grid of points. What is the greatest subset of these *N*^{2} points such that *no three of them* form an isosceles triangle? (Note: Degenerate triangles, or triangles with zero area, *do* count as triangles here, as long as their three vertices are distinct.)

As shown below, the largest anti-isosceles set in a 2×2 grid has two points; for a 3×3 grid, the largest anti-isosceles set has four points. (Note: For both grids, there are multiple sets with these maximum numbers of points.)

How many points are in the largest anti-isosceles set for a 4×4 grid?

*Extra credit:* What about a 5×5 grid, a 6×6 grid, or even larger square grids? Can you find an expression (or bounds) for the size of the largest anti-isosceles set for the general *N*×*N* grid? (If you figure out anything about the general case, Jordan would love to hear about it!)

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

Congratulations to 👏 Erik Voigt 👏 of New York, winner of last week’s Riddler Express.

Last week, you and your infinitely many friends were sharing a cake, and you came up with two rather bizarre ways of splitting it.

For the first method, Friend 1 took half of the cake, Friend 2 took a third of *what remained*, Friend 3 took a quarter of what remained after Friend 2, Friend 4 took a fifth of what remained after Friend 3, and so on. After your infinitely many friends took their respective pieces, you got whatever was left.

For the second method, your friends decided to save you a little more of the take. This time around, Friend 1 took 1/2^{2} (or one-quarter) of the cake, Friend 2 took 1/3^{2} (or one-ninth) of *what remained*, Friend 3 took 1/4^{2} of what remained after Friend 3, and so on. Again, after your infinitely many friends took their respective pieces, you got whatever is left.

How much of the cake did you get using each of these methods?

For both of these problems, a clever math move looked at how much of the cake remained after each friend took their portion. So for the first method, 1/2 remained after Friend 1. Friend 2 took 1/3 of what was left, meaning they left behind 2/3 of the 1/2, or 2/3 × 1/2 of the original cake. Similarly, after Friend 3, there was 3/4 · 2/3 · 1/2 of the cake left.

Reversing the order of the fractions in these products, the amount left after *N* friends took their pieces was 1/2 × 2/3 × 3/4 × 4/5 × 5/6 × 6/7 × … × *N*/(*N*+1). The denominator of each fraction canceled out with the numerator of the next fraction, so that the overall product was just 1/(*N*+1). In the limit of infinitely many friends, this product went to **zero**, meaning you got no cake using the first method!

The second method looked more promising. This time, after N friends took their respective pieces, the amount left was (2^{2}−1)/2^{2} × (3^{2}−1)/3^{2} × (4^{2}−1)/4^{2} × … × [(*N*+1)^{2}−1]/(*N*+1)^{2}. At first, this might have seemed rather tricky to evaluate. However, as noted by solver Elaine H. of Tampa, Florida, differences of squares like *a*^{2}−1 can always be factored as (*a*+1)(*a*−1).

Using this identity, we can rewrite that heftier product as (1×3)/(2×2) × (2×4)/(3×3) × (3×5)/(4×4) × (4×6)/(5×5) × … × [*N*×(*N*+2)]/[(*N*+1)×(*N*+1)]. This time around, the two factors in each denominator canceled out with factors in the numerators in both the preceding and following fractions. All that was left at the end was 1/2 × (*N*+2)/(*N*+1). In the limit of infinitely many friends, the fraction (*N*+2)/(*N*+1) approached 1, which meant you got exactly **half** the cake using the second method. That’s a half-decent portion right there!

For extra credit, you removed every other term from the second method’s product, leaving you with (2^{2}−1)/2^{2} × (4^{2}−1)/4^{2} × (6^{2}−1)/6^{2} × (8^{2}−1)/8^{2}, and so on. Because terms were removed from the second product, this new product had to be greater than a half. It turned out this was the reciprocal of the Wallis product, which meant you got **2/𝜋**, or about 63.7 percent of the cake. For more on the Wallis product, and a way to prove it using complex analysis, check out Laurent Lessard’s write-up.

Congratulations to 👏 Dallas Trinkle 👏 of Urbana, Illinois, winner of last week’s Riddler Classic.

Last week, three students in puzzle submitter Matt Yeager’s fourth-grade class — Players A, B and C — were engaged in a game of *veinte*. In each round, players took turns saying numbers in order (Player A, then B, then C, then A again, etc.). The first player to go said the number “1.” Each number had to be either one, two, three or four more than the number said by the previous player. When someone said “20,” the round was over and the *next* person was eliminated, with the following person beginning the subsequent round. For example, if Player A said “20,” then Player B was eliminated, while Player C began the next round by saying “1.” At no point could anyone say a number greater than 20.

All three players wanted to be the winner (i.e., the only player remaining) after the two rounds. But if they realized they couldn’t win, they then prioritized making it to the second round.

Player A started things off by saying “1.” Which player won?

Solver Madeline Argent of Launceston, Tasmania, Australia, worked backwards, starting from what would happen after one of the players was eliminated. In the two-player variation, the second player could always win by saying the next multiple of 5. So the first player would say 1, after which the second player would say 5. Then, no matter what numbers the first player said next, the second player would say 10, followed by 15, followed by 20, at which point the second player would be victorious.

So knowing that the second player to say a number would always win between two players, what happened with *three* players?

Again, working backwards was key — this time, with higher numbers. Suppose Player A was the first to say 20. That would eliminate Player B, after which Player C would say 1 in a two-player game with A. In other words, A would win, followed by C in second place and B in third! Similarly, if B or C said 20, then they would win, respectively.

Next, what if A was the first to say 19? Then B would have no choice but to say 20, and so B would win. In fact, if A said any number between 16 and 19 (inclusive), then B would say 20 next and win.

You could continue working backwards in the following way: Suppose a player said a number *k*. That meant the next player could say *k*+1, *k*+2, *k*+3 or *k*+4. This next player would pick the optimal result for themselves among those four numbers so that they won (or if not, so that they made it to the next round of play). And so that was exactly the result that would happen if the original player said *k*. Proceeding backwards in this fashion gave you the resulting table:

First, second and third-place finishes (1st-2nd-3rd) when each player says a certain number in three-player *veinte*

Number | When A says number |
When B says number |
When C says number |
---|---|---|---|

1 | B-A-C | C-B-A | A-C-B |

2 | A-C-B | B-A-C | C-B-A |

3 | C-B-A | A-C-B | B-A-C |

4 | C-B-A | A-C-B | B-A-C |

5 | C-B-A | A-C-B | B-A-C |

6 | C-B-A | A-C-B | B-A-C |

7 | B-A-C | C-B-A | A-C-B |

8 | B-A-C | C-B-A | A-C-B |

9 | B-A-C | C-B-A | A-C-B |

10 | B-A-C | C-B-A | A-C-B |

11 | A-C-B | B-A-C | C-B-A |

12 | C-B-A | A-C-B | B-A-C |

13 | C-B-A | A-C-B | B-A-C |

14 | C-B-A | A-C-B | B-A-C |

15 | C-B-A | A-C-B | B-A-C |

16 | B-A-C | C-B-A | A-C-B |

17 | B-A-C | C-B-A | A-C-B |

18 | B-A-C | C-B-A | A-C-B |

19 | B-A-C | C-B-A | A-C-B |

20 | A-C-B | B-A-C | C-B-A |

For example, if Player B said the number 8, then the final result would be that C came in first, B came in second and A came in third.

Since you were told in the puzzle that Player A said 1, then to determine the eventual winner we need only look at the first cell in the table. **Player B** was the eventual winner, followed by A, who made it to the second round, and finally C, who was eliminated in the first round.

For extra credit, you looked at the four-player variation of the game, with Players A, B, C and D, all of whom wanted to make it through as many rounds of the game as possible. Again, Player A started things off by saying “1.” Now which player won?

Again, suppose Player A said 20 in the first round. Then Player B would be eliminated, and C would kick off a three-way contest among C, D and A. Based on the above analysis, we already know what happens in the three-player variation: The person who goes first comes in second, the person who goes second comes in first and the person who goes third comes in last. So if A said 20, the final results would be that D came in first, C came in second, A came in third and B came in fourth.

Filling out a table just as we did in the three-person variation yielded the following:

First, second, third and fourth-place finishes (1st-2nd-3rd-4th) when each player says a certain number in four-player *veinte*

Number | When A says number |
When B says number |
When C says number |
When D says number |
---|---|---|---|---|

1 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

2 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

3 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

4 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

5 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

6 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

7 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

8 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

9 | A-D-B-C | B-A-C-D | C-B-D-A | D-C-A-B |

10 | D-C-A-B | A-D-B-C | B-A-C-D | C-B-D-A |

11 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

12 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

13 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

14 | C-B-D-A | D-C-A-B | A-D-B-C | B-A-C-D |

15 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

16 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

17 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

18 | B-A-C-D | C-B-D-A | D-C-A-B | A-D-B-C |

19 | A-D-B-C | B-A-C-D | C-B-D-A | D-C-A-B |

20 | D-C-A-B | A-D-B-C | B-A-C-D | C-B-D-A |

Since A went first and said 1, the eventual winner of the four-player game was **Player C**, followed by B, D and A.

Solver Stefano Perfetti of Zurich, Switzerland, looked at variations with increasing numbers of players. I don’t know about you, but I don’t see much of a pattern in these results as the number of players increases.

But what I *do* see is that poor Player A doesn’t win when there are anywhere from two to 30 players. Player A’s best option is to play on their own. My sympathies to that fourth grader.

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{18} 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.

You and your infinitely many friends are sharing a cake, and you come up with two rather bizarre ways of splitting it.

For the first method, Friend 1 takes half of the cake, Friend 2 takes a third of *what remains*, Friend 3 takes a quarter of what remains after Friend 2, Friend 4 takes a fifth of what remains after Friend 3, and so on. After your infinitely many friends take their respective pieces, you get whatever is left.

For the second method, your friends decide to save you a little more of the take. This time around, Friend 1 takes 1/2^{2} (or one-quarter) of the cake, Friend 2 takes 1/3^{2} (or one-ninth) of *what remains*, Friend 3 takes 1/4^{2} of what remains after Friend 3, and so on. Again, after your infinitely many friends take their respective pieces, you get whatever is left.

Question 1: How much of the cake do you get using the first method?

Question 2: How much of the cake do you get using the second method?

*Extra credit:* Your friends are feeling rather guilty for not saving enough of the cake for you, so they try one more method. This time, they only take the fractions with even denominators from the second method. So Friend 1 takes 1/2^{2} of the cake, Friend 2 takes 1/4^{2} of what remains, Friend 3 takes 1/6^{2} of what remains after Friend 2, and so on. After your infinitely many friends take their respective pieces, how much of the cake do you get?

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

From Matt Yeager comes a game that is immensely popular with his fourth-grade class:

Three of Matt’s students — Players A, B and C — are engaged in a game of *veinte*. In each round, players take turns saying numbers in order (Player A, then B, then C, then A again, etc.). The first player to go says the number “1.” Each number must be either one, two, three or four more than the number said by the previous player. When someone says “20,” the round is over and the *next* person is eliminated, with the following person beginning the subsequent round. For example, if Player A says “20,” then Player B is eliminated, while Player C begins the next round by saying “1.” At no point can anyone say a number greater than 20.

All three players want to be the winner (i.e., the only player remaining) after the two rounds. But if they realize they can’t win, then they will prioritize making it to the second round.^{19}

Player A starts things off by saying “1.” Which player will win?

*Extra credit:* Instead of three players, now suppose there are four — Players A, B, C and D — all of whom want to make it through as many rounds of the game as possible. Again, Player A starts things off by saying “1.” Which player will win?

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

Congratulations to 👏 Art Morris 👏, winner of last week’s Riddler Express.

Last week, in a puzzle I described as being “hopefully not too complex,” you had to find three distinct numbers such that the second was the square of the first, the third was the square of the second, and the first was the square of the third.

If you went hunting for numbers, you likely didn’t have much success. Squaring the first number gave you the second number, squaring the second number gave you the third, and squaring the third gave you the first number *again*. But when you square any number greater than 1, the results get bigger and bigger, while squaring a number between 0 and 1 gives you a smaller result. Meanwhile, squaring a negative number always gives you a positive number. So how could there possibly be three distinct numbers that met the criteria?

If we call the numbers *a*, *b* and *c*, then they had to satisfy the equations *b* = *a*^{2}, *c* = *b*^{2} and *a* = *c*^{2}. Putting these equations together meant that *a*^{8} = *a*, and dividing both sides by *a* meant that *a*^{7} = 1. While it was true that *a* = 1 was one solution to this equation, that didn’t help you find three distinct values for *a*, *b* and *c*.

At this point, many readers caught on to the clue from the beginning — “hopefully not too complex” was in fact suggesting that you explore complex numbers. The equation *a*^{7} = 1 has *seven* distinct solutions — the seven roots of unity. Solver Venk Natarajan wrote these out in standard form as cos(2𝜋*k*/7) + *i*·sin(2𝜋*k*/7), where *i* is the imaginary unit and *k* was any whole number less than 7 (i.e., 0, 1, 2, 3, 4, 5 or 6). Go ahead — pick a value for *k* and raise that complex number to the seventh power. You’ll get a value of 1!

Most solvers opted to write these complex roots of unity in polar form: exp(2𝜋*ik*/7). Squaring a complex number on the unit circle is equivalent to doubling its argument, or the angle it makes with the positive real axis. The number exp(2𝜋*i*/7) was one-seventh of the way around the unit circle; squaring gave you a number that was two-sevenths of the way around; squaring *that* gave you a number that was four-sevenths around; and squaring *that* gave you a number that was eight-sevenths around, which was equivalent to one-seventh — the original number!

So three distinct numbers that were squares of each other were **exp(2𝜋***i***/7), exp(2𝜋***i***·2/7) and exp(2𝜋***i***·4/7)**.

For extra credit, you had to find three other numbers with the same property. Solvers among The Hewitt School Problem Solving & Posing Class recognized that the answer above involved only three of the roots of unity — but what about the others?

The root exp(2𝜋*i*·3/7) was three-sevenths of the way around the unit circle; squaring gave you a number that was six-sevenths around; squaring *that* gave a number that was twelve-sevenths around, which was equivalent to five-sevenths; and squaring *that* gave you a number that was ten-sevenths around, which was equivalent to three-sevenths — the original number. That meant the other three distinct numbers that were squares of each other were **exp(2𝜋 i·3/7), exp(2𝜋i·6/7), and exp(2𝜋i·5/7)**.

Until the next time complex numbers make a surprise appearance in this column, remember to keep it real!

Congratulations to 👏 Emma Knight 👏 of Toronto, Ontario, Canada, winner of last week’s Riddler Classic.

Last week, Martina and Olivia each secretly generated their own random real number, selected uniformly between 0 and 1. Starting with Martina, they took turns declaring (so the other could hear) who they thought probably had the greater number until the first moment they agreed. Throughout this process, their respective numbers did not change. So for example, their dialogue might have gone as follows:

Martina: My number is probably bigger.

Olivia: My number is probably bigger.

Martina: My number is probably bigger.

Olivia: My number is probably bigger.

Martina: Olivia’s number is probably bigger.

They were playing as a team, hoping to maximize the chances they correctly predicted who had the greater number.

For any given round with randomly generated numbers, what was the probability that the person they agreed on really did have the bigger number?

There was more than one way to interpret this puzzle. One way was to assume that Martina and Olivia were truthful with each other, always using “probably” to indicate what they believed had at least a 50 percent chance of being true. For now, let’s follow this interpretation and see where it led.

Suppose Martina’s number was *x* and Olivia’s was *y*, both chosen randomly, independently and uniformly between 0 and 1. The set of all possible ordered pairs (*x*, *y*) was defined as the unit square with vertices at (0, 0), (1, 0), (1, 1) and (0, 1). For which regions inside this square would Martina and Olivia correctly agree?

When *x* was greater than 0.5, Martina would say her number was probably bigger (which Olivia would understand to mean that *x* was greater than 0.5). Meanwhile, if *y* was less than 0.5, Olivia would correctly agree that Martina’s number was greater. Similarly, if *x* was less than 0.5 and *y* was greater than 0.5, the pair would correctly agree. Already, these two scenarios accounted for half the area of the unit square.

But what if *both* *x* and *y* were greater than 0.5? Again, Martina would initially say that she probably had the bigger number. If *x* and *y* were on either side of 0.75, they would agree correctly. If both *x* and *y* were greater than 0.75, they would disagree and continue with 0.875 (halfway between 0.75 and 1) as their next pivot.

However, if *x* and *y* were both between 0.5 and 0.75, then Martina would first say she probably had the bigger number, and then Olivia would agree, ending things right then and there. At this point, they both knew *x* and *y* were between 0.5 and 0.75. But without any additional information, each was equally likely to be the bigger number, meaning there was a 50 percent chance that their guess that Olivia had the bigger number was right.

Solver Austin Shapiro graphed the unit square, splitting it up into the cases we’ve been discussing (and using binary to indicate the values of the guesses). The top-right half shows when Olivia’s number is greater, while the bottom-left half shows when Martina’s number is greater. Meanwhile, squares and rectangles shaded green indicate when they agree Olivia’s number is probably bigger, while pink indicates when they agree Martina’s number is probably bigger.

The probability that they would guess correctly was the fraction of the area that was on the “correct” side of the diagonal dividing line. One way to calculate this was to add up the areas of the triangles on the “wrong” side. The two largest triangles in the middle each had area 1/32, while each next largest pair was four times smaller than the previous pair. Summing these areas gave you an infinite geometric series: 1/16 · (1 + 1/4 + 1/4^{2} + 1/4^{3} + …), which equaled 1/16 · 4/3, or 1/12. Again, this was the probability that Martina and Olivia would be *wrong*, which meant the probability they’d be right was **11/12**. Not bad, and way better than a coin flip!

As I said from the beginning, there was another way to interpret the puzzle: Martina and Olivia might not have always been truthful, but could have strategically lied along the way if it meant improving their chances of being right — if only they were allowed to agree upon this strategy beforehand. As observed by solver Laurent Lessard, Martina and Olivia might agree that their first statements told whether their respective numbers were greater than 0.01, the second as to whether their numbers were greater than 0.02, and so on. They could get unlucky if, say, both numbers were between 0.63 and 0.64. But with arbitrary precision, they could be as likely to succeed as they had the patience for.

For extra credit, you considered what would happen if Martina and Olivia stopped only when Olivia first agreed with Martina. It turned out that this didn’t affect the results. If Martina and Olivia were playing truthfully, their chances of being correct were still **11/12**. But if they were playing strategically, there was now a way for them to always be correct *without* any prior communication: Martina could reveal the binary representation of her number, one digit at a time, so Olivia would know exactly the right moment to agree.

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

]]>Karaoke is a microcosm of everything we haven’t been able to do since the pandemic began. Large groups of people singing and shouting at the top of their lungs and sharing mics, drinks and hugs in a small, windowless space with little ventilation. Can you imagine?

For decades, we sang our hearts out with little regard for the germs we might have been spreading. In fact, for some of us, it was a way of life. We spoke to some self-professed karaoke lovers about why they feel so passionately about these dingy bars, and what not singing together for over a year has meant for them.

Watch: https://abcnews.go.com/fivethirtyeight/video/dr-fauci-life-post-vaccine-bidens-approach-pandemic-74864671

Watch: https://abcnews.go.com/fivethirtyeight/video/best-organize-bookshelf-fivethirtyeight-debate-club-71479106

]]>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,^{20} 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.

Here’s a puzzle that hopefully isn’t too complex:

Can you find three distinct numbers such that the second is the square of the first, the third is the square of the second, and the first is the square of the third? Assuming you can, what are three such numbers?

*Extra credit:* Can you find three *other* numbers with the same property?

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

From Fred Blundun comes a simple game of definitive diffidence:

Martina and Olivia each secretly generate their own random real number, selected uniformly between 0 and 1. Starting with Martina, they take turns declaring (so the other can hear) who they think probably has the greater number until the first moment they agree. Throughout this process, their respective numbers do not change. So for example, their dialogue might go as follows:

Martina: My number is probably bigger.

Olivia: My number is probably bigger.

Martina: My number is probably bigger.

Olivia: My number is probably bigger.

Martina: Olivia’s number is probably bigger.

They are playing as a team, hoping to maximize the chances they correctly predict who has the greater number.

For any given round with randomly generated numbers, what is the probability that the person they agree on really does have the bigger number?

*Extra credit:* Martina and Olivia change the rules so that they stop when Olivia first says that she agrees with Martina. That is, if Martina says on her turn that she agrees with Olivia, that is not a condition for stopping. Again, if they play to maximizing their chances, what is the probability that the person they agree on really does have the bigger number?

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

Congratulations to 👏 Charlie 👏 of Boston, Massachusetts, winner of last week’s Riddler Express (and perhaps the eponym of a certain river?).

Last week, you were playing a game of cornhole with your friends, and it was your turn to toss the four bean bags. For every bean bag you tossed onto your opponents’ board, you got 1 point. For every bag that went through the hole on their board, you got 3 points. And for any bags that didn’t land on the board or through the hole, you got 0 points.

Your opponents had had a terrible round, missing the board with all their throws. Meanwhile, your team currently had 18 points — just 3 points away from victory at 21. You were also playing with a special house rule: To win, you had to score *exactly* 21 points, without going over. (Also, you were clearly playing a variation such that one team threw all four bean bags before the other, rather than alternating tosses.)

Based on your history, you knew there were three kinds of throws you could have made:

- An
*aggressive*throw, which had a 40 percent chance of going in the hole, a 30 percent chance of landing on the board and a 30 percent chance of missing the board and hole. - A
*conservative*throw, which had a 10 percent chance of going in the hole, an 80 percent chance of landing on the board and a 10 percent chance of missing the board and hole. - A
*wasted*throw, which had a 100 percent chance of missing the board and hole.

For each bean bag, you could have chosen any of these three throws. Your goal was to maximize your chances of scoring exactly 3 points with your four tosses. What was the probability that your team would finish the round with exactly 21 points and declare victory?

Suppose you started with an aggressive throw. Then there was a 40 percent chance the bag went in the hole, in which case you could have wasted your remaining three throws. But there was also a 30 percent chance the bag landed on the board, in which case you needed to score 2 more points in your remaining three throws. Conservative throws were the best way to get those points. There was a 64 percent chance your first two throws landed on the board, a 6.4 percent chance your first and third throws landed on the board with a miss in between and another 6.4 percent chance your first throw missed while your last two landed on the board. So when your first throw landed on the board, that contributed 30 percent times 76.8 percent — or 23.04 percentage points — to your overall chances of winning.

But wait! There was a 30 percent chance that your first throw missed the board entirely. In this case, you had to score 3 points with your remaining three throws. If you were aggressive, you still had a 40 percent chance of winning on your second throw. But if that throw landed on the board (30 percent), you could still get 2 more points with conservative throws (64 percent). And if that second throw missed as well (30 percent), you could stay aggressive so that either your third or fourth throw went in the hole (40 percent plus 12 percent, or 52 percent). So with an aggressive second throw, you still had a 74.8 percent chance of winning.

Alternatively, your second throw could have been conservative. If so, it had a 10 percent chance of going on. If it landed on the board, that meant each of your final three throws had to land on the board for a (0.8)^{3}, or 51.2 percent, chance of winning. But if your second throw missed (10 percent), you were then better off playing aggressively on your third throw, with a 52 percent chance of winning. So a conservative second throw gave you a 10 percent plus 51.2 percent plus 5.2 percent — or 66.4 percent chance — of winning. That was less than 74.8, so if you missed the board on your first throw, your second throw should also have been aggressive, contributing 30 percent times 74.8 percent — or 22.44 percentage points — to your overall chances of winning.

Putting all of this together, if your first throw was aggressive, then you would win with that throw 40 percent of the time, win by landing on the board 23.04 percent of the time and win despite missing the board 22.44 percent of the time. Overall, your chances of reaching exactly 21 points stood at 85.48 percent.

Now, what would have happened if you had kicked things off with a conservative throw, rather than an aggressive throw? First off, you would (accidentally) have scored 3 points 10 percent of the time. Meanwhile, for the 80 percent of the time your throw landed on the board, you needed 2 points in three throws, which we said had a 76.8 percent chance of happening. Finally, for the 10 percent of the time you missed the board, you needed 3 points in three throws, which we said had a 74.8 percent chance of happening. All together, this meant your probability of winning was 0.1 + (0.8)(0.768) + (0.1)(0.748), or 78.92 percent.

The last time I checked, 85.48 was greater than 78.92, which meant your best bet was to start with an aggressive throw. This gave you an **85.48 percent** chance of scoring exactly 21 points.

That was a lot of casework. Several solvers, like Chris Smith and David Diamondstone, organized their calculations into a table with the number of points needed and throws remaining. By solving simpler versions of the problem with smaller numbers, they worked their way up (via a dynamic programming approach) to the given problem, which called for 3 points in four throws.

By the way, if you need a reminder of what it’s like being outside playing cornhole, here’s a photo from a neighborhood cornhole tournament, courtesy of solver Betts Slingluff.

Congratulations to 👏 Praveen Anumolu 👏 of Smithtown, New York, winner of last week’s Riddler Classic.

Last week, you were presented with a crossword puzzle title “Systematic Solving.” It had math:

Here was the solution, courtesy of the puzzle’s creator, Elise Corbin:

There were several themed clues, including 24-Across, which was the equation *y* = *x*−3, and 52-Across, which was the equation *y* = 2*x*+1. These two lines were visible in the puzzle as the circled coordinates, assuming the dead center of the grid was the origin of the coordinate plane.

Meanwhile, these two lines intersected in the southwest corner of the grid (also known as Quadrant III of the coordinate plane), at the point (−4, −7). These coordinates were part of the clues 37-Down and 70-Across.

Since this was technically a Friday crossword, I’ll bet that quite a few readers had Feyer-esque solve times. I wonder how Dr. Fill would have done …

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{21} 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.

You’re playing a game of cornhole with your friends, and it’s your turn to toss the four bean bags. For every bean bag you toss onto your opponents’ board, you get 1 point. For every bag that goes through the hole on their board, you get 3 points. And for any bags that don’t land on the board or through the hole, you get 0 points.

Your opponents had a terrible round, missing the board with all their throws. Meanwhile, your team currently has 18 points — just 3 points away from victory at 21. You’re also playing with a special house rule: To win, you must score *exactly* 21 points, without going over.

Based on your history, you know there are three kinds of throws you can make:

- An
*aggressive*throw, which has a 40 percent chance of going in the hole, a 30 percent chance of landing on the board and a 30 percent chance of missing the board and hole. - A
*conservative*throw, which has a 10 percent chance of going in the hole, an 80 percent chance of landing on the board and a 10 percent chance of missing the board and hole. - A
*wasted*throw, which has a 100 percent chance of missing the board and hole.

For each bean bag, you can choose any of these three throws. Your goal is to maximize your chances of scoring exactly 3 points with your four tosses. What is the probability that your team will finish the round with exactly 21 points and declare victory?

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

From Elise Corbin comes something new for The Riddler: a crossword puzzle! This one is titled “Systematic Solving,” and it may or may not have a mathematical twist. Enjoy!

For this puzzle, you can submit your solution by taking a picture or screenshot of your completed puzzle and uploading it via the submission form.

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

Congratulations to 👏 Dean Ballard 👏 of Seattle, Washington, winner of last week’s Riddler Express.

Last week, after you had intended to make a perfectly circular pancake, the batter spread out and filled every last corner of your square pan. (It was unclear why you were using a square pan in the first place.)

To salvage your breakfast, you planned to slice off the corners of your square pancake (an example of which is shown below), giving you something closer to a circle. Each slice had to be straight, and no slice could pass through the inscribed blue circle that represented your original desired pancake.

Of course, there was a catch — you could make at most five slices. If the blue circle had a radius of 1 unit, what was the minimum possible area your pancake can have after five slices?

For any number of cuts less than five, this problem was more straightforward. To remove as much area as possible, each of those cuts would have been tangent to the circle while making a 45 degree angle with the sides of the square. The total area of the original square pancake was 4, and each of these cuts removed (2−√2)^{2}/2 — or 3−2√2 — units of area.

Had the problem asked for exactly four cuts, the result would have been a regular octagon with an area of 4 − 4(3−2√2), which simplified to 8√2−8, or about about 3.3137 — not too far from 𝜋, the area of the ideal circular pancake.

That fifth cut really mucked up the works, however. As noted by Emma Sona, the pigeonhole principle dictated that having a fifth cut meant that one of the four corners would have *two* cuts.

Solver Ian Rhile recognized that to lop off as much area from the square as possible, both cuts had to be tangent to the circle. Ian parameterized the two tangent points by the angles they made within the circle — 𝜃_{2} and 𝜃_{3} in the diagram below — and then calculated the area of the remaining top right quarter-square after both cuts were made. This turned out to be tan(𝜃_{2}) + tan(𝜃_{3}) + tan(45°−𝜃_{2}−𝜃_{3}).

From there, you could have muddled through some more trigonometric identities, or you could have graphed the function to locate its minimum. This minimum occurred when both 𝜃_{2} and 𝜃_{3} equalled 15 degrees.

Whether you subtracted the areas of the sliced corners or counted up the area of the resulting nonagon itself, the smallest possible area of the pancake was **6√2−3√3 **units, or about 3.289 units — a smidge closer to 𝜋!

Alas, no one attempted to make such a nonagonal pancake in real life. I bet it would have been delicious.

Congratulations to 👏 Maxim Wang 👏 of St. Louis, Missouri, winner of last week’s Riddler Classic.

Last week, Riddler Nation’s neighbor to the west, Enigmerica, was holding an election between two candidates, A and B. Every person in Enigmerica voted randomly and independently, and that the number of voters was very, very large. Moreover, due to health precautions, 20 percent of the population decided to vote early by mail.

On election night, the results of the 80 percent who voted on Election Day were reported out. Over the next several days, the remaining 20 percent of the votes were then tallied.

What was the probability that the candidate who had fewer votes tallied on election night ultimately won the race?

Suppose *m* people voted by mail. Since they represented 20 percent of the voting population, that meant 4*m* people voted on Election Day. Since there were many, many voters, and each voter’s probability distribution was binomial (i.e., randomly split between candidates A and B), the central limit theorem dictated that the probability distribution for the percent of the vote each candidate received would be approximately normal.

Solver Emma Knight worked through the specific details. Call *x* the number of votes A received on Election Day minus the number of votes B received on Election Day, and let *y* be a corresponding variable for the mail-in ballots. The probability distribution for *x* was *P*(*x*) = exp[−*x*^{2}/(8*m*)]/√(8𝜋*m*), and the probability distribution for *y* was *Q*(*y*) = exp[−*y*^{2}/(2*m*)]/√(2𝜋*m*). Since the Election Day and mail-in votes were independent, meaning *x* and *y* were independent, the overall probability distribution was *R*(*x*,*y*) = *P*(*x*)*Q*(*y*).

This was a two-dimensional normal distribution. But what was it that you actually needed to compute?

For the candidate who had fewer votes tallied on election night to ultimately win the race, you needed *x* and *y* to be opposite in sign, and you also needed the magnitude of *y* to be greater than the magnitude of *x*. In mathematical terms, that meant either *y* > −*x* > 0 or *y* < −*x* < 0, both of which were regions that each made up one-eighth of the coordinate plane.

The figure below shows this graphically. The yellow region shows where the probability distribution is greater, and the blue region shows where it is smaller. The area between the two white lines represents our regions of interest.

All that was left to do was integrate the probability distribution over those regions. (Yikes!) Emma was up to the task, finding the surprisingly concise closed-form solution **tan**^{−1}**(1/2)/𝜋**, or about 14.7 percent. Several readers came up with an answer that was half this value, accounting for the case when A had fewer votes tallied on election night but ultimately won, and missing the case when B had fewer votes tallied on election night but ultimately won.

Meanwhile, many solvers went the computational route, simulating thousands or even millions of Enigmerican (that’s a word, right?) elections to see how often the candidate with fewer votes tallied on election night ultimately won. For example, Andrew Nicholls ran 100 million simulations that assumed Enigmerica had a population of 100 billion, which gave an answer of 14.75 percent — not too far from the analytical result!

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{22} 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.

After you intended to make a perfectly circular pancake, the batter has spread out, filling every last corner of your square pan. (It is unclear why you were using a square pan in the first place.)

To salvage your breakfast, you plan to slice off the corners of your square pancake, giving you something closer to a circle. The image below shows one such slice you might make. Each slice must be straight, and no slice can pass through the inscribed blue circle that represents your original desired pancake.

Of course, there’s a catch. You can make at most five slices. If the blue circle has a radius of 1 unit, what is the minimum possible area your pancake can have after five slices?

(Kudos to anyone who actually constructs an edible pancake with this minimum area.)

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

From Aaron Wilkowski comes a rather enigmatic election:

Riddler Nation’s neighbor to the west, Enigmerica, is holding an election between two candidates, A and B. Assume every person in Enigmerica votes randomly and independently, and that the number of voters is very, very large. Moreover, due to health precautions, 20 percent of the population decides to vote early by mail.

On election night, the results of the 80 percent who voted on Election Day are reported out. Over the next several days, the remaining 20 percent of the votes are then tallied.

What is the probability that the candidate who had fewer votes tallied on election night ultimately wins the race?

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

Congratulations to 👏 Mark Jackson 👏 of Rochester, New York, winner of last week’s Riddler Express.

Last week, you were creating a variation of a Romulan pixmit deck. Each card was an equilateral triangle, with one of the digits 0 through 9 (written in Romulan, of course) at the base of each side of the card. No number appeared more than once on each card. Furthermore, every card in the deck was unique, meaning no card could be rotated so that it matched (i.e., could be superimposed on) any other card.

What was the greatest number of cards your pixmit deck could have?

The numbers here were small enough that you could have written out every last card. A more efficient way to count them was to use combinatorics. To construct a unique card, you first had to pick three digits from 0 through 9. There were precisely 10 choose 3 — or 120 — ways to do this.

But 120 was not the answer. As noted by solver Yolanda Chang, there were *two* distinct ways to assign any three digits to the three sides of a card. For example, if the three digits were 1, 2 and 3, then one way to arrange them was 1-2-3 as you moved clockwise around the triangle. This card was then equivalent to 2-3-1 and 3-1-2. Alternatively, you could arrange them 3-2-1 as you moved clockwise — an arrangement that was equivalent to 2-1-3 and 1-3-2.

Another way to think about this was that, given three digits to place on the card, the two arrangements were reflections of each other. As observed by Reece Goiffon, pairs of cards connected via reflection were analogous to enantiomers in organic chemistry. Neat!

That meant the greatest number of pixmit cards was twice 120, or **240**.

For extra credit, numbers were allowed to appear two or three times on a given card. Once again, no card could be rotated so that it matched any other card. Now what was the greatest number of cards your pixmit deck could have?

In addition to the 240 cards with three distinct digits we just counted, there were 10 cards that had three of the same digit appear on them — one card for each digit. Finally, you had to count cards with two distinct digits, in which case one digit appeared twice on the card, while another digit appeared once. There were 10 digits that could appear twice, and for each of these digits you then had nine remaining digits to choose from for the digit that appeared once. Multiplying these together, there were 90 unique cards with two digits. Putting this all together, your new pixmit deck had 240 + 10 + 90, or **340** cards.

Anyway, the next time you play pixmit, don’t read too much into the cards. It’s not like the fate of the galaxy depends on it.

Congratulations to 👏 Rolfe Petschek 👏 of Kennebunkport, Maine, winner of last week’s Riddler Classic.

Last week marked the beginning of Ramadan in the U.S. The month-long observance traditionally begins with the sighting of a crescent moon, or hilal, after a new moon. For last week’s Classic, you took a closer look at a waxing crescent moon.

After a new moon, the crescent appears to grow slowly at first. At some point, the moon will be one-sixth full by area, then one-quarter full, and so on. Eventually, it becomes a half-moon, at which point its growth begins to slow down. The animation below provided some insight into what was happening here:

How many times faster was the area of the illuminated moon growing when it was a half-moon versus a one-sixth moon?

To keep the puzzle *relatively* simple, you were allowed to make a few assumptions, including:

- The moon is a perfect sphere.
- The moon’s orbit around Earth is a perfect circle.
- The moon orbits the Earth much faster than the Earth orbits the sun.
- The sun is very, very far away.

With these simplifying assumptions — and several more that Rolfe kindly listed but which I won’t get into here — you could think of the moon as having an illuminated hemisphere that was rotating with a constant angular speed. You were a distant observer who happened to be in position to see the full moon, new moon and all the phases in between.

From here, one approach to solving was to find an expression for the illuminated area *A* as a function of time *t*. You could then use calculus to find the derivative, or instantaneous rate of change, in *A*(*t*) when the moon was half-full and one-sixth full.

Here’s a helpful diagram from Thomas Stone, looking down at the moon from atop its axis of rotation and looking at the moon from Earth:

For an angle 𝜔*t* from a new moon, where 𝜔 was the angular velocity of the lunar terminator, the area of the crescent was the area of the half-moon — 𝜋*R*^{2}/2 — minus the area of a semi-ellipse. The semi-major axis of the ellipse was *R*, while the semi-minor axis was *R*cos(𝜔*t*). The semi-ellipse therefore had an area of 𝜋*R*^{2}cos(𝜔*t*)/2. That meant the crescent, which was the difference between the semicircle and the semi-ellipse, had area *A*(*t*) = 𝜋*R*^{2}[1−cos(𝜔*t*)]/2.

By the way, since the answer worked for any value of *R* and 𝜔, you could simplify this expression by choosing an *R* that canceled the 𝜋/2 and setting 𝜔 equal to 1. This gave you the simpler function *A*(*t*) = 1−cos(*t*).

Now that you had a formula for the area, you just needed to know for *which* values of *t* the moon was one-sixth full and half-full. Our simplified function *A*(*t*) oscillated between 0 and 2, meaning the moon was half-full when *A*(*t*) = 1, and one-sixth full when *A*(*t*) = 1/3. Setting 1−cos(*t*) equal to 1 meant that *t* was 90 degrees, or 𝜋/2 radians. And setting 1−cos(*t*) equal to 1/3 meant that cos(*t*) was equal to 2/3, or that *t* was cos^{−1}(2/3), or about 48.2 degrees.

Finally, to compare the *rates* at which the crescent was growing, you needed the derivative of *A*(*t*), which was *A*’(*t*) = sin(*t*). When *t* was 𝜋/2 radians for a half-moon, *A*’(*t*) was 1. (Note that the moon’s illuminated area grows the fastest at the half-moon!) Meanwhile, when *t* was cos^{−1}(2/3), *A*’(*t*) was sin[cos^{−1}(2/3)]. With the help of some trigonometry — that is, drawing a right triangle with one leg of length 2 and a hypotenuse of 3, whose *other* leg was then √5 — you were able to simplify this expression to √5/3. That meant the half-moon was growing **3/√5** — or about 1.34 — times faster than the one-sixth moon.

For those of you observing, Ramadan Mubarak!

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

]]>Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{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.

From Curtis Karnow comes a puzzle that boldly goes where no human has gone before:

You are creating a variation of a Romulan pixmit deck. Each card is an equilateral triangle, with one of the digits 0 through 9 (written in Romulan, of course) at the base of each side of the card. No number appears more than once on each card. Furthermore, every card in the deck is unique, meaning no card can be rotated so that it matches (i.e., can be superimposed on) any other card.

What is the greatest number of cards your pixmit deck can have?

*Extra credit:* Suppose you allow numbers to appear two or three times on a given card. Once again, no card can be rotated so that it matches any other card. Now what is the greatest number of cards your pixmit deck can have?

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

This past Monday marked the beginning of Ramadan in the U.S. The month-long observance traditionally begins with the sighting of a crescent moon, or hilal, after a new moon. For this week’s Classic, let’s take a closer look at a waxing crescent moon.

After a new moon, the crescent appears to grow slowly at first. At some point, the moon will be one-sixth full by area, then one-quarter full, and so on. Eventually, it becomes a half-moon, at which point its growth begins to slow down. The animation below provides some insight into what’s happening here:

How many times faster is the area of the illuminated moon growing when it is a half-moon versus a one-sixth moon?

(Some simplifying assumptions you might make for this problem are that the moon is a perfect sphere, that its orbit around Earth is a perfect circle, that the moon orbits the Earth much faster than the Earth orbits the sun and that the sun is very, very far away. If you make additional assumptions, feel free to include them in your response.)

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

Congratulations to 👏David Olsho 👏 of Seattle, Washington, winner of last week’s Riddler Express.

Last week, you had two 16-ounce cups — cup A and cup B. Both cups initially had 8 ounces of water in them.

You took half of the water in cup A and poured it into cup B. Then, you took half of the water in cup B and poured it back into cup A. You did this again. And again. And again. And then many, many, many more times — always pouring half the contents of A into B, and then half of B back into A.

When you finally paused for a breather, what fraction of the total water was in cup A?

A good first step was to work out what was happening with the first few pours. Writing the volumes of the cups as the ordered pair (*A*, *B*), the initial state was (8, 8). After the first pour from A to B, it was (4, 12); pouring half from B back to A resulted in (10, 6). Another pour from A to B resulted in (5, 11), and then back from B to A gave you (10.5, 5.5).

If you worked out a few more iterations, you noticed that with every subsequent pour from A into B, the volume of cup B appeared to be twice that of cup A. Similarly, with every pour from B into A, the volume of cup A appeared to be twice that of cup B. After many, many pours, with a final pour from B to A, cup A had **two-thirds** of the total water volume.

For extra credit, you had to consider a more general version of the problem, when both cups initially had somewhere between 0 and 8 ounces of water in them. You didn’t know the precise amount in each cup, but you knew that both cups were not empty. Again, you repeated the process of pouring half the water from cup A into cup B, and then half from cup B back to A. When you paused for a breather, what fraction of the total water was in cup A?

Surprisingly, no matter what the initial volumes in A and B were, the answer was always the same. The following animation shows 15 different starting ratios between the volumes in A and B. After just eight pours (i.e., four from A to B and another four from B to A), all of these cases were nearly the same, with **two-thirds** of the total volume in cup A.

Up to this point, we have merely *observed* that cup A appears to have two-thirds of the total volume. A number of solvers went a step further and proved this fact using a variety of different approaches, with many venturing into linear algebra, eigenvalues/eigenvectors and stationary points.

Meanwhile, solver David Kravitz offered an elegant little proof. Initially, suppose the initial fraction of water in cup A was 2/3+*x*, which meant the fraction in cup B was 1/3−*x*. Note that every ratio between the volumes in A and B can be described using some value of *x*, which could take on any value between −2/3 and 1/3.

As an ordered pair, the initial volumes were (2/3+*x*, 1/3−*x*). After the first pour from A to B, you had (1/3+*x*/2, 2/3−*x*/2). Then, after pouring from B back to A, you had (2/3+*x*/4, 1/3−*x*/4).

What was the significance of this? With every pair of pours, the difference between the volume in cup A and two-thirds of the total volume decayed by a factor of four, as did the difference between the volume in cup B and one-third of the total volume. So no matter the initial state of the cups, their volumes exponentially approached a 2:1 ratio.

I’ll drink to that!

Congratulations to 👏Mike Strong 👏 of Mechanicsburg, Pennsylvania, winner of last week’s Riddler Classic.

Last week, in an effort to make Riddler City friendlier for pedestrians and cyclists, the mayor decreed that all streets should be one-way. Meanwhile, the civil engineer overseeing this transition was not particularly invested in the project and was going to randomly assign every block of each street a particular direction.

For your daily commute to work, you drove a car two blocks east and two blocks south, as shown in the diagram below. What was the probability that, after each block was randomly assigned a one-way direction, there would still be a way for you to commute to work while staying within this two-by-two block region (i.e., sticking to the 12 streets you see in the diagram)? Here was one such arrangement of one-way streets that let you commute to work:

And no, you couldn’t get out of your car to hop on a bike or walk. I mean, you *could have*, but not in this puzzle.

Now, because this puzzle had 12 one-way streets that could all be one of two states — either left/right or up/down — that meant there were a total of 2^{12}, or 4,096, cases to consider in this problem. Yikes!

Solver Rohan Lewis realized that all viable paths had to be either 4, 6 or 8 blocks long. However, these paths were not mutually exclusive — some of the 4,096 cases included *multiple* paths, making the counting exercise even more difficult.

In the end, virtually everyone who solved this turned to their computers for some assistance. (A few brave souls nevertheless did their work by hand. I see you, Guy D. Moore and Ethan Rubin!) Of the 4,096 cases of one-way streets we have been discussing, 1,135 resulted in a viable commute to work. That meant your probability of being able to commute was 1,135/4,096, or about **27.71 percent**.

A few solvers went even further. Jezlyn Schaa and Peter Ji both determined that the probability you could make a *round trip* (i.e., to work and back) along the one-way streets was lower: 170/4,096, or about 4.15 percent. Meanwhile, others looked beyond the two-by-two block scenario, at three-by-three (~20 percent), four-by-four (~15 percent), five-by-five (~12 percent), etc.

If you work even farther from home, you might want to consult this handy chart from Maxim Wang:

Finally, it’s worth mentioning that Vikash Gayah, a professor of civil and environmental engineering at Penn State University, pointed out that “two-way streets are in fact safer for pedestrians and cyclists, and are even preferred for non-motorized roads too.” I’ll be sure to pass along this message to the Riddler City mayor!

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

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