From Quoc Tran comes a curious case of centrifugation:

Quoc’s lab has a microcentrifuge, a piece of equipment that can separate components of a liquid by spinning around very rapidly. Liquid samples are pipetted into small tubes, which are then placed in one of the microcentrifuge’s 12 slots evenly spaced in a circle.

For the microcentrifuge to work properly, each tube must hold the same amount of liquid. Also, importantly, the center of mass of the samples must be at the very center of the circle — otherwise, the microcentrifuge will not be balanced and may break.

Quoc notices that there is no way to place exactly one tube in the microcentrifuge so that it will be balanced, but he can place two tubes (e.g., in slots 1 and 7).

Now Quoc needs to spin exactly seven samples. In which slots (numbered 1 through 12, as in the diagram above) should he place them so that the centrifuge will be balanced?

*Extra credit:* Assuming the 12 slots are distinct, how many different balanced arrangements of seven samples are there?

From Oliver Roeder, who knows a thing or two about riddles, comes a labyrinthine matter of lexicons:

One of Ollie’s favorite online games is Guess My Word. Each day, there is a secret word, and you try to guess it as efficiently as possible by typing in other words. After each guess, you are told whether the secret word is alphabetically before or after your guess. The game stops and congratulates you when you have guessed the secret word. For example, the secret word was recently “nuance,” which Ollie arrived at with the following series of nine guesses: naan, vacuum, rabbi, papa, oasis, nuclear, nix, noxious, nuance.

Each secret word is randomly chosen from a dictionary with exactly 267,751 entries. If you have this dictionary memorized, and play the game as efficiently as possible, how many guesses should you expect to make to guess the secret word?

Congratulations to Seth Cohen of Concord, winner of last week’s Riddler Express.

Last week, you were helping the folks from Blacksburg, Greensboro and Silver Spring, who were getting together for a game of pickup basketball. Each week, anywhere from one to five individuals showed up from each town, with each outcome equally likely.

Using all the players that showed up, they wanted to create exactly two teams of equal size. Everyone was wearing a jersey that matched the color mentioned in the name of their town. To avoid confusion, they agreed that the residents of two towns should combine forces to play against the third town’s residents.

What was the probability that, on any given week, it was possible to form two equal teams with everyone playing, where two towns were pitted against the third?

Since there were five possibilities for the number of players from each of the three towns, that meant the total number of outcomes to consider here was 5×5×5, or 125. Some solvers, like the Highlands Latin School statistics class in Louisville, Kentucky, analyzed all 125 cases, finding the number of cases in which two towns combined to have the same number of players as the third. The probability was then *that* number divided by 125.

Meanwhile, many solvers worked backwards, starting with all the different ways to have two numbers add up to the third, and then counting up the corresponding number of permutations. For example, the towns could have had one, three and four players respectively, since 1+3=4. There were then six total ways to assign these numbers to the three towns: 1/3/4, 1/4/3, 3/1/4, 3/4/1, 4/1/3 and 4/3/1.

In total, there were six ways for two whole numbers between 1 and 5 so to add up to another number between 1 and 5. Here they are, along with how many ways each set of numbers could be assigned to the three towns:

- 1+1=2, three ways
- 1+2=3, six ways
- 1+3=4, six ways
- 1+4=5, six ways
- 2+2=4, three ways
- 2+3=5, six ways

Adding these up, there were 30 total ways for two towns to be fairly matched up against the third. Since there were 125 total outcomes to consider, the probability of a fair match was 30/125, or **24 percent**.

For extra credit, you looked at a broader version of the puzzle, in which each town had anywhere from one to *N* players, rather than just one to five. Just as the denominator in the original riddle was 5^{3}, here it was *N*^{3}. But finding the numerator was trickier work.

Solver Nicholas Robbins (from Blacksburg, Virginia!) and Alberto Rorai both supposed that Blacksburg happened to have the most players among the three towns. If Blacksburg had two players, there was only one way for the other two towns to make a fair match (1+1). If Blacksburg had three players, there were two ways (1+2 and 2+1). If Blacksburg had four players, there were three ways (1+3, 2+2 and 3+1). This pattern continued all the way up to when Blacksburg had *N* players, when there were *N*−1 ways. Counting these all up gave you 1+2+3+…+(*N*−1), or *N*(*N*−1)/2.

But wait! That was only when Blacksburg had the most players. What about Greensboro and Silver Spring? To account for them, Alberto multiplied by three (since there were *three* towns), which meant there were 3*N*(*N*−1)/2 ways to have two numbers add up to the third number.

That was the numerator for our probability, while the denominator was *N*^{3}. Dividing them gave a final answer of **3( N−1)/(2N^{2})**. Sure enough, this checked out for the case when

When the towns move on to softball, they’ll definitely need a new system for assigning teams.

Congratulations to Mikolaj Franaszczuk of New York, New York, winner of last week’s Riddler Classic.

Last week marked the return of the Tour de FiveThirtyEight. For every mountain in the bicycle race, the first few riders to reach the summit were awarded “King of the Mountain” points.

You were racing against three other riders up one of the mountains. The first rider over the top would get 5 points, the second rider would get 3, the third rider would get 2, and the fourth rider would get 1.

All four of you were of equal ability — that is, under normal circumstances, you all had an equal chance of reaching the summit first. You were riding for Team A, one of your opponents was riding for Team B, but *two* of your competitors were both on Team C, meaning they could work together, drafting and setting a tempo up the mountain. Whichever teammate happened to be slower on the climb would get a boost from their faster teammate, and the two of them would both reach the summit at the faster teammate’s time (minus a very small fraction of a second).

As a lone rider, the odds were stacked against you. How many points were you expected to win on this mountain?

First off, if there hadn’t been any teams (i.e., all four riders were on their own), then the 5+3+2+1 points, or 11 points, would be evenly split among the four riders, on average. That meant you would have expected to get 2.75 points.

But the fact that there was a *team* threw a wrench into this analysis. Fortunately, like last week’s Riddler Express, this could be solved by working through a few cases. If you were Rider A, the other solo rider was B, and the two team riders were C, then you could concisely write the outcome of a race like ABCC (i.e., you came in first, then Rider B, then the two teammates). However, an outcome like CABC would turn into CCAB, since the slower teammate C would catch up to their partner.

Without further ado, here are the 12 total outcomes had the teammates *not* worked together, along with how they then finished by working together. In parentheses are the number of points you earn for each case.

- ABCC → ABCC (5 points)
- ACBC → ACCB (5 points)
- ACCB → ACCB (5 points)
- BACC → BACC (3 points)
- BCAC → BCCA (1 points)
- BCCA → BCCA (1 points)
- CABC → CCAB (2 points)
- CACB → CCAB (2 points)
- CBAC → CCBA (1 points)
- CBCA → CCBA (1 points)
- CCAB → CCAB (2 points)
- CCBA → CCBA (1 points)

Averaging these together meant you could expect 29/12, or about **2.417 points** on average. Several solvers, like Emma Beer and Sandeep Narayanaswami, grouped some of these 12 cases together (e.g., one-fourth of the time you came in first, regardless of how the teammates did) for an even more efficient calculation.

Meanwhile, Phil Rauscher noticed that this result was exactly one-third less than 2.75, the expected number of points when there was no team. Phil took this a step further, showing that whenever there are *N* racers, you can expect to finish one-third places further back on average when two of your opponents are teammates. This is because your placement will only be affected when you would otherwise have finished *between *them (not when you’re ahead of both of them or behind both of them), which happens one-third of the time.

The Tour de FiveThirtyEight continues to be a grueling test of endurance and cleverness. I’ll see you at the next stage!

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

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>From Zack Beamer comes a baffling brain teaser of basketball, just in time for the NBA playoffs:

Once a week, folks from Blacksburg, Greensboro, and Silver Spring get together for a game of pickup basketball. Every week, anywhere from one to five individuals will show up from each town, with each outcome equally likely.

Using all the players that show up, they want to create exactly two teams of equal size. Being a prideful bunch, everyone wears a jersey that matches the color mentioned in the name of their city. However, since it might create confusion to have one jersey playing for both sides, they agree that the residents of two towns will combine forces to play against the third town’s residents.

What is the probability that, on any given week, it’s possible to form two equal teams with everyone playing, where two towns are pitted against the third?

*Extra credit:* Suppose that, instead of anywhere from one to five individuals per town, anywhere from one to *N* individuals show up per town. Now what’s the probability that there will be two equal teams?

This month, the Tour de France is back, and so is the Tour de FiveThirtyEight!

For every mountain in the Tour de FiveThirtyEight, the first few riders to reach the summit are awarded points. The rider with the most such points at the end of the Tour is named “King of the Mountains” and gets to wear a special polka dot jersey.

At the moment, you are racing against three other riders up one of the mountains. The first rider over the top gets 5 points, the second rider gets 3, the third rider gets 2, and the fourth rider gets 1.

All four of you are of equal ability — that is, under normal circumstances, you all have an equal chance of reaching the summit first. But there’s a catch — two of your competitors are on the same *team*. Teammates are able to work together, drafting and setting a tempo up the mountain. Whichever teammate happens to be slower on the climb will get a boost from their faster teammate, and the two of them will both reach the summit at the faster teammate’s time.

As a lone rider, the odds may be stacked against you. In your quest for the polka dot jersey, how many points can you expect to win on this mountain, on average?

Congratulations to Brendan Hill of Edmond, Oklahoma, winner of last week’s Riddler and the new ruler of Riddler Nation!

Last week was the fifth Battle for Riddler Nation, and things were a little different this time around.

In a distant, war-torn land, there were 13 castles — three more than the usual 10 from prior battles. There were two warlords: you and your archenemy. Each castle had its own strategic value for a would-be conqueror. Specifically, the castles were worth 1, 2, 3, …, 12, and 13 victory points. You and your enemy each had 100 soldiers to distribute, any way you liked, to fight at any of the 13 castles. Whoever sent more soldiers to a given castle conquered that castle and won its victory points. If sent the same number of troops as your opponent, you split the points. You didn’t know what distribution of forces your enemy had chosen until the battles began. Whoever won the most points won the war.

I received a total of 970 battle plans. Of those, I excluded ones that were not valid, including any that had in excess of 100 troops, or, like the strategy submitted by Lowell Vaughn, tried to sneak in 101 troops to Castles 2 through 13 by having -1,112 (yes, a negative number) troops at Castle 1. Also, to keep things fair, whenever anyone submitted multiple strategies, I only counted the *last *strategy they submitted. In the end, there were 821 valid strategies.

Next, I ran all 336,610 one-on-one matchups, awarding one victory to each victor. In the event of a tie, both warlords were granted half a victory. Brendan Hill was the overall winner, tallying 630 wins against just 186 losses and 4 ties. Here’s a rundown of the 10 strongest warlords, along with how many soldiers they deployed to each castle:

The top 10 finishers in FiveThirtyEight’s Battle for Riddler Nation, with their distribution of soldiers for each castle and overall record

Soldiers per castle | Record | ||||||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Name | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | W | T | L | |

1 | Brendan Hill | 0 | 0 | 5 | 7 | 9 | 11 | 13 | 2 | 20 | 1 | 3 | 27 | 2 | 630 | 4 | 186 |

2 | Carl Schwab | 2 | 2 | 3 | 5 | 7 | 9 | 13 | 1 | 1 | 24 | 27 | 3 | 3 | 613 | 8 | 198 |

3 | Alex Conant | 0 | 1 | 1 | 2 | 2 | 11 | 3 | 16 | 16 | 16 | 3 | 3 | 26 | 601 | 21 | 198 |

4 | Fivey The Swing Voter | 0 | 0 | 1 | 1 | 3 | 12 | 13 | 3 | 18 | 2 | 27 | 1 | 19 | 592 | 22 | 206 |

5 | Kyle P. | 0 | 0 | 0 | 1 | 1 | 10 | 13 | 16 | 2 | 3 | 3 | 28 | 23 | 599 | 4 | 217 |

6 | Jonathan Siegel | 0 | 0 | 0 | 1 | 2 | 2 | 12 | 15 | 15 | 22 | 2 | 27 | 2 | 592 | 14 | 214 |

7 | Eric V. | 0 | 0 | 0 | 1 | 1 | 1 | 12 | 14 | 16 | 21 | 3 | 28 | 3 | 593 | 5 | 222 |

8 | David Zhu | 0 | 2 | 2 | 2 | 9 | 13 | 14 | 15 | 6 | 3 | 28 | 3 | 3 | 593 | 4 | 223 |

9 | Jonathan Hawkes | 0 | 2 | 3 | 5 | 9 | 7 | 11 | 16 | 3 | 3 | 24 | 8 | 9 | 590 | 8 | 222 |

10 | Matthew Altman | 2 | 1 | 3 | 3 | 5 | 3 | 11 | 16 | 21 | 25 | 3 | 1 | 6 | 589 | 7 | 224 |

In previous battles, when there were just 10 castles, there were 55 points in play. As long as you won more than half them — that is, at least 28 points — you were guaranteed a victory. The top strategies clustered soldiers into a small number of castles worth exactly 28 points. It took at least four castles to achieve 28 points, and there were several ways to do it: 4+5+9+10, 3+6+9+10, etc.

This time around, with 13 castles, there were 91 points in play, which meant you needed at least 46 points to secure a victory. Two-time Battle of Riddler Nation victor Vince Vatter was the one who suggested increasing the number of castles to 13, since there was only a single way to reach 46 points by winning exactly four castles: 10+11+12+13. Vince was curious whether that strategy would prevail or instead a strategy that targeted more castles would win the day.

Our winner, Brendan, placed at least five soldiers at a whopping seven castles. Adding the values of these castles gave 3+4+5+6+7+9+12, which was indeed exactly 46 points.

Vince did fine, by the way, coming in 69th place with 532 wins against 266 losses and 22 ties. Our previous champion, David Love, was a little lower down, coming in 335th with 438 wins, 352 losses and 30 ties.

The complete data set of strategies will be posted in the coming weeks. In the meantime, the following graph summarizes all the strategies. Each column represents a different castle, while each row is a strategy, with the strongest performers on top and the weakest on the bottom. The shading of a cell indicates the number of soldiers placed. It’s a lot to take in, but at the very least you might see a few “bands” — for example, the 10+11+12+13 strategies are clustered together in a few places, since they were similarly (somewhat) successful.

Finally, I was delighted to see that there was a fierce competition in Iowa’s Sheldon Community School District. Three classrooms — Sheldon Middle School Advanced Math, Sheldon Middle School TAG, and Sheldon High School STEM — submitted strategies. Among these, Sheldon Middle School TAG was the strongest, coming in 282nd place, with 458 wins, 347 losses and 15 ties. Definitely a *talented* group, there.

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

Email Zach Wissner-Gross at riddlercolumn@gmail.com

**CORRECTION (Sept. 11, 2020, 3:36 p.m.):** In an earlier version of this article, the table about Riddler Nation’s strongest warlords transposed the columns for ties and losses.

But this week is special. The time has flown, and somehow it’s been a year since I (peacefully) took over this column from my predecessor, Oliver Roeder.

It is only fitting that we continue one of Ollie’s (and my) favorite traditions here at The Riddler: the Battle for Riddler Nation. In order to have a chance at winning and becoming the next ruler, I need to receive your battle plans before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Some readers may be familiar with the first, second, third and fourth Battles for Riddler Nation. If you missed out, you may want to consult the thousands of attack distributions from these previous contests.

I am pleased to say that this week marks the *fifth* such competition — but this time, the number of castles has changed!

In a distant, war-torn land, there are 13 castles. There are two warlords: you and your archenemy. Each castle has its own strategic value for a would-be conqueror. Specifically, the castles are worth 1, 2, 3, …, 12, and 13 victory points. You and your enemy each have 100 soldiers to distribute, any way you like, to fight at any of the 13 castles. Whoever sends more soldiers to a given castle conquers that castle and wins its victory points. If you each send the same number of troops, you split the points. You don’t know what distribution of forces your enemy has chosen until the battles begin. Whoever wins the most points wins the war.

Submit a plan distributing your 100 soldiers among the 13 castles. Once we receive all your battle plans, we’ll adjudicate all the possible one-on-one matchups. Whoever wins the most wars wins the battle royale and is crowned ruler of Riddler Nation!

Who can steal the crown from David Love of Ambler, Pennsylvania, who currently sits atop the throne? Also, don’t count out Vince Vatter of Gainesville, Florida, a two-time champion eager to take back what was rightfully his.

Will *you* defeat them?

Do you have the cunning and logic to be the next ruler of Riddler Nation?

The results to this Riddler can be found in the following week’s column.

Congratulations to Dan Speirs of Newtown Square, Pennsylvania, winner of last week’s Riddler Express.

Last week, you had a giant sheet that lay flat on the ground, covering the Earth (assumed to be a perfect sphere with a radius of 6,378 kilometers).

You wanted to raise the sheet so it was instead always 1 meter off the ground. To make it so, how much did you have to increase the area of the sheet?

You were essentially being asked to find the difference in surface area between two spheres — one with the radius of the Earth, or 6,378,000 meters, and another whose radius was one meter longer, or 6,378,001 meters.

Most readers knew the formula: A sphere with radius *r* has a surface area of 4𝜋*r*^{2}. So one approach was to precisely calculate the areas of the two spheres and subtract them.

You could also have worked through a little algebra before plugging anything in. If *R* is the radius of the Earth, then the difference in surface area was 4𝜋(*R*+1)^{2}−4𝜋*R*^{2}. After canceling out the squared terms, this difference became 8𝜋*R*+4𝜋 square meters, or approximately 160.3 million square meters. Since there are a million (not a thousand!) square meters in one square kilometer, this was equivalent to **160.3 square kilometers**.

In the grand scheme of things, the answer was a pretty tiny fraction of the Earth’s surface — a shade over 0.00003 percent, to be precise.

For extra credit, you had to identify a city, country, land mass or body of water whose area was very close to the answer. As far as countries went, Liechtenstein, with an area of 160 square kilometers, was the closest.

So when it comes time to grow our Earth-covering quilt, I think the most prudent thing to do would be to head to Liechtenstein and host a quilting bee. I’ll see you there!

Congratulations to Richard Guidry Jr. of Baton Rouge, Louisiana, winner of last week’s Riddler Classic.

Last week, you wanted to play a very special game of War.

War is a two-player game in which a standard deck of cards is first shuffled and then divided into two piles with 26 cards each — one pile for each player. In every turn of the game, both players flip over and reveal the top card of their deck. The player whose card has a higher rank wins the turn and places both cards on the bottom of their pile. In the event that both cards have the same rank, the rules get a little more complicated, with each player flipping over additional cards to compare in a mini “War” showdown.

Assuming a deck was randomly shuffled before every game, how many games of War would you expect to play until you had a game that lasted exactly 26 turns, with no mini “Wars?”

As you might have guessed, such a “perfect” game of war is very rare. So rare, in fact, that trying to simulate it was a fruitless exercise. Instead, your best bet was to work out an analytical solution.

A good first step was to determine the probability *p* of having a perfect game of War (either for you or for your opponent — there was no such distinction in the problem). If we knew the value of *p*, then the probability of achieving one perfect game in exactly one attempt would be *p*, in exactly two attempts it would be (1−*p*)*p*, in exactly three attempts it would be (1−*p*)^{2}*p*, and so on. By combining these probabilities, the expected number of games until having a perfect one was then *p* + 2(1−*p*)*p* + 3(1−*p*)^{2}*p* + 4(1−*p*)^{3}*p* + …, an infinite arithmetico-geometric series whose sum was simply 1/*p*.

So all you had to do was find the probability of having a perfect game of War, and then compute the reciprocal of that number. Alas, determining this probability was easier said than done.

You could have tried a back-of-the-envelope calculation. If we put aside mini “Wars” for a moment, pretending that each player has a 50 percent chance of winning each turn outright, what’s the probability of your winning a perfect game? You would have to win all 26 turns, meaning your chances stood at 1/2^{26}, or approximately 1.49×10^{−8}. Pretty unlikely.

But that was still an estimate. For each turn, you actually had *less than* a 50 percent chance of winning outright, since there was a nonzero chance of a mini “War” in which their card had a matching rank, leaving you and your opponent to split the leftover probability. That meant 1.49×10^{−8} was an overestimate.

Finding the exact probability, it turned out, was an advanced exercise in combinatorics. Solver Laurent Lessard defined the probability as a function of the remaining number of cards of each rank. He then set up a recurrence relation and had his computer crunch the numbers via memoization.

Meanwhile, Peter Norvig similarly “abstracted” what a deck was, considering only how many cards there were of relative ranks (rather than their specific ranks), and then tracked the probability trees in which one player won all the rounds outright.

Both Laurent and Peter found that the probability of sweeping your opponent in 26 turns was approximately 3.1324×10^{−9} — not too far off from our back-of-the-envelope calculation. which meant you would expect to play about **319 million** games before winning in such a fashion. Solver Angela Zhou calculated this value more precisely, finding it was 29,908,397,871,631,390,876,014,250,000 divided by 93,686,147,409,122,073,121. Wow!

This was the expected number of games until *you* won in a sweep. The expected number of games until either you or your opponent won in a sweep was *half* as many (since the probability of either of you achieving this was twice as high), or about **159,620,171** games. The question was ambiguous as stated, so I gave credit for both answers.

Finally, if you enjoyed this riddle, you might try your hand at a few similar combinatorial challenges, courtesy of solver Eric Farmer. You’ll find boring snaps, Frustration Solitaire and prohibited subwords. Sounds like just another day at The Riddler.

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

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Suppose you have a rope that goes all the way around the Earth’s equator, flat on the ground. (For the entirety of this puzzle, you should assume that the Earth is a perfect sphere with a radius of 6,378 kilometers.)

You want to lengthen the rope just the right amount so that it’s 1 meter off the ground all the way around the Earth. How much longer did you have to make the rope?

If you’ve never heard this puzzle before, the answer is surprisingly small — about 6.28 (i.e., 2𝜋) meters. Also, spoiler alert! (Darn, I was one sentence too late.)

Now, instead of tying the Earth up with *rope*, you’ve moved on to covering the globe with a giant *sheet* that lies flat on the ground. If you want the sheet to be 1 meter off the ground (just like the rope), by how much would you have to increase the area of your sheet?

*Extra credit:* What city, country, land mass or body of water has an area that is very close to your answer?

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

From Duane Miller comes a riddle that is good for absolutely nothing:

Duane’s friend’s granddaughter claimed that she once won a game of War that lasted exactly 26 turns.

War is a two-player game in which a standard deck of cards is first shuffled and then divided into two piles with 26 cards each — one pile for each player. In every turn of the game, both players flip over and reveal the top card of their deck. The player whose card has a higher rank wins the turn and places both cards on the bottom of their pile. In the event that both cards have the same rank, the rules get a little more complicated, with each player flipping over additional cards to compare in a mini “War” showdown. Duane’s friend’s granddaughter said that for *every* turn of the game, she always drew the card of higher rank, with no mini “Wars.”

Assuming a deck is randomly shuffled before every game, how many games of War would you expect to play until you had a game that lasted just 26 turns with no “Wars,” like Duane’s friend’s granddaughter?

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

Congratulations to Jacob Herlin of Denver, Colorado, winner of last week’s Riddler Express.

Last week, you were analyzing some unusual signals from deep space, measured at many regular intervals. You computed that you heard zero signals in 45 percent of the intervals, one signal in 38 percent of the intervals and two signals in the remaining 17 percent of the intervals.

Your research adviser suggested that it may just have been random fluctuations from *two* sources. Each source had some fixed probability of emitting a signal that you picked up, and together those sources generated the pattern in your data.

Was your adviser right? Was it possible for your data to have come from two random fluctuations?

At first, this may have sounded like an astronomy question, but as many solvers noted, it was a probability question in disguise. Since there were two random signals, we can assign probabilities to them and see what happens.

Suppose that, for any given interval, the first source emitted a signal with probability *p*, and the second source emitted a signal with probability *q*. If you made the reasonable assumption that the sources were independent, then the probability you heard *two* signals was the product of these probabilities, *pq*. Since you heard two signals 17 percent of the time, that meant *pq* = 0.17. But that still wasn’t enough information to solve the problem.

Next, most solvers opted to analyze the 45 percent of intervals that had zero signals (although looking at the one-signal intervals was equally valid). The probability an interval had zero signals was the probability that the first source didn’t emit a signal, 1−*p*, times the probability the second source didn’t emit a signal, 1−*q*. In other words, (1−*p*)(1−*q*) = 0.45. Expanding the expression on the left gave you 1−*p*−*q*+*pq* = 0.45. Substituting the value 0.17 for *pq* and rearranging a little resulted in the equation *p*+*q *= 0.72.

At this point, there was good news and bad news. The good news was that we were looking for two values, *p* and *q*, knowing both their sum (0.72) and their product (0.17). Anytime that’s the case, we can turn to the quadratic formula. The values *p* and *q* were the solutions to the quadratic equation *x*^{2}−0.72*x*+0.17 = 0.

The bad news, as noted by solver Madeline Barnicle, was that this equation had no real solutions. In other words, **your adviser was mistaken**, and you had an opportunity to show them up. So maybe that was some good news after all.

Looking back, what were the possible probabilities of seeing zero, one or two signals? If we replaced the respective values of 0.45, 0.38 and 0.17 with *A*, *B* and *C* and performed the same analysis, the resulting quadratic equation was *x*^{2}−(1+*C*−*A*)*x*+*C* = 0. For this equation to have any real roots, it had to have a nonnegative discriminant, which meant (1+*C*−*A*)^{2} ≥ 4*C*. So only when that equation was true could you have attributed your data to two random sources.

It seemed like the case was closed.

But not so fast, said solver Daniel Taub, who read the puzzle with a different interpretation. Rather than assume each source emitted exactly zero or one signal, Daniel assumed that each source emitted signals randomly at some average rate — essentially a Poisson process — and then was baffled when no interval produced *three* signals. Solvers Reece Goiffon and Tyler James Burch ran with this interpretation, nevertheless proving that two Poisson processes cannot possibly explain the frequencies listed in the puzzle.

Well, if your data didn’t come from random noise, the truth must still be out there.

Congratulations to Patrick Boylan of Alexandria, Virginia, winner of last week’s Riddler Classic.

Last week, you were building a large pen for your pet hamster. To create the pen, you had several vertical posts, around which you were wrapping a sheet of fabric. The sheet was 1 meter long — meaning the perimeter of your pen could be at most 1 meter — and weighed 1 kilogram, while each post weighed *k* kilograms.

You also wanted your pen to be lightweight and easy to move between rooms. The total weight of the posts and the fabric you used could not exceed 1 kilogram.

For example, if *k* were 0.2, then you could have made an equilateral triangle with a perimeter of 0.4 meters (since 0.4 meters of the sheet would weigh 0.4 kilograms), or you could have made a square with perimeter of 0.2 meters. However, you couldn’t have made a pentagon, since the weight of five posts would already hit the maximum and leave no room for the sheet.

You wanted to figure out the shape that enclosed the largest area possible. What was the greatest value of *k* for which you should have used four posts rather than three?

Three posts weighed 3*k*, which meant that you had 1−3*k* meters to build your triangular pen. Importantly, given a fixed perimeter, the triangle with the greatest area was an equilateral triangle (like courtesy of Keith from the United Kingdom). That meant each of the three sides was one-third of the total perimeter, or (1−3*k*)/3. With a little trigonometry, you found that area of this triangle was then (1−3*k*)^{2}√3/36.

Meanwhile, four posts weighted 4*k*, leaving you with 1−4*k* meters for a quadrilateral pen. Now, the shape that maximized the area was a square with sides of length (1−4*k*)/4, and with area (1−4*k*)^{2}/16.

At this point, it was just a matter of finding the values of *k* for which the square’s area was greater than the triangle’s area. The areas were equal when *k* was (3−2^{4}√3)/(12−6^{4}√3), or approximately **0.08964. **This was the answer — the greatest value of *k* for which four posts was a better choice than three posts.

For extra credit, you looked at pens with five posts, six posts, seven posts and so on. For which values of *k* did each number of posts provide the greatest area for your pen?

To solve this generalized version of the polygon, a good first step was to derive a general formula for the area of polygon with *N* sides and perimeter 1−*Nk* (since having *N* sides meant there would be *N* posts with a total weight of *Nk*). Like the triangle and the quadrilateral, the *N*-gon with the maximum area was a *regular N*-gon, with equal side lengths and angles.

First off, the central angle of a regular *N*-gon is 360/*N* (for those of you who prefer to work in degrees rather than radians). Then, using the fact that each of the *N* sides has a length of (1−*Nk*)/*N*, the area of the triangle formed by the central angle and its corresponding side was half the side length — (1−*Nk*)/(2*N*) — times the altitude of the triangle — (1−*Nk*)/(2*N*)·cot(180/*N*). Yes, that is the cotangent function.

But wait! That was the area for just one of the triangles. Since there were *N* of them, that meant the total area of the polygon was *N* times greater — *N*·(1−*Nk*)/(2*N*)·(1−*Nk*)/(2*N*)·cot(180/*N*), or (1−*Nk*)2/(4*N*)·cot(180/*N*).

Armed with this general formula for the area of an *N*-gon, all that remained was plugging in different values of *N* and *k* and seeing, for each *k*, which *N* resulted in the greatest area. Solver Mike Seifert plotted the graphs for the first few values of *N*, finding that as *k* decreased, the optimal number of posts went up incrementally.

So the extra credit was really asking for the points of intersection between the topmost curves from Mike’s graph. The formula for these points was quite messy, but if you’d like to see it nevertheless, check out the write-ups of Laurent Lessard, who found a condition on these values, and David Zimmerman, who attempted to find a closed-form solution.

While the rodent itself may have been of a usual size, I hope you took comfort in the fact that at least the pen could be of an unusual size — depending on your value of *k*.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>From Josh Silverman comes a puzzle that’s out of this world:

When you started your doctorate several years ago, your astrophysics lab noticed some unusual signals coming in from deep space on a particular frequency — hydrogen times tau. After analyzing a trove of data measured at many regular intervals, you compute that you heard zero signals in 45 percent of the intervals, one signal in 38 percent of the intervals and two signals in the remaining 17 percent of the intervals.

Your research adviser suggests that it may just be random fluctuations from *two* sources. Each source had some fixed probability of emitting a signal that you picked up, and together those sources generated the pattern in your data.

What do you think? Was it possible for your data to have come from two random fluctuations, as your adviser suggests?

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

From Scott Ogawa comes a riddle about rodents of usual size:

Quarantined in your apartment, you decide to entertain yourself by building a large pen for your pet hamster. To create the pen, you have several vertical posts, around which you will wrap a sheet of fabric. The sheet is 1 meter long — meaning the perimeter of your pen can be at most 1 meter — and weighs 1 kilogram, while each post weighs *k* kilograms.

Over the course of a typical day, your hamster gets bored and likes to change rooms in your apartment. That means you want your pen to be lightweight and easy to move between rooms. The total weight of the posts and the fabric you use should not exceed 1 kilogram.

For example, if *k* = 0.2, then you could make an equilateral triangle with a perimeter of 0.4 meters (since 0.4 meters of the sheet would weigh 0.4 kilograms), or you could make a square with perimeter of 0.2 meters. However, you couldn’t make a pentagon, since the weight of five posts would already hit the maximum and leave no room for the sheet.

You want to figure out the best shape in order to enclose the largest area possible. What’s the greatest value of *k* for which you should use four posts rather than three?

*Extra credit:* For which values of *k* should you use five posts, six posts, seven posts, and so on?

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

Congratulations to Alban of Jacksonville Beach, Florida, winner of last week’s Riddler Express.

Last week, you had a large pile of squares that each had a side length of 1 inch. One square was blue, while all the other squares were white. You wanted to arrange several white squares so they covered part of the blue square but didn’t overlap with each other. (The entire blue square did not have to be covered, while the blue area that each white square covered had to be nonzero.)

What was the greatest number of white squares you could have placed?

First, a quick acknowledgment that this was very similar to a problem posed by Martin Gardner some years ago. Special thanks to reader Brian Kell for pointing this out!

Nevertheless, this still proved to be a very challenging express, with a lot of disagreement: Among the hundreds of submitted responses, 2 percent said the answer was four, 3 percent said the answer was five, 14 percent said the answer was six, 56 percent said the answer was seven, 7 percent said the answer was eight, and 4 percent said the answer was nine. (There was a smattering of other answers as well, including readers who creatively wanted to stack paper-thin squares in the third dimension.)

Nine was *not* the answer. Remember, all the squares were the same size. So if you placed a 3-by-3 grid of white squares over the blue one and then rotated the grid about its center, at most *five* of the white squares would have an overlapping area with the blue square.

The majority of readers thought the answer was seven — there must be something to that. Solver Michael Branicky’s five-year-old (!) daughter Lydia may have been the youngest to attempt this puzzle. She got seven white squares to overlap with the blue square by arranging them in a rotated honeycomb pattern:

At this point, we’ve seen that seven squares were possible, while nine squares were not possible. So what about eight squares?

Solver Daniel Thompson illustrated different examples, from one white square all the way up to eight:

It wasn’t as symmetric as the seven-square solution. But by tilting the various squares just right, it was possible to make room for that eighth white square in the middle.

While eight may not be a perfectly square number, this week it was a perfectly hip number.

Congratulations to Richard Guidry Jr. of Baton Rouge, Louisiana, winner of last week’s Riddler Classic.

Last week, the Riddler Manufacturing Company had an issue with their production of foot-long rulers. Each ruler was accidentally sliced at three random points along the ruler, resulting in four pieces. Looking on the bright side, that meant there were then four times as many rulers — they just happened to have different lengths.

On average, how long were the pieces that contained the 6-inch mark?

The problem as stated was slightly ambiguous. I have it on good authority that, for each ruler, the Riddler Manufacturing Company chose the three random points before doing any slicing. (If you assumed a slice was made after each point was selected, it was possible to get different and equally interesting results.)

Solver Quoc Tran declared that “math is for nerds,” before running 50,000 simulations of broken rulers like a total geek. He found that the length of the piece containing the 6-inch mark followed a rather curious distribution (shown below), with an average of approximately 5.621 inches.

Julian Gerez found a similar distribution, further noting that the unusual shape was due to the mashing together of two distinct cases, depending on whether the 6-inch mark was on one of the two middle pieces or on one of the two end pieces — but more on that in a moment!

So that was the computer geeks; now back to the math nerds. The most rigorous way to solve this is with calculus, integrating the product of each possible length multiplied by its relative probability — essentially calculating the mean value of the theoretical curve that Quoc Tran’s histogram approximates. Emma Knight fearlessly worked her way through those messy integrals, arriving at an answer of **5.625 inches**.

An alternative approach was to break the problem down into two cases, depending on which of the four pieces contained the 6-inch mark. If the 6-inch mark was on one of the two end pieces of the broken ruler, that meant the three random breaks were all between the 0- and 6-inch marks (which occurred with a one-in-eight probability) or all between 6- and 12-inch marks (which also occurred with a one-in-eight probability). In other words, 25 percent of the time the three random breaks were all on one side of the 6-inch mark, while the other 75 percent of the time there was a single break on one side of the 6-inch mark and two breaks on the other side.

When all three breaks were on one side, the length of the piece with the 6-inch mark was 6 inches plus the average distance between the 6-inch mark and nearest of the three breaks. Regular solvers of Riddler Classics may know a thing or two about order statistics, particularly the result that when you choose *N* random values from a uniform distribution between 0 and 1, the expected value of the smallest number is 1/(*N*+1), the expected value of the second smallest is 2/(*N*+1), and so on up to the largest number, which has an expected value of *N*/(*N*+1). So when the three random breaks all occurred (uniformly) over a 6-inch range, the average distance between the 6-inch mark and the nearest break was one-fourth (since four is one more than three) of the total range, or 1.5 inches. Putting the two pieces together, that meant 25 percent of the time the average length was 6+1.5, or 7.5 inches.

But what about the other 75 percent of the time? As we said, there were two random breaks on one side of the 6-inch mark and one break on the other side. The average distance between the 6-inch mark and nearest among two breaks was one-third (since three is one more than two) of the length, or 2 inches. And the average distance between the 6-inch mark and the single break on the other side was one-half (since two is one more than one) of the length, or three inches. Putting *these* two pieces together, that meant 75 percent of the time the average length was 2+3, or 5 inches.

No integrals required, and we almost have our answer! Using the linearity of expectation, as solver Steve Gabriel did, the average length was 0.25(7.5) + 0.75(5), or 5.625 inches — the same answer we got from calculus!

But the fun didn’t stop there. Several readers took this puzzle to new heights, looking at how the average length of the piece with the 6-inch mark changed with the number of break points, as well as the average length of pieces containing *other* points on the ruler like the 1- or 2-inch mark. Laurent Lessard combined these into a single generalization, finding the expected length of the piece a fraction *a* along a ruler of length *L* broken at *N* random points. By computing the distribution for the number of breaks on either side of the selected point and applying order statistics, he found this length to be (2−*a*^{N+1}−(1−*a*)^{N+1})*L*/(*N*+1).

One more thing: As *N* got very large — meaning there were many random breaks in the ruler — the average length Laurent found approached 2*L*/(*N*+1) for *any* value of *a*. In other words, as soon as you pick a point on the ruler and ask for the average length of the piece containing that point, the answer will be *twice* the length of the average piece. Again, that’s for any point you pick!

It’s a bizarre paradox, to be sure. My sense is this happens because you are looking for the next break in each of the *two* directions along the ruler. If anyone happens to make further headway on this, be sure to let Laurent (and me!) know.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Since we’ll be live blogging the Democratic and Republican National Conventions this week and next, we thought it would be a good idea to ask … why? In this episode, we argue about whether parties should even have political conventions. Before you watch the video, take a guess as to which side our politics editor, Sarah Frostenson, is on.

Whose side are you on? Be sure to weigh in on YouTube or Twitter. And while you’re there, let us know what you’d like us to debate in the next episode.

]]>From Dean Ballard comes a sneaky sorting of squares:

You have a large pile of squares that each have a side length of 1 inch. One square is blue, while all the other squares are white. You want to arrange several white squares so they cover part of the blue square but don’t overlap with each other.

For example, here’s how you could arrange four white squares so they each cover part of the blue square.

What is the greatest number of white squares you can place so that each covers part of the blue square without overlapping one another? (The entire blue square does not have to be covered, while the blue area that each white square covers must be nonzero.)

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

From Angela Zhou comes one riddle to rule them all:

The Riddler Manufacturing Company makes all sorts of mathematical tools: compasses, protractors, slide rules — you name it!

Recently, there was an issue with the production of foot-long rulers. It seems that each ruler was accidentally sliced at three random points along the ruler, resulting in four pieces. Looking on the bright side, that means there are now four times as many rulers — they just happen to have different lengths.

On average, how long are the pieces that contain the 6-inch mark?

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

Congratulations to Richard Dickerman of Dallas, Texas, winner of last week’s Riddler Express.

Last week, I was hiking on Euclid Island, which was perfectly rectangular and measured 3 miles long by 2 miles wide. I was especially interested in locating the point on the shore that was nearest to my position.

From where I had been standing, there were in fact *two* distinct points on the shore that were *both* the closest such points. It turned out the trail I was hiking along connected all such locations on the island — those with multiple nearest points on the shore.

What was the total length of this trail on Euclid Island?

The dead center of the island was a logical place for many solvers to start. It was just 1 mile along the width — in either direction. But before we looked for other such points, what did it mean, mathematically, for there to be multiple nearest points on the shore?

One way to think about it was to grow a circle (kind of like blowing up a balloon) centered at your current position. As the circle got larger and larger, it eventually touched (i.e., was tangent to) the shore. The moment the circle touched the shore, you had to look at *how many places* it touched the shore. If there was only one such point of tangency, then you weren’t on the trail. But if there were two or more such points, then you were indeed on the trail.

So what did Euclid Island’s trail actually look like? Rebekah Murphy noticed the trail included a segment of length 1 that ran east-west down the middle of the island. Every point on this segment was equidistant from the north and south shores. (Interestingly, the endpoints of this segment were equidistant from *three* distinct points along the shore.)

But the fun didn’t end there. Ishaan Bhatia noted that the trail also had four additional segments, which ran between the endpoints of the middle segment and the four corners of the island. (While the corners themselves were not part of the trail — after all, they were *on* the beach — the trail came infinitesimally close, which meant the exclusion of these points didn’t affect your length calculations.)

The animation below puts all five segments together, revealing the entire green hiking trail on Euclid Island. The circle that grows and shrinks demonstrates the points of tangency (i.e., the nearest spots on the beach) for each location on the trail.

At this point, we’re ready to answer the original question, which asked you to determine the total length of the trail. As we already said, the middle segment had a length of 1. Using the Pythagorean theorem (or, rather, the Babylonian Formula), the remaining four segments had a length of √2. That meant the total length of the trail was **1+4√2 miles**, or about 6.657 miles.

For extra credit, you looked at Al-Battani Island, which was elliptical rather than rectangular. Al-Battani Island’s major axis was 3 miles long, while its minor axis was 2 miles long. Like Euclid Island, Al-Battani Island had a hiking trail that connected all locations with multiple nearest points on the shore. What was the total length of this trail on Al-Battani Island?

As with Euclid Island, a good place to start was the dead center, which was 1 mile from the two endpoints of the minor axis. And once again, there was a central horizontal segment that ran along the major axis. The challenge here was figuring out just where the trail ended.

It didn’t cover the *entire* major axis, because when it got too close to the endpoints, there was only a single point of tangency. In other words, you had to find when the circle — centered on an ellipse’s major axis and internally tangent to the ellipse — transitioned between one and two points of tangency, as illustrated below:

And that’s where the math got hairy. If you assumed the ellipse was centered at (0, 0), then it was described by the equation *x*^{2}/1.5^{2} + *y*^{2} = 1. Meanwhile, the equation for a circle with radius *R *centered at *h* on the *x*-axis was (*x*−*h*)^{2} + *y*^{2} = *R*^{2}. From there, you had to find coordinates (*x*, *y*) that satisfied both equations, but you also needed to use calculus to ensure the circle and the ellipse had matching slopes at these points.

In the end, as long as the center of the circle was between (−5/6, 0) and (5/6, 0), the circle was tangent to the ellipse in two locations. That meant the distance between these two points — about **1.667 miles** — was the length of the trail and the answer to the extra credit.

Quite the challenging hike, if you ask me. And if you’d like to explore this problem further, check out an interactive version of this graph on Desmos, courtesy of solver Hypergeometricx.

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

Last week, inspired by Kareem Carr, you looked at an alternative definition for addition and followed where it led. To compute the sum of *x* and *y*, you combined groups of *x* and *y* nematodes and left them for 24 hours. When you came back, you counted up how many you had — and that was the sum!

It turned out that, over the course of 24 hours, the nematodes paired up, and each pair had one offspring 50 percent of the time. (If you had an odd number of nematodes, they still paired up, but one was left out.) So if you wanted to compute 1+1, half the time you’d get 2, and half the time you’d get 3. If you computed 2+2, 25 percent of the time you’d get 4, 50 percent of the time you’d get 5, and 25 percent of the time you’d get 6.

We also redefined exponentiation: Raising a sum to a power meant leaving that sum of nematodes for the number of days specified by the exponent.

With this number system, what was the expected value of (1+1)^{4}?

A good strategy here was to work your way up, one power at a time. First, what was the expected value of (1+1)^{1}? You initially had two nematodes that paired up, and after 24 hours there was a 50 percent chance they’d have one offspring. So the expected value of (1+1)^{1} was 2.5.

Next, what was the expected value of (1+1)^{2}? In other words, what was the expected number of worms after 48 hours? As we said, after 24 hours, there was a 50 percent chance there were two worms and a 50 percent chance there were three worms. After *another* 24 hours, the two-worm case had an expected value of 2.5. Meanwhile, the three-worm case resulted in two worms that paired up along with a third wheel. The pair of worms resulted in an expected value of 2.5, and adding the third worm gave an expected value of 3.5. Half the time there would be 2.5 worms, and the other half the time there would be 3.5 worms, which meant the expected value was an even 3.

Let’s do one more: What was the expected value of (1+1)^{3}? Digging through the previous paragraph, we just saw that after 48 hours there was a 25 percent chance of having two worms, a 50 percent chance of having three worms and a 25 percent chance of having four worms. In the two-worm case, the expected number of worms after 72 hours was again 2.5. In the three-worm case, the expected number of worms after 72 hours was 3.5. The four-worm case was more complicated — 25 percent of the time there would still be four worms after 72 hours, 50 percent of the time there would be five worms, and 25 percent of the time there would be six worms. Putting all these cases (and sub-cases) together, the expected value of (1+1)3 was 0.25·2.5 + 0.5·3.5 + 0.25·(0.25·4 + 0.5·5 + 0.25·6), or 3.625.

I’ll spare you all the casework for (1+1)^{4}. After 96 hours, there could be anywhere from two to nine worms. The expected value was **4.40625** worms. Andrew Heairet neatly visualized the breakdown of probabilities over the course of 96 hours:

Mark Girard went beyond 96 hours and found the expected number of nematodes more than two weeks later. This value appeared to increase exponentially over time, which is the perfect segue into the extra credit, which asked you how the expected value of (1+1)^{N} behaved as *N* got larger and larger. As many solvers correctly observed, the expression goes off to infinity. But the real question was *how*.

The challenge here related to the “odd worms out,” which sat out reproductive cycles. Were it not for them, the math would have been more straightforward. For each pair of worms, their number either stayed the same or increased by 50 percent over each 24-hour period. Averaging those two possibilities meant the expected number increased by 25 percent every 24 hours. But again, this reasoning didn’t apply, thanks to the occasional one worm who didn’t pair up.

To figure this out, some solvers fit exponential models directly to the data. For example, Hypergeometricx found the growth was intriguingly close to 1.56·(1.23456789)^{N}. Meanwhile, Josh Silverman showed that 2·(1.25)^{N−1} was in fact a pretty decent approximation.

But Rajeev Pakalapati took the cake, showing — analytically, mind you — that 2·(1.25)^{N−1} + 0.5 was the limiting behavior of nematode exponentiation. His work is shown below, and you can also follow along via Steve Gabriel’s helpful transcription.

Indeed, this was a fun and challenging foray into an alternate set of operations. Only one submitter felt threatened enough to compare this riddle to an Orwellian dystopia. Needless to say, they didn’t get very far with the math.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>The Riddler Isles are a chain of small islands on the Constant Sea. One of them, Euclid Island, is perfectly rectangular and measures 3 miles long by 2 miles wide. While walking across the island on a recent vacation, I was often interested in locating the point on the shore that was nearest to my current position.

One morning, I realized that from where I was standing there were *two* distinct points on the shore that were *both* the closest such points. I was excited by my discovery, only to realize it had been made years earlier. It turned out I was on a hiking trail that connected all such locations on the island — those with multiple nearest points on the shore.

What is the total length of this trail on Euclid Island?

*Extra credit:* Al-Battani Island is another of the Riddler Isles, but it’s elliptical rather than rectangular. Al-Battani Island’s major axis is 3 miles long, while its minor axis is 2 miles long. Like Euclid Island, Al-Battani Island has a hiking trail that connects all locations with multiple nearest points on the shore. What is the total length of this trail on Al-Battani Island?

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

This week’s Riddler Classic is inspired by Kareem Carr:

We usually think of addition as an operation applied to a field like the rational numbers or the real numbers. And there is good reason for that — as Kareem says, “Mathematicians have done all the hard work of figuring out how to make calculations track with reality. They kept modifying and refining the number system until everything worked out. It took centuries of brilliant minds to do this!”

Now suppose we defined addition another (admittedly less useful) way, using a classic model organism: the nematode. To compute the sum of *x* and *y*, you combine groups of *x* and *y* nematodes and leave them for 24 hours. When you come back, you count up how many you have — and that’s the sum!

It turns out that, over the course of 24 hours, the nematodes pair up, and each pair has one offspring 50 percent of the time. (If you have an odd number of nematodes, they will still pair up, but one will be left out.) So if you want to compute 1+1, half the time you’ll get 2 and half the time you’ll get 3. If you compute 2+2, 25 percent of the time you get 4, 50 percent of the time you’ll get 5, and 25 percent of the time you’ll get 6.

While we’re at it, let’s define exponentiation for sums of nematodes. Raising a sum to a power means leaving that sum of nematodes for the number of days specified by the exponent.

With this number system, what is the expected value of (1+1)^{4}?

*Extra credit:* As *N* gets larger and larger, what does the expected value of (1+1)^{N} approach?

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

Congratulations to David Daly of Glendale, Arizona, winner of last week’s Riddler Express.

Last week, you made it to the final round of the Riddler Rock, Paper, Scissors tournament.

The rules were simple: Rock beat scissors, scissors beat paper, and paper beat rock. Moreover, the game was “sudden death,” so the first person to win a single round was immediately declared the grand champion. (If both players chose the same object, then you simply played another round.)

Fortunately, your opponent was someone you had studied well. Based on the motion of their arm, you could tell whether they would (1) play rock or paper with equal probability, (2) play paper or scissors with equal probability or (3) play rock or scissors with equal probability. (Every round fell into one of these three categories.)

If you strategized correctly, what were your chances of winning the tournament?

Well, if you strategized like solver Carenne Ludeña, you won the tournament a whopping **100 percent** of the time. When your opponent played rock or paper with equal probability, you knew with certainty they wouldn’t play scissors, and that meant if you played paper you couldn’t lose. You also had a 50 percent chance of winning the round, i.e., whenever your opponent played rock.

Each of the three scenarios listed in the problem had a corresponding response. As we just said, when your opponent played rock or paper, you should play paper. When your opponent played paper or scissors, you should play scissors. And when your opponent played rock or scissors, you should play rock. With this strategy, you would win half the time, draw half the time, and lose … never.

At this point, many solvers recognized this meant you were guaranteed to win the tournament eventually, as the probability of drawing infinitely many games was vanishingly small. Solver Stephen Paisley formally demonstrated this with the geometric series 1/2 + 1/4 + 1/8 + 1/16 + … (your chances of having won in each successive round), which indeed sums to 1, or 100 percent.

Finally, as some readers observed, the original puzzle was slightly ambiguous as written. When you were told that your opponent would “play rock or paper with equal probability,” most solvers assumed that meant both probabilities were 50 percent, rather than being equal but less than 50 percent. While that was the intent of the puzzle, it wasn’t explicitly stated.

If these equal probabilities were close to 50 percent, then your strategy wouldn’t change (although your chances of winning the tournament would go down). For example, if you knew there was a 40 percent chance your opponent would play rock, a 40 percent chance they’d play paper and a 20 percent chance they’d play scissors, then your best bet would still be to play paper. But instead of always winning the tournament, your chances of victory would be 0.4 + 0.4^{2} + 0.4^{3} + …, or 2/3.

But once the equal probabilities dipped below one-third, you needed to adjust your strategy. Suppose you knew there was a 10 percent chance your opponent would play rock, a 10 percent chance they’d play paper and an 80 percent chance they’d play scissors. Now your best bet was to play rock, and your chances were 0.8 + 0.8(0.1) + 0.8(0.1)^{2} + 0.8(0.1)^{3} + …, or 8/9.

Any which way, it seems like you’re pretty good at rock, paper, scissors. My winning move against you would be not to play.

Congratulations to Rick of Cloverdale, California, winner of last week’s Riddler Classic.

Last week, the inaugural graduating class of Riddler High School, more than 100 students strong, received their diplomas. They were lined up along the circumference of the giant circular mosaic on the plaza in front of the school, while their principal, Dr. Olivia Rhodes, stood atop a tall stepladder in the middle.

Dr. Rhodes observed that every one of the *N* graduates — all except Val, the valedictorian — was standing in the wrong place. Moreover, Dr. Rhodes said that no two graduates were in the correct position *relative to each other*. (Two graduates would be in the correct position relative to each other if, for example, graduate A was supposed to be 17 positions counterclockwise of graduate B and was indeed 17 positions counterclockwise of graduate B — even if both graduates were in the wrong positions.)

But Val corrected Dr. Rhodes. Given the fact that there were *N* students, there must have been at least two who were in the correct position relative to each other.

Given that there were more than 100 graduates, what was the *minimum* number of graduates who were posing for the class photo?

That was a lot of information to take in. To summarize, there were *N* students arranged in a circle. Exactly one of them was in the correct position, while *N*−1 were not. Furthermore, this value of *N* *guaranteed* that at least two students were in the correct position relative to each other.

The puzzle’s submitter, Dave Moran, solved this by looking at how far apart each student was from their correct position (say, in the clockwise direction) as a function of their correct position, *x*. Let’s call this function *E*(*x*). So because Val was correctly in the first position, *E*(1) = 0, meaning she was zero spots away from her correct position. But *E*(2), *E*(3), and so on, all the way to *E*(*N*), were nonzero, since all the other students were in the wrong position.

But what if *no students* had been in the correct position *relative to each other*? Well, imagine if there had been two students with correct positions *a* and *b*, such that *E*(*a*) = *E*(*b*). That would have meant they were shifted the same number of spots clockwise from their correct positions, and so they would have been in the correct position relative to each other! To *avoid* this situation, no two students could have had the same value of *E*(*x*).

Now recall that there were *N* students. That meant *E*(*x*) could take on exactly *N* distinct values, from 0 to *N*−1. And because no two students could have the same value of *E*(*x*), every value from 0 to *N*−1 had to be taken by one and only one student.

We’re getting there — but first, a slight detour. Solver Eric Dallal looked at the sum of all the errors, *E*(1) + *E*(2) + *E*(3) + … + *E*(N). If you started with a correct arrangement of the entire class, this sum would be zero. You could also swap different pairs of students over and over again to get any desired overall arrangement of the class. But every time you swapped two students, the sum either didn’t change, or it increased or decreased by *N*. And that meant this sum had to be a multiple of *N*.

Okay, we’re in the home stretch. For there to have been *no students* who were in the correct position relative to each other, we needed the sum of the errors — that is, the sum of the numbers from 0 to *N*−1, or *N*(*N*−1)/2 — to be a multiple of *N*. This was true whenever *N* was odd.

To summarize, it was only possible to arrange students so that no two were in the correct position relative to each other when *N* was odd.

Want an example of such an arrangement? Well, you’re in luck, thanks to solver Inga. Consider a class with 101 students — an odd number. Val was in the correct position, but suppose everyone else was reflected across the line connecting Val and Dr. Rhodes. If the salutatorian was supposed to be directly clockwise from Val, they now found themselves in position 101, meaning *E*(2) was 101−2, or 99. Working your way around the circle, you’d see that each student indeed had a unique value of *E*(*x*).

At last, we return to the original question that was posed. What was the smallest class size such that at least two students *had* to have been in the correct relative position? That would be the smallest even number greater than 100, and so the answer was **102**.

In addition to this week’s winner, Val from the inaugural graduating class really knew her number theory. The other seniors, including Zach, need to step up their game.

Email Zach Wissner-Gross at riddlercolumn@gmail.com

]]>Congratulations, you’ve made it to the final round of the Riddler Rock, Paper, Scissors.

The rules are simple: Rock beats scissors, scissors beat paper, and paper beats rock. Moreover, the game is “sudden death,” so the first person to win a single round is immediately declared the grand champion. If there’s a tie, meaning *both* players choose the same object, then you simply play another round.

Fortunately, your opponent is someone you’ve studied well. Based on the motion of their arm, you can tell whether they will (1) play rock or paper with equal probability, (2) play paper or scissors with equal probability or (3) play rock or scissors with equal probability. (Every round falls into one of these three categories.)

If you strategize correctly, what are your chances of winning the tournament?

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

From Dave Moran comes a perplexing puzzle of pomp and circumstance:

The inaugural graduating class of Riddler High School, more than 100 students strong, is lined up along the circumference of the giant circular mosaic on the plaza in front of the school. The principal, Dr. Olivia Rhodes, stands atop a tall stepladder in the center of the circle, preparing to take the panoramic class photo.

Suddenly, she shouts, “Hey, class, every single one of you *N* graduates — except Val, our valedictorian — is standing in the wrong place! Remember, you’re supposed to stand in order of your class rank, starting with Val directly in front of me, and going counterclockwise all the way to Zach, who should then be next to Val.” (Poor Zach was ranked last in the graduating class at Riddler High.)

Dr. Rhodes continues, “Not only are almost all of you in the wrong positions, but no two of you are even in the correct position *relative to each other*.” Two graduates would be in the correct position relative to each other if, for example, graduate A was supposed to be 17 positions counterclockwise of graduate B and was indeed 17 positions counterclockwise of graduate B — even though both graduates were in the wrong positions.

But Val speaks up: “Dr. Rhodes, that can’t be right. There must be at least two of us who are in the correct position relative to each other.”

Dr. Rhodes looks carefully around the circle of graduates below her and admits, “Of course, our brilliant Val is quite right. I do now see that there are two of you who are correctly positioned relative to each other.”

Given that there are more than 100 graduates, what is the *minimum* number of graduates who are posing for the class photo?

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

Congratulations to Joe Maloney of Atlanta, Georgia, winner of last week’s Riddler Express.

Last week, Riddler Township was having its quadrennial presidential election. Each of the town’s 10 “shires” was allotted a certain number of electoral votes: two, plus one additional vote for every 10 citizens (rounded to the nearest 10).

The names and populations of the 10 shires are summarized in the table below.

Shire | Population | Electoral votes |
---|---|---|

Oneshire | 11 | 3 |

Twoshire | 21 | 4 |

Threeshire | 31 | 5 |

Fourshire | 41 | 6 |

Fiveshire | 51 | 7 |

Sixshire | 61 | 8 |

Sevenshire | 71 | 9 |

Eightshire | 81 | 10 |

Nineshire | 91 | 11 |

Tenshire | 101 | 12 |

Under this sort of electoral system, it was quite possible for a presidential candidate to lose the popular vote and still win the election.

With two candidates running for president of Riddler Township, and every citizen voting for one or the other, what is the *lowest* percentage of the popular vote that a candidate could get while still winning the election?

To win with as few votes as possible, a candidate needed a majority of the 75 electoral votes, meaning they needed at least 38 votes. They also needed just over half the popular vote in any shire they won, while losing the entire popular vote in the shires they lost.

Meanwhile, the less populous shires also offered greater leverage. For example, Oneshire, with its three electoral votes and a population of 11 people (i.e., a majority consisting of six people), offered 0.5 — three divided by six — electoral votes per supporter. Twoshire, with four electoral votes and a voting majority of 11 people, offered about 0.36 electoral votes per supporter. Meanwhile, at the other end of the spectrum, Tenshire offered less than 0.24 electoral votes per supporter.

That meant our unpopular winner wanted to win the less populous shires, accruing electoral votes without winning too many popular votes.

The fewest votes a candidate needed to win turned out to be 136 out of Riddler Township’s 560 total citizens, and so the smallest possible winning percentage of the popular vote was approximately **24.3 percent**.

There were several different ways to generate this electoral nightmare. For example, the winner might get six votes in Oneshire, 16 votes in Threeshire, 21 votes in Fourshire, 26 votes in Fiveshire, 31 votes in Sixshire and 36 votes in Sevenshire. In all, that is indeed 38 electoral votes against just 136 popular votes. This particular scenario was illustrated by Andrew Heairet:

That map of Riddler Township looks vaguely familiar, but I can’t quite place it.

But that wasn’t the only solution. Many solvers, like Nora Corrigan from Columbus, Mississippi and Caspian from Stockholm, Sweden found multiple ways to generate exactly 38 electoral votes and 136 popular votes. Here is the complete list of such scenarios:

- Slim victories in Oneshire, Threeshire, Fourshire, Fiveshire, Sixshire and Sevenshire
- Slim victories in Oneshire, Twoshire, Fourshire, Fiveshire, Sixshire and Eightshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fiveshire, Sixshire and Nineshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fiveshire, Sevenshire and Eightshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fourshire, Sixshire and Tenshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fourshire, Sevenshire and Nineshire

As it turned out, there were many ways for the electoral and popular votes to wildly disagree. I guess that’s the electoral college for you. Oh, and if you’d like your vote to *really* count in Riddler Township, it’s a good idea to live in Oneshire.

Congratulations to Steven Trautmann of Aurora, Colorado, winner of last week’s Riddler Classic.

Last week, you played a game of Riddler Pinball, which had an infinitely long wall and a circle whose radius was 1 inch and whose center was 2 inches from the wall. The wall and the circle were both fixed and never moved. A single pinball started 2 inches from the wall and 2 inches from the center of the circle.

To play, you flicked the pinball toward a spot of your choosing along the wall, specified by its distance *x* from the point on the wall that’s closest to the circle, as shown in the diagram below.

The goal of the game was simple: Get the ball to bounce as many times as possible.

If you aimed too far to the right (i.e., your value of *x* was too small), the pinball quickly bounced its way through the gap between the circle and the wall. But if you aimed too far to the left (i.e., your value of *x* is too big), the pinball quickly came back out the same side it went in.

Riddler Pinball was an unforgiving game — the slightest error tanked your chances of victory. But if you strategized *just* right, it was possible to do quite well.

What was the greatest number of bounces you could achieve? And, more importantly, what value of *x* got you the most bounces?

There was certainly a sweet spot that resulted in many, many bounces. For example, when *x* was 0.82248632494339, there were 43 bounces:

But before we go any further, let’s return to the first question that was posed: What was the greatest possible number of bounces?

As solver Phillip Bradbury observed, all the angles and points of impact changed monotonically with *x*. As noted above, smaller values of *x* made the pinball pass through, but larger values of *x* made the pinball bounce back out.

So then what happened *between* these two cases? Zooming in revealed a single point on the number line where the pinball neither passed through nor bounced back. But if it did *neither* of these things, what else could the ball possibly do? It would be stuck bouncing back and forth **infinitely many times**. And as *x* approached this point — whatever it was — the number of bounces went up, up, up. This was nicely illustrated by solvers Mark Girard and Dinesh Vatvani.

There were a few other ways to see how this was true. Jason Bellenger reasoned backwards, thinking about a ball that was bouncing straight up and down between the tip of the circle and the wall. If you perturbed it infinitesimally, but just right, it would ultimately bounce its way to the correct starting point. Playing these bounces in reverse would then give you an infinitely high-scoring pinball shot.

Laurent Lessard, meanwhile, was able to better visualize the multitude of bounces by plotting the logarithm of the *x*-coordinate, as shown below. Graphically, this had the effect of spreading out the bounces, leading him to correctly suspect the answer was infinity.

As it turned out, this reasoning was the easier part of the problem. *Finding* this precise value of *x* that got you infinitely many bounces was another matter entirely.

At this point, most solvers took a computational approach. Simulating the pinball as it bounced off the wall wasn’t too tricky — the ball simply reflected off the wall so that the angle it made with respect to the wall didn’t change. But simulating bounces off the circle was a little tougher. Again, the angles of incidence and reflection were equal, but now they were measured from the radius of the circle. A slight mess of trigonometry ensued.

Once you had a functioning pinball simulator up and running, then it was just a matter of trying different values of *x*, adding one digit at a time, and seeing which values gave you more bounces. Indeed, the answer was very close to **0.82248632494339** — the true value actually goes on for infinitely many decimal places.

Some solvers, like Dean Ballard and Joseph Wetherell, went to great lengths to crank out more of the solution’s decimal places. In fact, the answer is very, very close to:

0.82248632494339006569162637706984990078

134582582348831326968696909788932771367

According to Joseph, this precise value of *x* resulted in more than 200 bounces!

Meanwhile, solver Emma Knight boldly attempted a more analytic approach. However, ballooning floating point errors ultimately got in the way.

Finally, a few very clever readers noticed something glaring about this problem: There was a second solution! While the original statement of the riddle asked where *along the wall* the pinball should be aimed, it was also possible to hit the circle first and *still* achieve infinitely many bounces. Josh Silverman went as far as animating this alternate strategy:

When I previously teased Riddler Pinball, I asked what interesting mathematical questions could be posed about it. There were multiple calls to alter the geometry, such as with a noncircular bumper, or to scale it up to three dimensions. With all these great suggestions, I don’t think it will be long before we all play another round of Riddler Pinball.

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

]]>From Peter Mowrey comes an elegant electoral enigma:

Riddler Township is having its quadrennial presidential election. Each of the town’s 10 “shires” is allotted a certain number of electoral votes: two, plus one additional vote for every 10 citizens (rounded to the nearest 10).

The names and populations of the 10 shires are summarized in the table below.

Shire | Population | Electoral votes |
---|---|---|

Oneshire | 11 | 3 |

Twoshire | 21 | 4 |

Threeshire | 31 | 5 |

Fourshire | 41 | 6 |

Fiveshire | 51 | 7 |

Sixshire | 61 | 8 |

Sevenshire | 71 | 9 |

Eightshire | 81 | 10 |

Nineshire | 91 | 11 |

Tenshire | 101 | 12 |

As you may know, under this sort of electoral system, it is quite possible for a presidential candidate to lose the popular vote and still win the election.

If there are two candidates running for president of Riddler Township, and every single citizen votes for one or the other, then what is the *lowest* percentage of the popular vote that a candidate can get while still winning the election?

Riddler Pinball is a game with an infinitely long wall and a circle whose radius is 1 inch and whose center is 2 inches from the wall. The wall and the circle are both fixed and never move. A single pinball starts 2 inches from the wall and 2 inches from the center of the circle.

To play, you flick the pinball toward a spot of your choosing along the wall, specified by its distance *x* from the point on the wall that’s closest to the circle, as shown in the diagram below.

The goal of the game is simple: Get the ball to bounce as many times as possible.

(Note: This is a geometry problem, not a physics problem. In other words, assume the system is frictionless and that all collisions are perfectly elastic.)

Let’s take a look at some games to see how they play out.

If you aim too far to the right (i.e., your value of *x* is too small), the pinball will quickly bounce its way through the gap between the circle and the wall. That’s what happened in the game below, when *x* was 0.75 inches, resulting in a rather pedestrian four bounces.

But if you aim too far to the left (i.e., your value of *x* is too big), the pinball will quickly come back out the same side it went in. That’s what happened in the next game, when *x* was 0.9 inches, again yielding just four bounces.

As you can see, Riddler Pinball is an unforgiving game — the slightest error can tank your chances of victory. But if you strategize *just* right, it’s possible to do quite well.

What’s the greatest number of bounces you can achieve? And, more importantly, what value of *x* gets you the most bounces?

Congratulations to Rick Schubert of San Diego, California, winner of last week’s Riddler Express.

Last week, you set your sights on breaking baseball records, albeit in a shortened season. Your true batting average was .350, meaning you had a 35 percent chance of getting a hit with every at-bat. If you had four at-bats per game, what were your chances of batting at least .400 over the course of the 60-game season?

In a 60-game season, with four at-bats per game, there were 240 total at-bats. Forty percent of 240 was 96, so to bat at least .400 you needed at least 96 hits. Since each at-bat was independent, you could use the binomial distribution to determine the probability of each number of hits.

For example, the probability of getting exactly 96 hits in 240 at-bats was equal to 0.35^{96} (i.e., getting a hit in 96 at-bats) times 0.65^{144} (i.e., *not* getting a hit in the remaining 144 at-bats), times the number of ways you can order 96 hits among 240 at-bats (240 choose 96).

Again, that was just the probability of getting *exactly* 96 hits. To find the probability of *at least* 96 hits, you had to add up the probabilities of getting 96 hits, 97 hits, 98 hits, and so on, up to (the very unlikely) case of 240 hits. Alternatively, you could have added up the probabilities of getting between zero and 95 hits, and then subtracted this from 1. Whether you did this by hand, by spreadsheet or by computer code, the correct answer was **6.1 percent**.

It turned out that it was 16 times easier to bat .400 over 60 games than it was over a full slate of 162 games, where the probability was just **0.38 percent**.

The puzzle’s submitter, Taylor Firman, extended this further, looking at the probability of *anyone in the league* batting .400, based on each player’s own career batting average. For a 60-game season, Taylor found that this probability was about 3.4 percent, while it was just 0.002 percent for a 162-game season. So if no one catches Ted Williams this year, forget about it.

For extra credit, you were asked to find your chances of getting a hit in at least 56 consecutive games within the 60-game season, tying or breaking Joe DiMaggio’s record. It was helpful to first calculate the probability of keeping the streak alive — that is, getting at least one hit in a game. Again using the binomial distribution, this probability was 82.15 percent. But while it was likely you’d get a hit in any given game, pulling that off 56 times *in a row* was another story.

As Mark Girard explained, there were several distinct ways to achieve a streak of at least 56 games:

- Getting a hit in games 1 to 56
- Not getting a hit in game 1 and getting a hit in games 2 to 57
- Not getting a hit in game 2 and getting a hit in games 3 to 58
- Not getting a hit in game 3 and getting a hit in games 4 to 59
- Not getting a hit in game 4 and getting a hit in games 5 to 60

Each of these five cases was very unlikely. And together, they were still very unlikely. Overall, your chances of having a 56-game hitting streak in a 60-game season stood at just **0.0028 percent**. And remember, that was assuming you were a lifetime .350 hitter! Over a 162-game season, these chances improved by about tenfold, to **0.033 percent**. It would appear that Joe DiMaggio’s record is quite safe.

Finally, for *extra* extra credit, you had to find your chances of both batting at least .400 *and* getting a hit in at least 56 games. This was especially tricky, since these events were *not* independent. In other words, if you hit .400, you were more likely to have a very long hitting streak, and vice versa.

Paul Wright worked it out analytically, starting with streaks of 56, 57, 58, 59 or 60 games, each of which had their own corresponding probabilities of occurring. Then, for each streak, you knew that the games *within* the streak had one, two, three or four hits, while the remaining handful of games could also have zero hits. Putting the streak and non-streak games together, the probability of batting at least .400 *and* having a hitting streak of at least 56 games was about **0.0027 percent**. This was very close to the answer for the extra credit, meaning If you tied or broke Joe DiMaggio’s record in 60 games, then it was *very* likely that you’d also bat at least .400.

Finally, Angela Zhou calculated your chances of reaching both milestones over a 162-game season. It wasn’t likely.

Congratulations to Jim Boyce of Kensington, Connecticut, winner of last week’s Riddler Classic.

Last week, the tortoise and the hare were about to begin a 10-mile race along a “stretch” of road. The tortoise was driving a car that traveled 60 miles per hour, while the hare was driving a car that traveled 75 miles per hour. (For the purposes of this problem, you were asked to assume that both cars accelerated from 0 miles per hour to their cruising speed instantaneously.)

The hare did a quick mental calculation and realized if it waited until two minutes had passed, they would cross the finish line at the exact same moment. And so, when the race began, the tortoise drove off while the hare patiently waited.

But one minute into the race, after the tortoise had driven 1 mile, something extraordinary happened. The road turned out to be magical and instantaneously stretched by 10 miles! As a result of this stretching, the tortoise was now *2* miles ahead of the hare, who remained at the starting line.

At the end of every subsequent minute, the road stretched by 10 miles. With this in mind, the hare did some more mental math.

How long after the race began should the hare have waited so that both the tortoise and the hare crossed the finish line at the same exact moment?

At first, it might have seemed like neither the tortoise nor the hare would ever finish the race. How could they, when they drove a mile or so per minute, while the road stretched *10* miles per minute? The key was to realize that the road stretched *uniformly*, which meant it could carry each car along as it stretched.

To better understand this, let’s take a closer look at the tortoise over time. After one minute, it traveled 1 mile, or 10 percent of the total distance. Then, the road stretched to a length of 20 miles. But because the stretching was uniform, the tortoise was still 10 percent of the way across, meaning it was 2 miles down the road. After another minute, it was 3 miles down the road. Then, after the road stretched to a length of 30 miles, the tortoise was 4.5 miles down the road.

But rather than focusing on the distance the tortoise traveled, you should have zeroed in on the *fraction* of the total distance it covered each minute. In the first minute, the tortoise drove 1/10 of the total distance. In the second minute, it drove 1/20. In the third minute, it drove 1/30, In the fourth minute, it was 1/40. After *N* minutes, it drove 1/10 · (1/1 + 1/2 + 1/3 + 1/4 + … + 1/*N*). As many solvers noticed, that sum inside the parentheses is none other than the *N*^{th} harmonic number! Once this sum exceeded 10, the tortoise was guaranteed to have finished the race. This happened 12,367 minutes — more than eight days — into the competition.

Calculating this was no small feat. But alas, much like the idling hare, you still had your work cut out for you.

The tortoise didn’t finish *exactly *12,367 minutes into the race — the sum of the first 12,367 terms of the harmonic series is in fact *greater than* 10. The precise time turned out to be approximately 12,366.47 minutes.

But as solvers Hector Pefo and Josh Silverman cleverly observed, you didn’t *need* to calculate this exact time. As long as the hare started running the moment the tortoise completed 20 percent of the race, they’d finish together. That’s because the hare would then travel a distance that was 25 percent longer over the same amount of time, perfectly balancing the fact that the hare traveled 25 percent faster.

At this point, you just needed to determine when the tortoise had finished 20 percent of the race. After just four minutes, when the race was 40 miles long, the tortoise had traveled about 8.33 miles — just a shade over 20 percent of the way. Since the tortoise drove 1 mile per minute, it must have passed the 20 percent mark a third of a minute (i.e., 20 seconds) prior. And so the hare took off exactly **3 minutes and 40 seconds** into the race.

You can see this epic tie in all its glory, courtesy of Allen Gu:

Solver Hypergeometricx extended the puzzle further, looking at what would happen if the road stretched *continuously* over time, rather than discretely every minute, finding that the hare would then wait *e*^{2}−1 minutes (about 6 minutes and 23 seconds) into the race.

And the moral of all this? Slow and stretchy wins the race.

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

]]>From Taylor Firman comes an opportunity to make baseball history:

This year, Major League Baseball announced it will play a shortened 60-game season, as opposed to the typical 162-game season. Baseball is a sport of numbers and statistics, and so Taylor wondered about the impact of the season’s length on some famous baseball records.

Some statistics are more achievable than others in a shortened season. Suppose your true batting average is .350, meaning you have a 35 percent chance of getting a hit with every at-bat. If you have four at-bats per game, what are your chances of batting at least .400 over the course of the 60-game season?^{10} And how does this compare to your chances of batting at least .400 over the course of a 162-game season?

*Extra credit:* Some statistics are *less* achievable in a shortened season. What are your chances of getting a hit in at least 56 consecutive games, tying or breaking Joe DiMaggio’s record, in a 60-game season? And how does this compare to your chances in a 162-game season? (Again, suppose your true batting average is .350 and you have four at-bats per game.)

*Extra extra credit:* In a 60-game season, what are your chances of *both* batting at least .400 and getting a hit in at least 56 consecutive games?

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

From Jason Shaw comes a new twist on an old fable:

The tortoise and the hare are about to begin a 10-mile race along a “stretch” of road. The tortoise is driving a car that travels 60 miles per hour, while the hare is driving a car that travels 75 miles per hour. (For the purposes of this problem, assume that both cars accelerate from 0 miles per hour to their cruising speed instantaneously.)

The hare does a quick mental calculation and realizes if it waits until two minutes have passed, they’ll cross the finish line at the exact same moment. And so, when the race begins, the tortoise drives off while the hare patiently waits.

But one minute into the race, after the tortoise has driven 1 mile, something extraordinary happens. The road turns out to be magical and instantaneously stretches by 10 miles! As a result of this stretching, the tortoise is now *2* miles ahead of the hare, who remains at the starting line.

At the end of every subsequent minute, the road stretches by 10 miles. With this in mind, the hare does some more mental math.

How long after the race has begun should the hare wait so that both the tortoise and the hare will cross the finish line at the same exact moment?

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

Congratulations to Seth Weinberg of Columbus, Ohio, Ryan Quattlebaum of Holly Springs, North Carolina, Garland Castaneda of Modesto, California and Jake Hancock of Atlanta, Georgia, winners of last week’s Riddler Express.

Last week, you played the Riddler version of the 24 Game. Your goal was to make a numerical expression that equals 24, using each of four given numbers once, along with parentheses, addition, subtraction, multiplication, division and exponentiation — that last one being a fun wrinkle.

For example, if I had given you the numbers 1, 2, 3 and 8, then valid solutions would have been 8×3×(2−1) and (3+1)×(8−2), since they both equal 24. However, concatenation was not an allowed operation, which meant (32−8)×1 was *not* a solution — that is, you couldn’t smush the 3 and the 2 together to get 32.

Given the four numbers 2, 3, 3 and 4, how could you make 24?

As it turned out, there was no way to get 24 using just parentheses, addition, subtraction, multiplication and division. There were four different classes of solutions, and every single one of them involved exponentiation:

- (3
^{2}−3)×4, which was found by Seth. - (4−2)
^{3}×3, which was found by Garland. - (4÷2)
^{3}×3, which was found by Ryan. - 4
^{(3/2)}×3, which was found by Jake.

Each of these solutions could also be rearranged to find several more. For example, (4÷2)^{3}×3 could also have been written as 3×(4÷2)^{3}, or even as 3÷(2÷4)^{3}.

I honestly thought that the last one, 4^{(3/2)}×3, would stump most solvers. After all, noninteger exponents aren’t something most people encounter every day. But I was wrong — more than 10 percent of solvers answered with 4^{(3/2)}×3 or an equivalent expression. I should never have doubted Riddler Nation!

A few solvers had some extra fun and submitted their own, even more challenging sets of numbers. How many ways can you make 24 with each of the following numbers?

- 2, 3, 10 and 10 (courtesy of Emma Knight)
- 3, 3, 8 and 8 (also courtesy of Emma Knight)
- 2, 3, 6 and 6 (courtesy of Andrew Heairet)
- 0, 0, 2 and 5 (courtesy of João Coelho)

Finally, while this seemed to be a conventional pencil-and-paper sort of riddle, a few solvers turned to their computers, having them generate every possible combination of the four numbers, the allowed operations and parentheses. For example, João’s C++ code and Scott Carr’s MATLAB code ran through the thousands of possibilities, finding every last expression that equals 24.

Scott and João further found that 37 was the smallest number you could *not* make with 2, 3, 3 and 4. Sounds like the makings of another riddle…

Congratulations to Tobias Tapirello of Budapest, Hungary, winner of last week’s Riddler Classic.

Last week, you were playing with a toy that had five rings of different diameters and a tapered column. Each ring had a “correct” position on the column, from the largest ring that fit snugly at the bottom to the smallest ring that fit snugly at the top.

When placed on the column, each ring slid down to its correct position, if possible. Otherwise, it rested on what was previously the topmost ring.

For example, if you stacked the smallest ring first, then you couldn’t stack any more rings on top. But if you stacked the second-smallest ring first, then you could stack any one of the remaining four rings above it, after which you couldn’t stack any more rings.

For example, here were four different stacks you could make:

How many unique stacks could you create using at least one ring?

First off, there were 31 stacks that resulted in each ring being in its correct position. That’s because each of the five rings could be on the stack or off it — two possibilities. So the number of such stacks was 2×2×2×2×2, or 32. Subtracting the one stack with *no* rings on it meant there were 31 unique stacks.

So then what about stacks with rings that were *not* in their correct position, but rather lying atop smaller rings? Counting up all these stacks, while being sure not to miss or double count any, got you to the heart of this puzzle.

One approach was to draw them all out, as solver Allen Gu did (with the help of his computer):

That’s all of them! It turned out there were a grand total of **65** unique stacks.

For extra credit, you were asked to find a general solution, the number of unique stacks given *N* rings. In working this out, many solvers tried their hand at stacks with fewer rings:

- With one ring, there was one possible stack.
- With two rings, there were three possible stacks.
- With three rings, there were eight possible stacks.
- With four rings, there were 22 possible stacks.

A pattern wasn’t immediately obvious. As noted by solver Meghan O’Keefe, the key was to divide the original problem up into five cases, depending on the number of rings you ultimately wanted on the stack.

There were five ways to place exactly *one* ring on the stack — one way for each ring. That wasn’t too hard. If you wanted exactly *two* rings on the stack, the first ring you placed couldn’t have been the topmost ring, because you couldn’t place a second ring above it. That meant you had four choices for the first ring. And the second ring could have been any ring except the one you already placed — again, four choices. So in all, there were 4×4, or 4^{2}, unique stacks with two rings.

What if you wanted a stack with *three* rings? Then the first ring couldn’t have been one of the top two rings, meaning it had to be one of the bottom three rings. Which ring could you have placed second? It couldn’t have been the ring you just placed, nor could it have been the topmost ring, which again left you with three rings to choose from. Which ring could you have placed third? It couldn’t have been either of the two rings you just placed, but the other three were all fair game. So in all, there were 3×3×3, or 3^{3}, unique stacks with three rings.

This pattern continued for any number of desired rings — with each ring you placed, it worked out that you still had the exact same number of rings to choose from. Including the cases with four or five desired rings, the total number of stacks was 5^{1} + 4^{2} + 3^{3} + 2^{4} + 1^{5}, which, sure enough, equals 65.

In general, given *N* rings, there are *N*^{1} + (*N*−1)^{2} + … + 2^{(N−1)} + 1^{N} unique stacks. This number grows rather quickly with *N*. And yes, these numbers have their own OEIS sequence.

I’m pleased to say there was no shortage of creative solutions to this extra credit. Josh Silverman arrived at the solution via recursion and telescoping, while Lucas Fagan impressively solved it using the inclusion-exclusion principle.

Oh, and just for fun, here are all 209 unique stacks when there are six rings (again courtesy of Allen Gu):

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

]]>The 24 Game is a fun test of mathematical fluency. In the Riddler version of the game, your goal is to make a numerical expression that equals 24, using each of four given numbers once, along with parentheses, addition, subtraction, multiplication, division and exponentiation.

For example, if I gave you the numbers 1, 2, 3 and 8, then valid solutions would be 8×3×(2−1) and (3+1)×(8−2), since they both equal 24. However, concatenation is not an allowed operation, which means (32−8)×1 is *not* a solution — that is, you can’t smush the 3 and the 2 together to get 32.

Given the four numbers 2, 3, 3 and 4, how can you make 24?

*Extra credit*: Can you find *another* way to make 24?

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

From Austin Shapiro comes a story of stacking that may stump you:

Mira has a toy with five rings of different diameters and a tapered column. Each ring has a “correct” position on the column, from the largest ring that fits snugly at the bottom to the smallest ring that fits snugly at the top.

Each ring she places will slide down to its correct position, if possible. Otherwise, it will rest on what was previously the topmost ring.

For example, if Mira stacks the smallest ring first, then she cannot stack any more rings on top. But if she stacks the second-smallest ring first, then she can stack any one of the remaining four rings above it, after which she cannot stack any more rings.

Here are a four different stacks Mira could make:

This got Mira thinking. How many unique stacks can she create using at least one ring?

*Extra credit*: Instead of five rings, suppose the toy has *N* rings. Now how many unique stacks can Mira create?

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

Congratulations to Bram Carlson of Fort Lauderdale, Florida, winner of last week’s Riddler Express.

Last week, my swimming pool was opening soon, with its five swimming lanes (and no general swim area). It seemed likely that social distancing regulations would prevent swimmers from occupying adjacent lanes.

I wanted to know what would happen if a queue of swimmers arrived at the pool when it opened at 9 a.m. One at a time, each person randomly picked a lane among the lanes that were available (i.e., the lane had no swimmer already and was not adjacent to any lanes with swimmers), until no more lanes were available.

At this point, what was the expected number of swimmers in the pool?

This was a simplified version of a previous Riddler, which looked at the general case of *N* lanes — last week’s extra credit — but in the context of misanthropes and houses.

First off, it was helpful to label the lanes — say, 1, 2, 3, 4 and 5, in that order. Whenever the first swimmer chose lane 2, the second swimmer had to choose either lane 4 or 5. At this point, there would be no more available lanes, meaning there were exactly two swimmers in total. This same reasoning applied to lane 4 as well.

But if the first swimmer happened to choose lane 3, the second and third swimmers would choose lanes 1 and 5, meaning there would be exactly three swimmers in total.

A trickier case was when the first swimmer chose lane 1. The second swimmer could then choose from lanes 3, 4 and 5, each with a one-third probability. If the second swimmer chose lane 3 or lane 5, then there would be three swimmers in total (lanes 1, 3 and 5). But if the second swimmer chose lane 4, then once again there would be no more available lanes, meaning there would be two swimmers in total. And when the first swimmer chose lane 5, the math worked out the same, with the second swimmer choosing from lanes 1, 2 and 3. So when the first swimmer chose lane 1 or 5, there was a two-thirds chance of having three swimmers and a one-third chance of having two swimmers — an average of 8/3 swimmers.

Overall, that meant there was a 40 percent chance of two swimmers (when the first swimmer chose lane 2 or 4), a 20 percent chance of three swimmers (when the first swimmer chose lane 3) and a 40 percent chance of 8/3 swimmers (when the first swimmer chose lane 1 or 5). Putting these probabilities together, the average number of swimmers was **37/15**, or about 2.467.

And what if there are more than five lanes? In the best-case scenario, about half of the lines would be occupied — when swimmers are in alternating lanes. And in the worst-case scenario, about a third of lanes would be occupied — when swimmers have two empty lanes on either side of them. As previously explored on this column, the exact answer is **(1− e^{−2})/2**, or about 0.432, which falls neatly between the extremes.

To see where this limiting behavior comes from, check out the explanations from Josh Silverman and Enrique Treviño.

And if you ever happen to be first in line at my local pool, be a good sport and choose lane 3.

Congratulations to Josh Rosenbluth of Oxnard, California, winner of last week’s Riddler Classic.

Last week, you were arranging the stars on the American flag. The 50 stars on the current flag form two rectangles: The larger rectangle is 5 stars wide, 6 stars long; the smaller rectangle is embedded inside the larger and is 4 stars wide, 5 stars long. This square-like pattern of stars is possible because the number of states (50) is *twice* a square number (25).

Now that the House of Representatives has passed legislation that would make the District of Columbia the 51st U.S. state, a natural question was how to aesthetically arrange *51* stars on the flag.

One pleasing design had a star in the middle, surrounded by concentric pentagons of increasing side length. The innermost pentagon had five stars, and subsequent pentagons were made up of 10, 15 and 20 stars. All told, that’s 51 stars.

It just so happened that when *N* equaled 50, *N* was twice a square and *N*+1 was a centered pentagonal number. After 50, what was the next integer *N* with this property?

First off, any number that was twice a square could be written as 2*x*^{2}, where *x* is an integer. But what about centered pentagonal numbers? For example, the seventh centered pentagonal number is 1 + 5 + 10 + 15 + 20 + 25 + 30. Other than the 1, every other term in the sum is a multiple of 5. Factoring out that 5 gives us 1 + 5 ᐧ (1 + 2 + 3 + 4 + 5 + 6). Now that expression inside the parentheses is something we can work with — it’s the sixth triangular number! For many solvers, this connection between centered pentagonal and triangular numbers unlocked a formula: Centered pentagonal numbers can be written as 1 + 5*y*(*y*+1)/2, where y is an integer.

Now all you had to do was find integers *x* and *y* such that 2*x*^{2} was one less than 1 + 5*y*(*y*+1)/2, meaning 2*x*^{2} = 5*y*(*y*+1)/2, or 4*x*^{2} = 5*y*(*y*+1). Right around this point, most solvers started cranking through values on their spreadsheets or writing some code to do this for them. But there were a few brave souls who tried to solve this analytically.

Solver Grant Larsen noticed that the left side of that last equation was the square of *2x* — an integer — which meant 5*y*(*y*+1) was also a square. Since 5 is prime and *y* and *y*+1 are guaranteed to be relatively prime, there were two possible cases: either (1) *y* and 5(*y*+1) were both square numbers, or (2) *y*+1 and 5*y* were both square numbers.

The first value of *y* to satisfy one of these cases — the former, as it turned out — was 4, since 4 and 5(4+1) are both squares. But this corresponded to *N* = 50, the value given in the problem. Trying out different squares, Grant found that the next solution — which happened to satisfy the latter case — was when *y* was 80, since 5(80) and 80+1 are both squares. This corresponded to *N* = **16,200**.

You read that right. The next time our flag can go from its current square-like arrangement to a centered pentagonal number is when the 16,201st state joins the Union.

Here’s what the flag would look like:

Now it’s true that the above solution had an element of “guess and check.” Some solvers, like Gareth McCaughan and Laurent Lessard (admittedly out of his comfort zone with this puzzle) ventured deeper into number theory and into the realm of Pell equations. These are well studied, and Gareth and Laurent used them to find a closed-form solution for all such values of *N*. Solver Emma Knight further extended the problem, looking for square-like numbers that were one *greater* than a centered pentagonal number.

Oh, and in case you were curious, the next value of *N* that worked was 5,216,450. And after that, it was 1,679,680,800.

Even the United Federation of Planets doesn’t have a flag with that many stars.

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

]]>It’s summertime and my local swimming pool, which has exactly five swimming lanes (and no general swim area), may be opening in the coming weeks. It remains unclear what social distancing practices will be required, but it’s quite possible that swimmers will not be allowed to occupy adjacent lanes.

Under these guidelines, the pool could accommodate at most three swimmers — one each in the first, third and fifth lanes.

Suppose a queue of swimmers arrives at the pool when it opens at 9 a.m. One at a time, each person randomly picks a lane from among the lanes that are available (i.e., the lane has no swimmer already and is not adjacent to any lanes with swimmers), until no more lanes are available.

At this point, what is the expected number of swimmers in the pool?

*Extra credit*: Instead of five lanes, suppose there are *N* lanes. When no more lanes are available, what is the expected number of swimmers in the pool?

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

Just in time for the Fourth of July, this week’s Classic is about stars on the American flag:

The 50 stars on the American flag are arranged in such a way that they form two rectangles. The larger rectangle is 5 stars wide, 6 stars long; the smaller rectangle is embedded inside the larger and is 4 stars wide, 5 stars long. This square-like pattern of stars is possible because the number of states (50) is *twice* a square number (25).

Now that the House of Representatives has passed legislation that would make the District of Columbia the fifty-first US state — and renamed Washington, Douglass Commonwealth, in honor of Frederick Douglass — a natural question is how to aesthetically arrange 51 stars on the flag.

One pleasing design has a star in the middle, surrounded by concentric pentagons of increasing side length, as shown below. The innermost pentagon has five stars, and subsequent pentagons are made up of 10, 15 and 20 stars. All told, that’s 51 stars.

It just so happens that when *N* equals 50, *N* is twice a square and *N*+1 is a centered pentagonal number. After 50, what is the next integer *N* with these properties?

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

Congratulations to Josiah Kollmeyer of Baton Rouge, Louisiana, winner of last week’s Riddler Express.

Last week, you were driving north in Riddler City, whose streets run north-south and east-west. At every intersection, you randomly turned left or right, each with a 50 percent chance.

After driving through 10 intersections, what was the probability that you were still driving north?

One way to solve this was to look at what happened after the first few intersections. It wasn’t long before a pattern emerged.

After the first intersection, you had a 50 percent chance of turning east and a 50 percent chance of turning west. In the event you turned *east*, upon reaching the second intersection, you then had a 50 percent chance of turning north and a 50 percent chance of turning south. But in the event you turned *west*, you *still* had a 50 percent chance of turning north and a 50 percent chance of turning south. Putting these possibilities together, that meant that after the second intersection, you were definitely driving north or south, each with a 50 percent chance.

Following this reasoning, solver Lily Koffman realized that after every odd number of intersections, you were driving east or west with equal probability, and after every *even* number of intersections, you were driving north or south with equal probability. Since the given number of intersections, 10, was even, that meant there was a **50 percent** chance you were driving north in the end.

For extra credit, instead of just turning left or right, you now also had the option of driving straight — each with a one-third chance. After driving through 10 intersections, *now* what was the probability that you were still driving north?

This was certainly a trickier scenario. Mike Bourdaa solved it by looking at what would happen if there were fewer intersections, hoping to find a pattern. After *N* intersections, there were a total of 3^{N} equally likely sequences of turns you could make. Mike found that approximately 3^{N}/4 of these sequences — or, more precisely, the smallest whole number greater than 3^{N}/4 — resulted in you driving north.

But that certainly wasn’t the only approach that worked here. Juan Casaravilla solved it using linear algebra, by first lining up the probabilities of driving north, south, east or west into a vector, which was initially [1; 0; 0; 0]. Each intersection could be modeled as multiplying this vector by the transition matrix [1/3 0 1/3 1/3; 0 1/3 1/3 1/3; 1/3 1/3 1/3 0; 1/3 1/3 0 1/3], where the output of this multiplication was a new probability vector that revealed your updated chances of driving north, south, east or west.

But why would you ever want to go through the hassle of encoding an intersection as a matrix in the first place? Well, if each intersection was equivalent to multiplying by a matrix, then driving through 10 intersections was equivalent to multiplying by 10 identical copies of that matrix — or, better yet, multiplying by the matrix *raised to the 10th power*, an operation that any computer can do with ease.

It turned out that your chances of driving in each of the four directions rapidly approached 25 percent. After 10 intersections, your chances of driving north stood at precisely **4,921/19,683**, or about 25.00127 percent.

Just to be extra sure, a few solvers went ahead and checked their work via computer simulation. Daniel Silva-Inclan ran 1,000 trials and verified that after 10 intersections, the probabilities of driving in all four directions were very close to 25 percent.

There’s definitely a life lesson here. If you ever find yourself approaching an intersection and you really want to come out of it driving in a random direction — but you know that pulling a U-turn is illegal — then don’t lose heart. Instead, randomly drive straight or turn left or right at each intersection for a few minutes. Before you know it, you’ll be driving in a random direction.

Yes, that was definitely an important life lesson.

Congratulations to Eli Wolfhagen of Brooklyn, New York, winner of last week’s Riddler Classic.

Last week, Polly Gawn was playing “connect the dots.” She specifically wanted to connect six dots so that they formed the vertices of a hexagon. To her surprise, she found that there were many different hexagons she could draw, each with the same six vertices.

What was the *greatest* possible number of unique hexagons Polly could draw using six points?

If the question had ended there, then there would have been some ambiguity around the use of the word “hexagon.” If self-intersecting hexagons (meaning the edges coincided or crossed each other) were allowed, then all you had to do was count how many ways she could connect the six points, one at a time. She could pick any starting point and then choose among the remaining five points to connect it to, then pick four points, then three, then two, and finally one. But wait — each polygon was the same whether she traversed its points in one direction or the other, so she had double counted. Dividing by two, that meant there were 60 total hexagons Polly could have drawn.

But if you read the hint, you saw that Polly was only counting simple hexagons — that is, hexagons whose sides didn’t cross. This was decidedly harder, but was within the realm of possibility if you had a weekend to kill and a bountiful supply of scratch paper.

Many solvers came close, missing a few potential hexagons. In the end, it was three points that all lay within the triangle formed by the other three points that produced the most polygons. Solvers Laurent Lessard and Emma Knight both arrived at this arrangement by leaning on the work of Oswin Aichholzer. They looked at “untangled” complete graphs that had the minimal number of intersections, arguing that this would result in the maximal number of polygons.

Without further ado, here is one such arrangement of six points and all the polygons it produces:

Polly could draw at most **29 hexagons** — a far cry from the upper bound of 60.

For extra credit, you were asked to find the greatest possible number of unique heptagons Polly could draw using seven points. In this case, the upper bound was 420, suggesting this was beyond the realm of pencil, paper and patience. Indeed, the greatest possible number of heptagons was **92**.

As some solvers noted, there’s an OEIS sequence for that — sequence A063546, to be exact, which lists the “largest number of crossing-free Hamiltonian cycles of *n* points in the plane.” Or as Polly likes to call them, polygons. (Yes, *simple* polygons.)

According to the sequence, Polly could draw at most 339 octagons (drawn below, courtesy of solver Josh Silverman), 1,282 nonagons and 4,994 decagons.

While the precise pattern for this sequence remains unknown, it *is* known to grow exponentially. Erik Demaine has followed the literature on this problem and tracked the evolution of the upper and lower bounds. As of 2011, it was known that as the number of points *n* increases, the maximal number of polygons is somewhere between 4.642^{n} and 56^{n}.

Let’s not overlook that many of the polygons from the above animations would make awesome corporate logos. If your mountain biking startup is in need of some edgy graphic design, call me.

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

]]>It turns out, our staffers have very strong opinions on books — how many books one should own, what kinds of books, and most importantly, how to best organize those books on a bookshelf. With that in mind, we hope to answer that very question on this episode of Debate Club.

What’s your preferred organizational method? Be sure to tell us on YouTube or Twitter. And while you’re there, let us know what you’d like us to debate in the next episode.

]]>In Riddler City, the city streets follow a grid layout, running north-south and east-west. You’re driving north when you decide to play a little game. Every time you reach an intersection, you randomly turn left or right, each with a 50 percent chance.

After driving through 10 intersections, what is the probability that you are still driving north?

*Extra credit*: Now suppose that at every intersection, there’s a one-third chance you turn left, a one-third chance you turn right and a one-third chance you drive straight. After driving through 10 intersections, *now *what’s the probability that you are still driving north?

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

Polly Gawn loves to play “connect the dots.” Today, she’s playing a particularly challenging version of the game, which has six unlabeled dots on the page. She would like to connect them so that they form the vertices of a hexagon. To her surprise, she finds that there are many different hexagons she can draw, each with the same six vertices.

What is the *greatest* possible number of unique hexagons Polly can draw using six points?

(*Hint:* With four points, that answer is three. That is, Polly can draw up to three quadrilaterals, as long as one of the points lies inside the triangle formed by the other three. Otherwise, Polly would only be able to draw one quadrilateral.)

*Extra Credit*: What is the greatest possible number of unique heptagons Polly can draw using seven points?

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

Congratulations to Emma Knight of Toronto, Canada, winner of last week’s Riddler Express.

Last week, I had a coin with a sun on the front and a moon on the back. I claimed that on most days it was a fair coin. But once a year, on the summer solstice, the coin always came up the opposite side as the previous flip.

Of course, you were skeptical of my claim. You initially figured there was a 1 percent chance that the coin was magical and a 99 percent chance that it was just an ordinary fair coin. You then asked me to “prove” the coin was magical by flipping it some number of times.

How many successfully alternating coin flips would it have taken for you to think there was a 99 percent chance the coin was magical (or, more likely, that I had rigged it so it always alternated)?

Wow, there was a lot of disagreement about this problem! About 40 percent thought the answer was eight flips, 30 percent thought the answer was seven, 12 percent thought the answer was 15 and 8 percent thought the answer was 14. (The remaining 10 percent gave answers ranging from zero to 100.) But before revealing the correct result, let’s take a closer look at what was happening with these coin flips.

Suppose my first flip happened to be a sun, my second flip was a moon, and the outcome continued alternating with every subsequent flip. The more alternating flips I got in a row, the more confident you’d become that the coin was magical.

To quantify this level of confidence, you needed an assist from Bayes’ theorem: Given *N* alternating flips in the row, the probability that the coin was magical was equal to the probability that a magic coin would generate *N* alternating flips, multiplied by the prior probability that the coin was magical, then divided by the prior total probability of getting *N* alternating flips.

Some of these were easier to calculate than others. For example, the probability that a magical coin would generate *N* alternating flips was 1. (That’s what it meant for the coin to be magical; the flips would *always* alternate.) And the prior probability that the coin was magical — prior to any flipping to *prove* it was magical — was given in the statement of the problem: 1 percent, or 0.01. The challenge was to determine the final value to plug into Bayes’ theorem: the prior total probability of getting *N* alternating flips.

On the (1 percent) off chance the coin was magical, you would always get *N* alternating flips. But even if the coin wasn’t magical (the remaining 99 percent of the time), you could still get *N* alternating flips, with a probability that decreased with *N*. In fact, the probability was 1/2^{N−1}, since the first flip could be a sun or a moon, and the remaining *N*−1 flips each had a 50 percent chance of being different from the previous flip. All together, that meant the total prior probability of getting *N* alternating flips was 0.01+0.99/2^{N−1}.

At last, with all that deep thought behind us, it’s time to plug and chug. Given *N* alternating flips in a row, the probability that the coin was magical was 0.01/(0.01+0.99/2^{N−1}). Now your task was to find the first value of *N* where this probability exceeded 0.99, meaning you could be 99 percent sure the coin was magical.

Starting with the inequality 0.01/(0.01+0.99/2^{N−1}) ≥ 0.99, some careful rearrangement led to the simplified inequality 2^{N−1} ≥ 99^{2}. The smallest power of two that exceeded 99^{2}, or 9,801, was 2^{14}, or 16,384. While many thought the answer was 14, it was in fact *N*−1 that equalled 14, meaning the answer was **15 flips**.

And in calculating that tricky prior total probability of getting *N* alternating flips, if you forgot to account for the 1 percent of the time the coin was magic, you got the incorrect answer of eight flips (or seven, if you again forgot to add one to the exponent). Or you might have discarded the prior probability altogether and looked for the first power of 1/2 that was less than 0.01.

No doubt this was a challenging Express. And in case you were curious, yes, the coin really was magical.

Congratulations to Eric Widdison of Kaysville, Utah, winner of last week’s Riddler Classic.

Last week, King Auric was passing his most prized possession — perfect spheres of solid gold — to his three children. He had spheres with diameters of 1 centimeter, 2 centimeters, 3 centimeters, and so on. To be fair to his children, he wanted to give each an equal weight of gold.

After much trial and error, the king managed to divide his spheres into three groups of equal weight. He was further amused when he realized that his collection contained the *minimum* number of spheres needed for such a division. How many golden spheres did King Auric have?

An important detail from the puzzle was that the spheres were solid gold, meaning their weight was proportional to the cube of their diameter. In other words, you had to find a whole number *N* such that 1^{3}, 2^{3}, 3^{3}, … , *N*^{3} could be partitioned into three groups with the same sum.

A good place to start was to find the weights of each of those three groups. Summing the cubes from 1^{3} to *N*^{3} always equals *N*^{2}(*N*+1)^{2}/4, and since each group had a weight that was one-third of the total, the weight of each group was therefore *N*^{2}(*N*+1)^{2}/12.

You also knew that the largest sphere must have belonged to one of the three groups, and clearly its weight couldn’t be greater than the sum of its corresponding group. In other words, *N*^{3} had to be less than or equal to *N*^{2}(*N*+1)^{2}/12. With a little algebraic manipulation, this meant that *N* had to be 10 or greater.

Also, in order to evenly split the spheres into three groups, you knew that the total sum of the weights, *N*^{2}(*N*+1)^{2}/4, had to be divisible by three. That meant *N* had to either be a multiple of three or two more than a multiple of three. So possible values of *N* were now 11, 12, 14, 15, 17, 18, etc.

At this point, solver after solver turned to their computer for assistance. As this week’s winner Eric Widdison noted, this problem appeared to be a variant of the famous knapsack problem, which is NP-complete. So it was computers or bust.

Nevertheless, Eric was able to find that the smallest possible number of spheres was **23**. The fair division of the spheres was {1, 4, 7, 8, 12, 16, 20, 22}, {2, 5, 9, 11, 14, 15, 17, 23} and {3, 6, 10, 13, 18, 19, 21}. For all three groups, the sum of the cubes was 76,176. That’s a whole lotta gold!

Last week’s extra credit asked you to find the smallest number of spheres that could be evenly shared among other numbers of children, like two, four, five, six, etc.

Solver Allen Gu reported several of these results, along with how long it took his computer to generate each result. With two children, King Auric would need 12 spheres; with four or five children, he would need 24 spheres; with six children, he would need 35 spheres; with seven children, he would need 41 spheres, and with eight children, he would need 47 spheres. At this point, Allen’s computer failed him, and he upgraded to a “beefier” virtual machine. After several hours, he found that with nine children, King Auric would need 53 spheres. Allen is still waiting to hear back from his computer in the case of 10 children.

These results formed a rather unpredictable sequence of integers: 1, 12, 23 (this week’s answer), 24, 24, 35, 41, 47, 53, and so on. And now, thanks to the puzzle’s submitter, Dean Ballard, this sequence has been enshrined for all eternity in the On-line Encyclopedia of Integer Sequences as sequence A330212. While he was there, Dean apparently updated the related sequence, A240070, which looks at splitting up powers other than cubes into three equal groups.

As for the computational difficulty in cranking out these integer sequences, Emma Knight put it best, concluding that “[a]t this point I (and my poor CPU) would advise the royal couple to ease up on their procreation habit.”

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

]]>I have a coin with a sun on the front and a moon on the back. I claim that on most days, it’s a fair coin, with a 50 percent chance of landing on either the sun or the moon.

But once a year, on the summer solstice, the coin absorbs the sun’s rays and exhibits a strange power: It always comes up the opposite side as the previous flip.

Of course, you are skeptical of my claim. You figure there’s a 1 percent chance that the coin is magical and a 99 percent chance that it’s just an ordinary fair coin. You then ask me to “prove” the coin is magical by flipping it some number of times.

How many successfully alternating coin flips will it take for you to think there’s a 99 percent chance the coin is magical (or, more likely, that I’ve rigged it in some way so it always alternates)?

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

From Dean Ballard comes a riddle of radiant spheres and fatherhood, just in time for the summer solstice and Father’s Day:

King Auric adored his most prized possession: a set of perfect spheres of solid gold. There was one of each size, with diameters of 1 centimeter, 2 centimeters, 3 centimeters, and so on. Their brilliant beauty brought joy to his heart. After many years, he felt the time had finally come to pass the golden spheres down to the next generation — his three children.

He decided it was best to give each child precisely one-third of the total gold by weight, but he had a difficult time determining just how to do that. After some trial and error, he managed to divide his spheres into three groups of equal weight. He was further amused when he realized that his collection contained the *minimum* number of spheres needed for this division. How many golden spheres did King Auric have?

*Extra credit:* How many spheres would the king have needed to be able to divide his collection among other numbers of children: two, four, five, six or even more?

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

Congratulations to Christopher R. Green of Oxford, Mississippi, winner of last week’s Riddler Express.

Last week, you tried your hand at a technique for rolling dice called “bowling,” in which you placed your index finger and thumb on two opposite sides of the die and rolled it along the table. When done correctly, the die never landed on the faces on which you held the die, leaving you with a 25 percent chance of landing on each of the remaining four faces.

You had to apply this technique to optimize your chances of rolling a 7 or 11 in a game of craps. With a standard rolling technique, your chances were about 22.2 percent. But if you bowled the dice one at a time (i.e., you knew the outcome of the first die before rolling the second), what were your chances of rolling a 7 or 11?

If your first roll had been a 1, 2, 3 or 4, then the only way to win was to have the two rolls add up to 7 — 11 would have been out of reach, since the maximum roll was 6. But if your first roll was a 5, then you could win with a second roll that was a 2 or 6. And if your first roll was a 6, then you could win with a second roll that was a 1 or 5. So with the first roll, you wanted to get a 5 or 6, since that doubled your chances of winning.

At this point, it was helpful to write out the three possible strategies, which we’ll call A, B and C, keeping in mind that opposing faces of a standard die always add up to 7:

**A**: Place your fingers on the 1 and the 6, resulting in a 25 percent chance of rolling a 2, 3, 4 or 5.**B**: Place your fingers on the 2 and the 5, resulting in a 25 percent chance of rolling a 1, 3, 4 or 6.**C**: Place your fingers on the 3 and the 4, resulting in a 25 percent chance of rolling a 1, 2, 5 or 6.

As noted by solver Carolyn Phillips, strategy C gave you the best chances of rolling a 5 or 6, so that should have been your first roll. And the result of your first roll determined your strategy for the second roll:

- If you rolled a 1, then you needed a 6. Strategies B and C both had a 25 percent chance of resulting in a 6.
- If you rolled a 2, then you needed a 5. Strategies A and C both had a 25 percent chance of resulting in a 5.
- If you rolled a 5, then you needed a 2 or 6. Strategy C had a 50 percent chance of resulting in a 2 or 6.
- If you rolled a 6, then you needed a 1 or 5. Strategy C had a 50 percent chance of resulting in a 1 or 5.

Averaging these four equally likely cases together, your chances of winning were 3/8, or **37.5 percent**. That was quite the improvement over the 22.2 percent chance you had with a standard rolling technique.

For extra credit, you still wanted to roll a 7 or 11 (earning you a point), but you also wanted to avoid rolling a 2, 3 or 12 (losing you a point). With a standard rolling technique, your average score would have been one-ninth of a point. But if you “bowled” to maximize your expected score, what was that average?

Once again, you could determine the optimal strategy by determining your second roll based on your first. For example, if your first roll was a 6, you wanted your second roll to be a 1 or a 5 (giving you a total of 7 or 11), but not a 6 (giving you a total of 12). Your best bets for the second roll would have been strategies A or C; either netted you 0.25 points on average.

If you did this analysis for each of the first rolls, you found that if your first roll was a 1, you’d get 0 points on average. If your first roll was a 2, 3, 4 or 6, you’d get 0.25 points on average. And if you were fortunate enough to get a 5 on your first roll, you’d get 0.5 points on average.

Putting these all together, your best option for the *first *roll was Strategy A, netting you an average of **5/16 points**. That was about three times better than how you would have done with standard rolling.

Solver David Alpert took this problem one step further, wondering what would happen if you bowled non-standard dice where the numbers on opposing sides did *not* have to add up to 7. Based on his analysis, your average point total jumped up to 3/8 — the exact same result as the original problem.

If there’s a lesson to be learned in all this, it’s that it pays to cheat in dice games. No, that can’t be right.

Congratulations to Dan Upper of Corvallis, Oregon, winner of last week’s Riddler Classic.

Last week, you were studying a new strain of bacteria, *Riddlerium classicum*. Each *R. classicum* bacterium did one of two things: split into two copies of itself or die. There was an 80 percent chance of the former and a 20 percent chance of the latter.

If you started with a single *R. classicum* bacterium, what was the probability that it would lead to an everlasting colony (i.e., the colony would theoretically persist for an infinite amount of time)?

Sometimes a puzzle is so good that it’s worth solving twice. After posting this puzzle last week, I learned that a very similar variation — the extra credit, in fact, in which the 80 percent was replaced by probability *p* — previously appeared on the Riddler column. It’s also on page 83 of “The Riddler” book.

Nevertheless, it’s a truly excellent puzzle, and I would like to acknowledge a few solvers from this past week who had never seen it before.

First off, the first bacterium had a 20 percent chance of dying outright, which meant the answer was at most 80 percent. After that, things got hairy.

As with many Riddler Classics, it was tempting to start with simulations in order to gain some intuition for what was happening. However, how could a finite simulation truly determine whether a colony was “everlasting?” Solver Greg Y. and a team who identified as the “MassMutual Crew” both tried their hand at this, each running 1 million simulations and seeing how many times the colony lasted 100 generations. It turned out that approximately 750,000 of those 1 million colonies made it, suggesting the answer was close to 75 percent.

Meanwhile, Jason Ash modeled the problem as an absorbing Markov chain and assumed that once the colony reached a population of 500 it was effectively guaranteed everlasting survival. With this approach, along with some code to construct the transition matrix, Jason approximated the answer as 75.00000000000211 percent.

Emma Knight solved the problem analytically and head-on, directly solving for the probability that the colony dies out with each generation and then summing all those infinite probabilities. That sum is the total probability the colony dies out, so 1 minus the sum is the probability of survival.

But many solvers, including Alain Bruguières, Hector Pefo, Josh Silverman and the international team (from Manila, Philippines and Ottawa, Canada) of Erin and Nicole, discovered a roundabout method that got to the answer in far fewer steps. They first defined *x* as the probability of extinction, when starting from a single bacterium. Without yet knowing exactly what *x* was, you had enough information to determine the probability that *two* such bacteria would go both extinct. Since they were independent events, this probability was *x*^{2}.

Returning to our single bacterium, what was the probability it led to an everlasting colony? By definition, it was 1−*x*. But it was *also* the probability that it divided into two bacteria, which did not *both* go extinct. And that probability was 0.8(1−*x*^{2}). Setting those probabilities equal gave the equality 1−*x* = 0.8(1−*x*^{2}), which meant *x* was equal to 0.25 and 1−*x* was 0.75. Frankly, it was remarkable that any solvable equation popped out at all, given the infinities inherent to this puzzle. And sure enough, the answer turned out to be **75 percent**, just as the simulations predicted.

For extra credit, you were asked to solve the problem when the probability each bacterium divided was *p*, rather than 80 percent. The above reasoning worked just as well for this general case, leading to the equation 1−*x* = *p*(1−*x*^{2}). When *p* was between 0.5 and 1, the answer was 2−1/*p*. But when *p* was less than 0.5 (i.e., when the value of 2−1/*p* dipped negative), *x* was 1, meaning the colony never survived.

This was a truly bizarre result. Imagine two strains of bacteria, one with a 50 percent chance of dividing and the other with a 50.0000000001 percent chance of dividing. Only the latter has any chance of forming an everlasting colony. Until I apply some antibacterial soap, that is.

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

]]>From Dave Mattingly comes a high-rolling question of craps:

There’s a technique for rolling dice called “bowling,” in which you place your index finger and thumb on two opposite sides of the die and roll it along the table. If done correctly, the die will never land on the faces on which you were holding the die, leaving you with a 25 percent chance of landing on each of the remaining four faces.

You’d like to apply this technique to improve your chances of winning a simplified game of craps, in which your goal is simply to roll a 7 or 11 using two dice. With a standard rolling technique, your chances of rolling a 7 or 11 are 2/9, or about 22.2 percent.

Now suppose you’re using your bowling technique, and you roll the dice one at a time (i.e., you know the outcome of the first die before rolling the second). If you play to maximize your chances of rolling a 7 or 11, what will be your chances of winning?

*Extra credit:* Suppose you get one point for rolling a 7 or 11, but now you *lose* a point for rolling a 2, 3 or 12. With a standard rolling technique, you’d average 1/9 of a point. But if you “bowl” to maximize your expected score, what will that average be?

From Adam Wagner comes a curious case of colonies:

You are studying a new strain of bacteria, *Riddlerium classicum* (or *R. classicum*, as the researchers call it). Each *R. classicum* bacterium will do one of two things: split into two copies of itself or die. There is an 80 percent chance of the former and a 20 percent chance of the latter.

If you start with a single *R. classicum* bacterium, what is the probability that it will lead to an everlasting colony (i.e., the colony will theoretically persist for an infinite amount of time)?

*Extra credit:* Suppose that, instead of 80 percent, each bacterium divides with probability *p*. Now what’s the probability that a single bacterium will lead to an everlasting colony?

Congratulations to Paul Berger of Forest Hills, New York, winner of last week’s Riddler Express.

Last week, the astronomers on planet Xiddler, having recently invented the telescope, discovered a new planet in their solar system!

Like Earth, Xiddler orbits its star in a nearly circular path, with an average distance of 150 million kilometers. But *unlike* Earth, there weren’t any other known planets in the solar system … until now.

Moments after the Xiddlerian sun set below the horizon, three astronomers used their telescopes to find the new planet at the zenith of the evening sky. They then raced to Xiddler’s Grand Minister to deliver the news.

The first astronomer said that the newly discovered planet orbited their sun with a radius of 50 million kilometers; the second astronomer said the orbital radius was 300 million kilometers; the third astronomer said the orbital radius was 150 million kilometers.

Which astronomer should the Grand Minister have believed?

The keys to this riddle were the relative locations of Xiddler’s sun and the new planet. While one was near the horizon (the equator of the celestial sphere), the other was at the zenith (the north pole of the celestial sphere). That meant the angle between the sun, Xiddler and the new planet had to be 90 degrees. In other words, as noted by solver (and astrophysicist) Megan Pickett, the newly discovered planet was at quadrature.

If you’re not convinced, then here’s a helpful visual, courtesy of solver Travis Bishop:

On the left is the sun, while Xiddler is in the bottom right. Because it was sunset, that meant the astronomers were on Xiddler’s terminator, which was perpendicular to the line between Xiddler and the sun. The new planet was observed at the zenith, and so the sun, Xiddler and the new planet formed a right triangle.

The distance between Xiddler and the sun — 150 million kilometers — was one leg of this right triangle. Meanwhile, the distance between the new planet and the sun was the hypotenuse, which, by the Pythagorean theorem, had to be greater than 150 million kilometers.

The only astronomer who calculated an orbit greater than 150 million kilometers was **the second astronomer, who said the orbit was 300 million kilometers**. She was the one the Grand Minister should have believed.

By the way, this math works just as well in our own solar system as it does on planet Xiddler. For example, if you’ve ever wondered why Venus is called a “morning star” or “evening star,” it’s because Venus’s orbit lies within the Earth’s orbit, so it always appears relatively close to the sun in the sky. You can’t see any stars or planets when the sun is directly overhead — you can only see Venus when the sun is just below the horizon, at dawn or in the evening.

Anyway, the work of the Xiddlerian astronomers has only just begun. I’m sure we’ll be hearing from them again soon…

Congratulations to Harel Dor of Sunnyvale, California, winner of last week’s Riddler Classic.

Last week, you were filling in a sign’s letters by drawing horizontal lines with a marker. This marker had a flat circular tip with a radius of 1 centimeter, and you were holding the marker so that it was upright, perpendicular to the sign.

Since the diameter of the marker’s tip was 2 centimeters, you decided to fill in the letters by drawing lines every 2 centimeters. However, this was the pattern you got, with apparent gaps between the strokes:

Of course, if you drew many lines all bunched together, you’d have a more uniform shading.

But you didn’t have all day to make the sign. If the lines couldn’t overlap by more than 1 centimeter — the radius of the marker tip — what should this overlap have been, in order to achieve a shading that was as uniform as possible? And how uniform was this shading?

Before we even discuss the first solution, it’s worth taking a minute to better understand the problem. You might have thought that lines separated by the width of the marker would be nice and uniform. So why were there gaps between the strokes in the first place?

The answer came from the shape of the marker’s tip — a circle. Imagine moving a circular tip along a sign at a constant speed. For any given spot on the side, the amount of ink present is proportional to how long the marker was in contact with that point. Points along the midline of a stroke are in contact with the marker the longest, while points that are 1 centimeter away from the midline are only in contact for an instant.

If you looked at a vertical slice of the sign and measured the relative amount of ink along that line, you got a pattern of repeating semicircular “hills”:

As the strokes began to overlap and move closer together, the “valleys” between them combined until they overtook the hills, as shown in the animation below:

By the end of the animation, when the overlap was a full centimeter, the distribution of ink was once again quite bumpy. But somewhere along the way, the variance in the distribution of the ink was minimized. And if you thought understanding that there even was a minimum was a challenge, *finding *that minimum was even more challenging.

Solvers Emma Knight and David Zimmerman both used calculus to measure the standard deviation of the ink as a function of the overlap between strokes. Along the way, they both encountered some rather nasty elliptic integrals, turning to their computers for a solution. If the height of each original hill of ink was 1, then the minimum standard deviation of roughly 0.0859 occurred when the overlap between strokes was approximately **0.308 centimeters**.

Josh Silverman also found the same solution via a computational approach, while Jason Ash arrived at a similar result by (reasonably) assuming the sign had a finite number of horizontal strokes.

Instead of minimizing the standard deviation, some solvers instead minimized the *relative* standard deviation (also known as the coefficient of variation), which is the ratio of the standard deviation to the mean. This was minimized when the overlap between the strokes was approximately **0.319 centimeters**: the standard deviation was about 0.08615 and the mean was about 0.9345, which meant the relative standard deviation was roughly 0.0922. Given the ambiguity in the problem, I accepted this answer as equally correct. (Even if you subtracted either of these answers from 2, the diameter of the marker tip, I *still* counted it as correct.)

So what does this optimally uniform shading look like, in the end? Feast your eyes!

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

]]>Of course, it’s not *just* black celebrities or *just* black former presidents who have spoken out — lots of Americans across races and parties have voiced concern about Floyd’s killing and what it says about the nation and its policies. And part of the black reaction can probably be explained by partisanship. The general position of the Democratic Party is that Floyd’s race was a factor in his death and that the criminal justice system is biased against black people. So it’s not surprising that black Americans are taking that view, as about 90 percent of them have voted for Democratic candidates in recent national elections.

Another factor, however, likely explains the collective black response to what has been happening in America over the past two weeks: The overwhelming majority of black Americans view their racial identity as a core part of their overall identity, and this black identity and kinship with other black people has likely been heightened by Floyd’s killing and the resulting debate over the status of black people in the United States.

About 52 percent of non-Hispanic black Americans said they viewed being black as “extremely important” to how they thought about themselves, according to a Pew Research Center poll conducted last year. Another 22 percent said it was “very important.” These numbers were considerably lower for non-Hispanic Asian, non-Hispanic white and Hispanic Americans. (More on the story with Asian and Hispanic Americans in a bit — it’s complicated.)^{16}

Pew polling from 2016 and 2017 also showed that black people were significantly more likely than other demographic groups^{17} to say that their race was central to their identities.

Similarly, Democracy Fund + UCLA Nationscape polling from last December found that 75 percent of black Americans said their ethnicity and race was “very important to their identity,” significantly higher than the share of Hispanic Americans (58 percent), Asian Americans and Pacific Islanders (40 percent) and white Americans (30 percent) who said the same. Another 15 percent of black Americans said that their race was “somewhat important.”^{18}

This heightened sense of black identity does not appear to be a particularly recent phenomenon — or one that was inspired by the Black Lives Matter movement, which began to emerge in 2013. In 2012, about 70 percent of black Americans said that being black was either extremely or very important to their identity, about the same proportion as in 2016, according to surveys conducted as part of the American National Election Studies. In both years, black Americans expressed much greater ties to their identity than white or Hispanic Americans did.^{19}

*[Related: Do You Know How Divided White And Black Americans Are On Racism?]*

Part of the story here is about ethnic and racial groups other than black Americans — why aren’t an overwhelming majority of white, Hispanic or Asian Americans saying that their race or ethnicity is very important to their personal identities? This is not a simple question, and we won’t try to unpack it all here. Penn State political science and African American studies professor Candis Watts Smith, who has written extensively about identity, said that “Asian” and “Hispanic” aren’t really the identities that some people who fall under these groups associate themselves with. Hispanic Americans, she argued, might think of themselves as Cuban or Mexican but not embrace the broader Latino or Hispanic labels. Similarly, some Americans of Chinese or Japanese ancestry might not describe themselves as Asian or feel much attachment to that identity. White Americans, Smith said, tend not to think of themselves racially, she said, because “whiteness is viewed as normal by white people.”

Some scholars, most notably Duke University political scientist Ashley Jardina, emphasize that a significant number of white Americans *do* define themselves by their race, though still at lower rates than black Americans. Her research suggests that people with high levels of white identity tended to vote for Trump in 2016, and you can imagine more liberal-leaning white Americans would avoid talking about their pride as white people to avoid being cast as racist. Also, at least one poll, the 2016 Collaborative Multi-Racial Post-Election Survey, suggests that Asian Americans and Latino Americans express fairly similar views to black Americans in terms of having a positive view of their association with their racial or ethnic group.

That said, experts agree that black Americans express high levels of connection to their blackness. Karyn Lacy, a University of Michigan sociology professor who wrote a book on black middle-class people living in the Washington, D.C., suburbs, said that the people she interviewed for her research wanted their children and grandchildren to be close to the broader black community.

“There is a lot of joy in being black,” Lacy said of the people she interviewed. “This is a really important point. Most of the media coverage of black people is negative. Scholars have spent a lot of time documenting the racial discrimination blacks experience. We do need to know about how and why discrimination persists. But there is very little attention to all the good things about being black.”

“We’re left with the impression that black people wake up every morning thinking, ‘Ugh, I’ve got to be black today, and it’s going to be awful.’ None of the people I interviewed held that view,” Lacy added. “They take a lot of pride in being black and worry that their kids might not embrace being black with the same enthusiasm.”

*[Related: How The Police See Issues Of Race And Policing]*

The centrality of racial identity to black Americans is important to consider in a lot of contexts. We mentioned earlier that black attitudes about policing could be explained in part by partisanship, namely that the overwhelming majority of black people vote Democratic. But that skips over something that’s extremely important to understand: Why are black people so much more Democratic-leaning than other ethnic and racial groups? Part of the answer sits in the power of black identity — scholars argue that, to some extent, black Americans vote as a collective to defend the broader group and sometimes shame and discourage other black people from voting Republican and breaking with that collective.

“Nobody likes Kanye right now,” Smith joked, noting that many black people have become frustrated with Kanye West since he started associating himself with President Trump and making controversial comments on racial issues.

Black NBA players’ doing everything they could to embrace Obama when he was president and then largely shunning Trump is no doubt related to those two presidents’ divergent personas and political stands, as well as to partisanship. But it’s tied to identity too — black NBA players took pride in a fellow black man being president and were angry after Trump slammed NFL players who knelt during the national anthem to protest racial inequality in America. Winfrey, in her decades as a celebrity, has generally avoided partisan politics. But she was very vocal in backing Obama during his 2008 presidential campaign and Stacey Abrams in her 2018 Georgia gubernatorial run. Obama became America’s first black president; Abrams would have been the nation’s first-ever black female governor.

“Their identity stems from lived experiences with discrimination, bias, violence, inequality, broken promises, empty rhetoric,” said Rosalee Clawson, a political science professor at Purdue University who studies the politics of race, class and gender. “I think we would be shocked if blacks didn’t share a sense of linked fate with their racial group.”^{20}** **

*[Related: 1968 Isn’t The Only Parallel For This Political Moment]*

So it’s worth considering Floyd’s killing and the black community’s reaction to it in that context. Police in the U.S. pull over, arrest and shoot and kill black people at much higher rates than their 13 percent share of the U.S. population. So perhaps men like Jordan and Obama see what happened to Floyd as something that could happen to them.

And it could. But it’s also likely that these famous black men and women, like most black Americans, view being black as a big part of who they are, and so feel that they should speak out when an issue related to being black is all over the news.

“Most of them were not always celebrities. And they have [black] friends and neighbors. And black celebrities face some of the same denigrating things [based on their race] as an average black person,” said Smith.

Even rich black people think, “It could have been me, it could have been my family member or my neighbor or a member of my community,” said Smith.

Being black, she said, “is always in their consciousness.”

]]>The astronomers on planet Xiddler have made several remarkable discoveries. After inventing the telescope, they quickly discovered a new planet in their solar system!

Xiddler is very much like Earth. The planet orbits its star in a nearly circular path, with an average distance of 150 million kilometers, a period of one Earth year and a day that lasts 24 hours. But *unlike* Earth, there weren’t any other known planets in the solar system…until now.

Moments after the Xiddlerian sun set below the horizon, three astronomers happened to focus their telescopes at the zenith of the evening sky, all seeing the same new planet. In their excitement, the astronomers race to Xiddler’s Grand Minister to deliver the momentous news.

The first astronomer says that, by her calculations, the newly discovered planet orbits their sun with a radius of 50 million kilometers. The second astronomer says that, by *her* calculations, the planet in fact orbits their sun with a radius of 300 million kilometers. The third astronomer disagrees with the other two — by *her* calculations, the planet has a very similar orbit to Xiddler, with a radius of 150 million kilometers.

Which astronomer should the Grand Minister believe?

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

Some friends have invited you to a protest, and you’ll be making a sign with large lettering. You’re filling in the sign’s letters by drawing horizontal lines with a marker. The marker has a flat circular tip with a radius of 1 centimeter, and you’re holding the marker so that it’s upright, perpendicular to the sign.

Since the diameter of the marker’s tip is 2 centimeters, you decide to fill in the letters by drawing lines every 2 centimeters. However, this is the pattern you get:

The shading doesn’t look very uniform — each stroke is indeed 2 centimeters wide, but there appear to be gaps between the strokes. Of course, if you drew many, many lines all bunched together, you’d have a rather uniform shading.

But you don’t have all day to make this sign. If the lines can’t overlap by more than 1 centimeter — half the diameter of the marker tip — what should this overlap be, in order to achieve a shading that’s as uniform as possible? And how uniform will this shading be, say, as measured by the standard deviation in relative ink on the sign?

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

Congratulations to Brian Leet of Burlington, Vermont, winner of last week’s Riddler Express.

Last week, I was playing one of my favorite video games, The Legend of Zelda: Breath of the Wild. Within the game, there were hundreds of hidden “Korok Seeds,” which I was having an increasingly difficult time finding.

Fortunately, I obtained a special mask that made a sound any time I was within a certain distance of a Korok Seed. While playing, I marked nine distinct locations on the game map, forming the 3-by-3 grid shown below:

Each leaf symbol was within range of a Korok Seed, while the point in the middle was *not* within range of a Korok Seed. Given this arrangement, what was the minimum possible number of Korok Seeds I could have detected?

There certainly could have been eight Korok Seeds, each near one of the eight leaves. But rather than start with larger numbers, many solvers jumped straight to the smaller ones. Was it possible to generate this pattern if there had been just one Korok Seed?

Solver Kiera Jones supposed for a moment that there *was* just one Korok Seed. It couldn’t have been at the center of the grid, since that would have resulted in a leaf icon in the middle. In fact, this one Korok Seed had to have been closer to all eight leaves than it was to the center. But the moment you were closer to one of the leaves, you’re farther away from its diametric opposite. For example, if you were closer to the leaf on the right than the middle, you *must* have been closer to the middle than the leaf on the left.

In short, Kiera proved there had to be at least two seeds. And as it turned out, the answer was indeed *exactly* **two**.

No information was given in the problem about what that “certain distance” was, within which the mask was able to detect a Korok Seed. You were left to assume it could be anything, and that meant the seeds could be arbitrarily far away from the nine points in the grid.

By placing the two seeds in opposite directions *outside* of the 3-by-3 grid, it was possible to have the eight points marked by leaves all be within some distance of a seed, while the middle point was outside that range. Here was one such placement of the two seeds, courtesy of Matt Jenny:

The large black circles show the regions where you would detect the Korok Seed at the center of the circle. Sure enough, the middle green point didn’t fall within one of the large black circles, while all eight surrounding green points fell within one of the two large circles. Countless other arrangements were also possible, but the key idea was to split the eight surrounding locations into two groups of four, each of which was in range of a single Korok Seed.

Raul Perera arrived at the same answer, but without the aid of any graphing technology. Instead, Raul dropped a disk onto some graph paper!

And even if you were to split the locations as Matt and Raul did, there were still many possible locations for the seeds. In the diagram below, the red and blue regions represent possible locations for the two seeds:

Those Koroks certainly are pesky. It may take me a few more years, but someday I will have found them all.

Congratulations to John Bullock of Lafayette, Indiana, winner of last week’s Riddler Classic.

Last week, everyone in the U.S. (about 330 million people) joined a single Zoom meeting between 8 a.m. and 9 a.m. — to discuss the latest Riddler column, of course.

The attendees all followed the same steps in determining when to join and leave the meeting. Each person independently picked two random times between 8 a.m. and 9 a.m. — not rounded to the nearest minute, mind you, but *any* time within that range. They then joined the meeting at the earlier time and left the meeting at the later time.

What was the probability that at least one attendee was on the call with everyone else (i.e., the attendee’s time on the call overlapped with everyone else’s time on the call)?

You don’t typically see a problem jump straight to the case of 330 million people unless something funny is going on. To understand what was happening here, many solvers first tried the case in which only two people were joining the meeting. What were the chances they’d be on at the same time?

You could have solved this with calculus, but you also could have skipped all the integrals with a combinatorial approach. With two people, there were four total times being randomly picked: two starting times and two ending times. Suppose these four times, in order from earliest to latest, were A, B, C and D. Then, there were six ways to split these times among the two attendees — the first person could have picked times A and B, A and C, A and D, B and C, B and D or C and D, while the other person would have been left with the remaining two times. The only time the two people didn’t meet up was when the first person picked A and B (leaving the other person with C and D) or C and D (leaving the other person with A and B) — two out of the six cases. That meant the chances they *did* meet up stood at four out of six.

So when there were two people in the meeting, the probability at least one attendee saw all the others was 2/3. But what if there were three people, or four — or 330 million?

This general problem is more involved, as there were way more than six cases to consider. Jim Crimmins, the puzzle’s submitter, along with solver Allen Gu, both referenced a 1990 paper that addressed this very problem. But before revealing the answer, let’s check in with those who took a computational approach.

John Bullock simulated 330 million attendees in Python and ran a whopping 364,000 total simulations. Among these, 242,919 — very nearly two-thirds of the simulations — had at least one attendee who was on the call with everyone else. Peter Ji, Matt Lee and Benjamin Phillabaum all kindly shared their code as well, and all three generated results that were remarkably close to 2/3 as well. Could it be that the answer was always 2/3, whether there were two attendees, 330 million or anything in between?

Indeed, the answer was **2/3**. Solver Josh Silverman proved this result with some powerful combinatorial reasoning and by swapping meeting times among certain attendees so that one could be on the call with everyone else.

For extra credit, you were asked to find the probability that at least *two* attendees were on the call with everyone else. Once again, a beautiful result was found by both the analysts and those using a computational approach: **2/5**.

As Josh explains, the probability of having exactly *k* attendees on the call with everyone else turned out to be (*k*+1)!/(2*k*+1)!! (The two exclamation points denote a double factorial — multiplying up all the evens or odds up to that number. In this case, since 2*k*+1 is always odd, it means multiplying all the odds from 1 to 2*k*+1.) Here’s what that probability distribution looks like, courtesy of Allen Gu:

One very cool consequence of this result is that, in the limit when there were many, many attendees (e.g., when there are 330 million of them), the *average* number of attendees who were on the call with everyone else approached 𝜋/2, or about 1.57.

Sometimes riddles just can’t help but circle back to 𝜋.

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

]]>