# My Favorite One-Liners: Part 93

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This is a wisecrack that I’ll use in my probability/statistics classes to clarify the difference between $P(A \cap B)$ and $P(A \mid B)$:

Even though the odds of me being shot by some idiot wielding a gun while I teach my class are probably a million to one, I’ve decided, in light of Texas’  campus-carry law, to get my concealed handgun license and carry my own gun to class. This is for my own safety and protection; after all, the odds of *two* idiots carrying a gun to my class must be absolutely microscopic.

# Guns on university campuses

Here in Texas, public universities are trying to figure out how they’re going to comply with a recently enacted state campus-carry law so that licensed handgun owners can bring their firearms to campus. A small sampling of local news articles and websites on this topic:

And in the midst of this debate, I found the opportunity for a mathematical wisecrack.

I’ve used this wisecrack in my probability class to great effect, as the joke pedagogically illustrates the important difference between $P(A \mid B)$ and $P(A \cap B)$.

For what it’s worth, here’s the version of the joke as I first saw it (in the book Absolute Zero Gravity):

Then there was the statistician who hated to fly because he had nightmares about terrorists with bombs. Yes, he knew that it was a million to one chance, but that wasn’t good enough. So he took a lot of trains until he realized what he had to do.

Now, whenever he flies, he packs a bomb in his own suitcase. Hey, do you know what the odds are against an airplane carrying two bombs?

Two final notes:

1. For the humor-impaired, I’m not referring to all gun owners as idiots. The only people I’m calling idiots are me and those that would slaughter innocent people (and these two sets are disjoint).
2. Though it’s certainly an important issue, I have no interest in debating the wisdom of the campus-carry law on this blog. Rather, the point of this post was using current events to memorably illustrate mathematical ideas.

# Lessons from teaching gifted elementary school students (Part 4b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

What is the chance of winning a game of BINGO after only four turns?

When my class posed this question, I was a little concerned that the getting the answer might be beyond the current abilities of a gifted elementary student.  Still, what I love about this question is that it gave me a way to teach my class some techniques of probabilistic reasoning that probably would not occur in a traditional elementary school setting.

As discussed yesterday, for a non-standard BINGO game with 44 numbers, the answer is

$\displaystyle 4 \times \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41}$

For a standard BINGO board with 75 numbers, the denominators are instead 75, 74, 73, and 72.

Now, for the next challenge: getting my students to simplify this product. I’m always mystified when college students blindly multiply numerators and denominators together without bothering to attempt to cancel common factors. Fortunately, this class already understands how to simplify fractions, and so the next step was easy:

$\displaystyle 4 \times \frac{1}{11} \times \frac{3}{43} \times \frac{1}{21} \times \frac{1}{41}$

So I was ready for the next step: cancelling 3 from the numerator and denominator. To my surprise, this was a major stumbling block. I tried probing around to prod them to perform this cancellation, but no luck. Eventually, I guessed the issue that my class was facing: they were familiar with the mechanics of both adding and multiplying fractions and also with writing fractions in lowest terms, but they weren’t yet comfortable enough with fractions to cancel 3 from the numerator of one fraction and the denominator of a different fraction.

So, toward this end, I asked my class if it was OK to shuffle a couple of the numerators and rewrite this product as

$\displaystyle 4 \times \frac{1}{11} \times \frac{1}{43} \times \frac{3}{21} \times \frac{1}{41}$

It took a moment, but then they agreed that this was OK because the order of multiplication doesn’t matter, even volunteering the word commutative to explain their reasoning. (I’m going to try to remember this technique for future reference as a way to get students new to fractions more comfortable with similar cancellations.) Once they got past this conceptual barrier, it was straightforward to continue the simplification:

$\displaystyle 4 \times \frac{1}{11} \times \frac{1}{43} \times \frac{1}{7} \times \frac{1}{41}$

$= \displaystyle \frac{4}{135,751}$

So I explained that if a game of BINGO took one minute, we could play round the clock for 135,751 minutes (about 96 days) and expect to win in the minimal number of turns only four times. Not very likely at all. (Though I didn’t discuss this with my class, the answer is even smaller with a standard BINGO game with 75 numbers: you’d expect to win only once every 211 days.)

# Lessons from teaching gifted elementary school students (Part 4a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

What is the chance of winning a game of BINGO after only four turns?

I leave a thought bubble in case you’d like to think this. One way of answering this question appears after the bubble.

When my class posed this question, I was a little concerned that my class was simply not ready to understand the solution (described below), as it takes more than a little work to get at the answer. Still, what I love about this question is that it gave me a way to teach my class some techniques of probabilistic reasoning that probably would not occur in a traditional elementary school setting. Also, I was reminded that even these gifted students might need a little help with simplifying the answer. So let me discuss how I helped these young students discover the answer. I found the ensuing discussion especially enlightening, and so I’m dividing this discussion into several posts.

Here’s a non-standard BINGO board:

Using the free space in the middle, there are four ways of winning the game in four moves:

• Horizontally (11-12-13-14)
• Vertically (3-8-17-22)
• Diagonally (1-7-18-24)
• Diagonally (5-9-16-20)

A standard BINGO board has 75 possible numbers (B 1-15, I 16-30, N 31-45, G 46-60, O 61-75). However, the board that I was using with my class (which was being used for pedagogical purposes) only had 44 possibilities. So the solution below assumes these 44 possibilities; the answer for a standard BINGO board is obvious.

My class quickly decided to start by solving the problem for the horizontal case. I began by asking for the chance that the first number will be on the middle row; after some thought, the class correctly answered $\displaystyle \frac{4}{44}$.

Next, I asked the chance that the next number would also be on the middle row. To my surprise, this wasn’t automatic for my young but gifted students. They felt that they didn’t know where the first number was, and so they felt like they couldn’t know the chance for the second number. To get them over this conceptual barrier (or so I thought), I asked them to pretend that the first number was 11. Then what would be the odds that the next number fell on the middle row? After some discussion, the class agreed that the answer was $\displaystyle \frac{3}{43}$.

Once that barrier was cleared, then the class saw that the next two fractions were $\displaystyle \frac{2}{42}$ and $\displaystyle \frac{1}{41}$. I then explained that, to get the answer for the four consecutive numbers on the middle row, these fractions have to be multiplied:

$\displaystyle \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41}$

I didn’t justify why the fractions had to be multiplied; my class just accepted this as the way to combined the fractions to get the answer for all four events happening at once.

Then I asked about the other three possibilities — the middle column and the two diagonals. The class quickly agreed that the answer should be the same for these other possibilities, and so the final answer should just be four times larger:

$\displaystyle 4 \times \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41}$

At this point, I was ready to go on, but then a student asked something like the following:

Shouldn’t the answer be $\displaystyle \frac{1}{44} \times \frac{1}{43} \times \frac{1}{42} \times \frac{1}{41}$? I mean, we chose 11 to be the first number so that we can figure out the chance for the second number, and the chance that the first number is 11 is $\displaystyle \frac{1}{41}$.

Oops. While trying to clear one conceptual hurdle (getting the answer of $\displaystyle \frac{3}{43}$ for the second number), I had inadvertently introduced a second hurdle by making my class wonder if the first number had to be a specific number.

I began by trying to explain that the first number really didn’t have to be 11 after all, but that only seemed to re-introduce the original barrier. Finally, I found an answer that my class found convincing: Yes, the chance that the first number is 11, the second number is 12, the third number is 13, and fourth number is 14 is indeed $\displaystyle \frac{1}{44} \times \frac{1}{43} \times \frac{1}{42} \times \frac{1}{41}$. But there are other ways that all the numbers could land on the middle row:

• The first number could be 12, the second number could be 11, the third number could be 13, and the fourth number could be 14.
• Quickly, light dawned, and my class began volunteering other orderings by which all the numbers land in the middle row.
• We then enumerated the number of ways that this could happen, and we found that the answer was indeed 24.
• I then tied the knot by noting that $4 \times 3 \times 2 \times 1 = 24$, and so gives another explanation for the numerators in the answer.

Having found the answer, it was now time to simplify the answer. More on this in tomorrow’s post.

# 2048 and algebra (Part 10)

In this series of posts, I used algebra to show that 114,795 moves were needed to produce the following final board. This board represents the event horizon of 2048 that cannot be surpassed.

I reached after about four weeks of intermittent doodling. It should be noted that the above game board was accomplished in practice mode, and I needed perhaps a couple thousand undos to offset the bad luck of a tile randomly appearing in an unneeded place.

For what it’s worth, my personal best in game mode was reaching the 8192-tile. I’m convinced that, even with the random placements of the new 2-tiles and 4-tiles, the skilled player can reach the 2048-tile nearly every time and should reach the 4096-tile most of the time.  However, reaching the 8192-tile requires more luck than skill, and reaching the 16384-tile requires an extraordinary amount of luck.

So what are the odds of a skilled player reaching the event horizon in game mode, without the benefit of undoing the previous move? I will employ Fermi estimation to approach this question. Of the approximately 100,000 moves, I estimate that about 5% of the moves require a certain 2-tile or 4-tile to appear at a certain location on the board. For example, in the initial stages of the game, the board is wide open and really doesn’t matter a whole lot where the new tiles appear. However, when the board gets quite crowded, it’s essential that new tiles appear in certain places, or else even a highly skilled player will get stuck.

What is the probability of getting the right tile on each of these occasions? Usually it’s quite high (over 90%). But sometimes it’s necessary to get a 4-tile in exactly the right place when there are four blank spaces (estimated probability of 3%). So let’s estimate 10% to be the probability for getting the right tile for all of these occasions. Let’s also assume that the random number generator is indeed random, so that the tiles appear independently of all other tiles.

With these estimates, I can estimate the probability of reaching the event horizon in game as $\displaystyle \left( \frac{1}{10} \right)^{5000} = \displaystyle \frac{1}{10^{5000}}$. While this analysis isn’t foolproof, it sure beats playing the game about $10^{10,000}$ times and then dividing by the number of times the event horizon is reached by the total number of attempts!

How small is $\displaystyle \frac{1}{10^{5000}}$? Since $2^3 \approx 10$, this is approximately equal to $\displaystyle \frac{1}{2^{15,000}}$, and that’s a probability so small that it was reached (and surpassed) when the Heart of Gold spaceship activated the Infinite Improbability Drive in The Hitchhiker’s Guide to the Galaxy. By way of comparison:

• $\displaystyle \frac{1}{2^{276,709}}$ is the probability that someone stranded in the vacuum of space will be picked up by a starship within 30 seconds.
• $\displaystyle \frac{1}{2^{100,000}}$ is the probability of skidding down a beam of light… or having a million-gallon vat of custard appearing in the sky and dumping its contents on you without warning.
• $\displaystyle \frac{1}{2^{75,000}}$ is the probability of a person turning into a penguin.
• $\displaystyle \frac{1}{2^{50,000}}$ is the probability of having one of your arms suddenly elongate.
• $\displaystyle \frac{1}{2^{20,000}}$ is the probability of an infinite number of monkeys randomly typing out Hamlet.

These are the events to which the probability of reaching the event horizon in 2048 without any undos should be compared.

# A surprising appearance of e

Here’s a simple probability problem that should be accessible to high school students who have learned the Multiplication Rule:

Suppose that you play the lottery every day for about 20 years. Each time you play, the chance that you win is $1$ chance in $1000$. What is the probability that, after playing  $1000$ times, you never win?

This is a straightforward application of the Multiplication Rule from probability. The chance of not winning on any one play is $0.999$. Therefore, the chance of not winning $1000$ consecutive times is $(0.999)^{1000}$, which we can approximate with a calculator.

Well, that was easy enough. Now, just for the fun of it, let’s find the reciprocal of this answer.

Hmmm. Two point seven one. Where have I seen that before? Hmmm… Nah, it couldn’t be that.

What if we changed the number $1000$ in the above problem to $1,000,000$? Then the probability would be $(0.999999)^{1000000}$.

There’s no denying it now… it looks like the reciprocal is approximately $e$, so that the probability of never winning for both problems is approximately $1/e$.

The above calculations are numerical examples that demonstrate the limit

$\displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x$

In particular, for the special case when $n = -1$, we find

$\displaystyle \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = e^{-1} = \displaystyle \frac{1}{e}$

The first limit can be proved using L’Hopital’s Rule. By continuity of the function $f(x) = \ln x$, we have

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \ln \left[ \left(1 + \frac{x}{n}\right)^n \right]$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} n \ln \left(1 + \frac{x}{n}\right)$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \ln \left(1 + \frac{x}{n}\right)}{\displaystyle \frac{1}{n}}$

The right-hand side has the form $\infty/\infty$ as $n \to \infty$, and so we may use L’Hopital’s rule, differentiating both the numerator and the denominator with respect to $n$.

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \frac{1}{1 + \frac{x}{n}} \cdot \frac{-x}{n^2} }{\displaystyle \frac{-1}{n^2}}$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \displaystyle \frac{x}{1 + \frac{x}{n}}$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \frac{x}{1 + 0}$

$\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = x$

Applying the exponential function to both sides, we conclude that

$\displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n= e^x$

In an undergraduate probability class, the problem can be viewed as a special case of a Poisson distribution approximating a binomial distribution if there’s a large number of trials and a small probability of success.

The above calculation also justifies (in Algebra II and Precalculus) how the formula for continuous compound interest $A = Pe^{rt}$ can be derived from the formula for discrete compound interest $A = P \displaystyle \left( 1 + \frac{r}{n} \right)^{nt}$

All this to say, Euler knew what he was doing when he decided that $e$ was so important that it deserved to be named.