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Can You Save Riddler Headquarters From Laser Larry? Please?!

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! 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 link below. 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 👏 Mike Phelan 👏 of Clonmel, Ireland, 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 doomsday (both in theme and difficulty) puzzle that comes to us from a philosophy professor who wishes to remain anonymous.

The archvillain Laser Larry threatens to imminently zap Riddler Headquarters (which, seen from above, is shaped like a regular pentagon with no courtyard or other funny business). He plans to do it with a high-powered, vertical planar ray that will slice the building exactly in half by area, as seen from above. The building is quickly evacuated, but not before in-house mathematicians move the most sensitive riddling equipment out of the places in the building that have an extra high risk of getting zapped.

Where are those places, and how much riskier are they than the safest spots? (It’s fine to describe those places qualitatively.)

Extra credit: Get quantitative! Seen from above, how many high-risk points are there? If there are infinitely many, what is their total area?

Submit your answer
Need a hint? You can try asking me nicely. Want to submit a new puzzle or problem? Email me.

And here’s the answer to last week’s Riddler, concerning how to beat Roger Federer at Wimbledon. If you have a 1 percent chance of winning any given point, but Federer offers to start the match at any score you name, you give yourself the best shot by declaring yourself up two sets to none, with the third set tied 6-6, and yourself up 6-0 in the tiebreaker. Your chance of winning the match in that scenario is roughly 5.86 percent. (The individual scores you name for the first two sets won’t matter, as long as you’ve won them, but you do need to get to the tiebreaker before the fifth set — there are no fifth-set tiebreakers at Wimbledon.)

Essentially, you’re trying to maximize the number of match points at your disposal, making it as difficult as possible for Roger to come back. In the best scenario, you’ve bought yourself at least six match points. (The six opportunities to win the tiebreaker, which you start up 6-0.) A more “obvious” solution might be to declare yourself up two sets to none, and up 5-0 in the third set with the score at 40-love. However, that buys you just three guaranteed match points, and only about a 3 percent chance of victory.

Calculating your exact chance of victory is a bit tedious, but we can arrive at a very good approximation quite easily by focusing only on your chances of winning the tiebreaker. Roger wins each point 99 percent of the time. Therefore, your chance of winning any of the first six points, which is by far your best shot to win Wimbledon, is \(1-0.99^6\approx 0.0585\). Your chance of winning the tiebreaker if it gets to 6-6 is just a negligible touch better than your chances of winning two points in a row, or \(0.01^2=0.0001\). So your approximate chance of winning the title is 0.0586, or 5.86 percent.

Your chances of winning the match if you fail to win the tiebreaker are vanishingly small — your probability of winning a fresh set, for example, is \(2.35\times 10^{-39}\) — so we can safely ignore them for all practical purposes. But Laurent Lessard, completist and Riddler Hall of Famer that he is, isn’t satisfied by a good approximation. He walks us through the particulars of arriving at the exact answer. Adding a few more decimal places of precision reveals a 5.86159 percent chance of victory.

This week’s 🏆 Coolest Riddler Extension Award 🏆 goes to Matjaž Konvalinka, who linked the correct tiebreaker solution to the “obvious” solution where you’re up two sets and 5-0 and 40-love in the third set. He asked: For what individual point-winning percentage (the 1 percent in the original problem) would the odds of winning the match be the same in both situations? In other words, when is it worthwhile to buy yourself the additional match points by losing games to get into a tiebreaker? Via computation, he found that the answer was 41.4 percent. Tyler Silber, John Faben and Ken Levine also came to similar conclusions, independently. Matjaž, please share your shiny emoji trophy.

Zach Wissner-Gross earns honorable mention for showing just how sensitive the chances of winning a match are to changes in the chances of winning any single point:

And elsewhere in the puzzling world:

  • Two (belated) Fourth of July puzzles. [New York Times]
  • More (belated) Fourth of July puzzles. [Expii]
  • A summertime audio cookout puzzle, perfect for the patio. [NPR]
  • A logic puzzle on Brexit and European language interpreters. [The Guardian]
  • A $2 million treasure hunt, based on a 24-verse poem. [New York Times]

Have a great weekend!

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.