Tag Archives: FiveThirtyEight

FiveThirtyEight Riddler Express Solution

I’d not peeked at FiveThirtyEight’s The Riddler in a while when I saw Neema Salimi ‘s post about the June 22, 2018 Riddler Express   Neema argued his solution was accessible to Algebra 1 students–not always possible for FiveThirtyEight’s great logic puzzles–so I knew it was past time to return.

After my exploration, I’ve concluded this is DEFINITELY worth posing to any middle school learners (or others) in search of an interesting problem variation.

Following is my solution, three retrospective insights about the problem, a comparison to Neema’s solution, and a proposed alternative numeric approach I think many Algebra 1 students might actually attempt.

THE PROBLEM:

Here is a screenshot of the original problem posted on FiveThirtyEight (2nd item on this page).

If you’re familiar with rate problems from Algebra 1, this should strike you immediately as a “complexification” of D=R*T type problems.  (“Complexification” is what a former student from a couple years ago said happened to otherwise simple problems when I made them “more interesting.”)

MY SOLUTION:

My first thought simplified my later algebra and made for a much more dramatic ending!!  Since Michelle caught up with her boarding pass with 10 meters left on the walkway, I recognized those extra 10 meters as irrelevant, so I changed the initial problem to an equivalent question–from an initial 100 m to 90 m–having Michelle catch up with her boarding pass just as the belt was about to roll back under the floor!

Let W = the speed of the walkway in m/s.  Because Michelle’s boarding pass then traveled a distance of 90 m in W m/s, her boarding pass traveled a total \displaystyle \frac{90}{W} seconds.

If M = Michelle’s walking speed, then her total distance traveled is the initial 90 meters PLUS the distance she traveled in the additional 90 seconds after dropping her boarding pass.  Her speed at this time was (M-W)  m/s (subtracting W because she was moving against the walkway), so the additional distance she traveled was D = (M-W) \cdot 90, making her her total distance D_{Michelle} = 90 + 90(M-W).

Then Michelle realized she had dropped her boarding pass and turned to run back at (2M+W) m/s (adding to show moving with the walkway this time), and she had \displaystyle \frac{90}{W} - 90 seconds to catch it before it disappeared beneath the belt.  The subtraction is the time difference between losing the pass and realizing she lost it.  Substituting into D = R*T gives

\displaystyle 90 + 90(M-W)=(2M+W)* \left( \frac{90}{W} - 90 \right)

A little expansion and algebra cleanup …

\displaystyle 90 + 90M - 90W = 180 \frac{M}{W} - 180M + 90 - 90W

\displaystyle 90M = 180 \frac{M}{W} - 180M

\displaystyle 270M = 180 \frac{M}{W}

And multiplying by \displaystyle \frac{W}{270M} solves the problem:

\displaystyle W = \frac{2}{3}

INSIGHTS:

Insight #1:  Solving a problem is always nice, but I was thinking all along that I pulled off my solution because I’m very comfortable with algebraic notation.  This is certainly NOT true of most Algebra 1 students.

Insight #2:  This made me wonder about the viability of a numeric solution to the problem–an approach many first-year algebra students attempt when frutstrated.

Insight #3:  In the very last solution step, Michelle’s rate, M, completely dropped out of the problem.  That means the solution to this problem is independent of Michelle’s walking speed.

Wondering if other terms might be superfluous, too, I generalized my initial algebraic solution further with A = the initial distance before Michelle dropped her boarding pass and B = the additional time Michelle walked before realizing she had dropped the pass.

\displaystyle A + B(M-W)=(2M+W)* \left( \frac{A}{W} - B \right)

And solving for gives \displaystyle W = \frac{2A}{3B}.

So, the solution does depend on the initial distance traveled and the time before Michelle turns around, and it was simplified in the initial statement with A=B=90.  That all made sense after a few quick thought experiments.  With one more variation you can show that the scale factor between her walking and jogging speed is relevant, but not her walking speed.  But now it was clear that in all cases, Michelle’s walking speed is irrelevant!

COMPARING TO NEEMA:

My initial conclusion matched Neema’s solution, but I really liked my separate discovery that the answer was completely independent of Michelle’s walking speed.  In my opinion, those cool insights are not at all intuitive.

AN ALTERNATIVE NUMERIC APPROACH:

While this approach is just a series of special cases of the generic approach, I suspect many Algebra 1 students would likely get frustrated quickly by the multiple variable and attempt

Ignoring everything above, but using the same variables for simplicity, perhaps the easiest start is to assume the walkway moves at W=1 m/s and Michelle’s walking speed is M=2 m/s.  That means her outward speed against the walkway is (2-1) = 1 m/s.  She drops the pass at 90 meters after 90 seconds.  So the pass will be back at the start after 90 seconds, the additional time that Michelle walks before realizing her loss.

I could imagine many students I’ve taught working from this through some sort of intelligent numeric guess-and-check, adjusting the values of M and W until landing at \displaystyle W=\frac{2}{3}.  The fractional value of W would slow them down, but many would get it this way.

CONCLUSION:

I’m definitely pitching this question to my students in just a few weeks.  (Where did summer go?)  I’m deeply curious about how they’ll approach their solutions.  I”m convinced many will attempt–at least initially–playing with more comprehensible numbers.  Such approaches often give young learners the insights they need to handle algebra’s generalizations.

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Mistakes are Good

Confession #1:  My answers on my last post were WRONG.

I briefly thought about taking that post down, but discarded that idea when I thought about the reality that almost all published mathematics is polished, cleaned, and optimized.  Many students struggle with mathematics under the misconception that their first attempts at any topic should be as polished as what they read in published sources.

While not precisely from the same perspective, Dan Teague recently wrote an excellent, short piece of advice to new teachers on NCTM’s ‘blog entitled Demonstrating Competence by Making Mistakes.  I argue Dan’s advice actually applies to all teachers, so in the spirit of showing how to stick with a problem and not just walking away saying “I was wrong”, I’m going to keep my original post up, add an advisory note at the start about the error, and show below how I corrected my error.

Confession #2:  My approach was a much longer and far less elegant solution than the identical approaches offered by a comment by “P” on my last post and the solution offered on FiveThirtyEight.  Rather than just accepting the alternative solution, as too many students are wont to do, I acknowledged the more efficient approach of others before proceeding to find a way to get the answer through my initial idea.

I’ll also admit that I didn’t immediately see the simple approach to the answer and rushed my post in the time I had available to get it up before the answer went live on FiveThirtyEight.

GENERAL STRATEGY and GOALS:

1-Use a PDF:  The original FiveThirtyEight post asked for the expected time before the siblings simultaneously finished their tasks.  I interpreted this as expected value, and I knew how to compute the expected value of a pdf of a random variable.  All I needed was the potential wait times, t, and their corresponding probabilities.  My approach was solid, but a few of my computations were off.

2-Use Self-Similarity:  I don’t see many people employing the self-similarity tactic I used in my initial solution.  Resolving my initial solution would allow me to continue using what I consider a pretty elegant strategy for handling cumbersome infinite sums.

A CORRECTED SOLUTION:

Stage 1:  My table for the distribution of initial choices was correct, as were my conclusions about the probability and expected time if they chose the same initial app.

App1

My first mistake was in my calculation of the expected time if they did not choose the same initial app.  The 20 numbers in blue above represent that sample space.  Notice that there are 8 times where one sibling chose a 5-minute app, leaving 6 other times where one sibling chose a 4-minute app while the other chose something shorter.  Similarly, there are 4 choices of an at most 3-minute app, and 2 choices of an at most 2-minute app.  So the expected length of time spent by the longer app if the same was not chosen for both is

E(Round1) = \frac{1}{20}*(8*5+6*4+4*3+2*2)=4 minutes,

a notably longer time than I initially reported.

For the initial app choice, there is a \frac{1}{5} chance they choose the same app for an average time of 3 minutes, and a \frac{4}{5} chance they choose different apps for an average time of 4 minutes.

Stage 2:  My biggest error was a rushed assumption that all of the entries I gave in the Round 2 table were equally likely.  That is clearly false as you can see from Table 1 above.  There are only two instances of a time difference of 4, while there are eight instances of a time difference of 1.  A correct solution using my approach needs to account for these varied probabilities.  Here is a revised version of Table 2 with these probabilities included.

App4

Conveniently–as I had noted without full realization in my last post–the revised Table 2 still shows the distribution for the 2nd and all future potential rounds until the siblings finally align, including the probabilities.  This proved to be a critical feature of the problem.

Another oversight was not fully recognizing which events would contribute to increasing the time before parity.  The yellow highlighted cells in Table 2 are those for which the next app choice was longer than the current time difference, and any of these would increase the length of a trial.

I was initially correct in concluding there was a \frac{1}{5} probability of the second app choice achieving a simultaneous finish and that this would not result in any additional total time.  I missed the fact that the six non-highlighted values also did not result in additional time and that there was a \frac{1}{5} chance of this happening.

That leaves a \frac{3}{5} chance of the trial time extending by selecting one of the highlighted events.  If that happens, the expected time the trial would continue is

\displaystyle \frac{4*4+(4+3)*3+(4+3+2)*2+(4+3+2+1)*1}{4+(4+3)+(4+3+2)+(4+3+2+1)}=\frac{13}{6} minutes.

Iterating:  So now I recognized there were 3 potential outcomes at Stage 2–a \frac{1}{5} chance of matching and ending, a \frac{1}{5} chance of not matching but not adding time, and a \frac{3}{5} chance of not matching and adding an average \frac{13}{6} minutes.  Conveniently, the last two possibilities still combined to recreate perfectly the outcomes and probabilities of the original Stage 2, creating a self-similar, pseudo-fractal situation.  Here’s the revised flowchart for time.

App5

Invoking the similarity, if there were T minutes remaining after arriving at Stage 2, then there was a \frac{1}{5} chance of adding 0 minutes, a \frac{1}{5} chance of remaining at T minutes, and a \frac{3}{5} chance of adding \frac{13}{6} minutes–that is being at T+\frac{13}{6} minutes.  Equating all of this allows me to solve for T.

T=\frac{1}{5}*0+\frac{1}{5}*T+\frac{3}{5}*\left( T+\frac{13}{6} \right) \longrightarrow T=6.5 minutes

Time Solution:  As noted above, at the start, there was a \frac{1}{5} chance of immediately matching with an average 3 minutes, and there was a \frac{4}{5} chance of not matching while using an average 4 minutes.  I just showed that from this latter stage, one would expect to need to use an additional mean 6.5 minutes for the siblings to end simultaneously, for a mean total of 10.5 minutes.  That means the overall expected time spent is

Total Expected Time =\frac{1}{5}*3 + \frac{4}{5}*10.5 = 9 minutes.

Number of Rounds Solution:  My initial computation of the number of rounds was actually correct–despite the comment from “P” in my last post–but I think the explanation could have been clearer.  I’ll try again.

App6

One round is obviously required for the first choice, and in the \frac{4}{5} chance the siblings don’t match, let N be the average number of rounds remaining.  In Stage 2, there’s a \frac{1}{5} chance the trial will end with the next choice, and a \frac{4}{5} chance there will still be N rounds remaining.  This second situation is correct because both the no time added and time added possibilities combine to reset Table 2 with a combined probability of \frac{4}{5}.  As before, I invoke self-similarity to find N.

N = \frac{1}{5}*1 + \frac{4}{5}*N \longrightarrow N=5

Therefore, the expected number of rounds is \frac{1}{5}*1 + \frac{4}{5}*5 = 4.2 rounds.

It would be cool if someone could confirm this prediction by simulation.

CONCLUSION:

I corrected my work and found the exact solution proposed by others and simulated by Steve!   Even better, I have shown my approach works and, while notably less elegant, one could solve this expected value problem by invoking the definition of expected value.

Best of all, I learned from a mistake and didn’t give up on a problem.  Now that’s the real lesson I hope all of my students get.

Happy New Year, everyone!

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.

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.)

SOLVING THE CURRENT PUZZLER:

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

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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.

App1

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.

App2

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.

App3

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

FINAL ANSWER:

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.