A Pi Day Puzzle

The island of 𝜋-land has two beaches: a semicircular arc and a diameter. If you pick a random point on the island, how likely are you to be closer to the diameter than the arc?

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. 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 🎬.

This Week’s Fiddler

From Jason Zimba (a colleague of mine at Amplify), comes a puzzle that’s perfect for Pi Day:

You are planning a picnic on the remote tropical island of 𝜋-land. The island’s shape is a perfect semi-disk with two beaches, as illustrated below: Semicircular Beach (along the northern semicircular edge of the disk) and Diametric Beach (along the southern diameter of the disk).

If you pick a random spot on 𝜋-land for your picnic, what is the probability that it will be closer to Diametric Beach than to Semicircular Beach? (Unlike the illustrative diagram above, assume the beaches have zero width.)

This Week’s Extra Credit

From Jason Zimba also comes some Extra Credit:

Suppose the island of 𝜋-land, as described above, has a radius of 1 mile. That is, Diametric Beach has a length of 2 miles.

Again, you are picking a random point on the island for a picnic. On average, what will be the expected shortest distance to shore?

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 classroom activity I made specifically for Pi Day. It’s an interactive simulation of Buffon’s needle problem, which you can use to (rather inefficiently) approximate 𝜋.

There’s also a collaborative version of the activity, where everyone can get in on the fun. If you have a few minutes, help us approximate 𝜋 by tossing some virtual needles. (In particular, check out Screen 8, where you can toss an unlimited number of needles.) At the time of writing, more than 10,000 virtual needles have been tossed, as shown below.

Needles in red cross a line segment, while needles in blue do not. Based on these 10,000+ throws, we can approximate 𝜋 to be 3.12, which is off by only a fraction of a percent.

With your help, we can toss 100,000 or even 1 million needles! And after you’ve tossed a few needles, check out the other Pi Day activities from Amplify, including Drumming With Angles and Circle Relationships.

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 timely correct solutions to Fiddlers and Extra Credits are worth 1 point each. 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: 🎻 Mark Goodrich 🎻 from Oxford, Pennsylvania. I received 67 timely submissions, of which 50 were correct—good for a 75 percent solve rate.

Last week, you were placing many, many dominoes in a straight line, one at a time. However, each time you placed a domino, there was a 1 percent chance that you accidentally tipped it over, causing a chain reaction that tipped over all the dominoes you had placed. After each chain reaction, you started over again.

If you did this many, many times, what could you expect the median (note: not the average) number of dominoes placed when a chain reaction occurred (including the domino that caused the chain reaction)?

More precisely, if this median number was M, then you would have expected to have placed fewer than M dominoes at most half the time, and more than M dominoes at most half the time.

Let’s look at the probability of a chain reaction for each number of placed dominoes. Perhaps we’ll find a pattern!

The probability that you placed exactly one domino and a chain reaction (I mean, what’s a “chain reaction” with one domino, anyway?) occurred was 1 percent. Simple enough!

To have a chain reaction after placing two dominoes, you needed there to not be a chain reaction after the first domino (with a 99 percent chance), and then you needed a chain reaction upon placing the second domino (with a 1 percent chance). Multiplying the probabilities of these two independent events together, a chain reaction after exactly two dominoes happened with probability (0.99)·(0.01).

Extending this line of reasoning, you could show that the probability of a chain reaction after exactly N dominoes had been placed was 0.99^N−1·0.01.

Now that you knew the probability of a chain reaction occurring with exactly N dominoes, what was the probability it occurred when there were anywhere from 1 to N dominoes? The moment this cumulative probability surpassed 50 percent, you knew you had just passed the median.

So let’s find an expression for that cumulative probability in terms of N, and let’s call it p_N. Taking the sum, p_N = 0.01 + 0.99·0.01 + 0.99²·0.01 + … 0.99^N−1·0.01. Multiplying both sides by 100 cleaned this up a bit: 100p_N = 1 + 0.99 + 0.99² + … 0.99^N−1. The right hand side of this equation was a finite geometric series, for which there’s a handy-dandy formula. The sum was (1−0.99^N)/1−0.99), which simplified to 100·(1−0.99^N). That meant p_N = 1−0.99^N.

Now, remember what we said: The median M was equal to the first value of N such that p_N was at least 0.5. Plugging these conditions into our equation for p_N gave you 1−0.99^M ≥ 0.5, which could be rewritten as 0.99^M ≤ 0.5.

In other words, as you raise 0.99 to successively higher powers, eventually the resulting product dips below 0.5 .You can try this out on your calculator, or solve directly by taking the logarithm of both sides of the inequality, which gave you M·log(0.99) ≤ log(0.5). Dividing both sides by log(0.99), a negative quantity, resulted in a sign flip: M ≥ log(0.5)/log(0.99). Finally, a useful logarithmic identity was that, for all values of x > 0, log(x) = -log(1/x). That meant you could rewrite the inequality as M ≥ log(2)/log(100/99).

This inequality worked for any logarithmic base, so let’s work in base 10, where log₁₀(2) ≈ 0.30103, and log₁₀(100/99) ≈ 0.0043648. Taking the ratio of these two values meant that M ≥ 68.968. Since M had to be a whole number, the median was 69.

To make sure that’s right, let’s double check the math. If 69 really was the median, then the following two statements had to be true:

• The probability the number of dominoes placed was less than 69 was less than 50 percent.

• The probability the number of dominoes less than or equal to 69 was greater than 50 percent.

The probability that the number of dominoes placed was anywhere from 1 to 68 was 1−0.99⁶⁸, or roughly 49.511 percent. The probability that the number of dominoes placed was anywhere from 1 to 69 was 1−0.99⁶⁹, or roughly 50.016 percent—just on the other side of 50 percent. Sure enough, 69 was the correct answer.

Note that a median of a data set must be a value within the data set.¹ Here, that meant the answer had to be an integer. That said, I was feeling rather generous, so I also accepted answers equivalent to log(2)/log(100/99), or ~68.968.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Chuck Yu 🎻 from Silver Spring, Maryland. I received 45 timely submissions, of which 43 were correct—good for a 96 percent solve rate.

You were placing dominoes again, but this time the probability of knocking each domino over and causing a chain reaction wasn’t 1/100, but rather 10^-k, where k was a whole number. When k = 1, the probability of knocking over a domino was 10 percent; when k = 2, this probability was 1 percent; when k = 3, this probability was 0.1 percent, and so on.

Suppose the expected median number of dominoes placed that initiated a chain reaction was M. As k got very, very large, what value did M/10^k approach?