Money eating *gulp* *gulp*#
All the money spent using a payment card (eventually) gets back to the bank.
Yeah, it's one of those reality checks...
In a perfect world, this article wouldn't exist. My research began with a video[5] in which Andrew Tate mentioned how real (paper) money moves:
"If I have a \(\$50\) bill and I give it to the barber for my haircut, and then he goes and buys groceries with it from the grocer, and the grocer goes and gets his car washed, the \(\$50\) goes from place to place, and after \(20\) or \(30\) transactions, the \(\$50\) bill belongs to somebody and it's worth \(\$50\)." - Andrew Tate
He contrasts this to the way the banks receive a commision on every card transaction:
"If I pay by card, \(1.5\%\) goes to the bank. And then if [the barber] takes the money I've given him and pays for the groceries, \(1.5\%\) goes to the bank. And after the groceries' been paid for, when he goes to get his car washed, \(1.5\%\) goes to the bank. So after \(20\) or \(30\) transactions, the \(\$50\) [bill] is gone and the bank has it all." - Andrew Tate
This can't be true... Can it?
How much per transaction?#
It turns out that a POS setup is a complex endeavour, but, since we are only interested in the ammount we get charged per transaction, we can find this data easily[2][3]:
In the United States, it can cost anywhere between \(1.5\%\) and \(3.5\%\) per transaction, plus some additional flat fee (somehwere near \(10\)¢).
However, this is the Merchant Discount Rate (MDR). The banks receive an interchange fee which is included in the MDR[4]. The interchange fee depends on many factors, including (most importantly) the geographic region of the payment. In Europe, the interchange fee can be as low as 0.2%, whilst in the USA, the fee can reach 2% of the initial transaction.
In this article, we'll only consider the MDR, since we're more interested in how the money "randomly" disappears, rather than what's the bank's cut.
The math#
We will be using the data above later. For now, let's focus on the math.
Let's consider how the money moves. In our model, every transaction converts a sum of money \(x\) into a smaller sum, by cutting both a percentage fee \(\alpha\in(0,1)\) and a fixed fee \(\beta\in\mathbb{R}_+\). Thus, mathematically, this can be written as:
\[x\rightarrow(1-\alpha) x-\beta\]Consider the function \(f(x)=(1-\alpha) x-\beta\). By applying it \(n\in\mathbb{N}\) times, we essentially move the money through \(n\) transactions. For example:
\[\begin{split} f^{(1)}(x)&=f(x)=(1-\alpha) x-\beta\\ f^{(2)}(x)&=f(f(x))= (1-\alpha)\big[(1-\alpha) x-\beta\big]-\beta\\ f^{(3)}(x)&=f(f(f(x)))= (1-\alpha)\Big[(1-\alpha) \big[(1-\alpha) x-\beta\big]-\beta\Big]-\beta\\ \vdots \end{split}\]It can be easily proved by induction that:
\[\begin{split} f^{(n)}(x)&=(1-\alpha)^nx-\beta\left[(1-\alpha)^{n-1}+(1-\alpha)^{n-2}+\dots+1\right]\\ &=(1-\alpha)^nx+\beta\frac{(1-\alpha)^n-1}{\alpha} \end{split}\]For our future calculations, we might want to get a formula for \(n\) as well:
\[\begin{split} f^{(n)}(x)+\frac{\beta}{\alpha}&=(1-\alpha)^nx+\frac{\beta}{\alpha}(1-\alpha)^n\\ \Rightarrow (1-\alpha)^n&=\frac{f^{(n)}(x)+\dfrac{\beta}{\alpha}}{x+\dfrac{\beta}{\alpha}}=\frac{\alpha f^{(n)}(x)+\beta}{\alpha x+\beta} \end{split}\]Extracting \(n\) is as easy as:
\[ n=\log_{1-\alpha}\left(\frac{\alpha f^{(n)}(x)+\beta}{\alpha x+\beta}\right)=\frac{\ln\Big[\alpha f^{(n)}(x)+\beta\Big]-\ln(\alpha x+\beta)}{\ln(1-\alpha)} \]If we're talking about a greater initial sum of money, the formula can be simplified a lot by not taking into account the flat fee, since it is very small compared to the percentage fee.
However, I wanted to do a correct analysis instead!
Implementing the code#
We can use Python to aid our calculations:
import math
def f_n(n, alpha, beta, x0):
return ((1-alpha)**n)*x0 + beta*(((1-alpha)**n)-1)/alpha
def get_n(alpha, beta, x0, f0):
return (math.log(alpha*f0+beta)-math.log(alpha*x0+beta))/math.log(1-alpha) As you can tell, f_n represents \(f^{(n)}\), while get_n computes \(n\) based on \(f^{(n)}\).
Plugging in the data#
Considering:
-
An initial sum of money, let's say Tate's \(x=\$50\)
-
The percentage fee: \(\alpha=3\%\)
-
The flat fee: \(\beta=\$0.1\)
We would reach:
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\(50\%\) of the initial sum of money in \(n=21\) transactions
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\(25\%\) of the initial sum of money in \(n=40\) transactions
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\(10\%\) of the initial sum of money in \(n=61\) transactions
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\(1\%\) of the initial sum of money in \(n=87\) transactions
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\(0.1\%\) (essentially \(x\rightarrow0\)) of the initial sum of money in \(n=91\) transactions
The \(\$50\) figure is very representative, since it is very close to Amazon's \(\$47.31\) AOV[1].
If we were to move around only \(x=\$10\), we would have less than \(\$1\) in \(n=45\) transactions. The lower the initial sum, the quicker it gets swollen by the banks.
If we were to use Toast's \(\alpha=2.49\%\), \(\beta=\$0.15\)[3], we would end up with less than \(10\%\) of \(\$10\), \(\$20\), \(\$50\) and \(\$100\) in \(33\), \(47\), \(65\) and \(75\) transactions respectively.
Conclusion#
Andrew Tate might have been a little off when it came to the numerical values themselves - however, his point still stands!
Most money, when moved around, get back to the banks!