Can You Break the Bell Curve?

This week, you’ll look at a biased bean machine that favors one direction. For Extra Credit, you’ll produce a curve that is decidedly not bell-shaped.

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

Bean machines can famously produce bell-shaped curves. But today, we’re going to change all that!

Suppose you have a board like the one shown below. The board’s topmost row has three pins (and two slots for a ball to pass through), while the bottommost row has two pins (and three slots for a ball to pass through). The remaining rows alternate between having three pins and two pins.

But instead of the 12 rows of pins in the illustrative diagram, suppose the board has many, many rows. And at the very bottom of the board, just below the two bottommost pins, are three buckets, labeled A, B, and C from left to right.

Whenever a ball encounters one of the leftmost pins, it travels down the right side of it to the next row. And whenever a ball encounters one of the rightmost pins, it travels down the left side of it to the next row.

But this isn’t your garden variety bean machine. Whenever a ball encounters any of the other pins, it has a 75 percent chance of traveling down the right side of that pin, and a 25 percent chance of traveling down the left side of that pin.

A single ball is about to be dropped into the left slot at the top of the board. What is the probability that the ball ultimately lands in bucket A, the leftmost slot at the bottom?

This Week’s Extra Credit

Suppose you have the board below, which starts with a row with six pins (and five slots), followed by a row with five pins (and six slots), and then back to six pins in an alternating pattern. Balls are only allowed to enter through the middle three slots on top. This time around, the pins that aren’t on the far left or far right behave normally, meaning each ball is equally likely to go around it via the left side or the right side.

Your goal is to create a trapezoidal distribution along one of the rows with six slots, which have been highlighted with dotted lines in the diagram above. More precisely, the frequency of balls passing through the six slots from left to right should be x−y, x, x+y, x+y, x, x−y, for some values of x and y with x ≥ y.

Suppose the ratio by which you drop balls into the top three middle slots, from left to right, is a : b : c, with a + b + c = 1. Find all ordered triples (a, b, c) that result in a trapezoidal distribution in at least one row with six slots.

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 something a little different: a digital escape room created by one of my colleagues at Amplify, Sean Sweeney, in collaboration with his children, Scottie and Vivi.

Actually, make that two digital escape rooms. The first is called “The House,” and the second is “The Museum.”

I love a good escape room, and these hit the spot. I hope you enjoy them!

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: 🎻 Paul Juster 🎻 from Walnut Creek, California. I received 89 timely submissions, of which 84 were correct—good for a 94 percent solve rate. That was the greatest number of solvers for a puzzle dating back to February. Nice job, everyone!

In Season 2 of Squid Game, contestants played a game of “Mingle” (spoiler alert!). For each round of this game, the contestants all waited in a common area until a number was called. They then had to split up into groups of that number. Any contestants not in a group of the correct number after 30 seconds … well, let’s just say bad things happened to them. Here, we referred to contestants who made it through a round in a group of the correct size as having “survived” the round.

Last week, you started with N contestants.

In the first round, contestants had to form groups of 1. Okay, that wasn’t too hard; everyone survived. In the second round, contestants had to form groups of 2. This continued (with groups of k in round k) until the 38th round, when contestants had to form groups of 38.

What was the smallest number N such that there was at least one person who survived all 38 rounds?

First off, it was impossible to have exactly one person survive all 38 rounds. (Note that I intentionally asked for the smallest N for which there was “at least one” survivor.) In the final round, contestants formed groups of 38, so the headcount had to be a multiple of 38: 0, 38, 76, 114, etc. Since we wanted N to be small, the final count, which was required to be nonzero, was presumably the smallest positive multiple of 38, which was precisely 38.

One approach was to work your way up from smaller values of N and see how many rounds the contestants would make it through:

• 1 → 1 (groups of 1) → 0 (groups of 2). They’d last 1 round.

• 2 → 2 (groups of 1) → 2 (groups of 2) → 0 (groups of 3). They’d last 2 rounds.

• 3 → 3 (groups of 1) → 2 (groups of 2) → 0 (groups of 3). They’d last 2 rounds.

• 4 → 4 (groups of 1) → 4 (groups of 2) → 3 (groups of 3) → 0 (groups of 4). They’d last 3 rounds.

• 5 → 5 (groups of 1) → 4 (groups of 2) → 3 (groups of 3) → 0 (groups of 4). They’d last 3 rounds.

• 6 → 6 (groups of 1) → 6 (groups of 2) → 6 (groups of 3) → 4 (groups of 4) → 0 (groups of 5). They’d last 4 rounds.

• 7 → 7 (groups of 1) → 6 (groups of 2) → 6 (groups of 3) → 4 (groups of 4) → 0 (groups of 5). They’d last 4 rounds.

• 8 → 8 (groups of 1) → 8 (groups of 2) → 6 (groups of 3) → 4 (groups of 4) → 0 (groups of 5). They’d last 4 rounds.

• 9 → 9 (groups of 1) → 8 (groups of 2) → 6 (groups of 3) → 4 (groups of 4) → 0 (groups of 5). They’d last 4 rounds.

• 10 → 10 (groups of 1) → 10 (groups of 2) → 9 (groups of 3) → 8 (groups of 4) → 5 (groups of 5) → 0 (groups of 6). They’d last 5 rounds.

In looking at values of N from 1 to 10, a few things were apparent. The number of rounds generally increased with N, in a way that was monotonic, but not strictly monotonic. In other words, the number of rounds stayed the same or increased, but never decreased.

Also, increases didn’t necessarily occur for numbers with more factors. For example, 8 and 9 are both composite whereas 7 is prime, and yet all three values of N resulted in 4 rounds. This was because they all reduced to 6 survivors after the first two rounds and proceeded in the same fashion from there.

Since the growth pattern wasn’t immediately obvious (to me, at least), it didn’t surprise that many solvers kept going—some with the help of code or spreadsheets—until they found the first value of N that lasted 38 rounds. That first value of N turned out to be 454. To prove this, it sufficed to show that 453 people did not make it 38 rounds, while 454 people did:

• 453 → 453 (groups of 1) → 452 (groups of 2) → 450 (groups of 3) → 448 (groups of 4) → 445 (groups of 5) → 444 (groups of 6) → 441 (groups of 7) → 440 (groups of 8) → 432 (groups of 9) → 430 (groups of 10) → 429 (groups of 11) → 420 (groups of 12) → 416 (groups of 13) → 406 (groups of 14) → 405 (groups of 15) → 400 (groups of 16) → 391 (groups of 17) → 378 (groups of 18) → 361 (groups of 19) → 360 (groups of 20) → 357 (groups of 21) → 352 (groups of 22) → 345 (groups of 23) → 336 (groups of 24) → 325 (groups of 25) → 312 (groups of 26) → 297 (groups of 27) → 280 (groups of 28) → 261 (groups of 29) → 240 (groups of 30) → 217 (groups of 31) → 192 (groups of 32) → 165 (groups of 33) → 136 (groups of 34) → 105 (groups of 35) → 72 (groups of 36) → 37 (groups of 37) → 0 (groups of 38). They’d last 37 rounds.

• 454 → 454 (groups of 1) → 454 (groups of 2) → 453 (groups of 3) → 452 (groups of 4) → 450 (groups of 5) → 450 (groups of 6) → 448 (groups of 7) → 448 (groups of 8) → 441 (groups of 9) → 440 (groups of 10) → 440 (groups of 11) → 432 (groups of 12) → 429 (groups of 13) → 420 (groups of 14) → 420 (groups of 15) → 416 (groups of 16) → 408 (groups of 17) → 396 (groups of 18) → 380 (groups of 19) → 380 (groups of 20) → 378 (groups of 21) → 374 (groups of 22) → 368 (groups of 23) → 360 (groups of 24) → 350 (groups of 25) → 338 (groups of 26) → 324 (groups of 27) → 308 (groups of 28) → 290 (groups of 29) → 270 (groups of 30) → 248 (groups of 31) → 224 (groups of 32) → 198 (groups of 33) → 170 (groups of 34) → 140 (groups of 35) → 108 (groups of 36) → 74 (groups of 37) → 38 (groups of 38) → 0 (groups of 39). They’d last 38 rounds.

Boy, figuring this all out from N = 1 to N = 454 was a lot of work! That’s why many clever solvers, like Rui Viana and Tom Singer, worked backwards. If you assumed there were 38 survivors coming out of the 38th round, then you needed a multiple of 37 that was at least 38 coming out of the 37th round. The smallest such multiple was 2·37, or 74. Then, you needed the smallest multiple of 36 that was at least 74 coming out of the 36th round, which was 3·36, or 108. If you kept going in this fashion, you could work your way right back to 454. Solver 🎬 Thomas Chick 🎬 made an awesome video showing this backtracking approach. Also, kudos to to Thomas for recognizing that the solution was suspiciously close to 456. (I could have picked any number of rounds for the puzzle; why else would I have picked 38?)

By the way, here’s a graph showing the survivors by round when you started with 453 (in blue) vs. 454 (in red):

These graphs were fascinating in their own right, and they appeared to be a quadratic in nature. In round k, the number of contestants eliminated was on the order of k, so after k rounds it made sense that the total number of contestants eliminated was on the order of k².

Solver Josh Silverman explored the relationship between the number of rounds and the minimum number of contestants needed for survivors. Josh found that the latter was approximately 0.32·N², which was indeed quadratic!

Finally, the answer of 454 and the sequence shown Josh’s graph have already been logged in the OEIS. Apparently, Erdös proved the sequence behaves like N²/𝜋, which is pretty darn close to Josh’s result. What was even more surprising, perhaps, was that the sequence has been named, “Smallest number of stones in Tchoukaillon (or Mancala, or Kalahari) solitaire that make use of n-th hole.” So there’s a mathematical connection between “Mingle” from Squid Game and Mancala. Awesome!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Lise Andreasen 🎻 from Valby, Copenhagen, Denmark. I received 47 timely submissions, of which 44 were correct—good for a 94 percent solve rate. And that was the greatest number of solvers for an Extra Credit dating back to February. Again, nice job, everyone!

There were N contestants about to play Mingle, where N was less than 500. This time, in each round, the number for group size could have been anywhere from 1 to 10, with each number equally likely to be called. Also, the numbers called in the rounds were all independent of each other. It appeared this version of Mingle would continue until there were no more survivors.

Before the game started, one of the contestants, Young-il, did some quick computation. He said to his companion, Gi-hun: “Oh no, it’s too bad we only have N people. If we had started with N+1 people instead, I’d expect the game to last about 10 more rounds, on average, than I’d expect when starting with N people.”

What was the value of N?