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.
Earlier this week, I had the pleasure of attending the MOVES conference in New York City, hosted by the National Museum of Mathematics. The opening keynote, “How to Invent Puzzles,” was delivered by puzzle master Scott Kim. In particular, his cuisenaire-rod puzzles got me thinking. …
A hexomino is a shape made by six identical, nonoverlapping squares that are connected by edges. Some hexominoes, like the one shown below, can be decomposed into an array of three squares, an array of two squares and an array of one square.
How many distinct hexominoes can you find that cannot be decomposed into arrays of three, two and one squares? For the purposes of this riddle, two hexominoes are considered equivalent if they can be turned into one another by rotation and/or reflection.
From Andrew Lin comes a game for getting yourself home:
You are stranded in a casino (lucky you!) and need to purchase a flight home. Flights cost $250, but you have only $100 at the moment. However, as I just said, you’re in a casino! Surely, you can gamble your way to $250.
The casino has a game called “Riddler’s Delight,” in which you can bet any amount of money in your possession for an even greater amount of money. You can even bet fractional (i.e., you can bet fractions of a penny), irrational or infinitesimal amounts if you so desire.
The catch is that the odds are not in your favor. In Riddler’s Delight, whenever you bet A dollars in an attempt to win B dollars (with B > A), your probability of winning is not A/B, which you would expect from a fair game. Instead, your probability of winning is always 10 percent less, or 0.9(A/B).
What should your betting strategy be to maximize your probability of getting home, and what is that probability?
Solution to last week’s Riddler Express
Congratulations to 👏 Michael Jackson 👏 of Grove City, Pennsylvania, winner of last week’s Riddler Express.
Last week, you were rolling a fair six-sided die on a ridged gnocchi board, such that two adjacent faces came up every time.
On average, what was the sum of the numbers shown on those two faces?
Each pair of adjacent faces were equally likely to be the two on top. Some solvers listed out all these pairs, added up the numbers of these corresponding faces and computed the average.
But there was another, more efficient approach, one used by solvers including Kiera Jones of Cincinnati, Ohio. Instead of listing pairs of faces, you could instead think about the edges between them since the face pairs and the edges exhibit a one-to-one correspondence.
Now, each cube has 12 edges. If you were to list these edges out, along with which two faces they’re between (for a total of 24 total slots), then each face would appear on the list four times. Why? Because by symmetry, no face could appear on the list any more or less often than the other faces.
So, computing the average sum of the face number pairs was equivalent to adding up all six face numbers four times and then dividing by 12. And that was equivalent to doubling the average of the six face numbers. That average was 3.5 (i.e., the sum of the whole numbers from 1 to 6, divided by 6), and doubling it gave you the answer, 7.
While this wasn’t technically extra credit from last week, solver Benjamin Dickman (as well as the puzzle’s original submitter, Michael Branicky) proposed an extension in which three adjacent faces (rather than two) were summed. In this case, by a similar symmetrical argument, the sum averaged to three times 3.5, or 10.5.
Perhaps the coolest part of this puzzle was that the precise arrangement of the numbers on the six faces turned out to be irrelevant. In the end, they all simply averaged out.
Solution to last week’s Riddler Classic
Congratulations to 👏 Christian Wolters 👏 of San Jose, California, winner of last week’s Riddler Classic.
Last week, Magritte the bowler was competing head-to-head against fellow bowler Fosse. However, rather than knocking down 10 pins arranged in a triangular formation, they were trying to knock down N2 pins (where N was some very, very large number) arranged in a rhombus, as shown below:
When Magritte rolled, he always knocked down the topmost pin. Then, if any pin got knocked down, it has a 50 percent chance of knocking down either of the two pins directly behind it, independently of each other. (If there was only one pin directly behind it, then it too has a 50 percent chance of being knocked over.)
Fosse was a stronger bowler than Magritte. Like Magritte, she always knocked down the topmost pin. But each pin that was knocked down then has a 70 percent chance (rather than Magritte’s 50 percent) of knocking down any of the pins directly behind it.
What were Magritte’s and Fosse’s respective probabilities of knocking down the bottommost pin in the rhombus formation?
At first glance (at least to some readers), this appeared to be relatively straightforward. Suppose each pin had a probability p of knocking over either pin behind it (i.e., p was 0.5 for Magritte and 0.7 for Fosse). And let’s further suppose the probability that two particular adjacent pins A and B in the same row got knocked over were a and b, respectively. What was the probability that the pin C directly behind both A and B would get knocked over? It seemed that the probability C was knocked down by A while B remained standing was a(1−b)p; the probability C was knocked down by B while A remained standing was (1−a)bp; and the probability C was knocked down along with both A and B was ab(2p−p2). Adding up these probabilities allowed you to iteratively compute each pin’s probability of being knocked down based on all the probabilities of the pins that came before it. Right?
Wrong! The fault in this logic was that it incorrectly assumed that pins A and B were independent. In reality, if A had been knocked down, that meant it had to have been knocked down by one of the pins in front of it, including the preceding pin it shared with B, which in turn meant B was more likely to have been knocked down. And because knocking down A and B were dependent events, that meant you couldn’t simply multiply their probabilities as we did in the preceding paragraph.
As it turned out, this is an unsolved problem2 from a branch of statistical math and physics known as percolation theory, so named because we can think of fluids (or, in this case, bowling pins) percolating through a medium. While I didn’t receive any solutions that appeared to solve this unsolved problem, several readers simulated Magritte’s and Fosse’s bowling to closely approximate their chances of knocking down the last pin.
Now, when p was small, the cascade of falling pins inevitably terminated. But when p exceeded what’s known as the percolation threshold, there was a chance the last pin could have been knocked down. As noted by solver Laurent Lessard (in an excellent write-up!), the percolation threshold for this particular problem is not yet known; however, it is known that the threshold lies between 0.5176 and 0.6298 (respectively proven in 1957 and 1982).
Since p was 0.5 for Magritte, which was definitely below the percolation threshold, that meant the probability Magritte knocked down the last pin was zero. Meanwhile, p was 0.7 for Fosse, which was definitely greater than the percolation threshold. So Fosse had a chance of knocking down that pin! Below are 15 Fosse simulations that Laurent illustrated when N was 100. Of these 15, 11 appeared to result in sustainable percolations. If you look closely at them, you can see surprising structure — streaks of pins that weren’t knocked over in darker blue and percolating streams of knocked over pins around them.
Laurent ultimately found that Fosse’s probability of knocking down the last pin was approximately 0.5782. Solver Peter Ji ran 10,000 simulations when N was 100 (i.e., very large), finding a probability of about 0.58, while solver Pradeep Niroula found it was about 0.5828. In the end, I accepted all answers for Fosse close to 0.58.
A few solvers, like Hernando Cortina, plotted the probability of knocking down the last pin as a function of p:
Sure enough, there appeared to be a percolation threshold between 0.5176 and 0.6298 where the probability was no longer zero.
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 firstname.lastname@example.org.