## Posts tagged ‘distribution’

### Constant Change

I’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?

The 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:

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:

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%.

### Stolen Truck Solution

Last week, I posted a modified version of Marilyn Burn’s horse problem, and I asked you to submit your answers. I received 331 responses; thanks for participating! An analysis of the submitted responses appears below, but first, a few comments and several different solutions.

My friend Jeane Joyner of Meredith College uses this problem in teacher and parent workshops. She said:

I have folks move to corners of the room — makes money, loses money, and breaks even. Then each groups selects an ambassador to go to the other groups to see if they can persuade folks to move. Fun!

Solution #1: He spent 600 + 800 = \$1,400, and he received 700 + 900 = \$1,600. That’s a profit of \$200.

Solution #2: Assume the man started with \$1,000 in his bank account. He bought the truck for \$600, so he had \$400 left. He then sold it for \$700, so his account increased to \$1,100. He bought it back for \$800, so he had \$300 left. When he sold it for \$900, his account increased to \$1,200. Since he started with \$1,000 and ended with \$1,200, he made a profit of \$200.

Solution #3: Use a number line to show how his amount of money changed.

After the four transactions, he is at +200 on the number line, so he made a profit of \$200.

Solution #4: Some people find it confusing that he buys and sells the truck twice. It might be easier to think of him doing these transactions with two different vehicles. For instance, what if he bought a truck for \$600 then sold it for \$700, and then bought a car for \$800 and sold it for \$900? It might be easier to wrap your head around that.

Approached that way, his actions represent two separate events. The first time he bought and sold the truck, he paid \$600 and sold it for \$700. That’s a profit of \$100. The second time, he bought it for \$800 and sold it for \$900. That’s another \$100 profit. In total, he made \$200.

After I posted the problem, a friend on Facebook asked, “Huh? How is that supposed to be hard?” Edward Early of St. Edwards College responded:

Sadly, I’ll only be surprised if there is a strong consensus for the correct answer. I’ve been teaching math too long to expect that to happen.

Of the 331 respondents, only 325 submitted usable responses. (Chalk this up to bad phrasing on my part. I asked folks to “enter a negative number if he lost money, 0 if he broke even, or a positive number if he made money.” I was meaning for people to enter the amount he lost or made, but some respondents entered a positive number that wasn’t possible in the context of the problem, such as 1 or 12. I think they thought they should enter any positive number to indicate that he made a profit. And one wiseacre responded, “a positive number.” Sorry, not even partial credit for that response!)

Of the 325 usable responses, 228 (70.1%) were correct. Answers from the other 28.9% ranged from ‑600 to 900, with 0 (28 responses) and 100 (47 responses) chosen most often. The chart below shows the distribution of incorrect responses. (Data for 200 has been removed since it overwhelms the others; its bar would be more than four times the height of the next highest bar.)

The vast majority of respondents (276) were 16 years of age or older. The 49 responses from people age 15 or younger looked like this:

 Age Responses Correct Responses 6 1 0 7 1 1 8 2 1 9 4 1 10 4 1 11 1 1 12 4 3 13 15 11 14 8 7 15 9 7

Interestingly, the under‑16 crowd, with 67.3% correct responses, did almost as well as the over‑16 crowd, which had 70.6% correct responses. And for the 11- to 15‑year old subset, an astounding 78.4% of the responses were correct.

From this, we can conclude that these youngsters are smarter than the rest of the population… but of course we already knew that, because teens and pre-teens know everything, right?

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.

## MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.