Archive for August, 2016

Constant Change

Coin PileI’m frustrated.

I’m also old, cranky, and cynical. Whatever.

My frustration is not the my-flight-was-delayed-three-times-then-eventually-cancelled-and-there-are-no-more-flights-to-Cleveland-till-tomorrow-morning type. It’s not even the can’t-believe-my-boss-is-making-me-go-to-Cleveland kind of frustration. More like the why-aren’t-there-the-same-number-of-hot-dogs-and-buns-in-a-pack variety. So it’s a First World problem, to be sure, but still annoying. I’ll explain more in a moment.

But first, how ’bout a math problem to get us started?

If you make a purchase and pay with cash, what’s the probability that you’ll receive a nickel as part of your change?

Sure, if you want to get all crazy about this, then we can take all the fun out of this problem by stating the following assumptions:

  • You only pay with paper currency. If you paid with coins, then the distribution of coins you’d receive as change would likely vary quite a bit.
  • You never use 50¢ coins. Honestly, they’re just too obscure.
  • Transaction amounts are uniformly distributed, so that you’re just as likely to receive 21¢ as 78¢ or any other amount.
  • Cashiers don’t round because they dislike pennies. So, if you’re supposed to get 99¢ change, the cashier doesn’t hand you a dollar and say, “Don’t worry about it.” Instead, you actually get 99¢ change.

But stating assumptions is a form of mathematical douchebaggery, isn’t it? (As an aside, check out the definition of douchey that’s returned when you do a search. Sexist, anyone?) I prefer problems with no assumptions stated; let folks make their own assumptions to devise a model. If you and I get different answers because of different assumptions, no worries. Maybe we both learn something in the process.

Anyway, where was I? Oh, yeah…

Understanding the solution to that problem is a precursor to the issue that’s causing me frustration. I’ll give the solution in a minute, so pause here if you want to solve it on your own, but let me now allow the proverbial cat out of its bag and tell you why I’m frustrated.

At our local grocery store, there’s a coin counting machine that will count your change, sort it, and spit out a receipt that you can take to the customer service desk to exchange for paper currency. Walk in with a jar full of change, walk out with a fistful of fifties. Pretty nifty, right? Except the machine charges a ridiculous 8.9% fee to perform this service. No, thank you.

My bank used to have a similar coin counting machine, and if you deposited the amount counted by the machine into your account, there was no fee. The problem is that everyone was doing this to avoid the grocery store fee, so the machine broke often. The bank finally decided the machine wasn’t worth the maintenance fees and got rid of it. Strike two.

Which brings me to my current dilemma. One Saturday morning every month, we now spend 30 minutes counting coins and allocating them to appropriate wrappers. Which is fine. The problem, however, is that we run out of quarter and penny wrappers way faster than we run out of nickel or dime wrappers. Which brings me to the real question for the day:

Since pennies, nickels, dimes, and quarters are not uniformly distributed as change, why the hell does every package of coin wrappers contain the same number for each coin type?

Coin WrappersThe Royal Sovereign Assorted Coin Preformed Wrappers is the best-selling collection of coin wrappers on Amazon, and it provides 54 wrappers for each coin type. They also offer a 360‑pack with 90 wrappers for each coin type; Minitube offers a 100‑pack with 25 wrappers for each coin type; and Coin-Tainer offers a 36‑pack with 9 wrappers for each coin type. But what no one offers, so far as I can tell, is a collection of coin wrappers with a distribution that more closely resembles the distribution of coins that are received as change.

Whew! It feels good to finally raise this issue for public consideration.

So, the question that I really wanted to ask you…

Given the distribution of quarters, dimes, nickels, and pennies that are received in change, and given the number of coins needed to fill a coin wrapper — 40 quarters, 50 dimes, 40 nickels, and 50 pennies — how many of each wrapper should be sold in a bundled collection?

To answer this question, I determined the number of coins of each type required for every amount of change from 1¢ to 99¢. The totals yield the following graph:

Change - Theoretical

The number of pennies is nearly five times the number of nickels. And there are nearly twice as many quarters as dimes.

But I realize that’s a theoretical result that may not match what happens in practice, since this assumes that the amounts of change from 1¢ to 99¢ are uniformly distributed (they aren’t) and that cashiers don’t round down to avoid dealing with pennies (they do). In fact, when I made a purchase of $2.59 yesterday, instead of getting one penny, one nickel, one dime, and one quarter as change, the cashier gave me one penny, three nickels, and one quarter, in what was clearly a blatant attempt to skew my data.

So for an experimental result, I counted the pennies, nickels, dimes, and quarters in our home change jar. The results were similar:

Change - Experimental

The ratio of pennies to nickels is closer to three, but the ratio of quarters to dimes is still roughly two.

Using a hybrid of the theoretical and experimental results, and accounting for the fact that only 40 quarters and nickels are needed to fill a wrapper whereas 50 pennies and dimes are needed, it seems that an appropriate ratio of coin wrappers would be:

quarters : dimes : nickels : pennies :: 17 : 8 : 6 : 19

Okay, admittedly, that’s a weird ratio. Maybe something like 3:2:1:4, to keep it simple. Or even 2:1:1:2. All I know is that 1:1:1:1 is completely insane, and this nonsense has got to stop.

Hello, Royal Sovereign, Minitube, and Coin-Tainer? Are you listening? I’ve completed this analysis for you, free of charge. Now do the right thing, and adjust the ratio of coin wrappers in a package accordingly. Thank you.


Wow, that was a long rant. Sorry. If you’ve made it this far, you deserve some comic relief.

How many mathematicians does it take to change a light bulb?
Just one. She gives it to a physicist, thus reducing it to a previously solved problem.

If you do not change direction, you may end up where you are heading. – Lao Tzu

The only thing that is constant is change. – Heraclitus

Turn and face the strange ch-ch-ch-changes. – David Bowie

A Buddhist monk walks into a Zen pizza parlor and says, “Make me one with everything.” The owner obliged, and when the pizza was delivered, the monk paid with a $20 bill. The owner put the money in his pocket and began to walk away. “Hey, where’s my change?” asked the monk. “Sorry,” said the owner, “change must come from within.”


As for the “probability of a nickel” problem that started this post, here’s my solution.

For change amounts from 1¢ to 25¢, there are ten values (5‑9 and 15‑19) for which you’ll receive a nickel as part of your change.

This pattern then repeats, such that for change amounts from 25n + 1 to 25n + 25, where n is the number of quarters to be returned, you’ll receive a nickel when the amount of change is 25n + k, where k ∈ {5, 6, 7, 8, 9, 15, 16, 17, 18, 19}. For 0 ≤ n < 4, there are 40 different amounts of change that will contain a nickel, so the probability of getting a nickel as part of your change is 40/100, or 40%.

August 30, 2016 at 5:10 am 3 comments

Math of the Rundetaarn

RundetaarnAs we were exiting the Rundetaarn (“Round Tower”) in Copenhagen, Denmark, I noticed a man wearing a shirt with the following quotation:

Find what you love, and let it kill you.

The only problem is that the shirt attributed the quotation to poet Charles Bukowski, when apparently it should have been attributed to humorist Kinky Friedman. For what it’s worth, my favorite Friedman quote is, “I just want Texas to be number one in something other than executions, toll roads, and property taxes.” But this ain’t a post about Kinky Friedman, or even Charles Bukowski. So, allow me to pull off the sidewalk and get back on the boulevard.

Whoever said it, the quotation hit me as drastically appropriate. I suspect that math will someday kill me… likely as I cross the street while playing KenKen on my phone, oblivious to an oncoming truck. As I exited the Rundetaarn, I was thinking about all the math that I had seen inside — much of which, I suspect, would not have been seen by many of the other tourists.

The Rundetaarn, completed in 1642, is known for the 7.5-turn helical ramp that visitors can walk to the top of the tower and, coincidentally, to one helluva view of the city. Rundetaarn Cross SectionThat leads to Question #1.

Along the outer wall of the tower, the winding corridor has a length of 210 meters, climbing 3.74 meters per turn. What is the (inside) diameter of the tower?

Above Trinitatis Church is a gift shop that is accessible from the Rundetaarn’s spiral corridor. The following clock was hanging on the wall in that little shop:

Rundetaarn clock

I have no idea who the bust is, but the clock leads to Question #2.

What sequence of geometric transformations were required to convert a regular clock into this clock?

And to Question #3.

Do the hands on this clock spin clockwise or counterclockwise?

And to Question #3a.

What is the “error” on the clock?

A privy accessible from the spiral corridor in the Rundetaarn has been preserved like a museum exhibit. Sadly, I have no picture of it to share, but a sign next to the privy implied that the feces deposited by a friar would fall 12 meters into the pit below.

That leads to Question #4.

What is the terminal velocity of a deposit when it reaches the bottom of the pit? (Or should that be “turd-minal velocity”?)

The first respondent to correctly answer all of these questions will earn inalienable bragging rights for perpetuity.

August 11, 2016 at 8:31 am Leave a comment

How Wide and How Deep?

In 2002, William Schmidt described the U.S. math curriculum as “a mile wide, an inch deep,” and it’s been bugging the sh*t out of me ever since.

I mean, I get what he and his co-authors were saying: The curriculum contains too many topics, so they can’t be covered with sufficient depth.

But if a mile is too wide and an inch is too shallow, then what dimensions would be appropriate?

One-inch wide and a mile deep would be problematic, too. That’d be like spending an entire year teaching kids to count to 10.

I suppose we could opt for a square curriculum instead. A curriculum that is a mile wide and an inch deep has an area of 5,280 × 1/12 = 440 square feet, so the conversion would look something like this, with the thin line representing a mile by an inch and the square representing 21 feet by 21 feet:

MileWideInchDeep.png

Sorry, it’s not to scale because of space limitations.

The square curriculum doesn’t feel quite right, either. The only way I know to make this problem tractable is to look at data.

In the late 1990’s, I was a standards weenie. I was fascinated by the variety from state to state. Because I didn’t have a girlfriend (and ostensibly didn’t want one, either), I would read state standards documents for fun. At the time Schmidt coined his phrase, Florida had more than 80 standards in each grade, and Utah subjected students to over 130 standards each year. As I recall, the average state had more than 100 standards at each grade level.

Today, Common Core represents a significant reduction in the number of standards. There are approximately 30 standards per grade for K‑8, and closer to 40 standards per course in high school.

Which means that if the curriculum used to be a mile wide, then the current curriculum is closer to ⅓ × 5,280 = 1,760 feet wide.

But if it’s ⅓ as wide, then it needs to be 3 times as deep. Which means the current curriculum is 1,760 feet wide by 3 inches deep, so it looks something like this:

1760ft_by_3in.png

Doesn’t feel like much of an improvement, does it? And the phrase “3 inches deep” doesn’t inspire confidence that the curriculum now has the depth it needs.

So, I give up. I don’t know what the proper dimensions ought to be. I just know that Schmidt’s phrase was hyperbole for dramatic effect, and it worked.

What do you think are the proper dimensions of the math curriculum?

Here’s a puzzle about width and depth:

How much dirt is in a hole that measures 4¾ feet × 5¼ feet?

And I know a joke about width, but you need to be able to read CSS:

.yomama {
width: 99999999px;
}

 

August 4, 2016 at 4:11 am 3 comments


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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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