A few days ago, Joey deVilla wrote a very nice blog post about using algebra (a system of two linear equations in two unknowns) to figure out how many people rented *The Interview* as opposed to seeing it in a movie theater. I posted my own note about how you can still analyze this example algebraically even if you’re allergic to equations. Joey then added some extremely clear pictures illustrating what I was talking about. After seeing his drawings, I thought it might be illuminating to reconcile the two approaches a little more explicitly.

To review the set up, The New York Times told us that the movie made $15M over Christmas weekend, from both rentals and ticket sales to a total of 2M people, where each person either rented for $6 or bought a ticket for $15. We want to know how many rented and how many bought tickets. If we write *r* = number of rentals, *s* = number of tickets sold, we know that

*r* + *s* = 2,000,000 (Eq. 1: total number of rentals and sales was 2M),

6*r* + 15*s* = 15,000,000 (Eq. 2: total revenue from rentals and sales was $15M).

If all 2M people rent — put another way, if the people counted by *r* AND the people counted by *s* each pay $6 — the total revenue would be $12M. We can write an equation for this by multiplying Eq. 1 by 6, obtaining

*6r* + 6*s* = 12,000,000. (Eq. 3)

Similarly, if the people counted by both *r* and *s* all pay $15 each (everyone buys a ticket at the theater), the total revenue would have been $30M:

*15r* + 15*s* = 30,000,000.

In reality, of course, the people counted by *r* paid $6 and the people counted by *s* paid $15. Eq. 2 represents that the resulting revenue was $15M, which is 1/6 of the way from $12M to $30M (picture by Joey):

From here, we can unlock the mystery of how many rented and how many bought tickets with one question: *What if we reanchor this picture at 0?* In other words, what if we slide the Seth Rogen sliding scale to the left so it starts at 0 rather at $12M?

This amounts to pretending that everything costs $6 less. In that case, a rental would be free, and a movie ticket would cost $9 rather than $15. The scale would run from 0 to $18M: if everyone rented, the total revenue would be 0, and if everyone bought a ticket, the total revenue would be $18M ( = 2M × $9). We didn’t change the spread between the cost of a rental and a ticket (still $9), so the difference between the two extremes of the scale is still the same: $18M. We also didn’t change the number of renters and ticket buyers, so the actual revenue based on *r* rentals and *s* sales is still 1/6 of the way across the (shifted) scale, or $3M. You could also say that letting each of 2M people pay $6 less reduces your total revenue by $12M — from $15M to $3M (again). If we work with equations, this is exactly what we’re doing when we subtract Eq. 3 from Eq. 2 to obtain 9*s* = 3,000,000. It may look like just an operation on equations, but it has real physical meaning!

Now we’re ready to read off the answer. Looking at things in terms of equations, you would divide 9*s* = 3,000,000 by 9 to obtain *s* = 1/3 M. Looking at things in terms of sliding scales, you would say that if rentals are free and tickets cost $9, the total revenue is $3M, which is 1/6 of what it would have been if everyone bought a ticket. So how many of the 2M people bought tickets? Clearly 1/6 of them, or 1/3 M again!

So we can think about and solve this problem in multiple ways, and relate the different approaches to each other. Does that matter, and if so, why?

It matters because *this is how math is supposed to work*. At its heart, math is not about learning procedures (long division, completing the square, elimination of variables), and it’s not about getting to an answer, though procedures can be useful tools and can also help increase our overall understanding, and answers are salutary, and sometimes necessary, outcomes. At its heart, math is about learning to order and make sense of the world, which means formulating good organizing principles and drawing logical conclusions from those principles. When you’re trying to order the world, having multiple ways to look at things isn’t a bug, it’s a feature. It means that if you’re looking at things one way and you’re stuck, you might make progress when you switch points of view. It means that if you like manipulating equations and I like averaging along a sliding scale, we can both figure out how many people bought tickets to see *The Interview*. And it means that when we compare our points of view, we can find connections between them and gain a deeper understanding.

When you walk into a research math department, you don’t (usually) find professors hunched over desks or computers, trying to find more digits of pi, or trying to solve rows of really complicated integrals. What you’ll usually find is people talking to each other, looking for connections. You might hear:

— Hey, haven’t seen you for a while, what have you been up to?

— Well, I found this funny formula, and I’m trying to figure out what it means. It reminds me of another formula I know that counts certain arrangements of hyperplanes in 4-dimensional space. You used to work on hyperplane arrangements, right? Think you could help me understand what it means?

— Sure, let me take a look. OK, this term here looks like it’s counting….

To paraphrase Joe Strummer: This is Math / This is how we feel.

But connections between math ideas don’t need to be limited to university math departments, or even to math blogs. Maybe you have a kid who’s learning how to multiply, divide, or solve equations differently from how you learned it. Maybe you’re a little frustrated: what’s wrong with the old way? The answer, in all likelihood, is nothing. But there’s probably nothing inherently wrong with the new way, either. And maybe if you and your kid sit down and try to reconcile the different approaches, you might find out that they’re related, and understand why they both get you to the same place, and gain new insights about each one.

If you’ve gotten this far, if you’ve managed to bear with Joey and me through multiple takes on who saw *The Interview* where, I bet you have a better feel for the underlying math than you did before you started — maybe even better than you imagined you could. Congratulations — that’s thinking like a math person. And now you too have every right in the world to make fun of the New York Times when it asks if algebra is necessary.

Your post reminds me a bit of a answer I posted to stackexchange: http://math.stackexchange.com/questions/478212/is-there-another-simpler-method-to-solve-this-elementary-school-math-problem/478357#478357

(see second highest rated answer)

There is one school of thought (I think it was said by a famous software guy) is that the best way of tackling a problem is brute force until you realize a better way. This might be software centric, as it is really easy to test out brute force methods, as computation is “free”. Sometimes the brute force solution is actually the fastest way to get from A to B – less thinking, more doing. Obviously, an engineering mindset!

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Quote is from Kenneth Thompson, godfather of C and Unix.

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