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.
You’re playing a game of cornhole with your friends, and it’s your turn to toss the four bean bags. For every bean bag you toss onto your opponents’ board, you get 1 point. For every bag that goes through the hole on their board, you get 3 points. And for any bags that don’t land on the board or through the hole, you get 0 points.
Your opponents had a terrible round, missing the board with all their throws. Meanwhile, your team currently has 18 points — just 3 points away from victory at 21. You’re also playing with a special house rule: To win, you must score exactly 21 points, without going over.
Based on your history, you know there are three kinds of throws you can make:
- An aggressive throw, which has a 40 percent chance of going in the hole, a 30 percent chance of landing on the board and a 30 percent chance of missing the board and hole.
- A conservative throw, which has a 10 percent chance of going in the hole, an 80 percent chance of landing on the board and a 10 percent chance of missing the board and hole.
- A wasted throw, which has a 100 percent chance of missing the board and hole.
For each bean bag, you can choose any of these three throws. Your goal is to maximize your chances of scoring exactly 3 points with your four tosses. What is the probability that your team will finish the round with exactly 21 points and declare victory?
From Elise Corbin comes something new for The Riddler: a crossword puzzle! This one is titled “Systematic Solving,” and it may or may not have a mathematical twist. Enjoy!
For this puzzle, you can submit your solution by taking a picture or screenshot of your completed puzzle and uploading it via the submission form.
Solution to last week’s Riddler Express
Congratulations to 👏 Dean Ballard 👏 of Seattle, Washington, winner of last week’s Riddler Express.
Last week, after you had intended to make a perfectly circular pancake, the batter spread out and filled every last corner of your square pan. (It was unclear why you were using a square pan in the first place.)
To salvage your breakfast, you planned to slice off the corners of your square pancake (an example of which is shown below), giving you something closer to a circle. Each slice had to be straight, and no slice could pass through the inscribed blue circle that represented your original desired pancake.
Of course, there was a catch — you could make at most five slices. If the blue circle had a radius of 1 unit, what was the minimum possible area your pancake can have after five slices?
For any number of cuts less than five, this problem was more straightforward. To remove as much area as possible, each of those cuts would have been tangent to the circle while making a 45 degree angle with the sides of the square. The total area of the original square pancake was 4, and each of these cuts removed (2−√2)2/2 — or 3−2√2 — units of area.
Had the problem asked for exactly four cuts, the result would have been a regular octagon with an area of 4 − 4(3−2√2), which simplified to 8√2−8, or about about 3.3137 — not too far from 𝜋, the area of the ideal circular pancake.
That fifth cut really mucked up the works, however. As noted by Emma Sona, the pigeonhole principle dictated that having a fifth cut meant that one of the four corners would have two cuts.
Solver Ian Rhile recognized that to lop off as much area from the square as possible, both cuts had to be tangent to the circle. Ian parameterized the two tangent points by the angles they made within the circle — 𝜃2 and 𝜃3 in the diagram below — and then calculated the area of the remaining top right quarter-square after both cuts were made. This turned out to be tan(𝜃2) + tan(𝜃3) + tan(45°−𝜃2−𝜃3).
From there, you could have muddled through some more trigonometric identities, or you could have graphed the function to locate its minimum. This minimum occurred when both 𝜃2 and 𝜃3 equalled 15 degrees.
Whether you subtracted the areas of the sliced corners or counted up the area of the resulting nonagon itself, the smallest possible area of the pancake was 6√2−3√3 units, or about 3.289 units — a smidge closer to 𝜋!
Alas, no one attempted to make such a nonagonal pancake in real life. I bet it would have been delicious.
Solution to last week’s Riddler Classic
Congratulations to 👏 Maxim Wang 👏 of St. Louis, Missouri, winner of last week’s Riddler Classic.
Last week, Riddler Nation’s neighbor to the west, Enigmerica, was holding an election between two candidates, A and B. Every person in Enigmerica voted randomly and independently, and that the number of voters was very, very large. Moreover, due to health precautions, 20 percent of the population decided to vote early by mail.
On election night, the results of the 80 percent who voted on Election Day were reported out. Over the next several days, the remaining 20 percent of the votes were then tallied.
What was the probability that the candidate who had fewer votes tallied on election night ultimately won the race?
Suppose m people voted by mail. Since they represented 20 percent of the voting population, that meant 4m people voted on Election Day. Since there were many, many voters, and each voter’s probability distribution was binomial (i.e., randomly split between candidates A and B), the central limit theorem dictated that the probability distribution for the percent of the vote each candidate received would be approximately normal.
Solver Emma Knight worked through the specific details. Call x the number of votes A received on Election Day minus the number of votes B received on Election Day, and let y be a corresponding variable for the mail-in ballots. The probability distribution for x was P(x) = exp[−x2/(8m)]/√(8𝜋m), and the probability distribution for y was Q(y) = exp[−y2/(2m)]/√(2𝜋m). Since the Election Day and mail-in votes were independent, meaning x and y were independent, the overall probability distribution was R(x,y) = P(x)Q(y).
This was a two-dimensional normal distribution. But what was it that you actually needed to compute?
For the candidate who had fewer votes tallied on election night to ultimately win the race, you needed x and y to be opposite in sign, and you also needed the magnitude of y to be greater than the magnitude of x. In mathematical terms, that meant either y > −x > 0 or y < −x < 0, both of which were regions that each made up one-eighth of the coordinate plane.
The figure below shows this graphically. The yellow region shows where the probability distribution is greater, and the blue region shows where it is smaller. The area between the two white lines represents our regions of interest.
All that was left to do was integrate the probability distribution over those regions. (Yikes!) Emma was up to the task, finding the surprisingly concise closed-form solution tan−1(1/2)/𝜋, or about 14.7 percent. Several readers came up with an answer that was half this value, accounting for the case when A had fewer votes tallied on election night but ultimately won, and missing the case when B had fewer votes tallied on election night but ultimately won.
Meanwhile, many solvers went the computational route, simulating thousands or even millions of Enigmerican (that’s a word, right?) elections to see how often the candidate with fewer votes tallied on election night ultimately won. For example, Andrew Nicholls ran 100 million simulations that assumed Enigmerica had a population of 100 billion, which gave an answer of 14.75 percent — not too far from the analytical result!
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.