Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.
Every Friday morning, I present mathematical puzzles intended to challenge and delight you. Most can be solved with careful thought, pencil and paper, and the aid of a calculator. The “Extra Credit” is where the analysis typically gets 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 puzzles are 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 🎬.
This Week’s Fiddler
This week’s puzzle is another collaboration between Fiddler on the Proof and Science News, where four math puzzles that are related were posted earlier this morning. The third and fourth among these are also appearing here as this week’s Fiddler and Extra Credit. Now—on to those puzzles!
You are participating in a holiday gift exchange with your classmates. You each write down your own name on a slip of paper and fold it up. Then, all the students place their names into a single hat. Next, students pull a random name from the hat, one at a time. If at any point someone pulls their own name from the hat, the whole class starts over, with everyone returning the names to the hat.
Once the whole process is complete, each student purchases a gift for the classmate whose name they pulled. Gifts are handed out at a big holiday party at the end of the year.
At this party, you observe that there are “loops” of gift-giving within the class. For example, student A might have gotten a gift for B, who got a gift for C, who got a gift for D, who got a gift for A. In this case, A, B, C and D would form a loop of length four. Another way to have a loop of length four is if student A got a gift for C, who got a gift for B, who got a gift for D, who got a gift for A. And of course, there are other ways.
If there are a total of five students in the class, how likely is it that they form a single loop that includes the entire class?
This Week’s Extra Credit
If there are N students in the class, where N is some large number, how likely is it that they form a single loop that includes the entire class, in terms of N? (For full credit, your answer should be proportional to N raised to some negative power.)
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. This week, I’m sharing a set of tracing puzzles I spotted with my kids at a local playground:
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.
Standings
I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. If you think you see a mistake in the standings, kindly let me know.
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Austin Calico 🎻 from Ashland, Kentucky. I received 58 timely submissions, of which 53 were correct—good for a 91 percent solve rate.
Last week, you and your assistant were planning to irrigate a vast circular garden, which had a radius of 1 furlong. However, your assistant was somewhat lackadaisical when it came to gardening. Their plan was to pick two random points on the circumference of the garden and run a leaky hose straight between them.
You were concerned that different parts of your garden—especially your prized peach tree at the very center—would be too far from the hose to be properly irrigated.
On average, how far could you have expected the center of the garden to be from the nearest part of the hose?
Wait—was this a puzzle about random chords in a circle? The premise here reminded several readers of Bertrand’s paradox, in which the manner of random chord selection is (intentionally) ill defined. Here, of course, I define it quite clearly: The two endpoints of the chord are randomly, uniformly selected from the circumference.
Now, suppose a chord had been chosen. What was the shortest distance between the center of the circle and that chord? This was the distance between the center of the circle and the midpoint of the chord. The line segment connecting these two points also happened to be perpendicular to the chord itself.
From there, the question became: What was the average distance between the center and the midpoint of a chord, when chords were chosen randomly as described?
To figure this out, let’s use a coordinate system where the circle had radius 1 and was centered at the origin, (0, 0). Let’s fix one end of the hose at the point (1, 0), while the other end was equally likely to be anywhere along the circumference. Let’s define 𝜃 as the counterclockwise angle between (1, 0), (0, 0), and the other end of the hose, so that 𝜃 was equally likely to be any value between 0 and 360 degrees (or, equivalently, between 0 and 2𝜋 radians).
Given a value of 𝜃, what was the distance between the circle’s center and the midpoint of the chord? With some trigonometry (assisted by the diagram below), this minimal distance was equal to cos(𝜃/2). Technically, since cosines can be negative, the distance was the absolute value of cos(𝜃/2). So, to keep things positive—and to take advantage of symmetry—we could instead consider 𝜃 between 0 and 𝜋 radians.
The average distance was then the integral of cos(𝜃/2) over this range, divided by 𝜋, which came to 2/𝜋, or approximately 0.63662. And that was the solution to the puzzle! On average, the peach tree was 2/𝜋 furlongs from the hose. Hopefully that was close enough to the hose to get some water! (But knowing anything about the size of trees, it probably wasn’t.)
With some more work, you could show that the cumulative probability distribution for the minimal distance x to the hose was p(x) = 2/𝜋·sin-1(x). This meant the median distance to the hose was sin(𝜋/4) = 1/√2, or about 0.7071, slightly greater than the mean. This result made sense, because precisely half of the time the two ends of the hose would lie within the same quarter of the circle, which was also when the distance to the hose was at most 1/√2.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Emanuele Macchi 🎻 from Galliate, Italy. I received 40 timely submissions, of which 28 were correct—good for a 70 percent solve rate.
As before, your assistant intended to pick two random points along the circumference of the garden and run a hose straight between them. This time, you decided to contribute to the madness yourself by picking a random point inside the garden to plant a second peach tree.
On average, how far could you have expected this point to be from the nearest part of the hose?
