Finding the percentage of people who lose money in a pyramid scheme

119 Views Asked by At

"In short, pyramid schemes that do not provide new members with a product that is fully worth the "membership fee" will result in a majority of all members losing their money. If we look at the 8-Ball Scheme, there are twice as many members in each new level. The bottom 3 levels always lose their money, and as shown below, this will be at least 7/8 ≈ 88% of the total of all participants."

(my work)->

n = levels in the pyramid

there are $2^n - 1$ members in the pyramid

at the bottom there are $2^{n-1}$ members

1 level from the bottom there are $2^{n-2}$ members

2 levels from the bottom there are $2^{n-3}$ members

Therefore

$\frac{Members losing money}{Total members}$ = $\frac{2^{n-1}+2^{n-2}+2^{n-3}}{2^n -1}$ =$\frac{2^{n-3}(2^2+2+1)}{2^n -1}$ =$\frac{2^{n-3}(7)}{2^n -1}$

I'm basing my calculations on a website, though mine are slightly different. On the site, they say, "If we neglect the quantity of 1 subtracted in the denominator we get the inequality

$\frac{Members losing money}{Total members}$ > $\frac{2^{n-1}+2^{n-2}+2^{n-3}}{2^n -1}$

With this they're able to factor out $2^{n-3}$ from the numerator and denominator and it cancels out completely leaving them with $\frac{7}{8}$.

I was hoping someone could explain why they're able to get rid of the -1 in the denominator and also why that turns it into an inequality.

https://mathmotivation.com/money/pyramid-scheme.html (website I based my calculations off of)