What mathematical rule would effectively eliminate .99 pricing?


In the year 2013, the government of Canada stopped minting pennies

Suppose that the United States government wanted to greatly reduce the number of pennies, nickles, and dimes manufactured by the mint.

Among other things, imagine that this future government the United States decides to impose hefty fines on some retail buissness if those bussinesses practice dot-nine-nine pricing.

An item which cost $\\\$ \text{19.99}$ in the past will cost $\\\$ \text{20.00}$ in the future.

Maybe the government wants to reduce the use of other small denominations as well as pennies.

Ideally, the only prices allowed are ones where robotic self check-out machines are unlikely to have dispense much change.

A price that was $\\\$49.75$ should now be $\\\$50.00$.

Beleive it or not, but I have met people for whom $25$ cents means the difference between buying stuff nearby and walking an hour away to a store with a lower price.

However, if you can afford to pay $\\\$ \text{49.75}$ , then you can afford to pay $\\\$ \text{50.00}$. Likewise, if you cannot afford to pay $\\\$50.00$, then you probably cant afford $\\\$49.75$ either. people who really care about $\\\$0.25$ don't have to worry about a shift in large prices.

If you have some qualms about the motivation, simply assume that you have been asked to devize some new price rules as some kind of mathematical puzzle.

Also, assume that some theoretical adversary wants to make life difficult.

If $\\\$$19.99 is a forbidden price, then the adversary will change the price to $\\\$$19.98.

If we forbid prices such as $\\\$$19.98, then the adversary might set the price to $\\\$$19.95.

The Question - Short Version

Can you provide an mathematical description of which prices are "simple" and which prices are not?

Prices with exactly one non-zero digit are usually simple. For example, \$10 or \$5,000.00

Prices with two non-zero digits are usually simple if the non-zero digits are consecutive (e.g. $\\\$ \text{25.00}$). AN exception occurs if the two digit price is very close to a one-digit price. For example, $\\\$\text{95}$ is close to $\\\$\text{100}$. So, $\\\$\text{100}$ is simple, but $\\\$\text{95}$ is not simple.

We define price with three nonzero digits to be simple only if ithe price is sufficiently large and not close to a two digit price. For example, $\\\$ \text{415,000.00}$ is a simple price. $\\\$ \text{415,000.00}$ is somewhat close to $\\\$ \text{410,000.00}$ and $\\\$ \text{420,000.00}$. However, for suffice tly large prices we do not require that 3 nonzero digits be rounded to two non-zero digits.

The Question - The long Version

We define a price to be any element of the set $\{x/100 \mid x \in \mathbb{N} \cup \{0\}\}$

Bob thinks he might know what it means for a price to a nice round number.

A set of prices $\mathrm{BPS} \subset \{x/100 \mid x \in \mathbb{N} \cup \{0\}\}$ is Bobain if and only if the following Bobain principles are satisfied:


    • INFORMAL: Most complex prices, such as $\\\$7,717.89$ and $\\\$4,769.81$ have lots and lots of non-zero digits. A house costing $\\\$835,000$ is easier to understand than $\\\$834,988$ in part because $\\\$835,000$ has only three significant figures.
    • FORMAL: If a set is Bobain, then every sufficiently small price in the set has at most $3$ non-zero digits.
    If $\\\$0.25$ is an element of a Bobain set, then $\\\$25.00$ must also be an element of the same set so must be $\\\$250,000.00$. For any price $bp \in \mathrm{BPS}$ and for any $k \in \mathbb{N}$, $bp*10^{k} \in \mathrm{BPS}$.


    • INFORMAL: $\\\$1,005$ is not a simple price, but $\\\$1,000$ definitely is a simple price.
    • FORMAL: All non-zero digits must appear consecutively.

    • INFORMAL: $\\\$0.99$ is bad. $\\\$4.97$ is also bad. $9$ million is okay though. The rule is almost that nines are bad, but there are "always exceptions to the rule."
    • FORMAL: $\mathrm{BPS}$ is a Bobain set of prices only if for all sufficiently small prices in $\mathrm{BPS}$, the price cannot contain $9$ as anything other than the most leftmost non-zero digit.
    • SUPER-FORMAL (read the super formal version only if you feel that the problem is not well-defined). $\exists B \in \mathbb{R}$ such that $B > 10$ and $\exists n \in \mathbb{N}$ such that $\forall p \in \mathbb{R}$ if $p < B$ and $p*100 \in \mathbb{N}$ then $9$ is not the most significant figure in price $p$. A corollary is that $\forall bp \in \mathrm{BPS}, bp - \lfloor bp \rfloor \neq 0.99$

    • INFORMAL: $\\\$10$ is a simple price. $\\\$5,000$ is also a simple price.
    • FORMAL: Bobain sets contain all prices having exactly one non-zero digit provided that the non-zero digit is a $1$ or a $5$.

    • INFORMAL: If $\\\$9$ million is a simple price, then that does not necessarily imply that $\\\$9.00$ is a simple price. $\\\$9.00$ should be $\\\$10.00$. As prices get larger an larger, we should see more prices allowed which used to be forbidden, like a tree branching out.
    • SEMI-FORMAL: There should be very large legal prices (for real estate, and more) which become illegal if you divide the price by $10$ a bunch of times. For example, we might allow $\\\$95,000$, but disallow $\\\$9.50$ (use \$10.00 instead of \$9.50)
    • FORMAL: set of prices $\mathrm{BPS}$ is Bobain only if $\begin{vmatrix} \left\{ bp \in \mathrm{BPS} \mid \exists z \in \mathbb{Z} \mid \dfrac{\lfloor bp * 10^{z} \rfloor }{100} \not\in BPS \right\} \end{vmatrix} = \begin{vmatrix} \mathbb{N} \end{vmatrix} $.

    • INFORMAL: You can't just write down a list of ten prices, say that those ten prices are simple, and quit there. There must be an infinite number of prices in a set in order for the set to be Bobain.
    • FORMAL: $\mid BPS \mid = \mid \mathbb{N} \mid$ and $\not\exists \text{ price } p : \forall sp \in \mathrm{BPS}, sp < p$

    • INFORMAL: $\\\$25$ and $\\\$0.75$ are okay. However, $\\\$14.70$ is not okay. Note that the right-most digit in the price $\\\$14.70$ is a $7$. Trailing $5\text{s}$ are nice.
    • FORMAL: For all sufficiently small prices, if the price contains more than $1$ non-zero digit, then the right-most nonzero digit must be a $5$.

    • INFORMAL: $2¢$ is a simple price, and a Normative economist might theoretically say that $2¢$ should not be outlawed (that was a joke). In all seriousness though, 0.02 is interesting because 0.02 is not a multiple of $1/8$ or a multiple of 5.
    • FORMAL: $\forall BPS \subseteq \begin{Bmatrix} x/100 \mid x \in \mathbb{N} \cup \{0\}\end{Bmatrix}$, if $\mathrm{BPS}$ is Bobain, then $0.02 \in \mathrm{BPS}$

What was the question again?

My question is, can you provide an example of a Bobain set?

Examples of non-Bobian Sets

Note that the set of whole dollar amounts is not Bobain.

$$\mathrm{SPS} = \mathbb{N}$$

Note that BOB'S TRINITY PRINCIPLE is violated.
$\\\$4,998$ is a whole-dollar amount, but the price contains more than $3$ digits.

BOB'S SCALE DOWN PRINCIPLE is also violated by the set of whole-dollar-amounts.

There should exist lots of large simple prices, such as 9.8 million USD, that when scaled down, are no longer simple.

The set of whole dollar amounts which are not multiples of a hundred dollars is the only set of prices which are not still in the set when scaled down. The set of multiples of one-hundre is finite.

It is tempting to look at multiples of $25$ cents.

$\\\$5.25$ isn't too bad of a price to understand.

$\\\$24.75$ should probably be $\\\$25.00$.

Note that it will not work to have prices always be a multiple of $\\\$0.125$. In the formula below, the abreviation $\mathrm{SPS}$ stands for either "simple prices" or "simple price set".

$$\mathrm{SPS} = \begin{Bmatrix}k/8 \mid k \in \mathbb{N} \cup \{0\}\ \end{Bmatrix}$$

$\mathrm{SPS}$ is not Bobain. $\\\$480.125$ is multiple of one-eighth of a dollar, but $\\\$480.125$ has more than $3$ non-zero digits. Bob's principle of trinity is not satisfied. $\left( \\\$480.125 \text{ has six non-zero digits} \right)$. If you are going to spend \$480, then \$0.125 differnce will not have significant impact on things. It is a bit like adding one more spec of dust to a train car already full of coal.

One last potential candidate I will mention as follows, but it also fails to have all of the Bobian properties:

$$\mathrm{SPS} = \begin{Bmatrix} \dfrac{ d*10^{n+3} + k*125}{100} \mid k \in \{0, 1, 2, \dots, 7\} \text{ and } n \in \mathbb{N} \cup \{0\} \text{ and } d \in \{0, 1, \dots, 9\} \end{Bmatrix} $$

  • 2
    $\begingroup$ I did not read this to its end - sorry, quite long - but it seems like you have an unclear economics question and a very weakly related math question? $\endgroup$
    – Giskard
    Commented Jul 18, 2022 at 6:39
  • 1
    $\begingroup$ On the econ front: I live in a EU country where they stopped using the smallest two coins (1,2). Prices still frequently end in 99. If you do not pay by card the cashier rounds the final sum to the closest number divisible by 5. $\endgroup$
    – Giskard
    Commented Jul 18, 2022 at 6:41
  • $\begingroup$ "any element of the set $\{x/100 \mid x \in \mathbb{N} \cup \{0\}\}$" seems to be a long-winded way to say "a whole number of cents" and since this whole thing is actually really informal I wonder why you bothered to formalize it. $\endgroup$ Commented Jul 18, 2022 at 18:06
  • $\begingroup$ This is far too long a "question"? $\endgroup$
    – BrsG
    Commented Jul 19, 2022 at 11:36
  • $\begingroup$ @BrsG I have a standing monitor, and on its 1920p high resolution the question is barely more than 2 pages long! $\endgroup$
    – Giskard
    Commented Jul 19, 2022 at 12:27


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