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Can You Break A Very Expensive Centrifuge?

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

Riddler Express

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.

Animation of a centrifuge with 12 slots spinning around.

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?

Submit your answer

Riddler Classic

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?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to рџ‘Џ Seth Cohen рџ‘Џ 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 53, here it was N3. 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 3N(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 N3. Dividing them gave a final answer of 3(N−1)/(2N2). Sure enough, this checked out for the case when N equals 5, giving the expected answer of 24 percent. But as N got bigger, the chances of a fair match went down.

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

Solution to last week’s Riddler Classic

Congratulations to рџ‘Џ Mikolaj Franaszczuk рџ‘Џ 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!

Want more riddles?

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

Want to submit a riddle?

Email Zach Wissner-Gross at


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.