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.
Scoring in bowling can be a tricky matter. There are 10 frames, and in each frame you get two chances to knock down as many of the 10 pins as you can. Each pin knocked down is worth 1 point, and the pins are reset after each frame. Your score is then the sum of the scores across all 10 frames.
If only it were that simple. There are special rules for spares (when you’ve knocked down all 10 pins with the second ball of a frame) and strikes (when you knock down all 10 pins with your first ball). Whenever you bowl a strike, that frame is scored as 10 plus the scores of your next two rolls. This can lead to some dependency issues at the end of the game, which means the final frame has its own set of rules that I won’t go into here.
For example, if you bowled three strikes and missed every subsequent shot (i.e., they were gutter balls), your third frame would be worth 10 points, your second frame would be worth 20 and your first frame 30. Your final score would be 30 + 20 + 10, or 60 points.
This week, Magritte is going bowling, and the bowling gods have decided that he will bowl exactly three strikes in three randomly selected frames. All the other frames will consist of nothing but gutter balls. Magritte also lacks patience for bowling’s particular rules. If one of his three strikes comes on the 10th and final frame and he is prompted to bowl further to complete the game, he will bowl gutter balls out of frustration.
What score can Magritte expect to achieve? (That is, with three randomly placed strikes, what is his average score?)
The solution to this Riddler Express can be found in the following column.
From Ben Edelstein comes a crash course in probability:
Eight two-way roads all converge at a single intersection, as shown in the diagram below.
Two cars are heading toward the single intersection from different directions chosen at random. Upon reaching the intersection, they both turn in a random direction (where proceeding straight is a possible “turn”) — however, neither car pulls a U-turn (i.e., heads back the way it came).
In some cases, the paths of the cars can be drawn so that they do not cross. In this case, all is well.
However, in other cases, the paths must cross. In this event, the cars will crash.
What is the probability the cars will crash? (If both cars head off in the same direction, that also counts as a crash.)
Extra credit: As the number of two-way roads converging at the intersection approaches infinity, what value does the probability of crashing approach?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to ÐÐ¯Ð¡Ð Al Shaheen ÐÐ¯Ð¡Ð of New Haven, Indiana, winner of last week’s Riddler Express.
Last week, you were hired to design a new logo for Riddler City. The mayor was a little eccentric and had requested that the logo have at least two lines of symmetry that intersected at an angle of precisely 1 radian, or 180/ÐÐÐ¬Ð (approximately 57.3) degrees.
What sorts of logos met the mayor’s requirements? (You were specifically asked to give an example or describe what all possible logos had in common.)
One way to solve this was to start with a few points and see what happened. Because the logo was symmetric across two lines, that meant that for any point you drew, there had to be a corresponding point on the other side of the line. (Otherwise the figure would not be symmetric!)
If you happened to start with three points and alternated reflecting them across either of the two lines, here was the resulting pattern:
Several readers recognized the inevitability of this result. As noted by solver Pramod Mathai, with each reflection, the distance between a point and the lines’ intersection was preserved. Meanwhile, because of the irrationality of the angle measure at the intersection (i.e., it was 1/(2ÐÐÐ¬Ð) of a complete rotation), each subsequent reflection resulted in a brand new point. The points in the resulting locus were all the same distance away from the intersection. In other words, they formed a circle!
By the way, you might have noticed that each ring in the animation was a little “clumpy” at first. In fact, before the animation speeds up, each ring appears to have 22 clumps. Meanwhile, one well-known approximation for ÐÐÐ¬Ð is 22/7 — again, the number 22! Could it be a coincidence?
On a separate note, before anyone shoots off an angry email to me (or to Riddler City’s mayor, for that matter), the logo didn’t technically have to be a complete circle. For those ready to venture into a little set theory, read on.
A circle on the real coordinate plane consists of ℵ1 points, since the points can be put in a one-to-one correspondence with the real numbers. Meanwhile, the countable process in the animation above, if allowed to run forever, will contain ℵ0 points. This set of points will be dense on the actual circle, but would make up an infinitesimal amount of the circle, similar to the set of rational numbers on a real number line.
Since the mayor likes continuous logos, concentric circles will do!
Solution to last week’s Riddler Classic
Congratulations to ÐÐ¯Ð¡Ð Peter Exterkate ÐÐ¯Ð¡Ð of Sydney, Australia, winner of last week’s Riddler Classic.
Earlier this month, Adam Kotsko asked the Twitterverse to choose four contiguous U.S. states for a new breakaway nation:
This got Philip Bump wondering about states you could pick such that the remaining (previously) contiguous states were no longer contiguous, but rather broken up into two near-halves by area. Treating the Upper Peninsula of Michigan as distinct from the Lower Peninsula, Philip hypothesized that removing Illinois, Missouri, Oklahoma and New Mexico would create two near-equal halves.
Last week, you were asked to remove a set of states (not necessarily four) so that you had two distinct contiguous regions among the lower 48 states, where the larger region had area A and the smaller region had area B. What states should you have removed to maximize area B?
It was ambiguous whether you should have used states’ land area or total area. It turned out these gave different answers!
But before we get to that, let’s see how Philip’s hypothesis did. By removing Illinois, Missouri, Oklahoma and New Mexico, Philip created eastern and western halves of the country. The eastern half was smaller, consisting of 35.39 percent of the land area and 37.19 percent of the total area. Not bad, considering only four states were selected.
When it came to land area, solver Jenny Mitchell was among those with the largest smaller region (i.e., the greatest value of B):
By removing Idaho, Wyoming, Nebraska, Missouri, Arkansas and Louisiana, the western half made up 42.65 percent of the land area — remarkably close to the 43.12 percent of the eastern half.
When it came to total area, solver Shawn Feils removed New Mexico, Oklahoma, Missouri, Iowa and Minnesota. This was very similar to Philip’s original idea, but resulted in a larger eastern half that produced a better balance. Here, the eastern half was again the smaller of the two, but it now made up 42.73 percent of the total area.
Again, due to the ambiguity of the problem, I received a number of solutions between 40 and 45 percent that appeared to make slightly different assumptions (e.g., land area vs. total area, whether or not to include Washington, D.C., etc.). So if you are convinced that you beat the results above, that’s totally cool. Share your map like Jenny did!
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 email@example.com.