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 🎬.

## This Week’s Fiddler

From Andrew E. Love comes a puzzle that’s sure to hook your attention:

A weaving loom set comes with a square with equally spaced hooks along each of its sides, as well as elastic bands that can be attached to the hooks.

Suppose a particular weaving loom has *N* hooks on each side, evenly spaced from one corner to another (i.e., there are two hooks on the two corners and *N*−2 hooks between them). Let’s label the hooks along one side *A*_{1} through *A _{N}*, the hooks on the next clockwise side

*B*

_{1}through

*B*(with

_{N }*A*and

_{N}*B*

_{1}denoting the same hook), the hooks on the third clockwise side

*C*

_{1}through

*C*, and the hooks on the final side

_{N}*D*

_{1}through

*D*.

_{N}Next, let’s use a whole bunch of elastic bands to connect hooks *A*_{1} and *B*_{1}, *A*_{2} and *B*_{2}, *A*_{3} and *B*_{3}, and so on, up to *A _{N}* and

*B*. When

_{N}*N*is 100, here’s what the loom looks like:

As *N* increases, what is the shape of the curve formed by the edges of the bands? Your answer can be a single word or a mathematical equation.

## Extra Credit

Let’s quadruple the number of bands placed on the weaving loom. In addition to the band connecting *A*_{1} and *B*_{1}, you also place bands connecting *B*_{1} and *C*_{1}, *C*_{1} and *D*_{1}, and *D*_{1} and *A*_{1}. You do this for all the sets of hooks from 1 through *N*, so that a total of 4*N* bands have been placed.

When *N* is 100, here is what the loom looks like:

As *N* increases, what fraction of the loom’s area lies between the four sets of bands? In other words, what fraction of the square above does the central white region make up?

## 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 an incredible fact I wasn’t aware of until recently.

Start with any four-digit number. Arrange the digits in decreasing order and call this value *x*. (If multiple digits are the same, that’s fine—just sequence them next to each other.) Next, arrange the digits in increasing order and call this value *y*, being sure to place any zeros at the front. Compute *x*−*y*. That’s your new number, and repeat the process over again.

For example, if your starting number is 3076, then *x* is 7630 and *y* is 0367. Computing the difference gives you your next number in the sequence: 7263. For this next iteration, *x* is 7632 and *y* is 2367, and so on. Eventually, you start to cycle through one or more numbers.

Depending on your initial four-digit number, if you repeat this process long enough, what numbers will you end up cycling through?

For spoilers—and more on the history of the math at play here—give this article a read.

## Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Vishal Doshi 🎻, from Glenview, Illinois. I received 120 timely submissions, of which 102 were correct—good for an 85 percent solve rate. That also set a new record for the number of solvers for a single puzzle here at The Fiddler. Woohoo!

Last week, you were betting on the final two football games of the season for your home team, the Fiddle-delphia Eagles. Thanks to the wonders of time travel, you happened to know that the Eagles would win one game and lose the other. Unfortunately, you couldn’t remember in which order they did so. Maybe the Eagles won the first game and lost the second, or maybe they lost the first and won the second.

You had $100, of which you could bet any amount (including fractions of pennies) that the Eagles would win. So if you had bet *x* dollars on the first game and the Eagles won, you had 100+*x* dollars to bet on the second game. But if the Eagles lost that first game, you were left with 100−*x* dollars to bet on the second game.

You wanted to implement a betting strategy that *guaranteed* you’d have as much money as possible after both games. If you did so, then after the two games, how much money were you guaranteed to have? (**Note:** By “guaranteed to have a certain amount of money,” I meant that you always had at least that much, no matter which specific order of wins and losses occurred.)

Let’s tackle this head on. (See what I did there?) If you had bet *x* dollars on the first game and the Eagles won, you then had 100+*x* dollars. Next, because they had won the first game, the Eagles were destined to lose the second game. So you wouldn’t bet anything for the second game, which meant you retained your 100+*x* dollars.

However, if the Eagles had lost that first game, you were left with 100−*x* dollars. The good news was that they were destined to win the second game. Putting all 100−*x* dollars on the line meant you finished with 2(100−*x*) dollars.

Now, you wanted to *guarantee* you’d have as much money as possible after both games, regardless of whether the Eagles won or lost that first game. The graph below shows how much money you would have finished with if the Eagles won (red line) or lost (blue line) the first game, for different values of *x*.

When *x* was less than $33.33 (really that’s 33.333… dollars, but for brevity I’ll omit that last third of a cent), you finished with less than $133.33 when the Eagles won their first game. In other words, you didn’t bet enough. And when *x* was greater than $33.33, you finished with less than $133.33 when the Eagles lost their first game. In other words, you bet too much.

But when *x* was $33.33 (and a third of a cent) on the nose, you were guaranteed to win $133.33 (and a third of a cent). As noted by Huzaifa Abbasi, an efficient way to solve for that optimal value of *x* was to set the quantities 100+*x*, your total when the Eagles *won* the first game, and 2(100−*x*), your total when the Eagles *lost* the first game, equal to each other. Sure enough, that gave you a value of $33.33, with which you could win a guaranteed **$133.33**—the *most* you could have had in guaranteed winnings.

## Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Michael Coffey 🎻, from Melbourne, Australia. For the Extra Credit, I received 53 timely submissions, 39 of which were correct—good for a 74 percent solve rate.

For the Extra Credit, there were now 12 games remaining in the season rather than two, and you knew the Eagles would win exactly eight of the 12 games. But, as before, you had no idea *which* eight of the 12 games they would win specifically.

Moreover, instead of only betting on the Eagles to win in each game, you could now bet on either team winning—the Eagles or their opponent.

Once again, you wanted to implement a betting strategy that *guaranteed* you’d have as much money as possible after the 12 games were complete. If you did so, then after the 12 games how much money were you guaranteed to have?

While the numbers of wins and losses in this puzzle were specific, several solvers cleverly attacked a generic version of this problem. Instead of eight wins and four losses, suppose the Eagles had *m* wins and *n* losses remaining in their schedule. Solver Sam Miner defined the function *f*(*m*, *n*) as the amount of money (divided by your starting amount of money) you could optimally generate off these remaining games. When either *m* or *n* was zero (meaning the Eagles won out or lost out), you could double your money each round. Mathematically, that meant *f*(*m*, 0) = 2* ^{m}* and

*f*(0,

*n*) = 2

*.*

^{n}Next, Sam found a recursive relationship. Suppose you already knew the values of *f*(*m*−1, *n*) and *f*(*m*, *n*−1) for specific values of *m* and *n*. Could you use these values to somehow calculate *f*(*m*, *n*)? Absolutely!

Here’s how: Suppose *m* is greater than or equal to *n*, and that you bet a fraction *x* of your money on the Eagles to win the next game. If they do win, and you continue to bet optimally, your final total will be (1+*x*)·*f*(*m*−1, *n*). If they lose, your final total will be (1−*x*)·*f*(*m*, *n*−1). As with last week’s Fiddler, you could maximize your *guaranteed* earnings by setting these two expressions equal to each other, so that (1+x)·*f*(*m*−1, *n*) = (1−*x*)·*f*(*m*, *n*−1). Solving for *x* gives you *x* = [*f*(*m*, *n*−1) − *f*(*m*−1, *n*)] / [*f*(*m*, *n*−1) + *f*(*m*−1, *n*)]. Finally, plugging this value of *x* back into either of the two expressions for your earnings meant your guaranteed final total was *f*(*m*, *n*) = 2 / [1/f(*m*−1, *n*) + 1/f(*m*, n−1)]. If *m *had been less than *n*, then you would have instead bet *against* the Eagles, resulting in the same formula.

While this recursive formula wasn’t pretty, it was enough to calculate *f*(8, 4), which solvers like Joe Kelley did via spreadsheet. Computing the function gave a result of 4096/495, or roughly 8.2747. Since you started with $100, that meant your guaranteed total after the 12 games was 100·8.2747…, or **about $827.47**.

Several solvers like Joel Lewis found a nicer, closed-form expression for the function: *f*(*m*, *n*) = 2^{m}^{+}* ^{n}*/(

*m*+

*n*choose

*m*). Note that

*f*(1, 1) = 2, rather than the 1.33… from last week’s Fiddler, since the Extra Credit allowed you to bet on the Eagles winning or losing, rather than just on them winning.

By the way, solvers Emilie Mitchell and Andrew Love, Jr. both observed how this puzzle is related to the Kelly criterion, which apparently gives the exact same answer. If (like me) you hadn’t heard of this previously, it makes for a fascinating read.

## 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.

For those planning to make future bets based on last week's Extra Credit:

The optimum fraction of your money to bet before each game is:

x = |m-n|/(m+n)

or, the difference in wins remaining for the two teams, divided by the total number of games remaining. Much simpler than I'd expect given the complexity of the problem.