# Great Probability Problems

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 drops you need to ensure that you can determine exactly the highest story from 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 probabilities that 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.

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 $\displaystyle \frac{5}{25}$ chance the siblings choose the same app and head to dinner after the first round.  The expected length of that round is $\frac{1}{5} \cdot \left( 1+2=3=4+5 \right) = 3$ minutes.

That means there is a $\displaystyle \frac{4}{5}$ 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 $\frac{1}{25} \cdot \left( 8*1+6*2+4*3+2*4 \right) = 1.6$ 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 $\displaystyle \frac{4}{20}=\frac{1}{5}$ 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 $\frac{1}{20} \cdot \left( 7*1+5*2+3*3+1*4 \right) = 1.5$ 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 $\frac{4}{5}$ 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 $\frac{1}{5}$ chance the choosing sibling ends the experiment with no additional time, and a $\frac{4}{5}$ 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.

$\displaystyle N = \frac{1}{5} \cdot 1 + \frac{4}{5} \cdot N \longrightarrow N=5$

Number of Rounds – Starting from the beginning, there is a $\frac{1}{5}$ chance of ending in 1 round and a $\frac{4}{5}$ chance of ending in an average 5 rounds, so the expected number of rounds of app choices before the siblings simultaneously end is

$\frac{1}{5} *1 + \frac{4}{5}*5=4.2$ rounds

Time until Eating – In the first choice, there is a $\frac{1}{5}$ chance of ending in 3 minutes.  If that doesn’t happen, there is a subsequent $\frac{1}{5}$ 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 $1.6+5*1.5=9.1$ minutes.  So the total average time until both siblings finish simultaneously will be

$\frac{1}{5}*3+\frac{4}{5}*9.1 = 7.88$ 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.

### 9 responses to “Great Probability Problems”

1. Steve Earth

Out of curiosity, I wrote a quick code snippet to perform the simulation (I did this before I read your response which showed that Nick Brown had already done something similar). Linked below is my graph (similar to Nick’s,including the second “hump” in the pdf which I don’t think is a fluke). In summary, my data resulted in a mean of about 9 (so I think 7.88 is too low).
http://tinyurl.com/phones538

• This is excellent. Your initial reactions got me worried about my answer and eventually led to the correction in the next post. Thanks.

I’m also intrigued by the secondary bump at 7ish. I’ve not played with many models like this one where there are actually two different pdfs at play. Here, there was the pdf of the first selections, and the pdf for any necessary additional selections. I suspect the “bump” is due to this transition between pdfs. I wonder how we would go about determining if this is real or an artifact of your simulations.

Another curiosity I noticed was the rather dramatic and “not smooth” fall off in frequencies immediately after 5. I don’t think it a coincidence that this happens directly after the domain of the first pdf. Again, I wonder how to prove this.

2. Steve Earth

Incidentally, I ran the simulation several times, and each time the mean was slightly more than 9 (9.005, 9.007, 9.003 etc, but never less), the mode was always at 5 (probability .153…) and there was always that secondary spike at 7… mysterious!

3. Steve Earth

belay my last comment… just got a mean of 8.9996 🙂

4. Steve Earth

ah, just realized something: it’s not that there’s a “spike” at 7, but rather that 6 is lower than your intuition would suggest. Definitely an intriguing problem with lots to explore!

5. P

The expected number of minutes for each task is 3, and the expected number of tasks before dinner is 3, so the expected number of minutes before dinner is 9.

In general, given tasks of length 1 to n minutes, we expect it will take (n+1)/2 minutes for each task and (n+1)/2 tasks to get to dinner, so a total of (n+1)^2/4 minutes. Note: (5+1)^2/4 = 36/4 = 9.

• Thanks for the reply. Ultimately, I believe you determined the correct mean number of minutes for the siblings to achieve “mutual clarity”, but your reasoning was flawed. Also, your conclusion that “the expected number of tasks before dinner is 3” is incorrect.

The mean time to complete a single task chosen from tasks of length 1 to n minutes is indeed (n+1)/2, and that is true for both siblings. Unfortunately, you reported (n+1)/2 tasks to get to dinner, and that is not true. The repeated factor comes from TWO people attempting clarity, not one person’s mean time multiplied by the number of tasks.

As I noted above and in the next post, the mean number of tasks completed (number of app choices made) in this specific scenario is 4.2, not the 3 you report. I’m quite certain that number is correct.

In the end, you spurred my thinking and ultimately helped me locate a few errors in my approach. I hope my suggestions help you, too. Thank you.

6. Steve Earth

Not sure that it helps compute the mean or not, but I found a (no doubt overly complicated) recursive formula for finding the pdf. Specifically, it yields the following numerator values for the probabilities of minutes 1 thru 10, where the denominators are 5^(mins+1): 1, 7, 49, 343, 2401, 4380, 22608, 110892, 509496, and 2165796, (I’m pretty sure the next is 8715072, though I might have made an arithmetic error on that one). I checked, and these very closely match all the simulated values. Observe that the numerators all go up by the powers of 7 until the 6th minute, when it takes a sudden dip (4380 instead of 16807), which matches the graph: this can be attributed to the fact that moments of clarity are no longer achievable by one of the sisters running just a single app.

7. Pingback: Mistakes are Good | CAS Musings