What Are The Odds World Cup Teams Play Each Other Twice?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Quick announcement: Have you enjoyed the puzzles in this column? If so, I’m pleased to tell you that we’ve collected many of the best, along with some that have never been seen before, in a real live book! It’s called “The Riddler,” and it will be released in October — just in time for loads of great holidays. It’s a physical testament to the mathematical collaboration that you, Riddler Nation, have helped build here, which in my estimation is the best of its kind. So I hope you’ll check out the book, devour the puzzles anew, and keep growing our nation by sharing the book with loved ones.

And now, to this week’s puzzles!

## Riddler Express

From Sam Kaplan, a chance double-take sporting encounter:

Assuming we don’t know anything about the strengths of the teams in the tournament, what are the chances that any pair of teams in a 32-team World Cup plays each other twice? (Given the way the World Cup bracket works, their first encounter would be in the round-robin, three-game group stage, with their second encounter in the final or the third-place game.)

Extra credit: What would those chances have been given FiveThirtyEight’s pre-tournament odds for this year’s field?

## Riddler Classic

From Jan-Willem Tel, a puzzle of efficient breaking and entering:

You have a gate that requires a passcode to open. There are 10 buttons: the digits 0 to 9. You have forgotten the passcode, but you remember it has four digits. You have no choice but to try them all.

Since there are $$10^4$$ = 10,000 four-digit passcodes, you might think this would take you 40,000 button presses to guarantee an opened gate. However, this gate’s keypad never resets: The gate opens as soon as the last four buttons you’ve pressed are the correct code, so you can be more efficient. For example, you can try two different codes by pressing just five buttons: The combination “12345” tries both “1234” and “2345.” Of course, pressed for time, you want to press as few buttons as possible while still trying different codes and eventually opening the gate.

So the question is: What’s the smallest number of buttons you need to press to make sure you open the gate — i.e., that you’ve tried every possible four-digit combination?

Extra credit: How do things change if you didn’t remember the passcode’s length?

## Solution to last week’s Riddler Express

Congratulations to 👏 Rebecca Gilbert 👏 of Atlanta, winner of last week’s Riddler Express!

Last week we met Michelle, who was exercising in the airport before her flight by walking the wrong way on a moving walkway. After walking 100 meters closer to her departure gate, she dropped her boarding pass onto the walkway. She didn’t notice at first and continued walking toward her gate. After walking for another 90 seconds, she finally realized she had dropped her boarding pass and immediately turned around and jogged in the direction of the walkway’s movement. Her jogging pace was exactly twice as fast as her walking pace. She caught up with her boarding pass 10 meters from the start of the moving walkway. How fast does the walkway move?

It moves at ⅔ meters per second, or about 1.5 mph.

How do we get there? Call the speed of the walkway w and call Michelle’s walking speed m. Michelle will end up 90(m-w) + 90w meters away from her boarding pass after the 90 seconds. This simplifies to 90m. Therefore, the time it takes to chase down her boarding pass after she realizes she dropped it is 90m/(2m + ww), or 90m/2m, or 45 seconds. Now we know that the boarding pass traveled 90 meters in 90+45 seconds. 90/(90 + 45) = 90/135 = ⅔, and we’re done!

A typical real-world moving walkway moves at about 1.4 mph — the Riddler, as ever, imitates life.

## Solution to last week’s Riddler Classic

Congratulations to ✌ Kevin Qiao ✌ of Philadelphia, winner of the inaugural Riddler Rock-Paper-Scissors Tournament and newly crowned Rochambeau Raja of Riddler Nation!

Last week, you were invited to participate in a Rock-Paper-Scissors battle royale series of best-two-out-of-three matches. To enter, you submitted two things: 1) the probabilities that you played rock, paper or scissors to start a match and 2) the conditional probabilities that you played rock, paper or scissors during the next throw in response to a play of rock, paper or scissors by your opponent.

I received 2,157 strategy submissions, 1,648 of which were valid — i.e., the relevant probabilities summed to 1 and so forth. Here are the best performers, along with their strategies:

##### All hail the scissor king!

The top five finishers in the inaugural Riddler Rock-Paper-Scissors Tournament, along with their first-throw choices

Name Hometown Rock Paper Scissors Winning percentage
Kevin Qiao Philadelphia 0% 0% 100% 57.5%
Linder Stillwater, MN 0 75 25 57.4
Josh Strauss Merrick, NY 0 70 30 57.3
Andrew Sydney 0 0 100 57.2
Tucker Belton Mineral, VA 0 60 40 57.2

Our winner, Kevin, went all in on scissors for his first throw. He then always played paper in response to rock, always played scissors in response to paper, and always played rock in response to scissors. Notably, none of the top performers ever featured rock on their first throw.

This strategy was effective for an apparently simple reason: Riddlers favored paper! On the first throw of a match, it was the selection more than 40 percent of the time, compared with less than 28 percent of the time for scissors. After the first throw, players tended to favor rock, on average, regardless of what they saw their opponent play on the throw before. Kevin’s response of paper to rock worked out well there, too.

##### Riddler Nation loves paper

Average percentage plays in the inaugural Riddler Rock-Paper-Scissors Tournament

In response to …
Choice First throw Rock Paper Scissors
Rock 31.9% 39.7% 35.7% 36.5%
Paper 40.2 31.1 34.9 30.8
Scissors 27.9 29 29.4 32.7

Many responses, not surprisingly, were equal or close-to-equal mixes of the three options: one-third rock, one-third paper, one-third scissors. This corresponds to the game-theory ideas behind the contest and the concept of a mixed-strategy Nash equilibrium, and obviously Riddler Nation knows all about that. These strategies did perfectly fine in the overall results! But they weren’t able to effectively exploit the small biases in Riddler Nation’s submissions as a whole. In this case, a scissor-heavy gamble paid off.

And Kevin guessed right, explaining his strategy thusly: “I just picked what I thought people would throw the most (paper).” Gotta get that paper.

Without naming names, the worst performer won just over 40 percent of his matches. He always, and I mean always, played rock. “Good ol’ rock. Nothing beats that!” he wrote. However, in a land of paper, rock is not king.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.

## Footnotes

1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. Eastern time on Sunday. Have a great weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.

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