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
Frankie the frog is hopping on a large, packed grid of lily pads, shown below. The pads are circular and each is a distance 1 from its nearest neighbors. (More concretely: Each pad has a diameter of 1 and they are arranged in a hexagonal lattice.) Frankie starts at (0, 0), the center of the pad labeled A. Then she hops due east to pad B at (1, 0), and from there she hops to pad C at (1.5, √(3)/2).
She wants to continue hopping in a counterclockwise, spiral-like pattern. Each of her jumps is to the center of a neighboring pad, a net distance of 1. But there are two rules her spiral must follow:
Each next pad must be in a more counterclockwise direction (relative to spiral’s origin at pad A) than the previous pad.
Each pad must be farther from A than the previous pad.
After a number of hops spiraling around, Frankie realizes she is, once again, due east from A. What is the closest to A she could possibly be? That is, what is the minimum possible distance between the center of the pad she’s currently on from the center of pad A?
This Week’s Extra Credit
Frankie has stored all of her food on lily pad A. However, her food has a tendency to “fly” away. Every second, the food that’s on every lily pad splits up into six equal portions that instantaneously relocate to the six neighboring pads.
At zero seconds, all the food is on lily pad A. After one second, there’s no food on pad A, and 1/6 of the food is on each of the surrounding six pads. After two seconds, 1/6 of the food is again on pad A, while the rest of the food is elsewhere.
After how many seconds N (with N > 2) will pad A have less than 1 percent of its original amount?
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 fun article (with colorful but not-to-scale diagrams) from The New York Times, titled “Infinite Pancakes, Anyone?”
It chronicles some recent mathematical research that builds off a classic puzzle that asks how many regions you can divide an infinite plane (or circle) into using N straight cuts. But things take a turn when these cuts are no longer straight, or semi-infinite rather than infinite.
For example, suppose your cuts are rays, extending forever in one direction but not the other. Or, better yet, suppose your cuts are two rays connected by a segment, forming an A shape whose bottom extends forever. Given N of these “cuts,” what is the greatest number of distinct regions you can carve out of the infinite plane?
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: 🎻 Rob Niederman 🎻 from Woodbridge, Virginia. I received 63 timely submissions, of which 49 were correct—good for a 78 percent solve rate.
A game of bingo typically consists of a 5-by-5 grid with 25 total squares. Each square (except for the center square) contains a number. When a square’s number is called, you place a marker on that square. The goal is to get “bingo,” which is five squares in a row, either across, down, or along one of the two long diagonals. The center square, which doesn’t have a number, is labeled “Free,” and begins with a marker on it before any numbers are called. Here’s an example of a winning 5-by-5 grid in which 10 squares (other than the “Free” square) have been marked:
Last week, you considered a smaller version of the game with a 3-by-3 grid: a “Free” square surrounded by eight other squares with numbers. Each of these eight squares was equally likely to be called, and without replacement (i.e., once a number was called, it didn’t get called again).
On average, how many markers did you have to place until you got “bingo” in this 3-by-3 grid? (The “Free” square didn’t count as one of the markers—it was “free”.)
Before calculating the average, let’s find some bounds. What were the fewest markers you could have placed to get a bingo, and what were the most?
The fewest was two—you could place one marker anywhere, and as long as the next marker was diametrically opposed from that first one you had a bingo. Meanwhile, the greatest number was five. When placing four markers (again, not counting the “free” space), it was possible to avoid having any bingos, as shown below. But no matter where you placed the fifth marker, you were guaranteed to have a bingo.
With these bounds, the number of markers placed upon getting a bingo had to be two, three, four, or five, each with some probability. Thus, the average was greater than two and less than five—but what was it exactly?
To get a good estimate, solver 🎬 Leonardo Palomera Tapia 🎬 simulated 100,000 games and calculated the average number of markers needed to win, which he found to be around 3.47 or 3.48. Leonardo’s resulting histogram for the number of markers placed is shown below (along with the histogram for the Extra Credit, which we’ll get to shortly). Sure enough, the minimum number of markers needed was two, while the maximum was five.
Of course, this was still just an approximation. To find the exact result, you could map out the various ways markers were placed on the grid. Technically, there were 8!, or 40,320 orders in which markers could have been placed. That said, the order in which markers were placed didn’t matter so much as the state of the board itself, of which there were 28, or 256, possibilities. That said, if you strategically considered states to be equivalent when you transform from one to another via rotation or reflection, you could reduce this to just 13 distinct non-bingo states.
Here, these 13 states (labeled A through M) are in a single diagram, along with the probabilities for transitioning from one state to another with the next marker:
When the transition probabilities emanating from a state didn’t add up to 1, that was because the remaining probability resulted in a bingo.
From this transition diagram, you could delineate all the possible paths to a bingo and sum their respective probabilities … or you could work backwards (from right to left in the diagram) and compute the average number of counters needed for each state. Let’s do the latter! For simplicity, we’ll use the variables A through M to indicate the expected number of markers needed for bingo from the corresponding state.
M = 1
L = 2/5·(1+M) + 3/5·1 = 7/5
K = 2/5·(1+M) + 3/5·1 = 7/5
J = 1/5·(1+M) + 4/5·1 = 6/5
I = 1/5·(1+M) + 4/5·1 = 6/5
H = 1
G = 1/6·(1+I) + 1/6·(1+J) + 1/6·(1+K) + 1/6·(1+L) + 2/6·1 = 28/15
F = 1/6·(1+H) + 2/6·(1+I) + 1/6·(1+K) + 2/6·1 = 9/5
E = 2/6·(1+J) + 1/6·(1+L) + 3/6·1 = 49/30
D = 1/6·(1+H) + 1/6·(1+I) + 1/6·(1+J) + 3/6·1 = 47/30
C = 2/7·(1+D) + 2/7·(1+F) + 2/7·(1+G) + 1/7·1 = 262/105
B = 2/7·(1+D) + 2/7·(1+E) + 2/7·(1+G) + 1/7·1 = 257/105
A = 4/8·(1+B) + 4/8·(1+C) = 243/70
Thus, the expected number of markers to get bingo was 243/70, or about 3.471. This agreed nicely with Leonardo’s simulations. As long as you provided an exact answer or a reasonable decimal approximation (e.g., 3.47), I awarded full credit.
A neat fact from the analysis above was this: If your first marker was on one of the four corner spaces (rather than along one of the four edge spaces), it then required slightly fewer markers on average to get a bingo. That made intuitive sense because, corners could be part of three-in-a-row across or down (like edge spaces), but corners could additionally be part of three-in-a-row diagonally.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Adnan Haque 🎻 from New York, New York. I received 39 timely submissions, of which 37 were correct—good for a 95 percent solve rate.
For Extra Credit, instead of a 3-by-3 grid, you returned to the original 5-by-5 grid.
On average, how many markers did you need to place until you got “bingo”? (As before, the “Free” square didn’t count as one of the markers—it was “free”.)
