Can You Make It Home in Time for Pancakes?

Three siblings travel home (for pancakes, of course) at different speeds. How quickly can they make it home, with piggybacking and some clever teamwork?

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 Nis Jørgensen comes a pernicious puzzle of pancakes:

Three siblings are at a playground: Alice, Bob, and Carey. Alice, the oldest, gets a call from their dad—their pancake dinner is ready! But they won’t get to eat until all three kids are home.

They each walk home at a different constant speed. Alice can walk home in 10 minutes, Bob can do it in 20, and Carey in 30. Fortunately, any of the kids can carry any of the others on their back without reducing their own walking speed. (However, none of them can carry a kid who is, in turn, carrying another kid.) Assume that they can pick someone up, set someone down, and change direction instantaneously.

What is the fastest they can all get home? (Your answer should be in minutes.)

Extra Credit

Nis Jørgensen also offered the following Extra Credit:

Suppose Alice, Bob, and Carey have a fourth sibling named Dee, who is with them at the playground. Dee is the slowest, needing 60 minutes to walk home. As before, any kid can carry any other kid, and they won’t start eating until everyone is home.

Once again, what is the fastest they can all get home? (Your answer should be in minutes.)

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 literary classic from John Conway (seriously!) that I only learned about recently. I’ll paraphrase below:

Suppose you have an equilateral triangle whose side lengths are infinitesimally greater than the whole number n. What is the minimal number of “unit equilateral triangles” (i.e., equilateral triangles of side length 1) you need to completely cover the larger one, in terms of n? Note that these unit triangles are allowed to overlap.

When the sides of the unit triangles are required to be parallel to those of the larger one, the answer was proven (spoiler alert!) in 2023. If the sides aren’t required to be parallel, I think this remains an open problem.

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.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Nigel 🎻 from Sanderstead, England. I received 81 timely submissions, of which 60 were correct—good for a 74 percent solve rate. I was pleased to see a very healthy increase in the number of solvers as compared to two weeks ago. Way to bounce back, everyone!

Last week, two people were sitting at a table together, each with their own bag of six “DnD dice”: a d4, a d6, a d8, a d10, a d12, and a d20. Here, “dX” referred to a die with X faces, numbered from 1 to X, each with an equally likely probability of being rolled.

Both people randomly picked one die from their respective bags and then rolled them at the same time. For example, suppose the two dice selected were a d4 and a d12. The players rolled them, and both rolls come up as 3. What luck!

What was the probability of something like this happening? That is, what was the probability that both players rolled the same number, whether or not they happened to pick the same kind of die?

A good first step was to work out the probability of rolling each number. For example, what was the probability of rolling a 20?

To figure this out, many readers counted up the faces on all the dice. The d4 had 4 faces, the d6 had 6 faces, and so on. In total, there were 60 faces. Only one of these was a 20, which meant the probability of rolling a 20 was 1/60. Right?

Not quite. Why? By this logic, you had a 4-in-60 chance of picking the d4, a 6-in-60 chance of picking the d6, and so on. However, the puzzle said that each player picked a die at random, implying that each die had the same 1-in-6 chance of being selected. So the probability of rolling a 20 was not in fact 1/60. First, you needed to pick the d20 (a 1-in-6 chance) and then, given that you were rolling the d20, you needed to actually roll a 20 (a 1-in-20 chance). Combining these meant the probability of rolling a 20 was in fact (1/6)·(1/20), or 1/120. The probability was the same for rolling 19, 18, 17, 16, 15, 14, and 13.

Meanwhile, your chances of rolling a 12 were greater, since there were two dice, the d20 and the d12, that could have resulted in a 12. As before, you had a 1-in-20 chance of picking the d20 and a 1-in-20 chance of then rolling the 12. But this time, you also had a 1-in-6 chance of picking the d12 and a 1-in-12 chance of then rolling a 12. Combining these cases meant the probability of rolling a 12 was (1/6)·(1/20) + (1/6)·(1/12), which simplified to 1/120 + 1/72, or 1/45. The probability was the same for rolling 11.

Continuing in this fashion (i.e., adding up the respective probabilities of the dice that could have resulted in each roll), you got the following:

• The probability of rolling 20, 19, 18, 17, 16, 15, 14, or 13 was 1/120.

• The probability of rolling a 12 or 11 was 1/45.

• The probability of rolling a 10 or 9 was 7/180.

• The probability of rolling an 8 or 7 was 43/720.

• The probability of rolling a 6 or 5 was 7/80.

• The probability of rolling a 4, 3, 2, or 1 was 31/240.

At this point, I’d like to quickly note that you were more likely than not to roll a 1, 2, 3, or 4, as this collective probability was approximately 51.7 percent. I found that a little surprising.

Okay, back to the puzzle. So far, we’ve worked out all the probabilities for one person’s roll. With two people rolling, you could determine the probability of each pair of rolls by creating a grid in which each row was one player’s roll and each column was the other player’s roll. Since the rolls were independent, the probability of the pair was the product of two respective probabilities.

Here’s what that grid looked like, with likelier pairs shown in yellow:

The cells outlined in black along the main diagonal of the grid were precisely when both players rolled the same value. And so, the answer to the puzzle was the sum of these 20 cells, also known as the trace of the grid (or “matrix”).

The exact answer was 4(31/240)² + 2(7/80)² + 2(43/720)² + 2(7/180)² + 2(1/45)² + 8(1/120)², which simplified to 3/32, or 9.375 percent.

Several solvers, like Sameer G. of Ithaca, New York, neatly arrived at the same result with a little less computation. Sameer worked out the probabilities with which the players had the same roll, for each possible pair of dice they picked. For example, if both players picked a d4, the probability they had the same roll was 1/4. If one picked a d4 and the other picked a d6, the probability was now 1/6. In general, if the two dice were dX and dY, the rolls were the same with probability 1/max(X, Y). Computing the average probability across all 36 possible pairs for X and Y, Sameer arrived at the same result of 3/32. 

In the end, despite having six different dice to choose from, the probability that both players rolled the same number wasn’t too much less than it would have been if they had both rolled a standard die (i.e., d6). This was largely thanks to the fact that every die had the numbers 1 through 4, which accounted for the majority of the time the players had the same roll.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Bogdan Lataianu 🎻 from Saskatoon, Canada. I received 39 timely submissions, of which 29 were correct—good for a 74 percent solve rate, matching the rate from last week’s Fiddler.

For Extra Credit, instead of two people sitting at the table, there were three. Again, all three players randomly picked one die from their respective bags and rolled them at the same time.

For example, suppose the three dice selected were a d4, a d20, and a d12, and that the d4 came up 4, the d20 came up 13, and the d12 came up 4. In this case, there were two distinct numbers (4 and 13) among the three rolls.

On average, how many distinct numbers would you have expected to see among the three rolls?