Can You Separate The World Cup Fans?
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.
Maryam is playing billiards on a 1 meter by 1 meter square table. She places the ball in one of the corners, aiming to strike the ball so that it travels as far as possible before hitting a wall for the third time. Note that the ball doesn’t necessarily have to hit three different walls of the table.
Assume the ball travels in a straight path and that it bounces off a wall as you’d expect.2 Also assume that it’s impossible for Maryam to hit the ball precisely in one of the corners of the table. Instead, it will hit both sides that are adjacent to the corner.
What is the farthest the ball can travel before hitting a wall for the third time?
Extra credit: What is the farthest the ball can travel before hitting a wall for the Nth time?
From Starvind comes a timely puzzle about the FIFA World Cup:
A certain hotel in Qatar is hosting 11 American fans and seven Dutch fans. Since no alcohol is available inside the stadiums, the fans spend the afternoon at the hotel bar before shuttle buses will take them to a match. Then, they haphazardly write their room numbers on a big board by the concierge desk.
To avoid any rowdiness between rival fans, shuttle bus drivers have been instructed to ferry American and Dutch fans separately. To ensure this, a shuttle pulls up in front of the hotel, and its driver calls out room numbers from the board, one by one at random. As long as they support the same team, each fan climbs aboard the bus and their room number is erased. Once the driver calls out the room number of a fan for the second team, the shuttle leaves with only the fans of the single team aboard. The next shuttle then pulls up and repeats the process.
What is the probability that the last shuttle ferries American fans?
Extra credit: On average, what is the expected number of shuttle buses needed to ferry all 18 fans?
Solution to the last Riddler Express
Congratulations to 👏 Michael Ringel 👏 of Jacksonville, Florida, winner of the last Riddler Express.
In the last Riddler Express, I was running a 5-kilometer “turkey trot” race. Before the race began, all the runners (including me) gathered behind the starting line in a random order. Once the race began, everyone started running at their own fixed pace.
I hadn’t run in several years, so I wasn’t sure how my pace would compare to that of the other racers. Nevertheless, once the race began, I found myself passing quite a few other runners — and being passed myself. It was safe to assume that if my pace was faster than that of any runner who started in front of me, I would pass them at some point during the race.
On average, what fraction of the other runners could I have expected to pass during the race?
First off, this was equivalent to asking what fraction of the runners in front of me were also slower than I was.
While the puzzle gave you no information about the distribution of paces among the different runners, the shape of this distribution didn’t matter. Suppose you ordered all the runners from fastest to slowest. Since I didn’t know how my pace compared to that of the other runners, you could assume (for the purposes of this puzzle) that I was equally likely to be anywhere in that ordering. And so, on average, I was faster than half of the runners and slower than the other half.
Now suppose I had been the first person in the queue at the starting line. In this case, I couldn’t possibly have passed anyone. If instead I had been the last person in the queue, I would have passed half of the runners on average.
But what if I had started somewhere else in the queue? As noted by Lisa of Los Altos, California, since the queue was random, I could expect to be faster than half the people in front of me, no matter where I started. Meanwhile, the number of people in front of me ranged from no one to everyone (minus me, of course — but it was safe to assume there were many runners in the race), which meant that, on average, half of the runners were in front of me.
To find the expected number of slower runners in front of me, all I had to do was multiply these two halves together, and one-half times one-half equals one-quarter.
Unfortunately, thanks to symmetry, I also expected to be passed by a quarter of the racers.
Solution to the last Riddler Classic
Congratulations to 👏 Sanandan Swaminathan 👏 of San Jose, California, winner of the last Riddler Classic.
In the last Riddler Classic, I had five kinds of fair Platonic dice: tetrahedra (whose faces are numbered 1-4), cubes (numbered 1-6), octahedra (numbered 1-8), dodecahedra (numbered 1-12) and icosahedra (numbered 1-20).3
When I rolled two of the cubes, there was a single most likely sum: 7. But when I rolled one cube and two tetrahedra, there was no single most likely sum — 8 and 9 were both equally likely.
Which whole numbers were never the single most likely sum, no matter which combinations of dice I picked?
As we already said, rolling two cubes had a most likely sum of 7. Similarly, rolling two tetrahedra had a most likely sum of 5, rolling two octahedra had a most likely sum of 9, rolling two dodecahedra had a most likely sum of 13 and rolling two icosahedra had a most likely sum of 21. (In general, adding two numbers chosen randomly, uniformly and independently from 1 to N had a most likely sum of N+1.)
From there, solver Michael Bradley of London, England, looked at what happened when you combined pairs of identical dice. For example, Michael found that if you rolled a pair of cubes and another pair of cubes, the most likely sum was 7 plus another 7, or 14. And if you rolled a pair of tetrahedra and a pair of icosahedra, the most likely sum was 5 plus 21, or 26.
Why did this work? Finding the sum of random variables from two different distributions is mathematically equivalent to convolving those two distributions. Convolving the symmetric triangular distributions you got from rolling two identical dice gave you more complicated distributions, but they always featured a unique central peak.
This meant you could pick pairs of dice to get any most likely number that was a linear combination of 5, 7, 9, 13 and 21.4 The only whole numbers that weren’t linear combinations of those five numbers were 1, 2, 3, 4, 6, 8 and 11.
Several readers stopped there, but 11 turned out to be a rather interesting case. If you had three tetrahedra and one cube, then the average sum was 2.5 + 2.5 + 2.5 + 3.5, or 11. But 11 also turned out to be the single most likely sum in this case, with a probability of 11/48. Meanwhile, the probabilities of 10 or 12 were both slightly lower, at 53/384.
As for 8, it might have been tempting to roll a cube and an octahedron. Here, the average sum was 3.5 + 4.5, which was indeed 8. However, 8 was not the single most likely sum, as 7, 8 and 9 were equally likely with a probability of 1/8.
In the end, the only single most likely sums that were not obtainable were 1, 2, 3, 4, 6 and 8.
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 email@example.com.