# Can You Defeat The TikTok Meme?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.

## Riddler Express

From Kyle Willstatter comes a puzzle that’s right on target:

You’re playing darts and trying to maximize the number of points you earn with each throw. You are deciding which sector to aim for. Your dart has a 50 percent chance of landing in that sector and a 25 percent chance of landing in one of the two neighboring sectors. Reading clockwise, the sectors are worth 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12 and 5 points, as shown below. (For the purposes of this puzzle, don’t worry about the bullseye, the outer ring that’s worth double or the inner ring that’s worth triple.)

Which sector should you aim for to maximize your expected score?

*Extra credit: *How would *you* “fairly” (by some definition of fair for you to define) assign the point values around a dartboard? Explain your thinking.

## Riddler Classic

From Angela Zhou comes a challenging meme analysis:

The #blindletterchallenge has recently taken TikTok by storm. In this challenge, you are presented with five letters, one at a time. Letters are picked randomly, but you can assume that no two letters are the same (i.e., letters are picked without replacement). As each letter is presented, you must identify which of five slots you will place it. The goal is for the letters in all five slots to be in alphabetical order at the end.

For example, consider an attempt at the challenge by Michael DiCostanzo. The first letter is X. Since this occurs relatively late in the alphabet, he puts this in the fifth slot. The second letter is U. He puts that in the fourth slot, since it also comes relatively late (and the fifth slot is already occupied). Next, the third letter is E. He takes a gamble, and places E in the first slot. The fourth letter is D. Since D comes before E alphabetically, but no slots prior to E are now available, Michael loses this attempt.

If you play with an optimal strategy, always placing letters in slots to maximize your chances of victory, what is your probability of winning?

## Solution to the last Riddler Express

Congratulations to 👏 Amy Leblang 👏 of Wayland, Massachusetts, winner of last week’s Riddler Express.

Last week, you were introduced to two friends who had birthdays on Feb. 9 and Nov. 18. When written numerically in MM/DD formatting, these dates were 02/09 and 11/18. Interestingly, the latter date included both the sum and the product of the values in the former date: 11 = 02 + 09 and 18 = 02 × 09.

How many pairs of dates were there such that one of the dates included both the product and the sum of the values in the other date (in either order)? Here, the order of the dates in the pair didn’t matter, so “02/09 and 11/18” was considered the same as “11/18 and 02/09.”

It turned out that there were quite a few such pairs. The key here was to carefully organize your counting so that you didn’t double-count any dates or miss any of them. Suppose the first date was A/B. Solver Jenny Mitchell distinctly considered when the second date was “sum-first” — A+B/A·B — or “product-first” — A·B/A+B. Sometimes, only one of these was possible. And occasionally, they were the same.

Here are the “sum-first” possibilities, organized by A:

- When A was 1, the first date could be anywhere from 01/01 to 01/11.
- When A was 2, the first date could be anywhere from 02/01 to 02/10.
- When A was 3, the first date could be anywhere from 03/01 to 03/09.
- When A was 4, the first date could be anywhere from 04/01 to 04/07.
- When A was 5, the first date could be anywhere from 05/01 to 05/06.
- When A was 6, the first date could be anywhere from 06/01 to 06/05.
- When A was 7, the first date could be anywhere from 07/01 to 07/04.
- When A was 8, the first date could be anywhere from 08/01 to 08/03.
- When A was 9, the first date could be anywhere from 09/01 to 09/03.
- When A was 10, the first date could be anywhere from 10/01 to 10/02.
- When A was 11, the first date had to be 11/01.
- A could not be 12.

In total, this accounted for 61 “sum-first” pairings. But what about when the second date was “product-first”?

- When A was 1, the first date could be anywhere from 01/01 to 01/12.
- When A was 2, the first date could be anywhere from 02/01 to 02/06.
- When A was 3, the first date could be anywhere from 03/01 to 03/04.
- When A was 4, the first date could be anywhere from 04/01 to 04/03.
- When A was 5, the first date could be anywhere from 05/01 to 05/02.
- When A was 6, the first date could be anywhere from 06/01 to 06/02.
- When A was 7, the first date had to be 07/01.
- When A was 8, the first date had to be 08/01.
- When A was 9, the first date had to be 09/01.
- When A was 10, the first date had to be 10/01.
- When A was 11, the first date had to be 11/01.
- When A was 12, the first date had to be 12/01.

This accounted for another 35 “product-first” pairings. In all, that meant you had 61 pairs plus another 35 pairs, which meant there were 96 possible pairs. Right?

Wrong. That’s because of the special case of 02/02. Its “sum-first” pair was 04/04, and its “product-first” pair was *also* 04/04. Since we counted this pair in both lists — rather than counting it once — the total tally was one too high. In the end, there were **95 pairs of dates**.

## Solution to the last Riddler Classic

Congratulations to 👏 Matt Frank 👏 of New York, New York, winner of last week’s Riddler Classic.

Last week, a restaurant at the center of Riddler City was testing an airborne drone delivery service against their existing fleet of scooters. The restaurant was at the center of a large Manhattan-like array of square city blocks, which the scooter had to follow.

Both vehicles traveled at the same speed, which meant drones could make more deliveries per unit of time. You could also assume that (1) Riddler City was circular in shape (2) deliveries were made to random locations throughout the city and (3) the city was much, much larger than its individual blocks.

In a (large) given amount of time, what was the expected ratio between the number of deliveries a drone could make to the number of deliveries a scooter could make?

This was equivalent to finding the ratio between the average distances — measured two different ways — from the center of Riddler City to a random location within Riddler City. For the drone, you simply needed the average Euclidean distance, or the straight-line distance.

For simplicity, let’s scale down Riddler City to a unit circle, centered at the origin and with radius 1. The distance between the origin and a point in the circle (*x*, *y*) was √(*x*^{2}+*y*^{2}). Since the city was circular, it actually made more sense to write this in polar notation: A point at (*r*, 𝜃) was a distance *r* from the origin. To find the *average* distance among all the points in the circle, you had to integrate *r* from 0 to 1, and 𝜃 from 0 to 2𝜋, using the area differential *r*d*r*d𝜃. Evaluating this integral gave you 2𝜋/3. Finally, you had to normalize by dividing by the total area of the circle, which was 𝜋. In the end, the average distance between a random point in a unit circle and the circle’s center was 2/3.

For our same point (*x*, *y*), the scooter traveled the “Manhattan distance,” or *x*+*y*. In polar form, this distance was |*r*cos𝜃| + |*r*sin𝜃|. After plugging this distance into the integral and separating variables, you were still integrating *r*^{2} from 0 to 1, which was 1/3. As for 𝜃, you were now integrating |cos𝜃|+|sin𝜃| from 0 to 2𝜋, which was 8. The total integral was the product of these two separated integrals, or 8/3. Normalizing by the area of the circle meant the average Manhattan distance was 8/(3𝜋).

At this point, you had the average Euclidean and Manhattan distances. All that was left was to find the ratio of these two values, which was 2/3 ÷ 8/(3𝜋), or 𝜋/4. Again, this was the ratio of the average *distances* traveled by the drone and scooter. To find the ratio of their expected number of deliveries, you needed to take the reciprocal, since distance and delivery rate were inversely related. In the end, the ratio of drone deliveries to scooter deliveries was **4/𝜋**, or about 1.273.

For extra credit, in addition to traveling parallel to the city blocks, scooters could also move diagonally from one corner of a block to the opposite corner of the block. With this additional motion in play, what was the expected ratio between the number of deliveries a drone could make and the number of deliveries a scooter could make?

Since the drone didn’t use these new paths, its average distance traveled in the unit circle remained 2/3. However, these new paths decreased the average distance for the scooter.

Consider points in the unit circle (*x*, *y*) with *x* > *y* > 0. (Note that these points make up one-eighth of the unit circle.) As shown below, the distance the scooter would travel to reach such a point was (x−y)+y√2.

In polar coordinates, this was *r*cos𝜃 + *r*sin𝜃(√2−1). Integrating and normalizing over the eighth of the unit circle resulted in an average distance of 16/(3𝜋)·(√2-1). And thanks to symmetry, this was the average for the other seven-eighths of the unit circle, meaning it was the average for the *entire *unit circle.

And so, with these diagonal routes now available, the ratio of drone deliveries to scooter deliveries was 8/𝜋(√2-1), or about 1.055. While the drone was still more efficient than the scooter, the additional diagonal paths impressively brought the scooter’s efficiency 80 percent closer to that of the drone.

## Want more puzzles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.