Probability of success: False uniqueness bias

It's very ideal to see that the path we choose, be it for success or anything else we try to achieve is unique, our own to follow. An extreme example I see is when a person models their own idea/business on someone else's yet have a false uniqueness bias. It made me wonder to what extent such a thing can be actually be true.

This blog post is a naive mathematical approach I have adopted to dispel the fact that even our ways to success are in fact not as unique as we may feel they may be,such that at-least one other person apart from us is following it; thus, trying to prove the false uniqueness bias.

Assuming the maximum number of unique path-ways to be successful is equal to the number of people in the world ,~7 billion (1 way to success for each person):
\(\mu=7,000,000,000\)
Let's take another assumption, that the path chosen is uniformly distributed across the \(\mu\) number of ways. Let \(success_i\) be the path of success the \(i^{th}\) person chooses, such that \(Pr\{success_i=r\}=1/\mu\) where \(i,r=1,2,3....\mu\). What are the chances that at least two people choose the same path in a group of \(k\) people ?

The probability that two people don't choose the same path is:
 \[1-\frac{1}{\mu} = \frac{999,999,999}{1,000,000,000} = 0.999999999\],
which makes sense, because this means that in 999,999,999 scenarios when comparing one person's path to another, they won't match. But with \(k\) people, it would mean \(\binom{k}{2}\) comparisons. That's a lot comparisons! And exponents have their own way of compounding! Iteratively applying the recurrence for \(k\) people choosing a different path is :
\[Pr\{A_k\}=1.(\frac{999,999,999}{1,000,000,000})(\frac{999,999,998}{1,000,000,000})....\]
\[Pr\{A_k\}=1.(\frac{\mu-1}{\mu}).(\frac{\mu-2}{\mu}).(\frac{\mu-3}{\mu})....(\frac{\mu-k+1}{\mu})\]
\[Pr\{A_k\}=1.(1-\frac{1}{\mu}).(1-\frac{2}{\mu})....(1-\frac{k-1}{\mu})\]
Note: \[1+x \leq e^x\]
\[Pr\{A_k\}\leq e^{\frac{-1}{\mu}}. e^{\frac{-2}{\mu}}. e^{\frac{-3}{\mu}}\]
\[ = e^{-\sum_{i=1}^{k-1} i/n}\]
\[ = e^{-k(k-1)/2n}\]
\[ \leq e^{-k(k-1)/2n}\]

So keeping the probability \(p(k)=1\) ,i.e, ensuring that a minimum of two people in a group of \$k\$ people follow the same path, we get solving for k:
\(k= 140,738\)

In a population of 7 billion, for a mere randomly selected 140,800, i.e, ~\(0.0002011\%\) of the population, min. 2 people will  always share the same path.

Conclusion : False uniqueness bias, hence proved. Even in the case where we choose from 7,000,000,000 options randomly, we still end up sharing common paths.