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
If you have young children (or if you’re still a child at heart), you probably have small blocks somewhere in your home.
I recently found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. Here’s a photo I took of this arrangement:
If you too have blocks at home (I mean, of course you do), see if you can make the same arrangement.
Now, if each of the four cubes has side length 1, then how far above the ground is the bottommost corner of the cube on top?
The solution to this Riddler Express can be found in the following column.
From Vince Vatter, one of the two reigning Wordsmiths Extraordinaire of Riddler Nation, comes a puzzle that is sure to please anyone who fondly remembers that time Spelling Bee appeared in this column:
You are a contestant on the game show Lingo, where your objective is to determine a five-letter mystery word. You are told this word’s first letter, after which you have five attempts to guess the word. You can guess any five-letter word, even one that has a different first letter.
After each of your guesses, you are told which letters of your guess are also in the mystery word and whether any of the letters are in the correct position. In the example below, T is in the correct position (remember, the first letter is provided to you), while A and C are in the mystery word but not in the correct positions.
For this example, here’s how you might have figured out the mystery word (TACOS) using all five guesses:
The mystery word and guesses can contain multiple instances of a letter. For example, the mystery word MISOS contains one O, so a guess with more than one O (like MOSSO) will only have the first O marked as correct (but in this case, in the wrong position).
As a contestant, your plan is to make a mockery of the game show by adopting a bold strategy: No matter what, before you are even told what the first letter of the mystery word is, you have decided what your first four guesses will be. Then, with your fifth guess, you will use the results of your first four guesses (and your encyclopedic knowledge of five-letter words!) to determine all remaining possibilities for the mystery word. If multiple mystery words are still possible, you will pick one of these at random.
Which four five-letter words would you choose to maximize your chances of victory? Assume that the mystery word is selected randomly from this word list, which is also the list your guesses must be chosen from.
Extra credit: For the four five-letter words you chose, what are your chances of victory?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Martin Miller 👏 of Mill Valley, California, winner of last week’s Riddler Express.
Last week, you reviewed some survey data that was randomly collected from the residents of Riddler City, which had a very large population.
Ten randomly selected residents were asked how many people (including them) lived in their household. As it so happened, their answers were 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Your job was to use this (admittedly limited) data to estimate the average household size in Riddler City. Your co-worker suggested averaging the 10 numbers, giving an answer of about 5.5 people. But you weren’t so sure.
Would your best estimate have been exactly 5.5, less than 5.5 or greater than 5.5?
On its face, it seemed like the answer should have been exactly 5.5 — the average of the 10 numbers from 1 through 10.
But Riddler Nation was not fooled. Among the 836 submissions I received before the midnight deadline on Monday, only 8.6 percent thought the best estimate was exactly 5.5, while another 19.5 percent thought the best estimate was greater than 5.5. Meanwhile, the overwhelming majority (71.9 percent) thought the best estimate was less than 5.5.
These solvers read the puzzle carefully, noting that it was residents — rather than households — that were randomly selected for the survey. If households had been randomly selected instead, then yes, your best estimate would have been 5.5 people per household. But because it was residents that were selected, solvers like Penelope Ackerman realized that a household with 10 people was 10 times more likely to have been chosen than a household with just one person. In other words, the sampling was biased toward larger households. For a household with one person and a household with 10 people to be selected with equal likelihood, there had to have been 10 times as many households with one person.
As it turned out, applying this reasoning to all 10 household sizes meant 2,520/7,381 of the households in Riddler City had one person, 1,260/7,381 had two people, 840/7,381 had three people, and so on, with 252/7,381 of the households having 10 people. The denominator was 7,381 because that was the sum of the smallest set of integers satisfying the ratios in the problem. These ratios were nicely illustrated by solver Reece J. Goiffon:
All this meant the average household size in Riddler City — calculated as 2,520/7,381·1 + 1,260/7,381·2 + 840/7,381·3 + … + 252/7,381·10 — was about 3.4 people. That is, less than 5.5.
Any readers who also happen to be pollsters in Riddler City should take notice. You have a pollster rating to maintain!
Solution to last week’s Riddler Classic
Congratulations to 👏 Ali Farhat 👏 of Dearborn, Michigan, winner of last week’s Riddler Classic.
Last week, you were competing in the finals of the Riddler Ski Federation’s winter championship! There was just one opponent left to beat, and then the gold medal would be yours.
Both of you were completing two runs down the mountain, and the times of your runs would be added together. Whoever skied in the least overall time would be the winner. Also, you knew that you and your opponent were evenly matched, and you both had the same normal probability distribution of finishing times for each run. And each skiing run was independent of all the others.
For the first run, your opponent went first. Then, as you crossed the finish line on your own first run, your coach excitedly signaled to you that you were faster than your opponent. Without knowing either exact time, what was the probability that you would still be ahead after the second run and earn your gold medal?
To no one’s surprise, many solvers simulated the skiing championship. Paulina Leperi and Harold Doran both wrote some code that simulated hundreds of thousands of pairs of skiing runs for both you and your opponent, finding that you emerged the winner just about 75 percent of the time. Could the answer have been 75 percent?
A few solvers, like Stergios Athanasoglou and David Ding, tackled the puzzle head on, working through the mathematics (and calculus) of the normal distribution.
But in a cruel (or clever?) twist of mathematics, it turned out that probability distribution of finishing times was irrelevant for this problem. As long as it was the same for both you and your opponent, it didn’t matter if this distribution was normal, uniform or even a Laplace distribution.
As John from Washington, D.C. observed, there was a 50 percent chance that you would be faster than your opponent on the second run, in which case you were guaranteed to be the overall winner. As for the other 50 percent of the time, when your opponent had the faster second run, it all came down to whether the time gap was greater in the first run or the second run — two cases that were equally likely, thanks to the symmetry in the problem. The 25 percent of the time the gap was greater in the first run, you won, while you lost the other 25 percent of the time. Putting it all together, your chance of victory was 50 percent plus 25 percent, or exactly 75 percent.
While it turned out that the normal probability distribution didn’t matter for the puzzle, the same could not be said for the extra credit, where you were asked to repeat the exercise in the case of 30 snowboarders (including you). Again, you were the last to complete the first run, and your coach signaled that you were in the lead at that point. What was the probability that you would win gold in snowboarding?
Even by Riddler standards, this problem was surprisingly hard to calculate. Forget 30 snowboarders with a normal probability distribution — even just three snowboarders with a uniform probability distribution was a challenge, requiring some hefty calculus and order statistics.
Everyone who solved this did so via computation, finding that you had between a 31.4 and 31.5 percent chance of winning the snowboarding championship. Josh Silverman took it one step further, looking at how your chances of winning depended on the number of competitors N. He found that this probability (the lower blue points in the graph below) appeared to be inversely proportional to the cube root of N (modeled by the higher orange points).
Of course, an even more general version of this puzzle was if there were N competitors and R runs in the competition, rather than just two runs. But I’ll save that riddle for our computer overlords to solve.
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.