# Can You Win The Penalty Shootout?

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.

## Riddler Express

From Irwin Altrows comes a rather sticky situation:

I have three dogs: Fatch, Fetch and Fitch. Yesterday, I found a brown 12-inch stick for them to play with. I marked the top and bottom of the stick and then threw it for Fatch. Fatch, a Dalmatian, bit it in a random spot — leaving a mark — and returned it to me. In her honor, I painted the stick black from the top to the bite and white from the bottom to the bite.

I subsequently threw the stick for Fetch and then for Fitch, each of whom retrieved the stick by biting a random spot. What is the probability that Fetch and Fitch both bit the same color (i.e., both black or both white)?

Submit your answer

## Riddler Classic

From Steven Pratt comes a probabilistic puzzle that you’ll get a kick out of:

Italy defeated England in a heartbreaking (for England) European Championship that came down to a penalty shootout. In a shootout, teams alternate taking shots over the course of five rounds. If, at any point, a team is guaranteed to have outscored its opponent after five rounds, the shootout ends prematurely, even if each side has not yet taken five shots. If teams are tied after five rounds, they continue one round at a time until one team scores and another misses.

If each player has a 70 percent chance of making any given penalty shot, then how many total shots will be taken on average?

Submit your answer

## Solution to last week’s Riddler Express

Congratulations to 👏 James Anderson 👏 of Charlotte, North Carolina, winner of last week’s Riddler Express.

Earlier this year, a new generation of Brood X cicadas had emerged in many parts of the U.S. This particular brood emerges every 17 years, while some other cicada broods emerge every 13 years. Both 13 and 17 are prime numbers — and relatively prime with one another — which means these broods are rarely in phase with other predators or each other.

Last week, you analyzed two broods of cicadas, with periods of *A* and *B *years, that had just emerged in the same season. However, these two broods could have also interfered with each other one year *after* they emerged due to a resulting lack of available food. For example, if *A* was 5 and *B* was 7, then *B*’s emergence in year 14 (i.e., 2 times 7) meant that when *A* emerged in year 15 (i.e., 3 times 5) there wasn’t enough food to go around.

If *A* and *B* were relatively prime and both less than or equal to 20, what was the longest stretch these two broods could have gone without interfering with each other’s cycle? (Remember, both broods just emerged *this* *year*.)

Had the broods only interfered during their years of emergence, then you would have had to find a pair of whole numbers (less than or equal to 20) with the greatest least common multiple (LCM). In that case, the answer would have been 19 and 20, whose least common multiple was 380.

However, that wrinkle of interference occurring one year post-emergence meant that if *A* were 19 and *B* were 20, then there wouldn’t have been enough food to go around in year 20. Thus, this puzzle was more than simply finding an LCM.

Since both *A* and *B* were between 1 and 20, that meant there were 20^{2}, or 400, total cases to check. (There were actually far fewer than 400 scenarios to check, since the puzzle stated that *A* and *B* had to be relatively prime.) Solvers like Rohan Lewis and Benjamin Dickman checked them all:

As you can see from Benjamin’s table, the latest possible interference occurred when *A* (the values along the horizontal axis) and *B* (the values along the vertical axis) were 17 and 19. The first few multiples of 17 were 17, 34, 51, 68, 85, 102, 119, 136 and 153. Meanwhile, the first few multiples of 19 were 19, 38, 57, 76, 95, 114, 133 and 152. The first time their multiples were within 1 of each other was when they were 153 and 152, meaning the answer was **153**. (I also accepted answers of 152.)

In general, consecutive odd values are always relatively prime and go a pretty long time without interfering. Given an odd value of *N*, when do *N* and *N*+2 first interfere? Since *N* is odd, that means *N*−1 and *N*+1 are both even, which in turn means that (*N*−1)/2 and (*N*+1)/2 are both integers. Multiplying *N*+2 by (*N*−1)/2 gives the expression (*N*^{2}+*N*)/2−1, while multiplying *N* by (*N*+1)/2 gives the expression (*N*^{2}+*N*)/2. Indeed, these two expressions differ by one, and the interference occurs after (*N*^{2}+*N*)/2 years. You can convince yourself it won’t happen sooner using modular arithmetic.

Should cicadas ever start interfering with each other like this, it wouldn’t surprise me if the brood cycles adapt over the course of the next million years or so.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Jordan Rader 👏 of Knoxville, Tennessee, winner of last week’s Riddler Classic.

Last week, the astronomers of Planet Xiddler were back!

This time, they had identified three planets that circularly orbited a neighboring star. Planet A was three astronomical units away from its star and completed its orbit in three years. Planet B was four astronomical units away from the star and completed its orbit in four years. Finally, Planet C was five astronomical units away from the star and completed its orbit in five years.^{2}

They reported their findings to Xiddler’s Grand Minister, along with the auspicious news that all three planets were currently lined up (i.e., they were collinear) with their star. However, the Grand Minister was far more interested in the three planets than the star and wanted to know how long it would be until the planets were next aligned.

How many years would it be until the three planets were again collinear (not necessarily including the star)?

First off, to all those who pointed out the non-physicality of this planetary system: Thanks, but seriously, read the footnotes.

Now, back to the puzzle. Like last week’s Express, it looked like this puzzle would involve LCMs. To find the next time all three planets returned to their original positions, meaning they were all collinear with their star, you had to compute the least common multiple of 3, 4 and 5, which was 60.

But wait! In *half* that time — or 30 years, the planets with periods of three years and five years would have returned to their original positions, while the planet with a period of four years would have been on the exact opposite side of the star from where it started. This meant they were all collinear after just 30 years.

But wait (again)! The puzzle said that only the *three planets* had to be on the same line. There was no need for the star to be collinear with them. That meant this was no longer a number theory puzzle involving LCMs but rather a dynamics puzzle involving trigonometry!

To see this in action, check out solver Tyler Barron’s animation of the three planets:

It appears that they lined up quite a few times before even hitting the 30-year mark.

There were a few different strategies to precisely solve for collinearity. The first step was always to write the coordinates of the three planets as a function of the number of years *t*: (3cos(2𝜋*t*/3), 3sin(2𝜋*t*/3)), (4cos(2𝜋*t*/4), 4sin(2𝜋*t*/4)) and (5cos(2𝜋*t*/5), 5sin(2𝜋*t*/5)).

From there, some solvers, like Frank van Biesen, wrote expressions for the distances between these three points over time, and then used Heron’s formula to calculate the area of the triangle whose side lengths were these three distances. Here was the resulting graph:

When the area was 0, that meant the triangle was degenerate and the three points were collinear. Sure enough, the graph first hits zero after approximately **7.77 years** — the correct answer.

Meanwhile, other solvers used dot products, cross products and even complex analysis, always arriving at the same result.

But wait (one more time)! The puzzle never specified that all three planets were orbiting in the same direction around the star (i.e., clockwise or counterclockwise). Solver Jason Weisman found that if only Planet A was moving in the opposite direction, the planets would be collinear after about 5.31 years. But if Planet C was the one moving the opposite way, collinearity occurred after just 1.63 years. Therefore, I accepted 1.63 years as a correct solution as well.

These Xiddlerian astronomers have been on a roll. I’m sure we’ll see them again soon — or perhaps they’ll see us!

## Want more riddles?

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 [email protected]