Fiddler on the Proof

Can You Spell Like a Queen Bee?

Zach Wissner-Gross — Fri, 28 Aug 2026 12:01:53 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

Every Friday morning, I present mathematical puzzles intended to challenge and delight you. Most can be solved with careful thought, pencil and paper, and the aid of a calculator. The “Extra Credit” is where the analysis typically gets hairy, or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after puzzles are released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

In Spelling Bee, a word game from The New York Times, you must create words using seven letters arranged in a honeycomb grid. Letters can be used more than once per word, but each word must be at least four letters long and include the central letter. You may recall a previous puzzle (RIP, FiveThirtyEight) based on Spelling Bee.

Each day, your goal is to score as many points as possible. Four-letter words are worth 1 point, longer words are worth the number of letters they contain, and “pangrams” (words containing every letter) provide 7 bonus points. But these specific details don’t matter for this week’s puzzle.

If you find all the words in a given day and therefore accrue the maximum number of points, you earn the “Queen Bee” ranking. Meanwhile, accruing smaller point totals earns you other rankings. In particular, the point cutoff for “Genius” is 70 percent of the maximum number of points, rounded to the nearest whole number. While the cutoff for “Genius” is clearly displayed in the puzzle, the “Queen Bee” point total is not readily shown.

Of course, this maximum total can be approximated by dividing the “Genius” cutoff by 0.7. Even so, the total may be ambiguous, since multiple “Queen Bee” values can result in the same “Genius” cutoff.

Suppose a given round of Spelling Bee has some very large, randomly chosen point total. What is the probability that this total can be precisely determined (i.e., without any ambiguity) from its point cutoff for “Genius”?

Submit your answer

This Week’s Extra Credit

Let’s put the rank of “Genius” aside. Here are some other ranks you can attain in Spelling Bee:

Amazing (if you get 50 percent of the maximum, rounded to the nearest whole number)
Great (40 percent)
Nice (25 percent)
Solid (15 percent)
Good (8 percent)
Moving Up (5 percent)
Good Start (2 percent)

Suppose a given round of Spelling Bee has some very large, randomly chosen point total. What is the probability that this total can be precisely determined from these cutoffs combined (i.e., from “Good Start” through “Amazing,” inclusive)?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing an old Fiddler column.

While dining out at a restaurant the other night, my family located a multi-player version of Shut the box!

Thanks to this puzzle from several years ago and its corresponding solution, I wound up winning most of our family’s games.

If you missed the original puzzle, give it a try! And if you’ve seen if before, think about how the strategy for solitaire might apply to a multi-player game, and what each player’s chances of winning are depending on the order of play.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. If you think you see a mistake in the standings, kindly let me know.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Nick Harper 🎻 from San Francisco, California. I received 42 timely submissions, of which 37 were correct—good for an 88 percent solve rate.

For her photography show, Frederica Fiddleria attached a light to a point on the circumference of a circular wheel with a radius of 1 meter. She pointed a camera at the wheel and, during a single long exposure, rolled the wheel for one revolution along the ground.

When she developed the film, she was curious about the path the light took as the wheel rolled. What was the length of this path?

First off, what was the path the light took as the wheel rolled? It was, by definition, a cycloid! But even if you recognized this fact, you still had to determine the length of a cycloid after one revolution of the wheel.

Every point on the path could be parameterized as (x, y) = (t − sin(t), 1 − cos(t)), where t varied between 0 and 2𝜋 radians (i.e., one complete revolution). That first t in the x-coordinate corresponded to the horizontal translation of the wheel’s center, while the other terms represented the motion around this center. Here was an illustration of this path (in yellow), courtesy of solver Rohan Lewis:

To find the length of this parametrically-defined curve, you could evaluate the following arc length integral:

The derivative of x with respect to t was 1 − cos(t), while the derivative of y with respect to t was sin(t). Squaring these derivatives and adding them together gave you 1 − 2·cos(t) + cos²(t) + sin²(t) = 2 − 2·cos(t) = 4·sin²(t/2), thanks to some trigonometric identities. The square root of this was 2·sin(t/2). (Technically it was the absolute value of this expression, since square roots are positive by definition. But that didn’t matter here, since the function was never negative between 0 and 2𝜋.)

At this point, you were integrating 2·sin(t/2) from t = 0 to t = 2𝜋. This was -4·cos(t/2) evaluated at t = 2𝜋 minus -4·cos(t/2) evaluated at t = 0, or 4 − (-4) = 8. Thus, in the end, the length of the light’s path was exactly 8 meters.

While this result was apparently well known to some solvers, I nevertheless found it remarkable. You had a point looping around a wheel that itself is rolling, with trigonometric functions and 𝜋 all over the place. But in the end, the length was a plain and simple whole number: 8.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Jason Weisman 🎻 from Wilton, Connecticut. I received 36 timely submissions, of which 28 were correct—good for a 78 percent solve rate.

For her next show, Frederica wanted to mix things up. Instead of placing the light on the circumference of the wheel, she picked a random point inside the circle (such that any two regions with the same area were equally likely to contain the point).

As before, she rolled the wheel for one revolution along the ground and captured the motion with a single long exposure on her camera. On average, what could she have expected the length of the path to be?

How Far Can You Roll?

Zach Wissner-Gross — Fri, 21 Aug 2026 12:02:44 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

For her photography show, Frederica Fiddleria attaches a light to a point on the circumference of a circular wheel with a radius of 1 meter. She points a camera at the wheel and, during a single long exposure, rolls the wheel for one revolution along the ground.

When she develops the film, she is curious about the path the light took as the wheel rolled. What is the length of this path?

Submit your answer

This Week’s Extra Credit

For her next show, Frederica wants to mix things up. Instead of placing the light on the circumference of the wheel, she will pick a random point inside the circle. (Before you ask, let me clarify what “random” means here: Any two regions with the same area are equally likely to contain the point.)

As before, she will roll the wheel for one revolution along the ground and capture the motion with a single long exposure on her camera. On average, what can she expect the length of the path to be?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing an introductory lesson on machine learning and neural networks for high school students that I’ve been developing over at Amplify. Any and all feedback (especially from current or former educators) is greatly appreciated!

The ask: Start by opening this survey and answering a few questions about your teaching experience and views on AI. The survey will then link you to the lesson. After spending about 15 minutes reviewing it, return to the survey to complete the remaining questions about the lesson and what would help you feel prepared to teach it.

And if you’re curious about the lesson but aren’t sure if you want to complete the survey, you can check it out here.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jeffrey Ling 🎻 from San Francisco, California. I received 15 timely submissions, of which 9 were correct—good for a 60 percent solve rate.

Yes, last week’s Fiddler was a little different. It was styled after the MIT Mystery Hunt (although, as you’ll see, there was still plenty of math involved) and ran in a game-centric special edition of the Financial Times. Here was the puzzle:

Camp Algebra

Camp Algebra held a competition in which campers were split into two teams: the Aces and the Ciphers. Various members of these teams competed in a triathlon of chess, basketball, and baseball. The results are shown below. To commemorate the event, the camp counselors designed a cuboidal trophy with dimensions that cleverly related to the results of the three events. Then, being Camp Algebra after all, they computed the square of the shortest path along the trophy’s surface from one corner to the opposite corner. After decoding that number, namely, what camp mascot did they carve into the trophy?

Chess

Occupatus, playing as white for the Aces, had Castor, playing as black for the Ciphers, on the ropes. But the industrious Castor clearly gave a dam. With what next move did Castor turn the tables?

Basketball

The Aces crushed the Ciphers in the second competition. While the box score showed all the players’ names in order, it neglected to tally the two teams’ scores bit by bit.

Statistics are minutes, rebounds, assists, steals, blocks, total field goals, 3-point field goals and free throws.

Baseball

In the final event, the Aces narrowly defeated the Ciphers, 5-4. But whatever did the scoreboard look like? Surely both line-ups had something to say about that.

Camp Algebra

Zach Wissner-Gross — Fri, 14 Aug 2026 12:03:40 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

This week’s Fiddler is a little different. It’s a puzzle in the style of the MIT Mystery Hunt rather than a mathematical brainteaser, although there’s certainly some math involved. The puzzle originally ran in a game-centric special edition of the Financial Times this past Saturday (edited by the one-and-only Ollie Roeder!). There will be no Extra Credit this week.

I am reproducing the puzzle below. You can also see it in its original form over at the Financial Times—but if you do, please refrain from looking up the solution.

Camp Algebra

Submit your answer

Chess

Occupatus, playing as white for the Aces, had Castor, playing as black for the Ciphers, on the ropes. But the industrious Castor clearly gave a dam. With what next move did Castor turn the tables?

Basketball

The Aces crushed the Ciphers in the second competition. While the box score showed all the players’ names in order, it neglected to tally the two teams’ scores bit by bit. (Statistics are minutes, rebounds, assists, steals, blocks, total field goals, 3-point field goals and free throws.)

Baseball

In the final event, the Aces narrowly defeated the Ciphers, 5-4. But whatever did the scoreboard look like? Surely both line-ups had something to say about that.

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a few other puzzles from the Financial Times’ special edition:

Killer grids: logical detective puzzles. Nail the killers at the picnic and the library.
Hexwords: a game of tactical vocabulary. Outspell and outflank in this alphabetical twist on a two-player, game-theory classic.
State of the nation: a jumbo American crossword. Solve this Texas-sized puzzle — then solve its nationwide meta-puzzle.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Eric Farmer 🎻 from Columbia, Maryland. I received 55 timely submissions, of which 43 were correct—good for a 78 percent solve rate.

The Fiddler Baseball League consisted of exactly two teams of equal skill: the Algebraists and the Geometers. Over the course of a season, these two teams played each other 162 times. Each team had an equal chance of winning each game, and the results of games were independent of one another.

At the end of the season, on average, how many games would you have expected the team with the better record to have won? (If the teams had the same record, i.e., 81 wins and 81 losses, then the “better” record was 81 wins.)

Suppose the teams were A and B. The number of games won by A was a binomial distribution. More specifically, the probability that team A won exactly k games was (162 choose k) / 2¹⁶². Thus, the average number of games won by A was the sum of k · (162 choose k) / 2¹⁶², for k ranging from 0 to 162. This sum came to … exactly 81. Of course! Because the teams were equally skilled, you’d expect A and B to each win 81 games on average.

But the problem wasn’t asking for the expected number of games won by either team, it was asking for the expected number of games won by the team that won more games. For example, in the event A won 71 games, which happened with probability (162 choose 71) / 2¹⁶², then the number of wins to include in the expected value calculation was 91 (the number of games won by B). Thus, you wanted the sum of max(k, 162−k) · (162 choose k) / 2¹⁶², for k ranging from 0 to 162.

At this point, solvers Michael Montuori and Mark Goodrich jumped to spreadsheets to compute the summation. That said, you could manipulate the expression inside it to remove the “max” function, giving you 81 · (162 choose 81) / 2¹⁶² plus the sum 2k · (162 choose k) / 2¹⁶², for k ranging from 82 to 162. With a little combinatorial cleverness, you could show that the latter sum was equal to 81, giving you 81 · (162 choose 81) / 2¹⁶² + 81, which was approximately 86.07.

Solver Eric Farmer wasn’t intimidated by the vast power of 2 in the denominator, and came up with a precise rational answer: 62895632625904371267814341812263929371665856338413 / 730750818665451459101842416358141509827966271488.

Meanwhile, solvers like “Maths ForFun,” Tom Singer, and Benjamin Phillabaum noted that 81 was a fairly large number, and so the first term could could be approximated by the limit of N · (2N choose N) /2²^N as N went to infinity, or √(N/𝜋). That meant a decent approximation was 81 +√(81/𝜋), or 86.078.

And so, on average, the better team recorded roughly 86.07 wins. Since this result was fairly close to a whole number, I accepted any solution between 86.0 and 86.1, inclusive.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Tyler D 🎻 from Los Angeles, California. I received 29 timely submissions, of which 26 were correct—good for a 90 percent solve rate.

After some expansion, the Fiddler Baseball League boasted 30 teams. Over the course of a season, each team played every other team five times. (Thus, each team played a total of 145 games.) As before, each team had an equal chance of winning each game, and the results of games were independent of one another.

At the end of the season, on average, how many games would you have expected the team with the best record to have won? (As before, if more than one team had the same best record, then any of them could be considered to have the “best” record.)

How Lucky Can a Baseball Team Get?

Zach Wissner-Gross — Fri, 07 Aug 2026 12:03:11 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

The Fiddler Baseball League consists of exactly two teams of equal skill: the Algebraists and the Geometers. Over the course of a season, these two teams play each other 162 times. Each team has an equal chance of winning each game, and the results of games are independent of one another.

At the end of the season, on average, how many games would you expect the team with the better record to have won? (If the teams have the same record, then you should include one of them in your calculation.)

Submit your answer

This Week’s Extra Credit

After some expansion, the Fiddler Baseball League now boasts 30 teams. Over the course of a season, each team plays each other team five times. (Thus, each team plays a total of 145 games.) As before, each team has an equal chance of winning each game, and the results of games are independent of one another.

At the end of the season, on average, how many games would you expect the team with the best record to have won? (If more than one team has the same best record, then you should include one of them in your calculation.)

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing another feature from Quanta Magazine (yes, Quanta is amazing). This time, it’s a snapshot of AI’s role at the frontier of mathematics, particularly why it’s been so readily applied to Erdős problems.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jade Nichols 🎻 from Kalamazoo Michigan. I received 54 timely submissions, of which 41 were correct—good for a 76 percent solve rate.

Last week, you analyzed a long vertical cylinder that had three narrow open rings, each of which wrapped around seven-eighths of the cylinder (leaving a one-eighth “gap”). The rings were evenly spaced vertically, but were otherwise randomly rotated about the cylinder’s central axis. For some orientations of the rings, there existed at least one vertical line down the cylinder’s surface that passed through each ring’s gap, as illustrated below.

What was the probability that at least one such vertical line existed?

First off, the fact that the rings were evenly spaced didn’t matter here—this fact was only relevant for the Extra Credit.

Let’s start with the ring at the top. You could always draw a vertical line that passed through its gap. But with the addition of a second ring, it wasn’t guaranteed that a line could pass through both. For this to occur, at least part of the rings’ gaps had to be overlapping. In other words, the center of the second ring’s gap had to be within a 1/8-rotation of the center of the first ring’s gap, which occurred 1/4 of the time. Thus, the probability you could draw a vertical line through both the first and second rings was 1/4.

The same was true for the third ring with respect to the first ring; the probability you could draw a vertical line through both the first and third rings was 1/4. Thus, you might have thought that the probability a line could pass through all three rings was (1/4)·(1/4) = 1/16. However, that wasn’t right. Why not? Because it was possible to draw a vertical line through the first and second rings, draw a vertical line through the first and third rings, but not be able to draw a line through all three rings, as illustrated below. So the probability wound up being less than 1/16.

To get ourselves back on track, let’s consider the angular overlap between the gaps in the first two rings. If they overlapped completely (both covering the same 1/8 of the circle around the cylinder), then the probability the third ring shared some of this overlap was 1/4. If they didn’t overlap at all, then the probability the third ring shared some of this (nonexistent) overlap was 0. And if their overlap was somewhere in between, say k (with 0 < k ≤ 1/8), then the probability the third ring shared some of this overlap was k + 1/8.

The probability that k was 0 was 3/4. The remaining quarter of the time, the overlap was uniformly distributed between 0 and 1/8. Thus, the probability the third ring shared some of this overlap (again, that was k + 1/8) was a uniform distribution between 1/8 and 1/4, so that the average probability was halfway between, or 3/16.

At this point, you had the following two facts:

The probability that the second ring’s gap had any overlap with the first ring’s gap was 1/4.
When there was some nonzero overlap between the first two rings’ gaps, the average probability that the third ring’s gap shared some of that overlap was 3/16.

Putting this all together, the probability that a vertical line could be drawn through all three rings was (1/4)·(3/16) = 3/64. Sure enough, this was slightly less than our prior naive calculation of 1/16.

This was the answer when the gap size g was 1/8 of the way around the circle. As long as g was less than 1/2, the probability the second ring’s gap had any overlap with the first ring’s gap was 2g. And when there was some nonzero overlap between the first two rings’ gaps, the average probability that the third ring’s gap shared some of that overlap was 3g/2. Putting these together, the probability that a vertical line could be drawn through all three rings was 3g². When g was at least 2/3, the probability a vertical line could pass through all three rings was 1. And when g was between 1/2 and 2/3 … well, that I shall leave as an exercise to the reader!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Japheth Wood 🎻 from Kingston, New York. I received 41 timely submissions, of which 29 were correct—good for a 71 percent solve rate.

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface was allowed. An example of such a helix passing through all three gaps is shown below.

What was the probability that there existed at least one such helix that could pass through each ring’s gap?

Can You Sneak Past the Rings?

Zach Wissner-Gross — Fri, 31 Jul 2026 12:01:32 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

From Randi Goldman comes a riddle of rings:

A long vertical cylinder has three narrow open rings, each of which wraps around seven-eighths of the cylinder (leaving a one-eighth “gap”). The rings are evenly spaced vertically, but are otherwise randomly rotated about the cylinder’s central axis. For some orientations of the rings, there exists at least one vertical line down the cylinder’s surface that passes through each ring’s gap, as illustrated below.

What is the probability that at least one such vertical line exists?

Submit your answer

This Week’s Extra Credit

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface is allowed. An example of such a helix passing through all three gaps is shown below.

What is the probability that there exists at least one such helix that can pass through each ring’s gap?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing Quanta Magazine’s features on the 2026 winners of the Fields and Abacus medals. In particular, I recommend the feature on Shayan Oveis Gharan and his work on the traveling salesperson problem.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Calvin Josenhans 🎻 from Madison, Wisconsin. I received 56 timely submissions, of which 50 were correct—good for an 89 percent solve rate.

Last week, Fiddler Nation made it to the semifinals of the World Cup. All four teams that had made it this far were equally matched in that they each possessed the same total amount of “energy.” In advance of each semifinal game, teams had to independently decide how much of their energy to allocate to the match; all remaining energy went toward the finals. The team that spent more energy in any given game won. The semifinals and finals occurred so close in time that teams couldn’t recuperate any of their energy in between.

You had heard that the managers for the other three teams were abysmal and had no idea how to allocate their teams’ energy. Each of the other managers independently picked a random percentage between 0 and 100 and allocated that portion of their team’s energy to the semifinal game; the remaining energy went toward the final.

Since you were the cleverest manager of the bunch, you could choose an optimal strategy that maximized Fiddler Nation’s probability of winning the World Cup. What was this optimal probability?

If you (naively) went with the same randomized strategy as all your opponents, then by symmetry you each had a 25 percent chance of winning the World Cup. Surely it was possible to improve upon these chances.

Suppose the fraction of energy you dedicated to the semifinal round was A, which meant you had 1−A energy for the final—if you made it that far. Let’s calculate your probability of winning the World Cup p as a function of A. Maximizing p(A) would then result in the optimal strategy.

You won the semifinal round as long as your opponent allocated less than A energy to the match. Since the probability distribution for their energy was uniform between 0 and 1, the probability that your energy was greater than theirs was just A. Thus, your probability of advancing to the final was similarly A.

However, the probability distribution for your opponent’s energy in the final was not uniform. Why? Because you knew for a fact that your finals opponent had won their semifinal match, which meant they had allocated more energy to their semifinal match than their semifinal opponent had. Therefore, their probability distribution for energy spent in the semifinal was that of the maximum of two uniformly chosen random variables.

Determining this probability distribution is a classic problem in order statistics, and you can work it out for yourself by considering the function max(x, y) over the region 0 < x, y < 1. The probability your opponent in the finals spent energy B to win their semifinal turned out to be 2B.

Because you faced the winner of the other semifinal, i.e., the one who had spent more energy in that match, they had less energy saved up for the final. That was good news for you, as it meant you had a greater chance of beating them. In particular, you needed 1−B (their remaining energy) to be less than 1−A (your remaining energy), or, equivalently, you needed B to be greater than A.

The probability distribution for B is shown below. For a given value of A, the probability that B was greater was the area of the purple trapezoidal region, or, equivalently, 1 minus the area of the blue triangular region. This was 1−(A)(2A)/2, or 1−A².

Putting this all together, your probability of winning your semifinal match was A, while your probability of winning the final (given that you won the semifinal) was 1−A². Your probability of winning the World Cup was the product of these, or A·(1−A²). Solver Chris Payne plotted this as a function of A:

To maximize this function, you could set its derivative, 1−3A², equal to zero, which occurred when A = 1/√3, or about 0.57753. Thus, it was optimal to dedicate about 58 percent of your energy to the semifinal match, saving the remaining 42 percent for the final. With this strategy, your probability of winning was 1/√3·(1−1/3), which simplified to 2/(3√3), or about 38.5 percent. That was a lot better than the 25 percent chance we naively started with!

Several readers submitted answers of 1/√3, the fraction of your energy you should have expended in the semifinal, which was not what the puzzle was asking for. However, given that these readers had essentially “solved” the puzzle, I awarded credit.

By the way, what were your three opponents’ respective chances of winning when you played optimally? To advance, your semifinal opponent had to allocate more than 1/√3 of their energy to the semifinal match, the probability of which was 1−1/√3. With some calculus, you could determine that their chances of winning in the finals was then (5−√3)/9. Putting this together, your semifinal opponent’s probability of winning the World Cup was (1−1/√3)·(5−√3)/9, or about 15.35 percent. That meant the remaining two teams each had a 23.08 percent chance of winning—not too far from 25 percent. Thus, by optimizing your strategy, you took greater advantage of your semifinal opponent than a potential finals opponent.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Aaron Slepkov 🎻 from Peterborough, Ontario, Canada. I received 27 timely submissions, of which 19 were correct—good for a 70 percent solve rate.

As it turned out, I spoke too soon. Fiddler Nation had made it to the quarterfinals of the World Cup rather than the semifinals. As before, teams had to allocate the same total amount of energy across up to three matches.

The managers for the other seven teams remained abysmal. Each manager independently picked a random percentage between 0 and 100 and allocated that amount of their team’s energy to the quarterfinal. If they won, they allocated a random amount of their remaining energy to the semifinal. And if they won that, the rest of their team’s energy went toward the final.

Fiddler Nation’s strategy had to be drawn up in advance, with no specific knowledge of the other teams’ strategies beyond what I already shared.

That said, as the cleverest manager of the bunch you could once again choose an optimal strategy that maximized Fiddler Nation’s probability of winning the World Cup. What was this optimal probability?

Can You Win the World Cup?

Zach Wissner-Gross — Fri, 17 Jul 2026 12:03:16 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

Note: Due to personal travel, the next edition of Fiddler on the Proof will be coming out on July 31, 2026.

This Week’s Fiddler

Congratulations to Fiddler Nation for making it to the semifinals of the World Cup! All four teams that made it this far are equally matched in that they each possess the same total amount of “energy.” In advance of each semifinal game, teams must independently decide how much of their energy to allocate to the match; all remaining energy goes toward the finals. The team that spends more energy in any given game will win. The semifinals and finals occur so close in time that teams can’t recuperate any of their energy in between.

You’ve heard that the managers for the other three teams are abysmal and have no idea how to allocate their teams’ energy. Each of the other managers will independently pick a random percentage between 0 and 100 and allocate that portion of their team’s energy to the semifinal game; the rest of that team’s energy will go toward the final.

Since you’re the cleverest manager of the bunch, you can choose an optimal strategy that will maximize Fiddler Nation’s probability of winning the World Cup. What is this optimal probability?

Submit your answer

This Week’s Extra Credit

As it turns out, I spoke too soon. Fiddler Nation has made it to the quarterfinals of the World Cup rather than the semifinals. My mistake. As before, teams must allocate the same total amount of energy across up to three matches.

The managers for the other seven teams remain abysmal. Each manager will independently pick a random percentage between 0 and 100 and allocate that amount of their team’s energy to the quarterfinal. If they win, they will allocate a random amount of their remaining energy to the semifinal. And if they win that, the rest of their team’s energy will go toward the final.

Fiddler Nation’s strategy must be drawn up in advance, with no specific knowledge of the other teams’ strategies beyond what I have already shared.

That said, as the cleverest manager of the bunch you can once again choose an optimal strategy that will maximize Fiddler Nation’s probability of winning the World Cup. What is this optimal probability?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing two puzzles from familiar sources:

July’s MoMath Monthly Mindbender asks you to draw 100 chords on a circle, with each chord generated by connecting two randomly selected points on the circumference. On average, how many intersections can you expect? This reminds me of a similar puzzle from my Riddler days…
July’s Ponder This from IBM has now apparently corrected its previous typos. If you’re in the mood for number theory and large numbers, this one’s for you!

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Ed Moores 🎻 from London, United Kingdom. I received 57 timely submissions, of which 48 were correct—good for an 84 percent solve rate.

Last week, the Tour de Fiddler was back!

This time, you analyzed a model for a cyclist’s speed v as a function of their pedaling power P, their mass m, and the ground’s angle of inclination 𝜃: v = P/(m·sin𝜃+10). For the purposes of this puzzle, you didn’t have to worry about the units for power, mass, or speed. (These are typically given in Watts, kilograms, and kilometers per hour or miles per hour, respectively.)

In cycling, roads are marked with a gradient g, which is a hill’s slope, typically expressed as a percentage. For example, an incredibly steep 45-degree incline has a gradient of 1, or “100 percent.”

You were asked to consider the following two riders:

A “climber,” who had a power of 300 and a mass of 60
A “sprinter,” who had a power of 325 and a mass of 80

At what gradient would the climber and sprinter have cycled at the same speed?

According to the formula, the climber’s speed was 300/(60·sin𝜃+10), while the sprinter’s speed was 325/(80·sin𝜃+10). Here was a graph of these two functions when 𝜃 was measured in degrees:

As their titles suggested, the sprinter (shown in blue) was faster on flatter terrain, while the climber (shown in red) was faster on steeper terrain. Somewhere around 3 degrees their speeds were equal.

To compute the exact speed at which this occurred, you had to set their speeds equal to each other, giving you the equation 300/(60·sin𝜃+10) = 325/(80·sin𝜃+10). Solving this gave you sin𝜽 = 1/18. Thus, 𝜃 = sin^-1(1/18), or about 3.185 degrees, consistent with our estimation from the graph.

However, the puzzle asked for the gradient, or slope, of the incline. A few readers thought the slope was the ratio between an angle and 45 degrees. However, that would have meant that a 90-degree angle had a gradient of 2—in reality, the slope of a 90-degree angle is infinite.

No, the slope of a hill is instead the tangent of its angle—just as the slope is “rise over run,” an angle’s tangent is “opposite over adjacent.” Therefore, if the angle of inclination was sin^-1(1/18), then the slope of the hill was tan(sin^-1(1/18)). You could find an exact value for this trigonometric expression using the following (not-to-scale!) diagram:

Here was an angle 𝜃 the sine of which was 1/18. By the Pythagorean theorem, the angle’s adjacent side had a length of √(18²−1²) = √(323). Thus, the tangent of this angle—and the slope of the hill—was 1/√(323), or approximately 0.0556, which could equivalently be written as 5.56 percent.

When 𝜃 is a small number of radians, all three of 𝜃, sin𝜃, and tan𝜃 are approximately equal. In this case, while tan𝜃 was 1/√(323) = 0.055641488…, 𝜃 was 1/18 = 0.0555555…, only about 0.1 percent smaller. Given how close these values were, I reluctantly awarded credit for answers that appeared to give either the gradient or the angle, i.e., anything between 0.055 and 0.056.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Ivor Traber 🎻 from Toronto, Canada. I received 31 timely submissions, of which 21 were correct—good for a 68 percent solve rate.

The climber and the sprinter were racing up a perfectly sinusoidal hill. They went from the base, where the gradient was 0 percent, to the peak, where the gradient was again 0 percent. For them to reach the top at the same time, what should the maximum gradient of the hill have been?

Importantly, the formula for v given above was for a rider’s speed along the ground. Thus, when the ground was inclined, the same speed would have covered less horizontal distance per unit time.

Can You Power up the Hill?

Zach Wissner-Gross — Fri, 10 Jul 2026 12:03:17 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

Every summer I try to run a cycling-related puzzle to coincide with the Tour de France. Previous puzzles have involved the shape of the peloton, catching the breakaway, and sprinting to the finish (not to mention three more puzzles from FiveThirtyEight). Man, I’ve been doing this for a while.

As it’s now July, the Tour de Fiddler is back!

This time, we’ll be looking at a model for a cyclist’s speed v as a function of their pedaling power P, their mass m, and the ground’s angle of inclination 𝜃:

(For the purposes of this puzzle, you needn’t worry about the units for power, mass, or speed. If you’re curious, they’re typically given in Watts, kilograms, and kilometers per hour or miles per hour, respectively.)

In cycling, roads are marked with a gradient g, which is a hill’s slope, typically expressed as a percentage. Thus, an incredibly steep 45-degree incline has a gradient of 1, or “100 percent.”

Consider the following two riders:

A “climber,” who has a power of 300 and a mass of 60
A “sprinter,” who has a power of 325 and a mass of 80

At what gradient will the climber and sprinter cycle at the same speed? (You can give your answer as a value between 0 and 1 or as a percentage.)

Submit your answer

This Week’s Extra Credit

The climber and the sprinter are racing up a perfectly sinusoidal hill. They go from the base, where the gradient is 0 percent, to the peak, where the gradient is again 0 percent. For them to reach the top at the same time, what should the maximum gradient of the hill be? (You can give your answer as a value between 0 and 1 or as a percentage.)

Importantly, note that the formula for v given above is for a rider’s speed along the ground. Thus, when the ground is inclined, the same speed will cover less horizontal distance per unit time.

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a post from Fawn Nguyen, my colleague over at Amplify. Fawn is a major advocate for non-routine mathematics in schools, and this post includes a nugget from the incomparable James Tanton plus a short, inspirational video from the Global Math Project.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Aslan Buluthan Gökbulut 🎻 from Erdek, Türkiye. I received 40 timely submissions, of which 40 were correct—good for a perfect 100 percent solve rate. Great work, everyone!

When designing her new nation’s flag, Retsy Boss wanted to compactly arrange some stars. These stars were positioned along a square grid, but she only wanted to include stars whose centers were at most two units away from some point on the plane.

For example, if she had centered the circle on a star itself, then she could have placed a total of 13 stars on the flag, as shown below:

What was the greatest number of stars Retsy could have placed on the flag?

The answer was at least 13, of course, thanks to the example given above. But was it possible to do any better? Indeed it was! Consider the circle placement shown below:

Sure enough, there were 14 stars inside this circle of radius 2. While this image was helpful, let’s prove that 14 stars fit inside the circle.

Fitting the stars was equivalent to circumscribing a circle around seven unit squares arranged as below:

When the circle was as tight as possible around the squares, it was tangent to them at four locations, including the two shown above. The center of the circle was a distance x above the bottom row of squares such that the two red radii in the diagram had the same length.

By the Pythagorean theorem, the squares of the radii were 0.5² + (2−x)² and 1.5² + (1+x)². Setting these equal and solving gave you x = 1/6, which meant the radii both had a length of √(65/18), or approximately 1.9003. Since this was less than 2, that confirmed that all the squares (and thus 14 stars) fit inside a circle of radius 2.

Finally, solver 🎬 Eli Wolfhagen 🎬 plotted how many stars were in the circle depending on where the circle’s center was within one of the unit squares on the grid. This value ranged from a minimum of 10 to a maximum of 14, as shown below. (Note that when the circle was centered exactly over a star, there were 13 stars in the circle, but this number dropped to 10 the moment the circle moved in any direction.)

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Amelie Zhou 🎻 from Richmond Hill, Ontario, Canada. I received 36 timely submissions, of which 32 were correct—good for an 89 percent solve rate.

After 250 years, the nation commissioned Retsy Boss VIII to design a new flag with one star for each of the nation’s current 58 states. As an homage to the original flag design, Retsy wanted to select 58 stars from the square grid that were all at most some distance R from a point on the plane.

What was the minimum distance R that Retsy could have used?

Can You Fit the Stars on the Flag?

Zach Wissner-Gross — Fri, 03 Jul 2026 12:03:24 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

For example, if she had centered the circle on a star itself, then she could have placed a total of 13 stars on the flag, as shown below:

What is the greatest number of stars Retsy could have placed on the flag?

Submit your answer

This Week’s Extra Credit

After 250 years, the nation has commissioned Retsy Boss VIII to design a new flag with one star for each of the nation’s current 58 states. As an homage to the original flag design, Retsy wants to select 58 stars from the square grid that are all at most some distance R from a point on the plane.

What is the minimum distance R that Retsy can use?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a reminder I received from my friend Scott Kominers about a neat date from last week:

June 28 was Tau Day, with “tau” being double pi, or approximately 6.28. (Tau Day is also “Perfect Number Day,” as the first two perfect numbers are 6 and 28.)

But that wasn’t all! 6/28/26 was a palindrome! For more, check out Scott’s article.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each.

The results from Q2 are in, and the finest of Fiddlers, who each solved all 26 puzzles, are:

👑 David Kong 👑 from Toronto, Ontario, Canada
👑 Jason Weisman 👑 from Wilton, Connecticut
👑 Peter Ji 👑 from Madison, Wisconsin

Congratulations to all who participated; the final standings are below. (If you think you see a mistake in the standings, kindly let me know.)

I will also be reaching out to the first-time top finishers so I can send along a fancy-schmancy t-shirt prize! (If you reside outside the US, I appreciate your patience on this.) In Q2, these top finishers were:

👕 Tom Keith 👕 from Toronto, Ontario, Canada
👕 Laurent Mandorla 👕 from Austin, Texas
👕 Mike Porter 👕 from Haddenham, United Kingdom

Finally, this week’s puzzles mark the beginning of the next quarter (Q3). If you’d like your shot at glory (and a t-shirt), become a paid subscriber today!

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Billy Mullaney 🎻 from Minneapolis, Minnesota. I received 32 timely submissions, of which 26 were correct—good for an 81 percent solve rate.

Last week, I was redoing my kitchen floor using rhombus-shaped tiles composed of two congruent equilateral triangles. One such tile is shown in blue below. How many distinct ways could I have used these to tile the outlined region below, which consisted of 24 equilateral triangles arranged in a regular hexagon?

You could have tried out various tilings by hand. Indeed, solvers like Kerry Brown and Mike Porter did exactly just that!

Another really neat approach involved recognizing that the three possible orientations of any placed rhombus corresponded to a face of a unit cube in an isometric projection. I think this made a litlte more sense with a visual: Below is one such tiling, courtesy of solver 🎬 Michael Schubmehl 🎬, along with a re-shading of the tilings by orientation.

In the shading on the right, it looks like a stack of two cubes shoved in the back corner with another cube next to them. Well, from one perspective, at least. Alternatively, it looks like a base layer of four cubes with one on top shoved in the back corner. (Maybe there’s another way to see it? My eyes are starting to hurt trying to find it.)

Either way, each isometric projection (from a given perspective) had a 1-to-1 correspondence with some packing of unit cubes within a 2-by-2-by-2 box. There were four vertical stacks of cubes, with each stack containing 0, 1, or 2 cubes. This gave an upper bound of 3⁴ = 81 projections, but not every projection was possible. In particular, if a was the height of the stack closest to the observer, d was the height farthest from the observer (i.e., the “back corner”), and the other two heights were b and c, then you needed a ≤ b, a ≤ c, b ≤ d, and c ≤ d. Indeed, in the above example, depending on your orientation, you had (a, b, c, d) equal either (0, 0, 1, 2) or (1, 1, 1, 2), both of which were valid ordered quadruples.

At this point, your challenge was to find the total number of such valid ordered quadruples. As it turned out, there were 20 of them:

(0, 0, 0, 0)
(0, 0, 0, 1)
(0, 0, 1, 1)
(0, 1, 0, 1)
(0, 1, 1, 1)
(0, 0, 0, 2)
(0, 0, 1, 2)
(0, 1, 0, 2)
(0, 1, 1, 2)
(0, 0, 2, 2)
(0, 2, 0, 2)
(0, 1, 2, 2)
(0, 2, 1, 2)
(0, 2, 2, 2)
(1, 1, 1, 1)
(1, 1, 1, 2)
(1, 1, 2, 2)
(1, 2, 1, 2)
(1, 2, 2, 2)
(2, 2, 2, 2)

And here were all 20 corresponding isometric projections, courtesy of solver 🎬 Jason Weisman 🎬:

Note that some of these tilings are equivalent under reflection and rotation. With this interpretation of the puzzle, there were 6 distinct tilings, shown below (along with how many tilings each corresponded to from the above list):

Thus, I awarded credit for answers of 20 or 6.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Emilie Mitchell 🎻 from Nashville, Tennessee. I received 20 timely submissions, of which 17 were correct—good for an 85 percent solve rate.

I was also redoing my patio using similar rhombus-shaped stones. How many distinct ways could I have tiled the outlined region below, which consisted of 54 equilateral triangles?

Can You Tile the Hexagon?

Zach Wissner-Gross — Fri, 26 Jun 2026 12:02:51 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

I’m redoing my kitchen floor using rhombus-shaped tiles composed of two congruent equilateral triangles. One such tile is shown in blue below. How many distinct ways can I use these to tile the outlined region below, which consists of 24 equilateral triangles arranged in a regular hexagon?

Submit your answer

This Week’s Extra Credit

I’m also redoing my patio, using similar rhombus-shaped stones. How many distinct ways can I tile the outlined region below, which consists of 54 equilateral triangles?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing—courtesy of reader Henry Huang—a story by Greg Egan that I haven’t yet in its entirety. But I got far enough to encounter some of the great puzzles that caught Henry’s attention, such as this one (among many delectable choices):

Can you find an integer … whose reciprocal has a recurring decimal with ten repeating digits, all of them different?

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Reid 🎻 from Kansas City. I received 50 timely submissions, of which 31 were correct—good for a 62 percent solve rate.

Last week, I was concerned that the random number generator on my calculator might have been malfunctioning.

Under normal conditions, it should have generated random numbers between 0 and 1. But my suspicion was that the calculator was “tanked,” meaning it only generated random numbers between 0 and some value 0 < a < 1. Beyond that, I had no knowledge regarding the value of a; it was equally likely to be any value from 0 to 1.

As an experiment, I asked the calculator to generate one random number. It produced a value of exactly 0.5. (While this was, admittedly, infinitely unlikely, I asked you to roll with it!)

Based on this result, what could I have expected the value of a to be, on average?

First off, whoever heard of “tanking” a calculator? That’s not a thing. “Bricking” would have made more sense. Ah, but this was a subtle nod to the famed German tank problem, a discrete problem that was analogous to last week’s continuous one.

In any case, we knew a couldn’t be less than 0.5. If it had been, then it would have been impossible for the calculator to produce a value of 0.5! Thus, we had the restriction that 0.5 ≤ a < 1.

At this point, it was tempting to conclude that any value of a within this range was equally likely, and that the answer was therefore 3/4, or 0.75. Indeed, quite a few readers submitted this answer. However, 0.75 wasn’t quite right.

Given a particular value of a, the calculator produced a random value x with 0 < x < a, which meant the uniform probability density in this region was p(x) = 1/a. Thus, the smaller the value of a, the greater the probability density function. Using a Bayesian approach, the probability for any particular value of a was then proportional to 1/a.

Let’s pause and think about this for a minute. Instead of saying the calculator gave you an exact value of 0.5, suppose it was within some very small range, like between 0.499 and 0.501. If a was 1, then the probability a random number would have fallen in this range was (0.501−0.499)/1, or 2/1000. But if a was smaller—say, 0.501—then the probability a random number would have fallen in this range was (0.501−0.499)/0.501, or 2/501. That was almost twice as likely!

So the probability for a given value of a was proportional to 1/a, and a could be anywhere between 0.5 and 1. To compute the expected value of a, you had to integrate this density, weighted by each value of a, and normalize:

The numerator was equal to 0.5, while the denominator was equal to ln(2). Thus, the average was 1/(2ln(2)), or approximately 0.72135. Sure enough, it was not 0.75; rather, it was less than 0.75, which made sense given how we expected smaller values of a to be favored.

If you’re still not convinced, you can confirm this result via simulations! As an illustration, I ran 1 million simulations in which I first picked a random value for a with 0 < a < 1 and then a random value for x with 0 < x < a. Among the 1 million simulations, 1378 resulted in a value of x between 0.499 and 0.501. The histogram below shows the distribution for a among these 1378 simulations:

You can clearly see the downward trend (although it’s harder to tell that this is hyperbolic rather than linear), and that values of a close to 1 were about half as common as values of a close to 0.5.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Christian Robert 🎻 from Paris, France. I received 38 timely submissions, of which 24 were correct—good for a 63 percent solve rate.

Frustrated with my old calculator, I tossed it in the ~~trash~~ recycling bin and bought a new one. But now I was concerned this second calculator was also “tanked.” As before, every value of a between 0 and 1 was equally likely (initially).

I asked my friend to generate one random number using this second calculator. My friend did so, and smirked. “I won’t tell you what the number is,” my friend said, “but it’s somewhere between 0 and 0.5.”

On average, what could I now have expected the value of a (for this second calculator) to be?

Can You Fix the Random Number Generator?

Zach Wissner-Gross — Fri, 19 Jun 2026 12:03:03 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

I think the random number generator on my calculator might be malfunctioning. Oh no!

Under normal conditions, it should generate random numbers between 0 and 1. But my suspicion is that the calculator is “tanked,” meaning it only generates random numbers between 0 and some value 0 < a < 1. Beyond that, I have no knowledge regarding the value of a. At the moment, it’s equally likely to be any value from 0 to 1.

As an experiment, I ask the calculator to generate one random number. It produces a value of exactly 0.5. (While this is, admittedly, infinitely unlikely, let’s roll with it!)

Based on this result, what can I expect the value of a to be, on average?

Submit your answer

This Week’s Extra Credit

Frustrated with my old calculator, I toss it in the trash and buy a new one. But now I’m concerned this second calculator is also “tanked.” As before, every value of a between 0 and 1 is equally likely at first.

I ask my friend to generate one random number using this second calculator. My friend does so, and smirks. “I won’t tell you what the number is,” my friend says, “but it’s somewhere between 0 and 0.5.”

On average, what can I expect the value of a (for this second calculator) to be?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing an old Substack note from Elliott Thornley, which I paraphrase below:

In front of you are two urns. The smaller urn contains exactly one white ball and one red ball. The larger urn contains 1 million white balls and two red balls.
You can choose one of the urns and remove exactly one ball from it at random. If the ball is white, you get $1. Two white balls are then added to the urn (the one you removed, and one additional white ball), and the game continues.
At any point, if the ball you remove is red, the game ends.
If your goal is to maximize your expected (i.e., average) earnings, which urn should you choose? And how much should you be willing to pay to play this game with that urn?

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Kenny Coven 🎻 from Chagrin Falls, Ohio. I received 46 timely submissions, of which 40 were correct—good for an 87 percent solve rate.

Semicircle Island is shaped like a perfect semicircle (or semidisk, technically), with a radius of 1 mile. It doesn’t have any permanent residents, but it’s a very popular destination for surfers.

Last week, rumor had it that a big wave was headed toward the island, but no one knew which direction it was coming from. This thin, straight wall of water never changed speed or direction. It would first make contact with the island at 10 a.m. and it would last be in contact with the island at 10:10 a.m.

What was the longest possible stretch of land that was directly under the wave at 10:05 a.m.?

At either 10 a.m. or 10:10 a.m. the wave was in contact with one of the island’s two sharp corners. For the puzzle, it didn’t matter which of these times it was, so let’s say this occurred at 10 a.m. Then, at 10:10 a.m., the wave was tangent to the island’s circular coast. The key was to realize that, to have the longest possible stretch of land under the wave at 10:05 a.m., you wanted the wave to pass over the other sharp corner of the island, as illustrated below:

Let’s set the center of the circle (if we had deigned to draw its bottom half) at the origin, O = (0, 0). The wave first hit the island at point A = (1, 0) and last made contact an angle 𝜃 around the circumference at the point B = (cos(𝜃), sin(𝜃)). At 10:05 a.m., the wave passed through both point C = (-1, 0) and the midpoint between A and B, which was point D = (1/2 + cos(𝜃)/2, sin(𝜃)/2).

Now, how to compute the length of that purple chord in the above diagram?

There were several ways to attack this, one of which was to recognize that line CD had to be parallel to the blue lines representing the wave’s initial and final positions. Meanwhile, the wave was perpendicular to the direction in which it propagated, along the line OB. Thus, the requirement was that CD was perpendicular to OB at their intersection, E.

Two vectors are perpendicular when their dot product equals zero. Vector CD was (3/2 + cos(𝜃)/2, sin(𝜃)/2), while vector OB was (cos(𝜃), sin(𝜃)). Their dot product was 3/2·cos(𝜃) + cos²(𝜃)/2 + sin²(𝜃)/2, which had to equal zero. Using the trigonometric identity sin²(𝜃) + cos²(𝜃) = 1, this gave you 3/2·cos(𝜃) + 1/2 = 0, or cos(𝜃) = -1/3. Angle 𝜃 was the measure of angle AOB in the diagram above, and if its cosine was -1/3, then the cosine of supplementary angle COB was 1/3. Since OC was a radius of the unit circle, it had length 1, which meant OE had length 1/3.

By the Pythagorean theorem, CE had length √(1−(1/3)²) = 2(√2)/3. Line segment CE was half the length of the purple chord, so the longest stretch of land under the wave was 4(√2)/3, or about 1.8856 miles. (I accepted any answer between 1.88 and 1.89 miles.)

Here’s an animation from solver 🎬 Jason Shaw 🎬 showing all the possible lengths. Sure enough, the maximum is approximately 1.8856 miles.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Vamshi Jandhyala 🎻 from London, United Kingdom. I received 35 timely submissions, of which 34 were correct—good for a 97 percent solve rate.

Another wave was approaching the island. Yet again, no one knew which direction it was coming from. For the moment, all directions were equally likely.

On average, what was the length of the stretch of land directly under the wave halfway between when the wave first and last made contact with the island?

Can You Catch the Longest Wave?

Zach Wissner-Gross — Fri, 12 Jun 2026 12:03:50 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

Rumor has it that a big wave is headed toward the island, but no one knows which direction it’s coming from. This thin, straight wall of water never changes speed or direction. It will first make contact with the island at 10 a.m. and it will last be in contact with the island at 10:10 a.m.

What is the longest possible stretch of land that is directly under the wave at 10:05 a.m.?

Submit your answer

This Week’s Extra Credit

Another wave is approaching the island, but again no one knows which direction it’s coming from—for the moment, all directions are equally likely. For example, it might come in the direction illustrated below:

On average, what is the length of the stretch of land directly under the wave halfway between when the wave first and last makes contact with the island?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m late to the party in sharing that a significant Erdős problem has been solved by AI. This one concerns unit distances:

Suppose you have N points in the plane. As N grows very, very large, how many pairs of points can you possibly have that are a unit distance apart? With a gridlike arrangement of points, the number of such pairs grows linearly with N. But the AI proved that a superlinear function of the form N^1+𝛿 (for some constant 𝛿 > 0) was in fact a lower bound. Wow.

By the way, this reminds me of another nugget involving unit distances:

Suppose every point in the plane is colored red, green, or blue. Find seven points in the plane such that, no matter how the points are colored, you will always have two points among the seven that are the same color and are a unit distance apart.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Alastair 🎻 from Cape Town, South Africa. I received 75 timely submissions, of which 73 were correct—good for a 97 percent solve rate.

Last week, three game show contestants had to work together to win a prize. They were shown a large bag that was initially empty, and then they saw three red hats and two white hats placed into the bag. At this point, the contestants were blindfolded. Each contestant then picked a hat at random from the bag and placed it on their head.

One at a time, their blindfolds were removed and they could see the hats on the others’ heads—but not the hat on their own head. If they could, with absolute certainty, identify the color of the hat on their own head, then the game was over and all three contestants won a prize! Otherwise, they skipped their turn, at which point the next contestant had their blindfold removed and the process continued. If all three contestants skipped their turn, then no prize was won. While blindfolded, contestants could still hear the decisions of other contestants and could identify who said what.

Did the contestants always win the prize?

Let’s approach this logically, by considering the contestants in order of blindfold removal. We’ll use “R” to mean a corresponding contestant had a red hat and “W” for white hat. There were eight potential cases to consider:

Since there were only two white hats in the bag, we could eliminate that last case right off the bat. So now we were down to seven cases:

Now imagine you were the first contestant looking at the other two hats. If you saw two white hats, you knew your hat was red. That’s because there were only two white hats, and equivalently because the case “RWW” was unique among the remaining cases in that ended with “WW.”

If that’s what the first contestant actually saw, then great! Otherwise, the first contestant wasn’t sure about their hat, and the game continued as one of the following six cases:

Now imagine you were the second contestant looking at the other two hats. As before, if you saw two white hats, you knew your hat was red. And again, one way to explain that was that “WRW” was the only remaining case that began and ended with a “W.”

But wait—there was another case with a unique beginning and end: “RRW.” If the second contestant saw a red hat on the first contestant’s head and a white hat on the third contestant’s head, they knew their own hat had to be red! Why? Suppose the second contestant had a white hat on. Then the first contestant would have seen two white hats and realized their own hat was red. But since the first contestant said nothing, this case could be eliminated.

So if the second contestant was able to identify their hat color, then great! Otherwise, the second contestant wasn’t sure about their hat, and the game continued as one of the following four cases:

Whenever the game reached the third contestant, their hat was guaranteed to be red. Thus, the contestants always won the prize. Interestingly, they always locked in their win when one of the contestants realized their hat was red, and never when it was white.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Aaron P. Moskalik 🎻 from Southfield, Michigan. I received 33 timely submissions, of which 30 were correct—good for a 91 percent solve rate.

Now, there were four game show contestants who had to work together. They were shown a large bag that was initially empty, and then they saw R red hats, W white hats, and B blue hats placed into the bag such that R, W, and B were each less than or equal to 4. Otherwise, the game was played as before.

For some triples (R, W, B), the contestants always won. Among these, what was the greatest value of R + W + B?

Can You Infer the Color of Your Hat?

Zach Wissner-Gross — Fri, 05 Jun 2026 12:03:28 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

From Alan Pierrot comes a great way to “cap” off your week:

Three game show contestants must work together to win a prize. They are shown a large bag that’s initially empty, and then they see three red hats and two white hats placed into the bag. At this point, the contestants are blindfolded. Each contestant then picks a hat at random from the bag and places it on their head.

One at a time, their blindfolds are removed and they can see the hats on the others’ heads—but not the hat on their own head. If they can, with absolute certainty, identify the color of the hat on their own head, then the game is over and all three contestants win a prize! Otherwise, they skip their turn, at which point the next contestant has their blindfold removed and the process continues. If all three contestants skip their turn, then no prize is won. While blindfolded, contestants can still hear the decisions of other contestants and can identify who said what.

Will the contestants always win the prize? If so, why? If not, why not?

Submit your answer

This Week’s Extra Credit

Now, there are four game show contestants who must work together. They are shown a large bag that’s initially empty, and then they see R red hats, W white hats, and B blue hats placed into the bag, such that R, W, and B are each less than or equal to 4. Otherwise, the game is played as before.

For some triples (R, W, B), the contestants can always win. Among these, what is the greatest value of R + W + B?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. Last weekend, I was a guest speaker at the American Regions Math League, one of the largest annual single-day math competitions in the world. The various rounds of problems have been posted. I tried my hand at the 10 individual problems (within the time limit of 10 minutes per pair of problems, mind you) and had, shall we generously say, “some” success.

But there was no way I was finishing Problem 8 in time. (It took me a few minutes to even understand what the problem was asking for.) Here it is, for you to try in the comfort of your own home:

The 60 students of the West Coast ARML team and the 120 students of the East Coast ARML team are at a party. A total of 1800 handshakes occur between a West Coast student and an East Coast student, with no pair of people shaking hands more than once and no other handshakes occurring. For each pair of students at the party, the coaches record how many students shook hands with both. Compute the minimum possible average of the 180 choose 2 numbers recorded by the coaches.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Mark Shepstone 🎻 from Norwich, England. I received 57 timely submissions, of which 57 were correct—good for a perfect 100 percent solve rate. Woohoo! (That’s only the fifth time that’s ever happened for a Fiddler.)

Last week, two sheep were at random points inside a square pen. They were munching grass and staring in two random directions. Each sheep had a field of view that was180 degrees.

What was the probability that they both saw each other?

Suppose the sheep were at any two locations in the square (let’s call these points A and B) and that they were looking in two particular directions. Without altering their positions, consider the following four related cases:

Both sheep were looking in the two directions as originally chosen.
Sheep A turned around 180 degrees and looked in the opposite direction, while sheep B’s orientation was unchanged.
Sheep B turned around 180 degrees and looked in the opposite direction, while sheep A’s orientation was unchanged.
Both sheep turned around 180 degrees and looked in the opposite direction from the ones originally chosen.

In the diagram below, these four cases were equivalent to picking one arrows on the left (red or blue) and one of the arrows on the right (red or blue). Each arrow represented a sheep and the direction it was facing.

In only one of these four cases were the two sheep ever able to see each other. (In the diagram above, this was when the blue arrow on the left and the red arrow on the right were chosen.) That’s because, for each sheep, you had to choose the arrow that was pointing within 90 degrees of the other sheep. For a given sheep, both arrows could never simultaneously do this, since they pointed in opposite directions.

In short, we took all the possible sheep position/orientation combinations and made groups of four equally likely scenarios, exactly one of which resulted in both sheep seeing each other. Thus, the probability that they saw each other was 1/4, or 25 percent.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Chris Connett 🎻 from New York, New York. I received 37 timely submissions, of which 31 were correct—good for an 84 percent solve rate.

Now, three sheep were at random points inside a square pen. They were munching grass and staring in three random directions. As before, each sheep had a field of view that was 180 degrees.

What was the probability that all three sheep saw each other?

Can You Spot the Sheep?

Zach Wissner-Gross — Fri, 29 May 2026 12:03:55 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

Two sheep are at two random points inside a square pen. They are munching grass and staring in two random directions. Each sheep has a field of view that’s 180 degrees.

What is the probability that they both see each other?

Submit your answer

This Week’s Extra Credit

Now, three sheep are at three random points inside a square pen. They are munching grass and staring in three random directions. As before, each sheep has a field of view that’s 180 degrees.

What is the probability that all three sheep see each other?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing the May puzzle from Peter Winkler and MoMath. Like last time, this one comes with an animation courtesy of Grant Sanderson.

This month’s puzzle concerns tying random ends of string together. I don’t know … something about plucking those random strings just sounds “harmonic” to me.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Rushil 🎻 from Charlotte, North Carolina. I received 57 timely submissions, of which 53 were correct—good for a 93 percent solve rate.

Last week’s puzzle was a very special collaboration between Fiddler on the Proof and Science News. Two of the four puzzles posted there served as last week’s Fiddler and Extra Credit. Now—on to those puzzles!

June the ant was on a cylinder. More specifically, she was on the edge of one of the cylinder’s two circular faces. Her dinner was on the edge of the opposite circular face, and all the way around on the other side of that face. The radius of the cylinder was 2 meters and its height was 2 meters.

Your job was to help June find the shortest path along the surface of the cylinder so that she could chow down as quickly as possible. What was the length of this shortest path?

It was tempting to head straight across the circular base and then down the height, as shown below. The length of this path was the diameter, 4 meters, plus the height, 2 meters, or 6 meters in total.

An alternative path was to spiral around the curved surface of the cylinder, as shown below. If you unraveled the curved surface to form a rectangle, this path was the hypotenuse of a right triangle with a vertical leg of length 2 meters and a horizontal leg of 2𝜋 meters, which meant the total distance was √(4+4𝜋²), or about 6.59 meters.

Thus, the shortest path was indeed 6 meters. If the cylinder had been taller, then eventually the curved path would have become more efficient than the straight path across the circle and down. In fact, you could calculate exactly when this transition occurred—it was when the cylinder’s height was 𝝅²/4−1, or about 1.4674, times the radius.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Nis Jørgensen 🎻 from Valsømagle, Denmark. I received 38 timely submissions, of which 28 were correct—good for a 74 percent solve rate.

Now, June was on a hollowed-out cylinder, also known as a “cylindrical shell.” The shell’s outer radius was 2 meters and its inner radius was 1 meter. The shell was 2 meters tall. June was on the outer edge of one of the cylinder’s two flat faces. Her dinner was on the opposite face, and all the way around on the other end of that face.

Your job was to help June find the shortest path along the surface of the shell so that she could chow down as quickly as possible. What was the length of this shortest path?

The Ant Goes Marching

Zach Wissner-Gross — Fri, 22 May 2026 12:03:19 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

This week’s puzzle is a very special collaboration between Fiddler on the Proof and Science News, where four math puzzles that are related were posted earlier this morning. The third and fourth among these are also appearing here as this week’s Fiddler and Extra Credit. Now—on to those puzzles!

June the ant is on a cylinder. More specifically, she is on the edge of one of the cylinder’s two circular faces. Her dinner is on the edge of the opposite circular face, and all the way around on the other side of that face. The radius of the cylinder is 2 meters and its height is 2 meters.

Your job is to help June find the shortest path along the surface of the cylinder so that she can chow down as quickly as possible. What’s the length of this shortest path?

Submit your answer

This Week’s Extra Credit

Now, June is on a hollowed-out cylinder, also known as a “cylindrical shell.” The shell’s outer radius is 2 meters and its inner radius is 1 meter. The shell is 2 meters tall. June is on the outer edge of one of the cylinder’s two flat faces. Her dinner is on the opposite face, and all the way around on the other end of that face.

Once again, your job is to help June find the shortest path along the surface of the shell so that she can chow down as quickly as possible. What’s the length of this shortest path?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing that I’ll be speaking to a group of high school students at the American Regions Math League later this month. To prepare, I snuck a peak at last year’s problems from the annual competition.

ARML’s crown jewel every year is its Power Round. Last year’s is titled “The Bases are Loaded!,” and it concerns integer partitions in which the parts are numbers whose only prime factors are 2 and/or 3. After some preliminary definitions, it gets rolling with a mix of computations and proofs. Enjoy!

After this year’s ARML, I’ll report back here with one or two of its latest and greatest problems.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Sean Garagan 🎻 from Halifax, Nova Scotia, Canada. I received 55 timely submissions, of which 39 were correct—good for a 71 percent solve rate.

Last week, I was playing hide-and-seek with my nephew. I started at point O, whereas my nephew could hide at point A, B, or C. I could walk from O to A in 2 minutes, from O to B in 3 minutes, from O to C in 4 minutes, and from B to C in 5 minutes. To get from A to B or from A to C, I had to pass through O.

My goal was to minimize the time it took to find my nephew, no matter how clever his strategy might have been. What was this optimal time?

In the words of Peter Exterkate, “This [was] an extremely poorly worded puzzle.” This was through no fault of the puzzle’s submitter, but rather my own. Therefore, I generously allowed for more than one interpretation.

The first interpretation was that of a “worst-case scenario.” What was the shortest path that guaranteed I’d look in all three hiding spots? There were six possible orders to consider:

Search A, then B, then C. This path took 2 + 2 + 3 + 5 = 12 minutes.
Search A, then C, then B. This path took 2 + 2 + 4 + 5 = 13 minutes.
Search B, then A, then C. This path took 3 + 3 + 2 +2 + 4 = 14 minutes.
Search B, then C, then A. This path took 3 + 5 + 4 + 2 = 14 minutes.
Search C, then A, then B. This path took 4 + 4 + 2 + 2 + 3 = 15 minutes.
Search C, then B, then A. This path took 4 + 5 + 3 + 2 = 14 minutes.

Thus, the most efficient path was to search A, then B, then C for a total of 12 minutes. This was a reasonable answer to this puzzle as I presented it. Also, since this was the worst-case scenario, I didn’t accept any answers greater than 12 minutes.

A second (and more challenging) interpretation was to minimize the average time it took to find my nephew, regardless of his hiding strategy. I still had the six distinct searching strategies listed above, whereas my nephew had three distinct hiding strategies (hide at A, hide at B, and hide at C). Thus, our game of hide-and-seek had the following payoff matrix:

To understand this table, consider the last row, where my searching strategy was C, then B, then A. If my nephew was hiding at C, I found him in 4 minutes; if he was hiding at B, I found him in 9 minutes; and if he was hiding at A, I found him in 14 minutes.

The question then became how I might mix strategies across the six rows so that the average seeking time across the three columns was minimized. If I mixed all six strategies, then if my nephew was hiding at A, I found him in an average of 8 minutes and 20 seconds. And the same went for B and C. Picking one of the six strategies at random was a fairly decent strategy. But it was possible to do better!

Instead of mixing all six strategies, suppose I mixed only two, each with a 50 percent chance. For example, suppose I equally mixed the strategies “A → B → C” and “C → A → B,” the first and fifth rows in the table above.

If my nephew was hiding at A, then the expected time to find him was (2+10)/2, or 6 minutes.
If my nephew was hiding at B, then the expected time to find him was (7+15)/2, or 11 minutes.
If my nephew was hiding at C, then the expected time to find him was (12+4)/2, or 8 minutes.

The greatest average among these was 11 minutes, when my nephew was hiding at B. So if I mixed those two strategies, it was possible for my nephew to adopt a strategy where it took me 11 minutes on average. Not great.

If you did a similar analysis for the various ways to equally mix two strategies, you got the resulting table:

There were two minima in this table. First, if I equally mixed “A → B → C” and “C → B → A,” then it took an average of 8 minutes to find my nephew, no matter where he was hiding. Alternatively, if I equally mixed “A → C → B” and “B → C → A,” then it again took an average of 8 minutes to find my nephew, no matter where he was hiding.

With a little work, you could convince yourself that other mixings of strategies could never result in a maximum average that was greater than 8 minutes. Solver 🎬 Michael Schubmehl 🎬 did this by plotting the regimes where each search strategy was optimal across the varying hiding strategies.

In the end, if I wanted to minimize the average time to find my nephew, this minimum average was 8 minutes.

Other answers submitted by readers made specific assumptions about where my nephew would hide (e.g., he was equally likely to hide in all three locations)—assumptions that were not in the spirit of the puzzle. And so, in the end, I only accepted answers of 8 and 12 minutes.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Srihari Narayanan 🎻 from Hayward, California. I received 20 timely submissions, of which 8 were correct—good for a 40 percent solve rate. (This was a tough one, and I’m sure I’ll receive at least one or two objections to the following solution.)

My nephew could no longer hide at C, and was instead limited to A and B. But this time, he had a teleporter that could instantaneously transport him from A to B or from B to A. He could use the teleporter as many times as he wanted. However, he couldn’t react to my approach, and he instead had to plan out his transport schedule ahead of time. That said, he did know the precise time when the game started.

My goal was to minimize the average time it took to find him, no matter how clever his strategy might have been. What was this optimal time?

Can You Play Hide-and-Seek?

Zach Wissner-Gross — Fri, 15 May 2026 12:01:20 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

From Nis Jørgensen comes a game wrapped in a puzzle (wrapped in an enigma?):

I am playing hide-and-seek with my nephew. I start at point O, whereas my nephew can hide at point A, B or C. I can walk from O to A in 2 minutes, from O to B in 3 minutes, from O to C in 4 minutes, and from B to C in 5 minutes. To get from A to B or from A to C, I must pass through O.

My goal is to minimize the time it takes to find him, no matter how clever his strategy might be. What is this optimal time?

Submit your answer

This Week’s Extra Credit

Also from Nis comes some Extra Credit:

My nephew can no longer hide at C, and is instead limited to A and B. But this time, he has a teleporter that can instantaneously transport him from A to B or from B to A. He can use the teleporter as many times as he wants. However, he can’t react to my approach, and must instead plan out his transport schedule ahead of time. That said, he does know the precise time when the game starts.

My goal is to minimize the average time it takes to find him, no matter how clever his strategy might be. What is this optimal time?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m belatedly sharing a puzzle from Ben Orlin, which was published in March over at Science News. This flower-themed puzzle certainly has a lot of “life” to it!

Oh, and word has it that another great math puzzle may be coming to Science News soon…

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 David Chasson 🎻 from Chandler, Arizona. I received 55 timely submissions, of which 40 were correct—good for a 73 percent solve rate.

Last week, I was preparing a mixture of “random-ade” using a large, empty pitcher and two 12-ounce glasses.

First, I filled one glass with some amount of lemon juice chosen randomly and uniformly between 0 and 12 ounces. I filled the other glass with some amount of water, also chosen randomly and uniformly between 0 and 12 ounces. Next, I poured an equal amount from each glass into the pitcher until one of the glasses was empty.

At this point, I refilled that empty glass with yet another random amount of the same liquid it previously contained. Once again, I poured an equal amount from each glass into the pitcher until one of the glasses was empty.

On average, how much random-ade could I have expected to prepare? (Note that all three random amounts in this problem were chosen independently of each other.)

To get started, let’s not worry about the refill. Suppose I just had the two glasses, each with a random initial amount, and I poured out both of them until one of the glasses was empty. How much lemonade was there at this point?

If the volumes of liquid in the two glasses were x and y, then I poured the minimum of these from each glass. Let’s write that minimum as min(x, y). Again, since this was the amount from each glass, the total volume of lemonade was double, or 2·min(x, y). Below is a 3D graph showing how 2·min(x, y) varied with x and y between 0 and 12.

The average amount of lemonade was the average height of the graph. To compute this, you could find the volume of the pyramid and divide it by the area of its base. A pyramid’s volume is 1/3 times the area of its base times its height, and dividing that by the area of the base left 1/3 times the height. The height of this pyramid was 24, the maximum amount of lemonade I could have made (when both glasses were full to the brim). One-third of this was 8, so I expected to have 8 fluid ounces after the first pouring—4 ounces from each glass.

Now let’s consider the refill. In the glass that hadn’t been emptied, the volume that remained was |y−x|. Meanwhile, the glass that was refilled had a new liquid volume z chosen randomly and uniformly between 0 and 12. How did z compare to |y−x|?

Consider the diagram below, showing how z could be anywhere between 0 and 12. The probability that z was less than |y−x| was the ratio between the segment left of |y−x| and the total length, or |y−x|/12. And the probability that z was greater than |y−x| was the ratio between the segment right of |y−x| and the total length, or 1 − |y−x|/12.

Next, let’s look at these two cases in greater detail:

First, suppose z was less than |y−x|, which occurred with probability |y−x|/12. In this case, z was poured into the pitcher from each glass, for an additional volume of 2z. Thus, the contribution to the expected value from this case was the volume times its corresponding probability, or 2z · |y−x|/12. Meanwhile, within this case, z was equally likely to be anywhere between 0 and |y−x|, so the average value of z was |y−x|/2. Plugging in this average gave you |y−x|²/12.

Second, suppose z was greater than |y−x|, which occurred with probability 1 − |y−x|/12. In this case, |y−x| was poured into the pitcher from each glass, for an additional volume of 2·|y−x|. This time, the contribution to the expected value was 2·|y−x| · (1 − |y−x|/12), or 2·|y−x| − |y−x|²/6.

Adding these two contributions together, the sum was |y−x|²/12 + 2·|y−x| − |y−x|²/6 = 2·|y−x| − |y−x|²/12. The 3D graph of this function was flatter than the one above, since we already averaged over all the different possible values of z:

This time, the maximum occurred when x was 12 and y was 0, or vice versa, since that meant that one of the two glasses was completely full after the refill.

In any case, computing the average value of this function over all x and y was trickier due to its curved surface. (We couldn’t just use the formula for the volume of a pyramid.) Instead, you needed some calculus. Using symmetry, let’s look at half the domain, such as x > y. The volume under the function in this region was given by the following double integral:

Evaluating this gave a value of 432. And to find the average value of the graph, you had to divide this result by the area of the region’s triangular base, which was 12²/2. In the end, the average was 6.

Adding up the average of the first pouring (8 ounces) and the average of the second pouring (6 ounces) gave a total average of 14 ounces. If you gave an answer of 7, forgetting to multiply by the two glasses involved in each of the pourings, I did not award credit.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 “Maths ForFun” 🎻 from the Pacific Northwest. I received 40 timely submissions, of which 32 were correct—good for an 80 percent solve rate.

I was once again preparing random-ade, but this time I had three 12-ounce glasses.

I filled the first glass with a random amount of lemon juice, the second glass with a random amount of lime juice, and the third glass with a random amount of water. As before, each amount was chosen uniformly between 0 and 12 ounces, and all amounts were independent. Next, I poured an equal amount from each glass into the pitcher until one of the glasses was empty.

At this point, I refilled that empty glass with yet another random amount of the same liquid it previously contained. Once again, I poured an equal amount from each glass into the pitcher until one of the glasses was empty.

Then I refilled that now-empty glass with yet another random amount of the same liquid it previously contained. Again, I poured an equal amount from each glass into the pitcher until one of the glasses was empty.

On average, how much random-ade could I have expected to prepare?

Can You Drink the “Random-ade”?

Zach Wissner-Gross — Fri, 08 May 2026 12:03:18 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

I’m preparing a mixture of “random-ade” using a large, empty pitcher and two 12-ounce glasses.

First, I fill one glass with some amount of lemon juice chosen randomly and uniformly between 0 and 12 ounces. I fill the other glass with some amount of water, also chosen randomly and uniformly between 0 and 12 ounces. Next, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.

At this point, I refill that empty glass with yet another random amount of the same liquid it previously contained. Once again, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.

On average, how much random-ade can I expect to prepare? (Note that all three random amounts in this problem are chosen independently of each other.)

Submit your answer

This Week’s Extra Credit

Once again I’m preparing random-ade, but this time I have three 12-ounce glasses.

I fill the first glass with a random amount of lemon juice, the second glass with a random amount of lime juice, and the third glass with a random amount of water. As before, each amount is chosen uniformly between 0 and 12 ounces, and all amounts are independent. Next, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.

At this point, I refill that empty glass with yet another random amount of the same liquid it previously contained. Once again, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.

Then I refill that now-empty glass with yet another random amount of the same liquid it previously contained. Again, I pour an equal amount from each glass into the pitcher until one of the glasses is empty.

On average, how much random-ade can I expect to prepare?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing IBM’s “Ponder This” puzzles. (Last month, some readers noted that IBM had posed a problem that was thematically similar to the Fiddler from two weeks ago.)

This month’s puzzle is called “The Powers of a Binary Matrix.” If I’m reading it correctly, it’s asking you to find permutation matrices that result in the identity matrix only when taken to maximally high powers. It’s a neat problem, but I think it could be grounded a little more. Here’s my attempt at doing so:

Imagine you shuffle a deck of cards exactly the same way every time. In the worst-case scenario, what’s the maximum number of shuffles needed to return to the original ordering of the deck? What if the deck has 10 cards? 50 cards? A million cards?

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Fred 🎻 from Kansas City. I received 72 timely submissions, of which 61 were correct—good for an 85 percent solve rate.

Last week, a frog was hopping around a chessboard, always from the center of one square to the center of another square. Each square had side length 1, but the board itself was not necessarily 8-by-8. Instead, it was N-by-N, where N was some large whole number.

Every jump the frog made was the same distance, which we called L. The frog wanted to make four jumps such that:

After the fourth jump, the frog returned to its starting square.
The frog visited a total of four distinct squares along the way, including the square on which it started (and also stopped).
The path the frog took was not a square loop.
The frog was never on a square that was diagonal (i.e., a bishop’s move away) or horizontal/vertical (i.e., a rook’s move away) from the starting square.

What was the smallest jumping distance L for which this was possible?

First off, I had neglected to specify the rook condition in that last bullet point when I originally posted the puzzle. Without that condition, the shortest distance the frog had to hop was √5, much like a knight in chess. Here was a four-jump path the frog could have taken:

But notice that no matter which square in this path the frog started from, two moves later it was on a square that was a rook’s move away. So with the rook condition, the puzzle was definitely trickier. Nevertheless, because of my erratum, I accepted answers of √5 that were submitted on May 1.

Now, instead of moving 2 squares in one direction and 1 square in a perpendicular direction (like a knight), suppose the frog jumped a squares in one direction and b squares in a perpendicular direction. If the frog started at the origin, then some possible four-jump paths to consider were:

(0, 0) → (a, b) → (2a, 0) → (a, -b) → (0, 0). However, like the path illustrated above, (2a, 0) was a rook’s move from (0, 0).
(0, 0) → (a, b) → (a+b, a+b) → (b, a) → (0, 0). However, (a+b, a+b) was a bishop’s move from (0, 0).
(0, 0) → (a, b) → (a−b, a+b) → (-b, a) → (0, 0). However, the shape of this loop was a square.

These were all dead ends, which meant something was up. And that something was that the frog had to be able to make two equally distant jumps in two substantially different directions. In other words, the frog had to be able to jump a squares in one direction and b squares in a perpendicular direction, or c squares in one direction and d squares in a perpendicular direction. Then a plausible path would be:

(0, 0) → (a, b) → (a+c, b+d) → (c, d) → (0, 0)

By the Pythagorean theorem, a jump that was a and then b in perpendicular directions had length √(a²+b²), while a jump that was c and then d in perpendicular directions had length √(c²+d²). For these lengths to be equal, you needed √(a²+b²) = √(c²+d²), or a²+b² = c²+d². In other words, you were looking for a number—the square of the jump length, to be precise—that could be written as the sum of two squares two different ways.

The smallest such number is 25, which equals both 4²+3² and 5²+0². However, paths with a jump length of √25 = 5 would have passed through a point like (5, 0), which was a rook’s move from the origin. Thus, we were looking for a number that could be written as the sum of two squares two different ways, such that neither square was 0.

The next candidate was 50, which equals both 7²+1² and 5²+5². However, paths with a jump length of √50 would have passed through a point like (5, 5), which was a bishop’s move from the origin. So in addition to neither square being zero, the two squares also couldn’t be equal.

And so, we came to the next candidate, 65, which equals both 8²+1² and 7²+4². Plugging these values into the path above gave one such valid loop:

(0, 0) → (8, 1) → (15, 5) → (7, 4) → (0, 0)

This path was the rhombus shown below, and there were also other similar loops you could have made with the same jumping distance. Thus, the smallest jumping distance L was √65.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Eli Wolfhagen 🎻 from Brooklyn, New York. I received 45 timely submissions, of which 34 were correct—good for a 76 percent solve rate.

The frog was jumping around the board with the same minimum distance L we had just found: √65.

But this time, the frog also wanted to be able to hop to every location on the chessboard. What was the minimum value of N for which this was possible?

Can You Hop Around the Chessboard?

Zach Wissner-Gross — Fri, 01 May 2026 12:01:46 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

This week’s puzzle was a collaboration with high-schooler Connor Hill, the First Place Winner from this year’s Regeneron Science Talent Search. Connor’s project involved an original, computer-assisted proof to find the complete list of noble polyhedra, in which all the faces are indistinguishable and all the vertices are indistinguishable. The five Platonic solids are perhaps the best known examples of noble polyhedra, but Connor documented all of them, including two infinite sets and 146 exceptional cases, among which were 85 new examples!

Here’s the puzzle from Connor—unrelated to his work—which is about a frog that makes very particular hops:

A frog is hopping around a chessboard, always from the center of one square to the center of another square. Each square has side length 1, but the board itself is not necessarily 8-by-8. Instead, it’s N-by-N, where N is some large whole number.

Every jump the frog makes must be the same distance, which we’ll call L. The frog wants to make four jumps such that:

After the fourth jump, the frog has returned to its starting square.
The frog visits a total of four distinct squares along the way, including the square on which it starts (and also stops).
The path the frog takes is not a square loop.
The frog is never on a square that is diagonal (i.e., a bishop’s move away) or horizontal/vertical (i.e., a rook’s move away) from the starting square.

What is the smallest jumping distance L for which this is possible?

Submit your answer

This Week’s Extra Credit

From Connor also comes some Extra Credit:

The frog is jumping around the board with the same minimum distance L you just found.

But this time, the frog also wants to be able to hop to every location on the chessboard. What is the minimum value of N for which this is possible?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing an online game that came my way courtesy of Bernardo Recamán Santos. It’s called “The Snake and the Hunter” and it’s billed as a “a strategic combinatorial game.” Well, that’s exactly what it is. Can you beat the computer?

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Zach Donofrio 🎻 from Brooklyn, New York. I received 86 timely submissions, of which 79 were correct—good for a 92 percent solve rate.

Last week, you studied an analog clock that included an hour hand, a minute hand, and 60 minute markers, 12 of which were also hour markers.

At a certain time, the hour hand and minute hand were both pointing directly at minute markers (either of which could also have been an hour marker). The hour hand was 13 markers ahead (i.e., clockwise) of the minute hand.

At what time did this occur? (You didn’t have to worry about a.m. vs. p.m. for this puzzle.)

Suppose the time was m minutes after h o’clock, where m was at least 0 and less than 60. So, for example, if h was 3 and m was 47, then the time was 3:47. Let’s figure out just which markers the hour and minute hands were pointing in terms of m and h.

The minute hand was the easier of the two. Because the time was m minutes after the hour, the minute hand was pointing to minute marker m.

Now for the hour marker. There were five minute markers for every hour marker on the clock, so the hour hand pointed directly at a minute marker every fifth of an hour, or every 12 minutes. Thus, m had to be one of 0, 12, 24, 26, or 48. Moreover, hour marker h was equivalent to minute marker 5h. That meant the hour hand pointed to minute marker 5h + m/12. Sure enough, you can see that this expression was an integer only when m was a multiple of 12.

For convenience, let’s define k = m/12. Since m/12 had to be an integer, so did k. More specifically, k had to be 0, 1, 2, 3, or 4.

At this point, you knew the minute hand pointed to minute marker m = 12k, while the hour hand pointed to minute marker 5h + m/12 = 5h + k. For what values of h and k was the hour hand 13 markers ahead of the minute hand? In other words, when was 5h + k equal to 12k + 13?

Setting these two expressions equal to each other gave you 5h + k = 12k + 13, or 5h = 11k + 13. For there to be integer solutions, both sides of this last equation had to equal the same multiple of 5. Let’s check the five possible values of k to see when this occurred:

k = 0: Then 11k + 13 = 11·0 + 13 = 13, which was not a multiple of 5.
k = 1: Then 11k + 13 = 11·1 + 13 = 24, which was not a multiple of 5.
k = 2: Then 11k + 13 = 11·2 + 13 = 35, which was a multiple of 5 (huzzah!). That gave you 5h = 35, or h = 7.
k = 3: Then 11k + 13 = 11·3 + 13 = 46, which was not a multiple of 5.
k = 4: Then 11k + 13 = 11·4 + 13 = 57, which was not a multiple of 5.

The only time k and h were both integers was when k = 2 (in which case m = 24) and h = 7. Thus, the time on the clock—and the answer to the puzzle—was 7:24. Here’s an illustration of this time, courtesy of solver Michael Schubmehl:

Sure enough, you can see that the hour hand was exactly 13 markers ahead of the minute hand.

But we weren’t quite done with this puzzle yet. We had neglected the possibility where the hour hand was 13 markers clockwise of the minute hand, but they were on opposite sides of the 12-hour (or 0-minute) marker. For this to occur—and for the hour hand to still point to a minute marker—you needed m = 48. For the hour hand to be 13 markers ahead, it had to point to m = 1, which corresponded to a time of 12:12, and not 12:48. Thus, no such time was possible.

With this check complete, 7:24 was indeed the unique answer to the puzzle.

By the way, the most common incorrect answer I received was 4:36, which was when the hour hand was 13 markers behind (i.e., counterclockwise to) the minute hand.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Jason Shaw 🎻 from Decorah, Iowa. I received 39 timely submissions, of which 29 were correct—good for a 74 percent solve rate.

At various times of day, the minute and hour hands form a right angle. But was there a time of day when the hour hand, minute hand, and second hand—each of which rotated in a continuous manner—together formed two right angles, with any of the hands in the middle? If you could find such a time or times, what were they?

If you couldn’t find any such times, suppose the measures of the two nearly right angles formed by the three hands measured A and B degrees. What time or times of day minimized the square error function f(A, B) = (A − 90)² + (B − 90)²?

Either way, your answer had to be precise to at least a thousandth of a second. (Again, you didn’t have to worry about a.m. vs. p.m. for this puzzle.)

Will You Be Right on Time?

Zach Wissner-Gross — Fri, 24 Apr 2026 12:03:19 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

From Steve Curry comes a puzzle that’s right on schedule:

A standard analog clock includes an hour hand, a minute hand, and 60 minute markers, 12 of which are also hour markers.

At a certain time, both the hour hand and minute hand are both pointing directly at minute markers (either of which could also be an hour marker). The hour hand is 13 markers ahead (i.e., clockwise) of the minute hand.

At what time does this occur? (Don’t worry about a.m. vs. p.m. for this puzzle.)

Submit your answer

This Week’s Extra Credit

At various times of day, the minute and hour hands form a right angle. But is there a time of day when the hour hand, minute hand, and second hand—each of which rotates in a continuous manner—together form two right angles, with any of the hands in the middle? If you can find such a time or times, what are they?

If you cannot find any such times, suppose the measures of the two nearly right angles formed by the three hands measure A and B degrees. What time or times of day minimize the square error function f(A, B) = (A − 90)² + (B − 90)²?

Either way, your answer should be precise to at least a thousandth of a second. (Again, don’t worry about a.m. vs. p.m. for this puzzle.)

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m a little late to the party—and kudos to Benjamin Dickman for posting about this—but I’m sharing a recent PROMYS for Teachers problem set, with materials prepared by Nate Stambaugh (screenshot below).

This one looks to be about geodesics and such. But as with any problem set, it’s wise to “[d]iscuss any fears before you proceed.”

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Emily Kelly 🎻 from Brookhaven, Georgia. I received 85 timely submissions, of which **buzz** were correct—good for a 93 percent solve rate.

I recently introduced my children to a game called “Buzz” (also known as “Fizz buzz”), in which players take turns reciting whole numbers in order. However, in one particular variant of the game, anytime a number is a multiple of 7 or at least one of its digits is a 7, the player must say “buzz” instead of that number.

For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.

Last week, I asked you to determine how many times “buzz” should be said in the first 100 turns of the game (including those mentioned above in the first 20 turns).

You could have counted all the numbers from 1 to 100 and checked which of them were buzzed. Alternatively, you could take an analytical approach.

As we said, every multiple of 7 was buzzed. Between 1 and 100, the greatest multiple of 7 was 98, which was 7·14. Thus, there were 14 multiples of 7 that were buzzed.

And how many numbers contained at least one digit that was a 7? There were 10 numbers with a 7 in the ones place (7, 17, 27, etc.) and 10 numbers with a 7 in the tens place (70, 71, 72, etc.). Importantly, we double counted 77, which has a 7 in the ones place and the tens place, so there were 10 + 10 − 1 = 19 such numbers.

So it seemed the answer was 14 + 19 = 33.

But wait! Some numbers were both multiples of 7 and had at least one digit that was a 7, and these were still being double counted across the two groups. There were three such numbers: 7, 70, and 77. Thus, the answer was three less than 33, or 30.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Jacob Doran 🎻 from New Orleans, Louisiana. I received 60 timely submissions, of which 46 were correct—good for a **buzz** percent solve rate.

As we just saw, in the first 20 turns of the game, 15 percent of the numbers were “buzzed.” And in the first 100 turns, 30 percent of the numbers were buzzed. As the game proceeded, an increasing frequency of numbers got buzzed.

In fact, there was a certain minimum number N such that, for the Nth turn in the game and for every turn thereafter, at least half the numbers up to that point had been buzzed. What was this value of N?

Have You Heard the Buzz?

Zach Wissner-Gross — Fri, 17 Apr 2026 12:03:19 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.

How many times should “buzz” be said in the first 100 turns of the game (including those mentioned above in the first 20 turns)?

Submit your answer

This Week’s Extra Credit

As we just saw, in the first 20 turns of the game, 15 percent of the numbers were “buzzed.” But as the game proceeds, an increasing frequency of numbers get buzzed.

There is a certain minimum number N such that, for the Nth turn in the game and for every turn thereafter, at least half the numbers up to that point have been buzzed. What is this value of N?

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing that The Daily Baffle, whose Arithmeglyphs puzzle I previously shared, is now available for download on various devices. Yay, more puzzling fun on the go!

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jason Donofrio 🎻 from Brooklyn, New York. I received 63 timely submissions, of which 56 were correct—good for an 89 percent solve rate.

Last week, you had four number cubes, where each face of each cube could display one numeric digit from 0 to 9. You could make various numbers by picking three faces on three distinct cubes and lining them up. For example, you could make the number “123” by lining up three cubes to show 1-2-3, and you could make the number “7” by lining up three cubes to show 0-0-7.

Importantly, any face with a “6” could also be used to display a “9” by flipping the cube around, and vice versa. However, no other pairs of digits were interchangeable in this way.

You could choose which digits to place on the various faces of the four cubes. Your goal was to be able to make all the whole numbers from 1 to N, without skipping any numbers in between. With optimal design, what was the greatest possible value of N?

Solver Nis Jørgensen noted that it was not possible to make all the whole numbers up to and including 777. Why? Because then you would have needed:

two 0s ( for 007, 070, 700, etc.)
three 1s (for 111)
three 2s (for 222)
three 3s (for 333)
three 4s (for 444)
three 5s (for 555)
three 6s (for 666)
three 7s (for 777)
two 8s (for 088, 188, etc.)

If you tallied all these up, that was 25 digits. But with four number cubes, each of which had six faces, you could only place a total of 24 digits on the cubes.

So while 777 was impossible, 776 was still in play. If you didn’t need to generate 777, then you only needed two 7s on the dice, reducing the total number of required digits to 24. Thus, 776 seemed likely to be the answer.

But to prove that 776 was indeed possible, you still had to assign a set of six digits to each of the four cubes in some valid way. Solver Jerome Socolof came up with one such assignment:

{0, 1, 3, 4, 5, 7}
{0, 2, 3, 4, 6, 8}
{1, 2, 3, 5, 6, 7}
{1, 2, 4, 5, 6, 8}

Sure enough, with these digits on the four dice, you could generate every whole number from 1 (i.e., 0-0-1) to 776.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Joe Kong 🎻 from Mississauga, Ontario, Canada. I received 33 timely submissions, of which 23 were correct—good for a 70 percent solve rate.

Now that you had determined (in last week’s Fiddler) that the greatest possible value of N was 776, how many distinct ways could you have assigned numbers to the four cubes that allowed you to generate all the whole numbers from 001 to 776?

Note that the cubes were not ordered in any way. And, very importantly, swapping digits between two faces on a single cube did not count as producing a distinct arrangement. In other words, you didn’t have to worry about the various ways to assign six given digits to the faces of a cube. (Also, “6” and “9” counted as the same digit here.)

Can You Number the Number Cubes?

Zach Wissner-Gross — Fri, 10 Apr 2026 12:03:17 GMT

Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.

This Week’s Fiddler

From John Torrey comes a puzzle any sufficiently clever Bond villain would appreciate:

You have four number cubes, where each face of each cube can display one numeric digit from 0 to 9. You can make various numbers by picking three faces on three distinct cubes and lining them up. For example, you can make the number “123” by lining up three cubes to show 1-2-3, and you can make the number “7” by lining up three cubes to show 0-0-7.

Importantly, any face with a “6” can also be used to display a “9” by flipping the cube around, and vice versa. However, no other pairs of digits are interchangeable in this way.

You can choose which digits to place on the various faces of the four cubes. Your goal is to be able to make all the whole numbers from 1 to N, without skipping any numbers in between. With optimal design, what is the greatest possible value of N?

Submit your answer

This Week’s Extra Credit

From John Torrey also comes some Extra Credit:

How many distinct ways can you assign numbers to the four cubes that result in this greatest possible value of N?

Note that the cubes are not ordered in any way. And, very importantly, swapping digits between two faces on a single cube does not count as producing a distinct arrangement. In other words, don’t worry about the various ways to assign six given digits to the faces of a cube. (Also, “6” and “9” count as the same digit here.")

Submit your answer

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing (for the last time, I promise) my March Madness predictions.

As a quick reminder, my methodology involved (1) ranking teams via regression using all the thousands of games in the season and (2) only advancing underdogs when their output from the regression was within some distance of the favorite’s.

My predictions did really well this year. Yes, I got lucky, but I was within a second (looking at you, Duke vs. UConn) of correctly identifying all eight Final Four teams across the men’s and women’s tournaments. I also correctly predicted both the men’s and women’s champions.

I’ll definitely have a similar post next year, but past results do not guarantee future performance.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jim R 🎻 from Cambridge. I received 39 timely submissions, of which 30 were correct—good for a 77 percent solve rate.

Last week, an alien—more specifically, an Eridian—named “Rocky” had to pass a solid xenonite crystal to his human friend in a neighboring spaceship. The crystal was shaped like a regular tetrahedron, as shown below, and all its edges had length 1.

To safely transport the crystal, Rocky created a long, cylindrical tunnel between the spaceships, and then oriented the tetrahedron so that it fit through the tunnel. (It was okay if the crystal fit snugly inside the tunnel—in this case, it could slide along without any friction.)

What was the minimum possible radius for the tunnel so that the crystal fit through it?

At first, it might have been tempting to push the tetrahedron down the tunnel with one of its four triangular faces leading the way. This meant the circle needed to circumscribe an equilateral triangle with side length 1. As shown in the diagram below, such a circle had a radius of 1/√3, or approximately 0.57735.

But Rocky could do even better. Instead of sliding the tetrahedron face-first down the tube, he could slide one of its edges along so it scraped the side of the tube, as shown below:

In this orientation, the tube wasn’t circumscribing an equilateral triangle, but rather the projection of a tilted equilateral triangle. With some 3D trigonometry, the radius of the tube worked out to be 3/(4√2), or approximately 0.53033. That was quite the improvement!

But Rocky could still do better. The image below shows how the tetrahedron could have been squeezed into an even narrower cylinder. One of its edges ran diametrically across the tube, while the opposite edge (i.e., the one between the remaining two vertices of the tetrahedron) also ran diametrically across the tube, but in the perpendicular direction.

Here’s what this would have looked like from Rocky’s perspective, peering down the tube:

Thus, the tube had to circumscribe a square with a diagonal of length 1, which meant the tube required a radius of 1/2, or 0.5. (If you mistakenly submitted the diameter instead of the radius, I still awarded credit.)

As a check, solver 🎬 Michael Schubmehl 🎬 looked at all possible orientations and determined what the resulting cylindrical radius would be for each. The orientations were parameterized by subsequent rotations about the z-axis (also the assumed axis of the cylinder) and then about the x-axis, where rotations of zero corresponded to the optimal solution above. Here are his results:

Sure enough, the cylinder couldn’t get any tighter than a radius of 0.5. What’s more, the radius of 1/√3 we found initially corresponded to the least efficient orientation for the tetrahedron!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Pierre Bierre 🎻 from Pleasanton, California. I received 23 timely submissions, of which 20 were correct—good for an 87 percent solve rate.

Next, Rocky wanted to transport a solid crystal shaped like a regular dodecahedron to his human friend. As before, each edge had length 1.

This time, the long tunnel between the space ships could have been any right prismatic shape, not necessarily a cylinder. Once again, Rocky needed the crystal to fit through the tunnel, and it was okay if that fit was snug.

What was the minimum possible cross-sectional area for the tunnel so that the crystal fit through it?