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 next week’s 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

Congratulations, you’ve made it to the final round of the Riddler Rock, Paper, Scissors.

The rules are simple: Rock beats scissors, scissors beat paper, and paper beats rock. Moreover, the game is “sudden death,” so the first person to win a single round is immediately declared the grand champion. If there’s a tie, meaning *both* players choose the same object, then you simply play another round.

Fortunately, your opponent is someone you’ve studied well. Based on the motion of their arm, you can tell whether they will (1) play rock or paper with equal probability, (2) play paper or scissors with equal probability or (3) play rock or scissors with equal probability. (Every round falls into one of these three categories.)

If you strategize correctly, what are your chances of winning the tournament?

The solution to this Riddler Express can be found in the following week’s column.

## Riddler Classic

From Dave Moran comes a perplexing puzzle of pomp and circumstance:

The inaugural graduating class of Riddler High School, more than 100 students strong, is lined up along the circumference of the giant circular mosaic on the plaza in front of the school. The principal, Dr. Olivia Rhodes, stands atop a tall stepladder in the center of the circle, preparing to take the panoramic class photo.

Suddenly, she shouts, “Hey, class, every single one of you *N* graduates — except Val, our valedictorian — is standing in the wrong place! Remember, you’re supposed to stand in order of your class rank, starting with Val directly in front of me, and going counterclockwise all the way to Zach, who should then be next to Val.” (Poor Zach was ranked last in the graduating class at Riddler High.)

Dr. Rhodes continues, “Not only are almost all of you in the wrong positions, but no two of you are even in the correct position *relative to each other*.” Two graduates would be in the correct position relative to each other if, for example, graduate A was supposed to be 17 positions counterclockwise of graduate B and was indeed 17 positions counterclockwise of graduate B — even though both graduates were in the wrong positions.

But Val speaks up: “Dr. Rhodes, that can’t be right. There must be at least two of us who are in the correct position relative to each other.”

Dr. Rhodes looks carefully around the circle of graduates below her and admits, “Of course, our brilliant Val is quite right. I do now see that there are two of you who are correctly positioned relative to each other.”

Given that there are more than 100 graduates, what is the *minimum* number of graduates who are posing for the class photo?

The solution to this Riddler Classic can be found in the following week’s column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Joe Maloney 👏 of Atlanta, Georgia, winner of last week’s Riddler Express.

Last week, Riddler Township was having its quadrennial presidential election. Each of the town’s 10 “shires” was allotted a certain number of electoral votes: two, plus one additional vote for every 10 citizens (rounded to the nearest 10).

The names and populations of the 10 shires are summarized in the table below.

##### Which way will Riddler Township swing?

Shire | Population | Electoral votes |
---|---|---|

Oneshire | 11 | 3 |

Twoshire | 21 | 4 |

Threeshire | 31 | 5 |

Fourshire | 41 | 6 |

Fiveshire | 51 | 7 |

Sixshire | 61 | 8 |

Sevenshire | 71 | 9 |

Eightshire | 81 | 10 |

Nineshire | 91 | 11 |

Tenshire | 101 | 12 |

Under this sort of electoral system, it was quite possible for a presidential candidate to lose the popular vote and still win the election.

With two candidates running for president of Riddler Township, and every citizen voting for one or the other, what is the *lowest* percentage of the popular vote that a candidate could get while still winning the election?

To win with as few votes as possible, a candidate needed a majority of the 75 electoral votes, meaning they needed at least 38 votes. They also needed just over half the popular vote in any shire they won, while losing the entire popular vote in the shires they lost.

Meanwhile, the less populous shires also offered greater leverage. For example, Oneshire, with its three electoral votes and a population of 11 people (i.e., a majority consisting of six people), offered 0.5 — three divided by six — electoral votes per supporter. Twoshire, with four electoral votes and a voting majority of 11 people, offered about 0.36 electoral votes per supporter. Meanwhile, at the other end of the spectrum, Tenshire offered less than 0.24 electoral votes per supporter.

That meant our unpopular winner wanted to win the less populous shires, accruing electoral votes without winning too many popular votes.

The fewest votes a candidate needed to win turned out to be 136 out of Riddler Township’s 560 total citizens, and so the smallest possible winning percentage of the popular vote was approximately **24.3 percent**.

There were several different ways to generate this electoral nightmare. For example, the winner might get six votes in Oneshire, 16 votes in Threeshire, 21 votes in Fourshire, 26 votes in Fiveshire, 31 votes in Sixshire and 36 votes in Sevenshire. In all, that is indeed 38 electoral votes against just 136 popular votes. This particular scenario was illustrated by Andrew Heairet:

That map of Riddler Township looks vaguely familiar, but I can’t quite place it.

But that wasn’t the only solution. Many solvers, like Nora Corrigan from Columbus, Mississippi and Caspian from Stockholm, Sweden found multiple ways to generate exactly 38 electoral votes and 136 popular votes. Here is the complete list of such scenarios:

- Slim victories in Oneshire, Threeshire, Fourshire, Fiveshire, Sixshire and Sevenshire
- Slim victories in Oneshire, Twoshire, Fourshire, Fiveshire, Sixshire and Eightshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fiveshire, Sixshire and Nineshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fiveshire, Sevenshire and Eightshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fourshire, Sixshire and Tenshire
- Slim victories in Oneshire, Twoshire, Threeshire, Fourshire, Sevenshire and Nineshire

As it turned out, there were many ways for the electoral and popular votes to wildly disagree. I guess that’s the electoral college for you. Oh, and if you’d like your vote to *really* count in Riddler Township, it’s a good idea to live in Oneshire.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Steven Trautmann 👏 of Aurora, Colorado, winner of last week’s Riddler Classic.

Last week, you played a game of Riddler Pinball, which had an infinitely long wall and a circle whose radius was 1 inch and whose center was 2 inches from the wall. The wall and the circle were both fixed and never moved. A single pinball started 2 inches from the wall and 2 inches from the center of the circle.

To play, you flicked the pinball toward a spot of your choosing along the wall, specified by its distance *x* from the point on the wall that’s closest to the circle, as shown in the diagram below.

The goal of the game was simple: Get the ball to bounce as many times as possible.

If you aimed too far to the right (i.e., your value of *x* was too small), the pinball quickly bounced its way through the gap between the circle and the wall. But if you aimed too far to the left (i.e., your value of *x* is too big), the pinball quickly came back out the same side it went in.

Riddler Pinball was an unforgiving game — the slightest error tanked your chances of victory. But if you strategized *just* right, it was possible to do quite well.

What was the greatest number of bounces you could achieve? And, more importantly, what value of *x* got you the most bounces?

There was certainly a sweet spot that resulted in many, many bounces. For example, when *x* was 0.82248632494339, there were 43 bounces:

But before we go any further, let’s return to the first question that was posed: What was the greatest possible number of bounces?

As solver Phillip Bradbury observed, all the angles and points of impact changed monotonically with *x*. As noted above, smaller values of *x* made the pinball pass through, but larger values of *x* made the pinball bounce back out.

So then what happened *between* these two cases? Zooming in revealed a single point on the number line where the pinball neither passed through nor bounced back. But if it did *neither* of these things, what else could the ball possibly do? It would be stuck bouncing back and forth **infinitely many times**. And as *x* approached this point — whatever it was — the number of bounces went up, up, up. This was nicely illustrated by solvers Mark Girard and Dinesh Vatvani.

There were a few other ways to see how this was true. Jason Bellenger reasoned backwards, thinking about a ball that was bouncing straight up and down between the tip of the circle and the wall. If you perturbed it infinitesimally, but just right, it would ultimately bounce its way to the correct starting point. Playing these bounces in reverse would then give you an infinitely high-scoring pinball shot.

Laurent Lessard, meanwhile, was able to better visualize the multitude of bounces by plotting the logarithm of the *x*-coordinate, as shown below. Graphically, this had the effect of spreading out the bounces, leading him to correctly suspect the answer was infinity.

As it turned out, this reasoning was the easier part of the problem. *Finding* this precise value of *x* that got you infinitely many bounces was another matter entirely.

At this point, most solvers took a computational approach. Simulating the pinball as it bounced off the wall wasn’t too tricky — the ball simply reflected off the wall so that the angle it made with respect to the wall didn’t change. But simulating bounces off the circle was a little tougher. Again, the angles of incidence and reflection were equal, but now they were measured from the radius of the circle. A slight mess of trigonometry ensued.

Once you had a functioning pinball simulator up and running, then it was just a matter of trying different values of *x*, adding one digit at a time, and seeing which values gave you more bounces. Indeed, the answer was very close to **0.82248632494339** — the true value actually goes on for infinitely many decimal places.

Some solvers, like Dean Ballard and Joseph Wetherell, went to great lengths to crank out more of the solution’s decimal places. In fact, the answer is very, very close to:

0.82248632494339006569162637706984990078

134582582348831326968696909788932771367

According to Joseph, this precise value of *x* resulted in more than 200 bounces!

Meanwhile, solver Emma Knight boldly attempted a more analytic approach. However, ballooning floating point errors ultimately got in the way.

Finally, a few very clever readers noticed something glaring about this problem: There was a second solution! While the original statement of the riddle asked where *along the wall* the pinball should be aimed, it was also possible to hit the circle first and *still* achieve infinitely many bounces. Josh Silverman went as far as animating this alternate strategy:

When I previously teased Riddler Pinball, I asked what interesting mathematical questions could be posed about it. There were multiple calls to alter the geometry, such as with a noncircular bumper, or to scale it up to three dimensions. With all these great suggestions, I don’t think it will be long before we all play another round of Riddler Pinball.

## 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 riddlercolumn@gmail.com.