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Can You Be Mediocre Enough To Win?

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

Riddler Express

This week, both puzzles come courtesy of Ben Orlin, author of the newly released “Math Games with Bad Drawings.” In addition to its terrible drawings, the book includes several games that have appeared in this column, including Tax Collector and Racetrack. Ben’s Express puzzle relates to tic-tac-toe.

In tic-tac-toe, you can rate a square’s “real estate value” by how many possible three-in-a-rows pass through it. Thus, the center scores 4, the corners score 3 and the edges score 2.

A tic-tac-toe board with numbers in each cell. Reading left-to-right, top-to bottom, these are: 3, 2, 3, 2, 4, 2, 3, 2, 3. The 3 in the top right also shows the 3 ways to get 3 in a row using that corner.

That means the center is the “most valuable” real estate, while the edges are the “least valuable.”

Now, let’s try three-dimensional tic-tac-toe, played on a 3-by-3-by-3 board. Which positions are the most and least valuable, and how much are they worth?

Extra credit: What about four-dimensional tic-tac-toe?

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

Riddler Classic

This week’s Classic also comes from Ben Orlin, although it’s relatively mediocre compared to the Express.

In the three-player Game of Mediocrity, you win by not winning too much.

Each round, every player secretly picks a number from 0 to 10. The numbers are simultaneously revealed, and the median number wins that number of points. (If two or more players pick the same number, then the winner is randomly selected from among them.)

Player A says 6, B says 3, and C says 2. B is the winner, having said the median, and earns 3 points.

After five rounds, the winner is whoever has the median number of points. (Again, if two or more players have the same score, then the winner is randomly selected from among them.)

With one round remaining, players A, B and C have 6, 8 and 10 points, respectively. Player A sighs and writes down “3,” but fails to do so in secret. Players B and C both see player A’s number (and both see that the other saw A’s number), and will take care to write their own numbers in secret. Assuming everyone plays to win, what numbers should B and C choose?

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

Solution to last week’s Riddler Express

Congratulations to 👏 Matt Garrett 👏 of Cardiff, Wales, winner of last week’s Riddler Express.

Multiplying polynomials can be challenging work. But every once in a while, there’s an expansion that is particularly elegant. Last week, I shared one of my favorites:


The ellipsis indicated that every letter was subtracted at some point. If you multiplied all these binomials together, what was the expansion in simplest form?

When you multiply two binomials, you get four resulting terms. When you multiply three binomials, you get eight terms. And when you multiply 26 binomials — as this riddle apparently asked you to do — you get 226 terms. That’s more than 67 million terms! Even if each term could be written or typed so it was just 1 millimeter wide, the entire expansion would still be 67 kilometers long. There was no reasonable way to submit such an answer.

But if you paid close attention to your calendar last week, you would have noticed it was April 1 — April Fools’ Day. Indeed, there was mischief afoot!

As noted by solver Sharon McCathern of Placentia, California, the 24th term in the factored expression was (xx), which was zero. And that meant the entire product was zero.

Surely the April Fools’ fun ended here, with the Express. Mischief has no place in a Classic puzzle … right?

Solution to last week’s Riddler Classic

Congratulations to 👏 Robert Bauer 👏 of Southbury, Connecticut, winner of last week’s Riddler Classic.

Last week, you tried your hand at the game of “Zactoll,” named for yours truly. To play the game, you started with the number 1 and then tried to reach different target numbers through a series of steps. For each step, you could always choose to double the number you currently had. However, if the number happened to be one more than an odd multiple of 3, you could choose to “reduce” — that is, subtract 1 and then divide by 3.

For example, to reach a target number of 5, you could double (1 → 2), double (2 → 4), double (4 → 8), double (8 → 16) and reduce (16 → 5). And to get a target number of 6, you could double (1 → 2), double (2 → 4), double (4 → 8), double (8 → 16), reduce (16 → 5), double (5 → 10), reduce (10 → 3) and double (3 → 6).

What was the smallest target whole number that you could not reach in the game of Zactoll?

Solver Ken Lin explored what would happen if we played Zactoll in reverse. When played in reverse, you halved numbers instead of doubling them. Also, when you had an odd number, you increased (multiplying by 3 and adding 1) rather than reduced (subtracting 1 and dividing by 3).

For example, suppose again that your target number was 6. Playing backwards to get to 1, you would halve (6 → 3), increase, (3 → 10), halve (10 → 5), increase (5 → 16), halve (16 → 8), halve (8 → 4), halve (4 → 2) and halve (2 → 1).

If this seemed familiar, that’s because Zactoll played in reverse was the Collatz Problem. (Most solvers also realized that “Zactoll” was merely an anagram of “Collatz.”) It has been conjectured that every starting whole number will eventually reach the number 1, though this has not yet been proven.

That meant there was no known solution to last week’s Classic. A few solvers got in on in the April Fools’ fun. This week’s winner, Robert, as well as Jenny Mitchell both jokingly claimed to have found numbers that didn’t reach 1, instead bouncing around like a hailstone. In submitting an answer, Robert remarked: “I have discovered a truly marvelous proof of this, which this text box is too small to contain.”

Solver Billy Mullaney made my week, writing: “APRIL FOOLS! lol it’s the collatz conjecture. Thanks for this, I literally laughed out loud, and then spent 20 minutes explaining the joke to my coworkers.” That’s what we’re all about, here at The Joker … errrr … The Riddler.

Want more puzzles?

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

Want to submit a riddle?

Email Zach Wissner-Gross at


  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.


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