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How Long Will You Be Stuck Playing This Bar Game?

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 👏 Nathan Rooy 👏 of Cincinnati, 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 pub game puzzle that comes to us from James Cherry, an electrical engineer from Waterloo, Ontario.

Consider a hot new bar game. It’s played with a coin, between you and a friend, on a number line stretching from negative infinity to positive infinity. (It’s a very, very long bar.) You are assigned a winning number, the negative integer -X, and your friend is assigned his own winning number, a positive integer, +Y. A marker is placed at zero on the number line. Then the coin is repeatedly flipped. Every time the coin lands heads, the marker is moved one integer in a positive direction. Every time the coin lands tails, the marker moves one integer in a negative direction. You win if the coin reaches -X first, while your friend wins if the coin reaches +Y first. (Winner keeps the coin.)

How long can you expect to sit, flipping a coin, at the bar? Put another way, what is the expected number of coin flips in a complete game?

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

Last week’s Riddler concerned an archvillain who was slicing the pentagonal Riddler headquarters in half with a powerful laser. To recap the details: In-house mathematicians were tasked with moving sensitive riddling equipment to the safer locations in the headquarters. Your job was to identify those parts of the building that were at especially high risk of being zapped and, for extra credit, to calculate their area. Our solution comes to us from an anonymous philosopher.

The solution put simply: Keep away from the center! The pentagon’s area bisectors intersect near that point, raising the odds that central regions will be zapped. The perimeter is safest, and if archvillain Laser Larry randomizes his laser attack uniformly over perimeter points rather than angle, the middles of the sides are the safest of the safe.

For an even-sided regular polygon, such as a square, all area bisectors intersect at the center point itself, making it the most dangerous place, and danger decreases with distance from it. But for an odd-sided polygon, such as our pentagon, the most dangerous places are still where bisectors intersect, but it is not true that all bisectors intersect at a single point. That’s where things get interesting.

When we examine all the possible paths along which Laser Larry might slice our building in half, some especially dangerous points emerge that take the shape of a curvy star. These are the points that are sliced by multiple possible laser bisectors. It looks like this:

Steer clear of the curvy star.

For much, much more on the deep math underlying this problem, and methods to calculate the area of the star-shaped danger zone, see the rest of the full solution provided by the anonymous philosopher. (Editor’s warning: The solution is exhaustive, to say the least, and corybantic, to say the most, and you certainly needn’t have worked out all its pieces to answer the problem correctly. But the puzzle is so cool and rich I’ve included it all there for the riddling record.) For some other takes, see these analyses from our winner, Nathan, and from Laurent Lessard.

Elsewhere in the puzzling world:

  • Find four fractions. [The New York Times]
  • Some tennis problems. [Expii]
  • A swarm of B’s. [NPR]
  • You thought this solution was long? Here’s the largest ever math proof. [Nature]

Have a super weekend! Keep away from lasers.

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.