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. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
At long last, Dakota Jones is close to finding the Lost Arc, a geometric antiquity buried deep in the sands of Egypt. Along the way, she discovered what she described as a “highly symmetric crystal” that’s needed to precisely locate the Arc. Dakota measured the crystal using her laser scanner and relayed the results to you. But nefarious agents have gotten wind of her plans, and Dakota and the crystal are nowhere to be found.
Locating the Arc is now up to you. To do that, you must recreate the crystal using the data from Dakota’s laser scanner. The scanner takes a 3D object, and records 2D cross-sectional slices along the third dimension. Here’s the looping animation file the scanner produced for the crystal:
What sort of three-dimensional shape is the crystal? No pressure — Dakota Jones, nay, the entire world, is counting on you to locate the Lost Arc and ensure its place in a museum!
Recent Riddlers have tackled Scrabble Superstrings and road trips through 48 states. For this week’s Riddler Classic, Max Maguire combines these two puzzles into one:
The challenge is to find the longest string of letters in which (1) every pair of consecutive letters is a two-letter state or territory abbreviation, and (2) no state abbreviation occurs more than once. For example, Guam, Utah and Texas can be combined into the valid four-letter string GUTX. Another valid string is ALAK (Alabama, Louisiana and Alaska), while ALAL (Alabama, Louisiana and Alabama) is invalid because it includes the same state, Alabama, twice.
For reference, the full list of abbreviations is available here, courtesy of the United States Postal Service.
Solution to last week’s Riddler Express
Congratulations to 👏Dan Anderson 👏 of Encinitas, California, winner of last week’s Riddler Express.
Last week, you and your friend played a game of “Acchi, Muite, Hoi” (also known as the “lookaway challenge”). In the first round, you randomly pointed up, down, left or right, while your friend randomly looked in one of those four directions. If your friend looked in the same direction you pointed, you won. Otherwise, as long as no one wins, you kept switching off who pointed and who looked.
Your chances of winning were exactly 4/7. There are a couple ways to go about finding that number.
As solver Alison Lynch observed, you can calculate the probabilities of winning each round and add them all up. You could only win the game when you were the one doing the pointing, which happened in the first round, the third round, the fifth round and all the other odd-numbered rounds. Furthermore, within each round, the pointer had a 1/4 chance of winning, since only one out of the four directions — up, down, left and right — matched what the looker did. Your chances of winning in the first round were therefore 1/4. To win in the third round, you must have failed to win in the first round (which had a 3/4 chance of happening), your opponent must have failed to win in the second round (again, with probability 3/4), and then you won in the third round (with probability 1/4). Multiplying these together, you had a 3/4 × 3/4 × 1/4 = 9/64 chance of winning in the third round, while in the fifth round your chances were 3/4 × 3/4 × 3/4 × 3/4 × 1/4. Continuing with this logic, adding up the infinitely many probabilities from all the odd-numbered rounds gave you a geometric series whose sum was exactly 4/7.
The students of The Hewitt School in New York City, meanwhile, thought about it another way. They called your chances of winning b. What are your opponent’s chances of winning? Well, first your opponent needs the game to proceed to the second round, which happens 3/4 of the time (the other 1/4 of the time you win in the first round). At this point, their chances of winning must also be b — their situation is identical to yours back in the first round. In fact, any time one of you is the designated pointer, that person’s probability of winning is b.
Since one of you must be the winner, you and your opponent’s chances of winning have to add up to 1. That means b + 3/4b = 1, which means, sure enough, that b = 4/7, or about 57 percent.
Just to be super-duper sure, solver Angelos Tzelepis simulated 10 million games, finding that whoever goes first wins a shade over 57 percent of the time.
Why isn’t the answer just 50 percent? Because pointing in the first round and having that very first opportunity to win offers you a slight edge. So who says it’s not polite to point? (I do. Don’t point.)
Solution to last week’s Riddler Classic
Congratulations to 👏Matt Friedrichsen 👏 of Overland Park, Kansas, winner of last week’s Riddler Classic.
Last week, you competed at the U.S. Open, where you were slightly better than the competition: Your chances of winning any given point were exactly 55 percent. What were your chances of winning a three-set match and a five-set match? And what were your chances of winning the whole tournament?
As solver Dennis Okon figured out, one way to solve this problem is to work backwards. Once you find the probabilities of winning from deuce, advantage-in and advantage-out, you can find the chances of winning a single game from lower and lower scores. For example, suppose you’re up 30-0 in a game. 55 percent of the time you’ll win the point, and the score will become 40-0, while the other 45 percent of the time you’ll lose the point, and the score will become 30-15. Your probability of winning the game at 30-0 turns out to be a weighted average of those other two scores: 55 percent of the probability you’ll win at 40-0 plus 45 percent of the probability you’ll win at 30-15.
You can use this technique to work all the way back to a score of 0-0, which represents your chances of winning a game at the start. If you do the same analysis for a set, you will find the probability of winning the set from different game scores, as shown in the table below.
What are your chances of winning the set?
The probability you’ll win a set of tennis given the number of games you and your opponent have won so far, assuming you have a 55 percent chance of winning each point
|Games your opponent has won|
|Games you’ve won||0||1||2||3||4||5||6||7|
The percentage when the score is 0-0 is equivalent to your chances of winning any given set: 81.5 percent. Your chances of winning two out of three sets are then 91 percent, and they are 95.3 percent for winning three out of five sets. Raising these probabilities to the seventh power gives you your chances of winning the entire U.S. Open: 51.7 percent in the women’s tournament and 71.4 percent in the men’s. As you can see, playing more sets gives stronger players a better chance at winning. So let the women play five sets, I say.
It’s a little shocking just how high your chances of winning are when you start with what seemed like a slight advantage on every point. Scoring in tennis is highly nonlinear: A tiny advantage in winning a point snowballs into a big probability shift over the course of an entire match. Riddler Nation Hall-of-Famer Diarmuid Early elegantly captured this nonlinearity in a single graph, where you can see how quickly your chances of winning a set and a match change as your probability of winning each point crosses 50 percent:
Sure enough, winning 55 percent of points does not make you “slightly better than the competition,” as the puzzle stated — instead, it puts you squarely in the pantheon of tennis’s all-time greats! Roger Federer, Rafael Nadal and Novak Djokovic have all won approximately 55 percent of their career points.
Finally, if you’d like to play around with these percentages for yourself, check out this tennis calculator by Mark Bennett. It will serve you well.
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 firstname.lastname@example.org.