# Correlation is not Causation (Part 4)

One of the standard topics in an undergraduate statistics course is the principle that two things that are highly correlated do not necessarily have a cause-and-effect relationship. Here is a hilarious example of this fallacy.

And, in case you’re wondering, here’s the rest of the story:

# Lessons from teaching gifted elementary school students: Index (updated)

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

# Another poorly written word problem (Part 3)

Textbooks have included the occasional awful problem ever since Pebbles Flintstone and Bamm-Bamm Rubble chiseled their homework on slate tablets while attending Bedrock Elementary. But even with the understanding that there have been children have been doing awful homework problems since the dawn of time (and long before the advent of the Common Core), this one is a doozy.

There’s no sense having a debate about standards for elementary mathematics if textbook publishers can’t construct sentences that can be understood by elementary students.

# Careers in Industry for Math Majors

I recently came across an excellent article promoting internships from math majors who would like to use their quantitative skills in an industrial setting (as opposed to an academic setting). The concluding paragraph:

Faculty will continue to train students for academic careers. Some will pursue tenure-track positions in the institutions of their choice, but an increasing number of our students will take positions very different from our own. Let’s learn about those options and share them with our students. Then, when a student takes a good job and enjoys a successful career, let’s call that a win.

# Teachers Who Are Funnier Than Their Students

I really enjoyed this, courtesy of BuzzFeed:

# My Mathematical Magic Show: Part 8c

This mathematical trick was not part of my Pi Day magic show but probably should have been… I’ve performed this for my Precalculus classes in the past but flat forgot about it when organizing my Pi Day show. The next time I perform a magic show, I’ll do this one right after the 1089 trick. (I think I learned this trick from a Martin Gardner book when I was young, but I’m not sure about that.)

Here’s a description of the trick. I give my audience a deck of cards and ask them to select six cards between ace and nine (in other words, no tens, jacks, queens, or kings). The card are placed face up, side by side.

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others.

I then announce that we’re going to some addition together… with the understanding that I’ll never write down a number larger than 9. For example, the 4 and 6 of spades are next to each other. Obviously, $4+6 = 10$, but my rule is that I’m going to write down a number larger than 9. So I’ll subtract 9 whenever necessary: $10-9 = 1$. Since 1 corresponds to ace, I place an ace about the 4 and 6 of spades.

Continuing in this way (and having the audience participate in the arithmetic so that this doesn’t get boring), I eventually get to this position:

Finally, I add the two cards at the top (and, in this case, subtract 9) to get $6+9-9 = 6$, and I dramatically turn over the last card to reveal a 6.

I’ll often perform this trick when teaching Precalculus, as the final answer involving Pascal’s triangle. As discussed yesterday, suppose that the six cards are $a$, $b$, $c$, $d$, $e$, and $f$. Forgetting for now about subtracting by 9, here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

Not surprisingly, the coefficients in the above chart involve the numbers in Pascal’s triangle. Indeed, the reason that I chose to use 6 cards (as opposed to any other number of cards) is that the bottom row has only 1, 5, and 10 as coefficients, and $10 \equiv 1 (\mod 9)$. Therefore, the only tricky part of the calculation is multiplying $b+e$ by $5$, as the final answer can then be found by adding the remaining four numbers.

My students usually find this to be a clever application of Pascal’s triangle for impressing their friends after class.

P.S. After typing this series, it hit me that it’s really easy to do this trick mod 10 (which means getting rids of only the face cards prior to the trick). All the magician has to do is subtly ensure that the second and fifth cards are both even or both odd, so that $b+e$ is even and hence $5(b+e)$ is a multiple of 10. Therefore, since $10c+10d$ is also a multiple of 10, the answer will be just $a+f$ or $a+f-10$.

(If the magician can’t control the placement of the second and fifth cards so that one is even and one is odd, the answer will be just $a+f+5$ or $a+f-5$.)

Henceforth, I’ll be doing this trick mod 10 instead of mod 9.

# My Mathematical Magic Show: Part 8b

This mathematical trick was not part of my Pi Day magic show but probably should have been… I’ve performed this for my Precalculus classes in the past but flat forgot about it when organizing my Pi Day show. The next time I perform a magic show, I’ll do this one right after the 1089 trick. (I think I learned this trick from a Martin Gardner book when I was young, but I’m not sure about that.)

Here’s a description of the trick. I give my audience a deck of cards and ask them to select six cards between ace and nine (in other words, no tens, jacks, queens, or kings). The card are placed face up, side by side.

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others.

I then announce that we’re going to some addition together… with the understanding that I’ll never write down a number larger than 9. For example, the 4 and 6 of spades are next to each other. Obviously, $4+6 = 10$, but my rule is that I’m going to write down a number larger than 9. So I’ll subtract 9 whenever necessary: $10-9 = 1$. Since 1 corresponds to ace, I place an ace about the 4 and 6 of spades.

Continuing in this way (and having the audience participate in the arithmetic so that this doesn’t get boring), I eventually get to this position:

Finally, I add the two cards at the top (and, in this case, subtract 9) to get $6+9-9 = 6$, and I dramatically turn over the last card to reveal a 6.

How does this trick work? This is an exercise in modular arithmetic (see also Wikipedia). Suppose that the six cards are $a$, $b$, $c$, $d$, $e$, and $f$. Forgetting for now about subtracting by 9, here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

Therefore, the top card will simply be $a+5b+10c+10d+5e+f$ minus a multiple of 9.

That’s a pretty big calculation for the magician to do on the spot. Fortunately, $9c + 9d$ is also a multiple of 9, and so the top card will be

$a+5b+10c+10d+5e+f - (9c + 9d)$ minus a multiple of 9, or

$5(b+e) + a + c + d + f$ minus a multiple of 9.

For the case at hand, $b = 6$ and $e =8$, so $5(b+e) = 70$. That’s still a big number to keep straight when performing the trick. However, since I’m going to be subtracting 9’s anyway, I can do this faster by replacing the 8 by $8 - 9 = -1$. So, for the purposes of the trick, $5(b+e) = 5 \times (6-1) = 25$, and I subtract $18$ to get $7$.

I now add the rest of the cards, subtracting 9 as I go along. For this example, I’d add the 2 first to get 9, which is 0 after subtracting another 9. I then add the remaining cards of 4, 3, and 8 (remembering that the 8 is basically $8-9 = -1$, yielding $4+3-1 = 6$. So the top card has to be 6.

The key point of this calculation is to subtract 9 whenever possible to keep the numbers small, making it easier to do in your head when performing the trick.

# My Mathematical Magic Show: Part 8a

This mathematical trick was not part of my Pi Day magic show but probably should have been… I’ve performed this for my Precalculus classes in the past but flat forgot about it when organizing my Pi Day show. The next time I perform a magic show, I’ll do this one right after the 1089 trick. (I think I learned this trick from a Martin Gardner book when I was young, but I’m not sure about that.)

Here’s a description of the trick. I give my audience a deck of cards and ask them to select six cards between ace and nine (in other words, no tens, jacks, queens, or kings). The card are placed face up, side by side.

After about 5-10 seconds, I secretly write a pull out a card from the deck and place it face down above the others.

I then announce that we’re going to some addition together… with the understanding that I’ll never write down a number larger than 9. For example, the 4 and 6 of spades are next to each other. Obviously, $4+6 = 10$, but my rule is that I’m going to write down a number larger than 9. So I’ll subtract 9 whenever necessary: $10-9 = 1$. Since 1 corresponds to ace, I place an ace about the 4 and 6 of spades.

Next, I consider the 6 of spades and 2 of diamonds. Adding, I get 8. That’s less than 9, so I pull an 8 out of the deck.

Next, $2+3 = 5$, so I pull out a 5 from the deck.

Next, $8+8=16$, and $16-9=7$. So I pull out a 7.

(To keep this from getting dry, I have the audience perform the arithmetic with me.)

On the the next row. The next cards are $1+8 = 9$, $8+5-9 = 4$, $5+2 =7$, and $2+7 = 9$.

On the the next row. The next cards are $9+4-9=4$, $latex$4+7-9 = 2\$, and $7+9-9 = 7$.

Almost there: $4+2 = 6$ and $2+7= 9$.

Finally, $6+9-9 = 6$, and I dramatically turn over the last card to reveal a 6.

Naturally, everyone wonders how I knew what the last card would be without first getting all of the cards in the middle. I’ll discuss this in tomorrow’s post.

# My Mathematical Magic Show: Part 7

This mathematical trick, which may well be the best mathematical magic trick ever devised, was not part of my Pi Day magic show. However, it should have been. Here’s a description of the trick, modified from the description at http://mathoverflow.net/questions/20667/generalization-of-finch-cheneys-5-card-trick:

The magician walks out of the room. A volunteer from the crowd chooses any five cards at random from a deck, and hands them to your assistant so that nobody else can see them. The assistant glances at them briefly and hands one card back, which the volunteer then places face down on the table to one side. The assistant quickly place the remaining four cards face up on the table, in a row from left to right. After all of this is completed, the magician re-enters the room, inspects the faces of the four cards, and promptly names the hidden fifth card.

In turns out that the trick is a clever application of permutations (there are $3! = 6$ possible ways of ordering 3 objects) and the pigeon-hole principle (if each object belongs to one of four categories and there are five objects, then at least two objects must belong to the same category). These principles from discrete mathematics (specifically, combinatorics) make possible the Fitch-Cheney 5-Card Trick.

Unlike the other tricks in this series, the Fitch-Cheney 5-Card Trick requires a well-trained assistant (or a smartphone app that plays the role of the assistant).

A great description of how this trick works can be found at Math With Bad Drawings. For a deeper look at some of the mathematics behind this trick, I give the following references: