Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world, including you, the readers. You’ll find this week’s puzzle below.
Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the form at the bottom. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday — have a great weekend!
Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 Si Ferrel 👏 of the United States Air Force Academy, our big winner. You can find a solution to the previous Riddler at the bottom of this post.
Now here’s this week’s Riddler, a puzzle about a misanthropic neighborhood that comes to us from Jim Ferry, a mathematician at Metron, a scientific consulting company in Reston, Virginia.
The misanthropes are coming. Suppose there is a row of some number, N, of houses in a new, initially empty development. Misanthropes are moving into the development one at a time and selecting a house at random from those that have nobody in them and nobody living next door. They keep on coming until no acceptable houses remain. At most, one out of two houses will be occupied; at least one out of three houses will be. But what’s the expected fraction of occupied houses as the development gets larger, that is, as N goes to infinity?
Extra credit: It’s been a while, so let’s offer up a 🏆 Coolest Riddler Extension Award 🏆. Complicate the neighborhood, make the misanthropes friendlier, build better fences to make better neighbors, or something even more creative. Submit your extension and its solution via the form below. The winner gets a shiny emoji trophy next week.
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And here’s the solution to last week’s Riddler, concerning the best strategy to win a space race to an alien artifact near Jupiter, which came to us from Robert Youngquist.
If you buy one Russian engine ($400 million, saving 200 days), three ion engines ($450 million total, saving 150 days) and three light payloads ($150 million, saving 75 days) you will spend your $1 billion and get there 425 days earlier. This seems like the right answer — and it would be if the question were how to get there fastest.
But the challenge was to get there first. If you go with the option above, a competitor could also buy the same items and you would have no advantage. Here’s one correct answer (there may exist variations on it that are also correct): Buy two Russian engines ($800 million, saving 300 days), one ion engine ($150 million, saving 50 days) and the rest of the world’s xenon (25,000 kilograms for $50 million). Basically, you want to hoard the xenon. You will get there only 350 days earlier, but the best anyone else can do is to buy one Russian engine and four payloads to get there 300 days earlier. You are guaranteed to win by at least 50 days.
Here’s an illustration of your various spaceship options, from Zach Wissner-Gross:
Fantastic hashtag. But it takes all manner of approaches to attempt a Riddler, of course, and Trevor Akin went the colorful-scrawl route:
While the team of Samson Perry and Ben Novacek went classic note card:
Xenon for days! Live long and prosper and, more immediately, have a wonderful weekend!