Can You Meet Me at the Mall?

Some friends arrive at the mall at random times within a one-hour window, each waiting 15 minutes. On average, what’s the maximum number of friends simultaneously at the mall?

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. 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. 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 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

This week’s puzzle is a very special collaboration between Fiddler on the Proof and Science News, where four math puzzles that are related were posted earlier this morning. The second and third among these are also appearing here as this week’s Fiddler and Extra Credit. Now—on to those puzzles!

You and two friends have arranged to meet at a popular downtown mall between 3 p.m. and 4 p.m. one afternoon. However, you neglected to specify a time within that one-hour window. Therefore, the three of you will be arriving at randomly selected times between 3 p.m. and 4 p.m. Once each of you arrives at the mall, you will be there for exactly 15 minutes. When the 15 minutes are up, you leave.

At some point (or points) during the hour, there will be a maximum number of friends at the mall. This maximum could be one (sad!), two, or three. On average, what would you expect this maximum number of friends to be?

This Week’s Extra Credit

Instead of three total friends, now suppose there are four total friends (yourself included). As before, you all arrive at random times during the hour and each stay for 15 minutes.

On average, what would you expect the maximum number of friends meeting up to be?

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 few other math puzzles that appeared earlier this year over at Science News, authored by none other than Ben Orlin!

• Imagine there’s no zero

• The Lesser Fool

• The conundrum of sharing

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: 🎻 Ted Carniol 🎻 from Vienna, Virginia. I received 86 timely submissions, of which 73 were correct—good for an 85 percent solve rate.

In bowling, you roll a large, heavy ball toward a triangular array of 10 pins many feet away. You’d think the objective is to just knock down as many pins as possible. But thanks to the sport’s byzantine scoring system, it’s not that simple.

Bowling consists of 10 “frames,” and in each frame you have up to two chances to knock down the 10 pins. Your score for the frame is however many pins you knock down, but there are bonuses if you knock down all 10 pins:

• If you knock down all 10 in one bowl, it’s called a “strike.” Your score for the frame is 10 plus however many pins you knock down in your next two bowls. Note that these next two bowls might occur in a single frame, or could be across the next two frames (if your very next bowl is again a strike).

• If it takes you two bowls to knock down all 10 pins, it’s called a “spare.” Your score for the frame is 10 plus however many pins you knock down in your next one bowl.

Something extra may be needed for the 10th and final frame. If you get a strike on the 10th frame, you get two extra rolls to determine your score for that frame. And if you get a spare, you similarly get a final roll to determine your score. The maximum score for a game is 300, earned from 12 consecutive strikes—10 strikes in the 10 frames, and then two additional strikes to compute the score for the 10th (and ninth) frame.

Now that we’ve reviewed how to score a game, let’s get to last week’s actual puzzle!

If you knocked down 100 pins over the course of a game, you were guaranteed to have a score that was at least 100. But what was the minimum total number of pins you needed to knock down such that you could attain a score of at least 100?

To get a high score without knocking over too many pins, you wanted to generate as many bonus points as possible. And the best way to get bonus points was to bowl consecutive strikes. Let’s start off with a few strikes and see what happens!

• Suppose you bowled one strike, and that was it. The remaining frames were all gutter balls, good for zero points. Your score was 10—a far cry from 100.

• Suppose you bowled two strikes, and that was it. You earned 10 points on the second frame, and 10 + 10 = 20 points on the first frame, for a total of 30—still a far cry from 100, but getting there.

• Suppose you bowled three strikes, and that was it. You earned 10 points on the third frame, and 10 + 10 = 20 points on the second frame, and 10 + 10 + 10 = 30 points on the first frame, for a total of 60—more than halfway to 100!

• Suppose you bowled four strikes, and that was it. You earned 10 points on the fourth frame, 10 + 10 = 20 points on the third frame, and 10 + 10 + 10 = 30 points on the second and first frames, for a total of 90—almost there!

• Suppose you bowled five strikes, and that was it. You earned 10 points on the fifth frame, 10 + 10 = 20 points on the fourth frame, and 10 + 10 + 10 = 30 points on the third, second, and first frames, for a total of 120. In knocking down 50 pins, you made it to 100, albeit with an excess of 20 points.

The answer was therefore at most 50. At this point, you wanted to see how many fewer pins you could knock down while still scoring at least 100. If you left a pin standing from any of the first three frames, the score for that frame went from 30 down to nine or less, meaning you no longer broke 100 points. You could have left some pins standing in the fourth frame, but your best option was to look at the fifth frame.

Instead of bowling a strike, suppose you knocked down N pins in the fifth frame. Your scores for the first two frames were still 30, but now your score for the third frame was 20 + N, your score for the fourth frame was 10 + N, and your score for the fifth frame was N. In total, your score was now 90 + 3N. For this to be at least 100, you needed 90 + 3N ≥ 100, which meant N ≥ 10/3. Since N had to be a whole number, the minimum value was 4.

In short, if you bowled four strikes in a row, and then knocked down four pins on the very next frame—a total of 44 pins knocked down—you scored 102 points. So 44 was the answer!

By the way, 102 points on 44 pins knocked down gave you an average of about 2.32 points per pin. That average wasn’t too far off from the figure for a perfect game, 2.5 points per pin. (Maximizing this average would have been an interesting puzzle in its own right. It turns out that you can do slightly better than the 2.5 points per pin from a perfect game.)

Finally, the following graph from solver 🎬 Peter Ji 🎬 shows the maximum attainable score as a function of pins knocked down. Sure enough, the score passed 100 at 44 pins. While the relationship was fairly linear, it became nonlinear at the extremes (i.e., zero and 120 pins).

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Tate Chutkow 🎻 from Bloomfield Hills, Michigan. I received 62 timely submissions, of which 42 were correct—good for a 68 percent solve rate.

In last week’s Extra Credit, you and an opponent each played a single game in which you both knocked down the same number of pins. However, your scores were quite different.

Your opponent remarked, “Given only the information that we knocked down the same number of pins in our two games, there’s no way the difference between our scores could have been any greater!”

What was this difference between your two scores?