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.


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.


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.


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.


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.


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.


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.


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!


7 responses to “Mistakes are Good

  1. Pingback: Great Probability Problems | CAS Musings

  2. Where do the 1/5, 1/5, and 3/5 come from? I count 20 possibilities with 4 leading to end, 6 not adding time, and 10 adding time.

  3. Never mind- forgot to weight them with the probability of being in each state.

  4. I love Chris’ flowcharts, and how they show a completely different approach to arriving at the answer! Note that your second flowchart, the one for N, already takes into account the 1/5 chance that the initial pick terminates at N=1; so I believe your last line computing a 4.2 expected number of rounds is superfluous, and that the N=5 already represents the desired quantity. A re-ran the simulation of the 2,000,000 trials (this time with the modification of having it track the rounds rather than total minutes) and it seems to confirm your N=5 result. http://tinyurl.com/538rounds

    • Hmmm….I need to think on that. Your simulation is compelling, but I’m trying to reason why I don’t need the last computation for Stage 1 when I did need it for Stage 2.

  5. I like how the “paradox of means” appears in the problem analysis: the average app length is 3 minutes, and the average number of apps run by both sisters before syncing is 5. Thus, one would think the average sum phone time for both sister is 15 mins. However, the mean sync time is 9 mins, indicating a mean sum of 18 mins of phone time for the sisters.

  6. oops, ignore my previous comment. The number of apps run by both sisters before syncing is actually SIX (not 5)… I forgot that Chris started counting his rounds at N=1 (even though the sisters actually ran TWO apps simultaneously that first round. Thus #apps = N+1). And since 3*6=18, and 18/2 = 9, this is perfectly aligned with the results in the previous blog post. (I had thought maybe some sort of Simpson’s paradox was rearing its head, but I was mistaken. Everything works out as expected).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s