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How Many Squares Can You Squeeze?

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 or send me an email.

Riddler Express

The winners of the 2023 Regeneron Science Talent Search were announced on March 14. (Full disclosure: I was a finalist in this very same competition exactly one eternity ago. You can find me among a gaggle of fellow New Yorkers, if you look closely.)

I am delighted that one of this year’s winners was able to share his favorite puzzle for this week’s column!

Hailing from Harrisonburg, Virginia, high schooler Max Misterka studied quantum calculus, also known as q-calculus, extending it to a version he calls “s-calculus.” This week, Max is putting the quantum aside and challenging you to a puzzle that may or may not be solvable with traditional calculus:

Max and I are playing a game in which we both pick a number in secret. Let’s call Max’s number m and my number z. After we both reveal our numbers, Max’s score is mz, while my score is zm. Whoever has the greater score wins.

When we played most recently, Max and I selected distinct whole numbers. Surprisingly, we tied — there was no winner! Which numbers did we pick?

Extra credit: Max and I play another round. This time, we both pick positive numbers that are not necessarily whole numbers. I tell Max my number without knowing his, at which point he tells me the game is once again a tie. “Ah,” I reply, “that means we must have picked the same number!” Which number did we both pick?

Submit your answer

Riddler Classic

From Ethan Rubin comes a matter of squeezing squares among the stars:

Ethan has been playing Star Battle, a sudoku-like game. In the five-star variant of the game, you are trying to fill a 21-by-21 grid with stars according to certain rules:

  • Every row must contain exactly five stars.
  • Every column must contain exactly five stars.
  • Every bold outlined region must contain exactly five stars.
  • No two stars can be horizontally, vertically or diagonally adjacent.

For example, here is a solved game board:

A 21 by 21 grid with 105 stars. Each row and column has exactly 5 stars. There are also dark borders forming 21 distinct regions, each of which also has exactly 5 stars.

After playing the game, Ethan noticed that the stars seemed to be rather evenly distributed throughout the board, although there were some gaps. Specifically, he wondered how many distinct two-by-two squares in the grid didn’t contain a star. Here’s the same game board in which all 20 empty 2-by-2 squares are highlighted:

The same 21 by 21 grid with 105 stars is shown. There are also 20 locations with empty 2 by 2 squares, which are shaded in red.

As you can see, some of these 2-by-2 regions overlap — even so, they still count as distinct.

In a solved board of Star Battle, what are the minimum and maximum possible numbers of empty 2-by-2 squares?

Submit your answer

Solution to the last Riddler Express

Congratulations to Ñб─Ñб÷âб─б≤Ðб▐ Thomas Stone Ñб─Ñб÷âб─б≤Ðб▐ of San Francisco, California, winner of last week’s Riddler Express.

I recently competed on “Jeopardy!” Heading into the Final Jeopardy! round, challenger Karen Morris was leading the way with $11,400, returning champion Melissa Klapper had $8,700 and I had $7,200. The Final Jeopardy! category was revealed as being “American Novelists,” and it was now time for all three of us to wager anywhere from $0 to the total amount we had for this final clue.

Despite the dramatic swings in the match, my assessment was that all three of us were somewhat evenly matched in terms of knowledge. Having studied my opponents, I was also confident that Karen would wager enough money to cover the most aggressive wager from Melissa, and that Melissa would wager enough to cover my most aggressive wager.

With these assumptions, it was logical for me to keep my own wager small, since my only chance at winning was if both Karen and Melissa guessed incorrectly. Not particularly liking the category, I chose to wager $0. What was the maximum dollar amount I could have wagered without affecting my chances of winning? (Again, you could assume Karen wagered enough to cover Melissa and Melissa wagered enough to cover me.)

If Melissa had bet everything she had and answered correctly, she would have doubled up, finishing with $17,400. To come out on top, Karen needed to finish with at least $17,401, which meant she’d have to wager at least $6,001. Similarly, in the event I had bet everything and answered correctly, I would have finished with $14,400. To finish with at least $14,401, Melissa needed to wager at least $5,701.

As I said earlier, I was hoping that both Karen and Melissa would get Final Jeopardy! wrong. In that case, Karen would have lost at least $6,001, so that her final total was at most $5,399. Similarly, Melissa would have lost at least $5,701, so that her final total was at most $2,999.

For me to have a shot at winning under these assumptions, I had to finish with more than the greater dollar amount of $5,399 and $2,999 (i.e., $5,399). To guarantee I had at most $5,400 by the end of the show, I should have wagered no more than $7,200 minus $5,400, or $1,800.

All the clues from my episode are available via J! Archive, which further provides wagering suggestions for Final Jeopardy! Sure enough, it recommends I not exceed $1,800. (It also recommends I wager at least $1,501 to cover a wager of $0 from Melissa, which would have been a good idea.)

In the end, Karen wagered $6,001, Melissa wagered $8,000, and I wagered $0 — all very reasonable bets, in my opinion. For extra credit, knowing these were the wagers we made, you had to further assume that all three of us had the same probability p of getting Final Jeopardy! correct, and that these three events were independent of one another. If the value of p was random and uniformly distributed between 0 and 1, what was my probability of winning the match?

Given these wagers, there were two ways I could have won: if all three of us whiffed on Final Jeopardy! (known as a “triple stumper”), which happened with probability (1−p)3, or if I was the only one to get Final Jeopardy! correct, which happened with probability p·(1−p)2. Adding these together gave you (1−p)2, i.e., the probability that both Karen and Melissa were wrong, since my answer didn’t matter. Since p was equally likely to be any value between 0 and 1, solver Paige Kester recognized that my probability of winning was the integral of (1−p)2 with respect to p from p = 0 to p = 1. By symmetry, this was the same as the integral of p2 from 0 to 1, which was 1/3. All things considered, I had a decent chance of pulling off the victory!

Solution to the last Riddler Classic

Congratulations to Ñб─Ñб÷âб─б≤Ðб▐Michael Bradley Ñб─Ñб÷âб─б≤Ðб▐ of London, England, winner of last week’s Riddler Classic.

It feels like there’s more parity in college basketball’s March Madness than ever, with lower-seeded teams advancing further in the tournaments at the expense of the favorites.

For last week’s Riddler Classic, you supposed that each team was equally likely to win any given game. What were the chances that the Sweet 16 consisted of exactly one of each seed?

The key to this puzzle was recognizing the inherent structure of the March Madness bracket. For example, in each of the four regions, the 1 seed plays against the 16 seed in the first round, and then the winner of that game plays the winner of the 8 seed vs. the 9 seed in the second round. That meant that exactly one of those four teams (1, 16, 8 and 9) could make it to the Sweet 16 out of each of the four regions. There were 44, or 256, ways to choose which of these seeds advanced to the Sweet 16. But there were only 4!, or 24, ways to have a 1 seed in one region, a 16 seed in another, an 8 seed in another and a 9 seed in the last. Therefore, the probability of having a 1 seed, a 16 seed, an 8 seed and a 9 seed in the Sweet 16 was 24/256, or 3/32.

Because of the bracket’s structure, the same was true for the 5, 12, 4 and 13 seeds, the 6, 11, 3 and 14 seeds, and the 7, 10, 2 and 15 seeds. For all four of these clusterings of seeds, the probability that one of each type of seed advanced to the Sweet 16 was 3/32. And because each clustering was independent of the other, that meant the probability of having all 16 seeds represented in the Sweet 16 was (3/32)4, which was 81/1,048,576, or about 0.0077 percent.

For extra credit, you now assumed that seed A would defeat seed B with probability 0.5 + 0.033·(BA). Again, what were the chances that the Sweet 16 consists of one of each seed?

To figure this out, let’s take a closer look at the 1 seed. To advance to the Sweet 16, it had to defeat the 16 seed in the first round, which occurred with probability 0.5 + 0.33·15, or 0.995. Then, it had to defeat either the 8 seed with probability 0.731 (the 53.3 percent of the time the 8 seed advanced to the second round) or the 9 seed with probability 0.764 (the 46.7 percent of the time the 9 seed advanced to the second round). All told, each 1 seed had a 74.27 percent chance of making it to the Sweet 16.

A similar analysis for the remaining seeds revealed that the 2 seed had a 65.47 percent chance of making it to the Sweet 16, the 3 seed had a 56.46 percent chance, and so on. As noted by solver Kiera Jones, to find the probability that each seed made it, you had to multiply all these probabilities together, but then multiply by (4!)4 to account for all the different ways these seeds could have come from the four regions. In the end, this probability turned out to be approximately 8.53×10-10.

Parity or no parity, it will be a very long time until we see all seeds 1-16 represented in the Sweet 16.

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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