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: If you want to be eligible for the shoutout, I need to receive your correct answer before midnight EST tonight. Speed is prized around here, but so is considered thought.

But before we get to the new puzzle, let’s dwell on last week’s! Congratulations to 👏 **Alana Christie** 👏 of Dallas, our big winner. You can find a full solution to the previous Riddler at the bottom of this post.

Now, here’s this week’s Riddler, which comes to us from **Olivia Walch**, a mathematics Ph.D. student and cartoonist:

You’ve just finished unwrapping your holiday presents. You and your sister got brand-new smartphones, opening them at the same moment. You immediately both start doing important tasks on the Internet, and each task you do takes one to five minutes. (All tasks take exactly one, two, three, four or five minutes, with an equal probability of each). After each task, you have a brief moment of clarity. During these, you remember that you and your sister are supposed to join the rest of the family for dinner and that you promised each other you’d arrive together. You ask if your sister is ready to eat, but if she is still in the middle of a task, she asks for time to finish it. In that case, *you* now have time to kill, so you start a new task (again, it will take one, two, three, four or five minutes, exactly, with an equal probability of each). If she asks you if it’s time for dinner while you’re still busy, you ask for time to finish up and she starts a new task and so on. From the moment you first open your gifts, *how long on average does it take for both of you to be between tasks at the same time so you can finally eat?* (You can assume the “moments of clarity” are so brief as to take no measurable time at all.)

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

*Note: Because of New Year’s, The Riddler will run on Tuesday next week (Dec. 29), as well. Happy holidays!*

Now, here is the full solution to last week’s Riddler, concerning the eruption of geysers, courtesy of Brian Galebach. In a slight uptick from the week before, 29.7 percent of you submitted a correct answer. *Note: 100 percent of you are awesome*.

Because it erupts precisely every two hours, we know that geyser A must erupt within the first two hours following our arrival. Because geyser B erupts every four hours and because we have no idea when it last erupted, the probability that B will erupt within the first two hours is 1/2. And, similarly, the probability that C, which erupts every six hours, will erupt within the first two hours is 1/3. (In other words, from our point of view, the eruption times for A, B and C are continuously uniformly distributed between the time we arrive and two, four and six hours, respectively, from the time we arrive.) There are four cases to consider:

- A, B, and C all erupt within two hours.
- A and B, but not C, erupt within two hours.
- A and C, but not B, erupt within two hours.
- Only A erupts within two hours.

In each of the four cases above, each of the geysers that can erupt within the first two hours will erupt first with equal probability. So for example, in Case 1, A, B and C all have a 1/3 chance to erupt first. In Case 4, A is guaranteed \((p(A)=1)\) to erupt first.

So to calculate the probability of A erupting first, we sum the probabilities of each case occurring multiplied by the probability of A erupting first in each case.

\[p(A) = \left(\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2}\right) + \left(\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2}\right) + \left(\frac{1}{2} \cdot \frac{2}{3} \cdot 1\right) = \frac{23}{36}\]

Likewise, we can calculate the probabilities of B and C using the same method. For B, we need only consider Cases 1 and 2, and for C, we only need consider Cases 1 and 3.

\[p(B) = \left(\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{2}\right) = \frac{8}{36}\]

\[p(C) = \left(\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2}\right) = \frac{5}{36}\]

**So the probabilities for A, B and C are 23/36, 8/36 and 5/36, respectively.**