Happy (Almost) New Year from The Fiddler!

Can you solve two puzzles about the number 2024?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

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

Importantly, due to the holidays, there will be no Fiddler next week. The next column will come out on Jan. 5, 2024. See you next year!

This Week’s Fiddler

The new year is almost upon us! From Dean Ballard comes an early celebration of the number 2024:

There are several rectangular prisms with integer edge lengths that have an internal diagonal of length 2024.

What is the greatest volume among these prisms?

Extra Credit

From Dean Ballard comes yet another puzzle to ring in 2024:

The Collatz Conjecture asserts that if you start with any positive integer and repeatedly apply a certain operation, the resulting sequence of numbers will always reach the number 1. The operation is: For even numbers divide by 2; for odd numbers, multiply by 3 and add 1.

For example, if you start with 11, the sequence goes: 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1.

The Collatz Conjecture is a well-known unsolved problem in mathematics. No proof has ever been found to show that it is true, nor has any counterexample been found to show that it is false.

What is the smallest starting number for which 2024 appears somewhere in its Collatz sequence?

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. Once again, I’m sharing Mathigon’s annual puzzle calendar, curated by my colleague Eda Aydemir. A new puzzle is being posted every day during the month of December.

This week, I’m sharing the puzzle from Dec. 16, which reminds me of the delightful geometric riddles of Catriona Agg:

A semicircle is placed inside a square of side length 1. One endpoint of the semicircle’s diameter coincides with the bottom-right corner of the square, while the other endpoint lies somewhere on the top edge of the square. Importantly, the semicircle is tangent to the left edge of the square. What is the semicircle’s diameter?

At first, it might seem like there’s not enough information to solve this puzzle—but indeed there is!

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Brad Slavens 🎻 from Atlanta, Georgia. I received 40 timely submissions (this was a toughie!), of which 33 were correct—good for an 82.5 percent solve rate.

Last week, Kyle and Julien were playing a game in which they each tossed their own fair coins. On each turn of the game, both players flipped their own coin once.

If, at any point, Kyle’s most recent three flips were Tails, Tails, and Heads (i.e., TTH), then he won. If, at any point, Julien’s most recent three flips were Tails, Tails, and Tails (i.e, TTT), then he won.

However, both players couldn’t win at the same time. If Kyle got TTH at the same time Julien got TTT, then no one won, and they continued flipping. They didn’t start over completely or erase their history, mind you—they merely continued flipping, so that one of them could conceivably win in the next flip or two.

What was the probability that Kyle won this game?

A good first step was to work out the different possible states this game could have been in, along with the probabilities of transitioning from one state to another. From there, you could have worked out and solved a system of equations or you could have plugged these probabilities into a Markov model.

So, what were these “states”? Solver 🎬 Laurent Lessard 🎬 considered the most recent three flips of both Kyle and Julien, for a total of 2⁶ (or 64) distinct states. Laurent even diagrammed all the possible transitions between these 64 states:

Meanwhile, solver Jeffrey Ling reduced the complexity by considering just nine distinct states, based instead on the lengths of Kyle’s and Julien’s respective current “tails streaks,” assuming neither Kyle nor Julien had yet won. Jeffrey defined the ordered pair (i, j) as Kyle having just flipped i tails in a row and Julien having just flipped j tails in a row. Both i and j could equal 0, 1, or 2.

Let’s define p(i, j) as the probability that Kyle won when Kyle’s tails streak was i and Julien’s was j. What was an equation we could use to relate p(0, 0) to some of these other probabilities? Well, there was a 1-in-4 chance they both flipped heads, in which case they returned to the state (0, 0). There was a 1-in-4 chance only Julien flipped tails, in which case they transitioned to the state (0, 1). There was a 1-in-4 chance only Kyle flipped tails, in which case they transitioned to the state (1, 0). And there was a 1-in-4 chance they both flipped tails, in which case they transitioned to the state (1, 1). That meant our equation was p(0, 0) = p(0, 0)/4 + p(0, 1)/4 + p(1, 0)/4 + p(1, 1)/4.

Working out the transition probabilities among all nine states gave you a total of nine equations with nine unknowns. Four of these equations (including the one we just derived) were relatively straightforward , since the next flip never resulted in Kyle or Julien winning:

• p(0, 0) = p(0, 0)/4 + p(0, 1)/4 + p(1, 0)/4 + p(1, 1)/4

• p(0, 1) = p(0, 0)/4 + p(0, 2)/4 + p(1, 0)/4 + p(1, 2)/4

• p(1, 0) = p(0, 0)/4 + p(0, 1)/4 + p(2, 0)/4 + p(2, 1)/4

• p(1, 1) = p(0, 0)/4 + p(0, 2)/4 + p(2, 0)/4 + p(2, 2)/4

The probabilities of winning from the remaining five states were a little trickier to work out, since they potentially resulted in Kyle’s or Julien’s victory. When Julien had a chance to win on the next flip (i.e., when j = 2), Kyle could only win when Julien’s next flip was heads, setting i back to 0:

• p(0, 2) = p(0, 0)/4 + p(1, 0)/4

• p(1, 2) = p(0, 0)/4 + p(2, 0)/4

And when Kyle had a chance to win on the next flip (i.e., when i = 2), that chance of victory showed up as a constant in the expression. The game only continued when Kyle flipped a tails, in which case i remained 2:

• p(2, 0) = 1/2 + p(2, 0)/4 + p(2, 1)/4

• p(2, 1) = 1/2 + p(2, 0)/4 + p(2, 2)/4

That left one state to consider—when both Kyle and Julien had a chance to win on the next flip (i.e., i = 2 and j = 2). At this point, there was a 1-in-4 chance they both flipped heads, in which case Kyle won. There was a 1-in-4 chance they both flipped tails, in which case Julien won. There was a 1-in-4 chance Kyle flipped tails and Julien flipped heads, in which case they returned to the state (2, 0). And there was a 1-in-4 chance Kyle flipped heads and Julien flipped tails. Since they couldn’t both win at the same time, they instead returned to the state (0, 2). Putting this together, the final equation was:

• p(2, 2) = 1/4 + p(0, 2)/4 + p(2, 0)/4

Whether you solved this system of equations by hand or with the aid of a computer, in the end you wanted the value of p(0, 0), which gave you Kyle’s probability of winning the game at the beginning, before any coins had been flipped. This probability turned out to be 45,667/71,106, or about 64.2 percent.

At first glance, the nine equations above looked suspiciously symmetric. So how could it be that Kyle had almost a two-thirds chance of winning? The asymmetry appeared in p(2, 0) and p(2, 1) as compared to p(0, 2) and p(1, 2). When Julien was one flip away from winning and needed just one more tails, getting a heads set i all the way back to 0. But when Kyle was one flip away from winning and needed a heads, getting a tails didn’t set j back to 0. Instead, j remained at 2, which meant Kyle still had a chance to win on his very next flip. It was this key difference that provided Kyle with his significant advantage.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 David Combs 🎻 from Palm Springs, California. I received 23 timely submissions, of which 21 were correct—good for a 91 percent solve rate!

Continuing their game, Kyle and Julien wrote down all eight possible sequences for three coin flips (HHH, HHT, HTH, THH, HTT, THT, THH, and TTT) on eight different slips of paper. They placed these slips into a hat and shook it.

They each randomly drew slips of paper out of the hat, after which they played the same game as previously described, but looking for the sequence specified on the slip of paper they each selected. Kyle drew first and looked at his slip of paper. After doing some calculations, he said: “Well, at this point, it’s about as fair a match as it could possibly be.” 

Which slip or slips of paper might Kyle have drawn? And what were his chances of winning at this point (i.e., before Julien selected his own slip of paper)?

In last week’s Fiddler, you found that TTH had a roughly 64.2 percent chance of defeating TTT. But for Extra Credit, you now had to consider how all eight sequences fared against each other, for a total of 56 scenarios (or 64 if Kyle and Julien picked with replacement—but more on that in a moment).

When you want to see how various possibilities all fare against each other, a table or grid is a great way to present the data. Solver Tom Keith created such a table:

Each row shows a different slip that Kyle might have picked, while each column shows a different slip that Julien might have picked. All the symmetry you’d expect when swapping heads with tails (and vice versa) is in there: For example, if Kyle picked HHH and Julien picked TTT, they each had a 50 percent chance of winning.

Kyle picked first, at which point his chances of winning were the average of the seven columns with values in the corresponding row. No slip had an average of 50 percent, but two of them—THT and HTH (a symmetric pair)—came very close. When Kyle picked either of these slips, he had roughly a 49.6 percent chance of winning the game.

If you interpreted Julien picking from the hat with replacement, meaning he could have picked the same slip as Kyle, then you averaged all eight columns together, with the eighth column having a value of 0.5 (since two identical slips were equally likely to beat each other). This only marginally increased Kyle’s chances of winning, and I still counted it as correct.

Want to Submit a Puzzle Idea?

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