**UPDATE: **Unfortunately, there are a couple errors in my computations below that I found after this post went live. In my next post, Mistakes are Good, I fix those errors and reflect on the process of learning from them.

**ORIGINAL POST:**

A post last week to the AP Statistics Teacher Community by David Bock alerted me to the new weekly Puzzler by Nate Silver’s new Web site, http://fivethirtyeight.com/. As David noted, with their focus on probability, this new feature offers some great possibilities for AP Statistics probability and simulation.

I describe below FiveThirtyEight’s first three Puzzlers along with a potential solution to the last one. If you’re searching for some great problems for your classes or challenges for some, try these out!

**THE FIRST THREE PUZZLERS:**

The first Puzzler asked a variation on a great engineering question:

You work for a tech firm developing the newest smartphone that supposedly can survive falls from great heights. Your firm wants to advertise the maximum height from which the phone can be dropped without breaking.

You are given two of the smartphones and access to a 100-story tower from which you can drop either phone from whatever story you want. If it doesn’t break when it falls, you can retrieve it and use it for future drops. But if it breaks, you don’t get a replacement phone.

Using the two phones, what is the

minimum number of dropsyou need to ensure that you can determineexactly the highest storyfrom which a dropped phone does not break? (Assume you know that it breaks when dropped from the very top.) What if, instead, the tower were 1,000 stories high?

The second Puzzler investigated random geyser eruptions:

You arrive at the beautiful Three Geysers National Park. You read a placard explaining that the three eponymous geysers — creatively named A, B and C — erupt at intervals of precisely two hours, four hours and six hours, respectively. However, you just got there, so you have no idea how the three eruptions are staggered. Assuming they each started erupting at some independently random point in history,

what are the probabilitiesthat A, B and C, respectively, will be the first to erupt after your arrival?

Both very cool problems with solutions on the FiveThirtyEight site. The current Puzzler talked about siblings playing with new phone apps.

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,

younow 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.)

**SOLVING THE CURRENT PUZZLER:**

Before I started, I saw Nick Brown‘s interesting Tweet of his simulation.

If Nick’s correct, it looks like a mode of 5 minutes and an understandable right skew. I approached the solution by first considering the distribution of initial random app choices.

There is a chance the siblings choose the same app and head to dinner after the first round. The expected length of that round is minutes.

That means there is a chance different length apps are chosen with time differences between 1 and 4 minutes. In the case of unequal apps, the average time spent before the shorter app finishes is minutes.

It doesn’t matter which sibling chose the shorter app. That sibling chooses next with distribution as follows.

While the distributions are different, conveniently, there is still a time difference between 1 and 4 minutes when the total times aren’t equal. That means the second table shows the distribution for the 2nd and all future potential rounds until the siblings finally align. While this problem has the potential to extend for quite some time, this adds a nice pseudo-fractal self-similarity to the scenario.

As noted, there is a chance they complete their apps on any round after the first, and this would not add any additional time to the total as the sibling making the choice at this time would have initially chosen the shorter total app time(s). Each round after the first will take an expected time of minutes.

The only remaining question is the expected number of rounds of app choices the siblings will take if they don’t align on their first choice. This is where I invoked self-similarity.

In the initial choice there was a chance one sibling would take an average 1.6 minutes using a shorter app than the other. From there, some unknown average N choices remain. There is a chance the choosing sibling ends the experiment with no additional time, and a chance s/he takes an average 1.5 minutes to end up back at the Table 2 distribution, still needing an average N choices to finish the experiment (the pseudo-fractal self-similarity connection). All of this is simulated in the flowchart below.

Recognizing the self-similarity allows me to solve for N.

**FINAL ANSWER:**

**Number of Rounds – **Starting from the beginning, there is a chance of ending in 1 round and a chance of ending in an average 5 rounds, so the expected number of rounds of app choices before the siblings simultaneously end is

rounds

**Time until Eating – **In the first choice, there is a chance of ending in 3 minutes. If that doesn’t happen, there is a subsequent chance of ending with the second choice with no additional time. If neither of those events happen, there will be 1.6 minutes on the first choice plus an average 5 more rounds, each taking an average 1.5 minutes, for a total average minutes. So the total average time until both siblings finish simultaneously will be

minutes

**CONCLUSION:**

My 7.88 minute mean is reasonably to the right of Nick’s 5 minute mode shown above. We’ll see tomorrow if I match the FiveThirtyEight solution.

Anyone else want to give it a go? I’d love to hear other approaches.