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 👏 **Silvia Adduci **👏 of Buenos Aires, 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 tough interplanetary war puzzle that comes to us from **Roberto Linares**, a Texas A&M graduate and data reconciliation aficionado at Pimsoft, in Houston.

A guardian constantly patrols a spherical planet, protecting it from alien invaders that threaten its very existence. One fateful day, the sirens blare: A pair of hostile aliens have landed at two random locations on the surface of the planet. Each has one piece of a weapon that, if combined with the other piece, will destroy the planet instantly. The two aliens race to meet each other at their midpoint on the surface to assemble the weapon. The guardian, who begins at another random location on the surface, detects the landing positions of both intruders. If she reaches them before they meet, she can stop the attack.

The aliens move at the same speed as one another. *What is the probability* that the guardian saves the planet if her linear speed is 20 times that of the aliens’?

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 solution to last week’s Riddler, concerning a bar game played with coin flips. A marker is placed at zero on a number line, and is moved one integer in a positive direction if the coin lands heads, and one in a negative direction if the coin lands tails. You win if the marker reaches \(-X\) first, and your friend wins if the marker reaches \(+Y\) first. You can expect this game to last for \(X\cdot Y\) flips of the coin.

Why? Let \(E(N)\) be the expected number of flips to end the game when the marker is sitting \(N\) places to the right of your winning number, \(-X\). We want to find \(E(X)\), as the game begins with the marker at zero. We already know \(E(0)=0\) and \(E(X+Y)=0\) — the game is already over in those cases, and no more flips are required.

For *other* values of \(N\), we can set up a recurrence relation. Specifically,

$$E(N) = 1 + \frac{1}{2} E(N-1) + \frac{1}{2} E(N+1)$$

We know we’ll have to flip the coin at least once, and then we’ll move the marker either one number higher or one number lower, with equal probability. Solving this equation for \(E(N)\) we get

$$E(N) = -N^2 + A\cdot N + B$$

Where \(A\) and \(B\) are two arbitrary numbers. But we know \(E(0)=0\), so we know \(B=0\). And we known \(E(X+Y)=0\), so we know \(A=X+Y\). Now we can find our solution, \(E(X)\):

$$E(X) = -X^2 + (X+Y)\cdot X$$

$$= X\cdot Y$$

Elsewhere in the puzzling world:

- Where should you tie your shoes in the airport? [The Guardian]
- New puzzles from Po-Shen Loh, coach of the victorious U.S. International Math Olympiad team. [Expii]
- Some word puzzles. [NPR]
- The authorities need your help! [The Wall Street Journal]
- “A 91-year-old woman is under investigation in Germany after filling in blank spaces on a crossword-themed artwork in a museum.” [BBC]

Have a great weekend!