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.
Riddler Express
From Dee Harley comes a devilish matter of dominos:
In a set of dominos, each tile has two sides with a number of dots on each side: zero, one, two, three, four, five or six. There are 28 total tiles, with each number of dots appearing alongside each other number (including itself) on a single tile.
Question 1: What is the probability of drawing a “double” from a set of dominoes — that is, a tile with the same number on both sides?
Question 2: Now you pick a random tile from the set and uncover only one side, revealing that it has six dots. What’s the probability that this tile is a double, with six on both sides?
The solution to this Riddler Express can be found in the following week’s column.
Riddler Classic
A certain 2-year-old is eating his favorite snack: an apple. But he eats it in a very particular way. When he first receives the apple, and every minute thereafter, he rotates the apple to a random position and then looks down. If there’s any skin of the apple left in the spot where he plans to take a bite, then he will indeed take that bite. But if there’s no skin there (i.e., he’s already taken bites at that spot), he won’t take a bite and will rotate the apple for another minute. Here’s a photo of what the apple might look like after a while:
Once he has bitten off all the skin of the apple, he’s done eating.
Suppose the apple is a sphere with a radius of 4 centimeters, and that each bite of the apple is a circle of the sphere whose radius, as measured along the apple’s curved surface, is 1 centimeter. On average, how many minutes will it take this 2-year-old to eat the apple?
The solution to this Riddler Classic can be found in the following week’s column.
Solution to last week’s Riddler Express
Congratulations to 👏 Peter Biggart 👏 of Los Angeles, California, winner of last week’s Riddler Express.
Last week, Dakota Jones had found a highly symmetric crystal in her pursuit of the Temple of Diametra. However, nefarious agents had gotten wind of her plans, and Dakota and the crystal were nowhere to be found.
It was up to you to recreate the crystal using the data from Dakota’s laser scanner, which takes a 3D object and records 2D cross-sectional slices along the third dimension. Here was the looping animation file the scanner produced for the crystal:
What sort of three-dimensional shape was the crystal?
While the cross sections in the animation were all rectangles, if you stacked them on top of each other, you found a solid with no apparent rectangles at all. That’s exactly what solver Jacob Kes did, finding that the resulting three-dimensional figure was a tetrahedron:
Peter Ji went so far as to build his own paper models and similarly arrived at a tetrahedral solution:
You would typically think of a tetrahedron as having triangular slices that shrink as you move along the altitude, from the base to the apex. But in this case, the cross sections were moving from one edge of the tetrahedron to the opposite edge, producing rectangles (and even a square) along the way.
Returning for a moment to the original puzzle, the crystal was described as “highly symmetric.” I have it on good authority that the crystal was in fact a regular tetrahedron, whose four faces were all equilateral triangles.
However, because the thickness of the slices in the 3D scan was never specified, slightly less symmetric answers were also possible. Solver Jenny Mitchell identified the solid as an “isosceles tetrahedron,” while Tim Tebbe called it a “disphenoid.” I didn’t pick hairs — as long as you said it was some kind of tetrahedron, I gave you credit!
And so Dakota Jones once again owes a debt of gratitude to Riddler Nation. Now it’s simply a matter of time before she successfully locates the Temple of Diametra. Stay tuned — she may need your help on her next adventure!
Solution to last week’s Riddler Classic
Congratulations to 👏 Gabe Martin 👏 of Hove, England, winner of last week’s Riddler Classic.
Last week, you were locked in a castle dungeon with three fellow prisoners, in separate cells with no means of communication. But it just so happened that all four of you were logicians.
To entertain themselves, the guards decided to give you all a single chance for immediate release. Each prisoner was given a fair coin, which could either have been fairly flipped one time or returned to the guards without being flipped. If all flipped coins came up heads, you would all be set free! But if any of the flipped coins came up tails, or if no one chose to flip a coin, you would all be doomed to spend the rest of your lives in the dungeon.
The only tools you and your fellow prisoners had to aid you were random number generators, which gave each prisoner a random number, uniformly and independently chosen between zero and one.
What were your chances of being released?
Since all four prisoners (including you) were “logicians,” but you couldn’t communicate with each other, it was fair to assume that you would all pursue the same optimal strategy. And without any further information, the best you could have done was pick some value p between zero and one — if your random number generator gave you a value less than or equal to p, you’d flip your coin and hope it came up heads. But if the generator gave you a number greater than p, you’d return to the coin to the guards and hope at least one other prisoner would flip a coin.
At this point, the entire puzzle boiled down to a single variable, p. And you immediately knew p shouldn’t be zero — then no one would flip a coin, and you’d have no chance at freedom. If p were one, then everyone would flip a coin, and the chances of exclusively getting heads would be small. So p was somewhere in between, and your task was to find the value that maximized your chances of freedom, presumably by keeping the number of flips close to one.
You can break the problem down into five cases, depending on how many prisoners decide to flip a coin:
- The probability that zero prisoners flipped coins was (1−p)4, and the resulting probability of freedom was zero.
- The probability that exactly one prisoner flipped a coin was 4p(1−p)3, and the resulting probability of freedom was 1/2 — when that one coin came up heads.
- The probability that exactly two prisoners flipped a coin was 6p2(1−p)2, and the resulting probability of freedom was 1/4 — when both coins came up heads.
- The probability that exactly three prisoners flipped a coin was 4p3(1−p), and the resulting probability of freedom was 1/8 — when all three coins came up heads.
- The probability that exactly four prisoners flipped a coin was p4, and the resulting probability of freedom was 1/16 — when all four coins came up heads.
Putting this all together, when you and your fellow prisoners flipped your coins with probability p, your chances of freedom were 2p(1−p)3 + 3/2p2(1−p)2 + 1/2p3(1−p) + 1/16p4.
To maximize this function, you had to take the derivative, 2 − 9p + 21/2p2 − 15/4p3, and set it equal to zero. This occurred when p was equal to about 0.342. If you plugged this value back into the original expression, you got approximately 0.285, which meant that your chances of freedom stood at about 28.5 percent. If you’re still not convinced, then I recommend checking out solver David Robinson’s millions of simulations, which gave the same result.
But that was just the case of four prisoners. Last week’s extra credit asked you to further solve the general case of N prisoners — and many readers were up to the challenge!
Rather than break the problem up into many cases, as we just did for N = 4, solvers Laurent Lessard and Josh Silverman both used the binomial theorem to more compactly write out your chances of freedom, which turned out to be (1−p/2)N − (1−p)N. Again taking the derivative and setting it equal to zero, Laurent found that the maximum occurred when p was equal to 1 − 1/(2N/(N−1)−1). Finally, plugging in this value for p meant that your probability of being set free was 1/(2N/(N−1)−1)N−1.
But what does all this algebra mean? Well, as you’d expect, when there were more prisoners (i.e., as the value of N increased), your chances at freedom went down — which makes sense, because it’s harder to guarantee that only one or two prisoners will flip their coins when there are many prisoners. But how your chances at freedom went down was arguably the most interesting part of this puzzle.
Here’s a graph of how the probability of freedom varied with p for different numbers of prisoners, courtesy of Jason Ash:
As the number of prisoners increased, the peak of the curves moved left, and your chances of freedom appeared to approach … 25 percent?!
Yes, it was 25 percent! As you plugged larger and larger values of N into the probability for freedom, it indeed approaches a value of 1/4. (Solver Emma Knight also offered some neat intuition as to why this happens.)
So if the guards ever planned on enacting this scheme in a dungeon with loads of imprisoned logicians, they would have been better off just flipping two coins themselves and seeing if they both came up heads. It certainly would have saved them a lot of time!
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.