How Long Is the All-Star Streak?
Many years from now, an all-star team is on a winning streak of at least one game. On average, how long a streak would you expect them to have?
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
The Fiddler Basketball Association’s All-Star Game consists of two teams: “East” and “West.” Every year these two teams play a game, each with a 50 percent chance of winning that’s independent of the outcomes of previous years.
Many, many years into the future, you look at the most recent results of the All-Star Game. On average, what is the longest current winning streak that one of the teams is on? (Here, having won only the most recent game still counts as a “streak” of one game.)
This Week’s Extra Credit
To spice up the All-Star Game, the commissioner of the FBA has decided that there will now be three teams competing in All-Star Games: “Stars,” “Stripes,” and “International.” Each year, two of the three teams play each other. If one year has Stars vs. Stripes, the next year has Stripes vs. International, the year after that has International vs. Stars, and then the cycle repeats with Stars vs. Stripes.
Many, many years after this new format has been adopted, you look at the most recent results of the All-Star game. On average, what is the longest current winning streak that one of the teams is on? (As before, having won one game counts as a “streak.” Also, note that the team with the longest winning streak might not be one of the two teams that played in the most recent All-Star Game.)
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 nugget from Jack Murtagh over at Scientific American:
“There are five children under 10 years old. Their ages are listed as whole numbers. …. The mean, median, mode and range of their ages are all equal. One of the children has a birthday and becomes a year older. The median and mode of the group remain equal, but both change value. What are the children’s ages before … the birthday?”
At first, you might think that the given information doesn’t provide enough constraints to solve for all five children’s ages. But indeed it does!
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.
Interestingly, both ChatGPT-5.1 Reasoning and Gemini 3 Pro missed both of last week’s puzzles. A win for us humans!
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Lise Andreasen 🎻 from Valby, København, Danmark. I received 50 timely submissions, of which 40 were correct—good for an 80 percent solve rate.
Frankie the frog was hopping on a large, packed grid of lily pads, shown below. The pads were circular and each was a distance 1 from its nearest neighbors. (More concretely: Each pad had a diameter of 1 and they were arranged in a hexagonal lattice.) Frankie started at (0, 0), the center of the pad labeled A. Then she hopped due east to pad B at (1, 0), and from there she hopped to pad C at (1.5, √(3)/2).
She wanted to continue hopping in a counterclockwise, spiral-like pattern. Each of her jumps was to the center of a neighboring pad, a net distance of 1. But there were two rules her spiral had to follow:
Each next pad had to be in a more counterclockwise direction (relative to the spiral’s origin at pad A) than the previous pad.
Each pad had to be farther from A than the previous pad.
After a number of hops spiraling around, Frankie realized she was, once again, due east from A. What was the closest to A she could possibly have been? That is, what was the minimum possible distance between the center of the pad she was currently on from the center of pad A?
Our goal was to spiral around once (i.e., from being due east of A to being due east of A again) while staying as close as possible to pad A. So when presented with multiple destination pads, we should consider hopping to pads that were as counterclockwise as possible and/or as close to A as possible. With this in mind, let’s take things one hop at a time.
Frankie’s next hop after C could have been northwest or northeast. Both landing pads were more counterclockwise (with respect to A) and farther from A than C was. However, the northwest pad was both more counterclockwise and closer to A than the alternative. So let’s call that pad D, which was a distance 2 from pad A.
From there, Frankie’s only option was northwest again. Jumping due west meant moving closer to A, while jumping northeast maintained the same angle with respect to A, meaning it wasn’t more counterclockwise. The situation was similar for her next jump as well (west was the same distance from A, while northeast was clockwise), so here was Frankie’s path from A to F:
At this point, Frankie could finally change direction and start traveling due west. But she was stuck traveling in that direction until a pad to the southwest was farther from A than her current pad. That first happened when she was 30 degrees north of west with respect to A.
Extending this reasoning, Frankie could change direction whenever she was 30 degrees north of east, due north, 30 degrees north of west, 30 degrees south of west, due south, or 30 degrees south of east from A. Here’s what her resulting path looked like:
When Frankie was due east, she was a distance 64 from pad A. How interesting that the answer wound up being a power of 2 (more specifically, it was 26).
It turned out that this was no coincidence! Frankie was changing direction every time she was another 60 degrees counterclockwise with respect to A. That meant you could break her path down into 30-60-90 right triangles, as shown below. The hypotenuse of any given triangle wound up being the shorter leg of the next larger triangle. And the ratio between a 30-60-90 triangle’s shortest leg and its hypotenuse is … wait for it … 2!
Thus, when Frankie was first due east of A, her distance from A was 1. When Frankie was 60 degrees north of east, the distance was 2. When she was 60 degrees north of west, the distance was 4. When she was due west, the distance was 8, and so on. Going from due east back to due east represented a rotation of 360 degrees, or six 60-degree turns. Thus, her distance from A at the end was 26 = 64.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Steve Curry 🎻 from Albuquerque, New Mexico. I received 39 timely submissions, of which 34 were correct—good for an 87 percent solve rate.
Frankie had stored all of her food on lily pad A. However, her food had a tendency to “fly” away. Every second, the food that was on every lily pad split up into six equal portions that instantaneously relocated to the six neighboring pads.
At zero seconds, all the food was on lily pad A. After one second, there was no food on pad A, and 1/6 of the food was on each of the surrounding six pads. After two seconds, 1/6 of the food was again on pad A, while the rest of the food was elsewhere.
After how many seconds N (with N > 2) did pad A have less than 1 percent of its original amount?
