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I apologize if this question might not sound feasible. I'm not a math major, but I have knowledge of basic math. I want to find this solution purely out of curiosity. So my thought process was: first assume that there are n numbers in the sequence, and the random number generator has a range of 1 to k (both inclusive). So, the probability of the first two numbers being different would be $1*\frac{(k-1)}{k}$. Then, the probability of the third number in the sequence being different would be $\frac{(k-2)}{k}$ as the first two numbers are defined and there are k-2 numbers left. Then, it would be $\frac{(k-3)}{k}$ for the fourth number, and so on. So for all the n numbers in the sequence, the probability that they would all be different would be:

$\frac{(k-1)(k-2)(k-3)...(k-n+1)}{k^{(n-1)}}$

Since I want the probability of repetition, I'll subtract this from 1, and if I multiply and divide the fraction with $k(k-n)!$ I get this as the final expression:

$1- [\frac{k!}{\{k^{n}(k-n)!\}}]$

Now for an infinite sequence, n and k both have to tend to infinity. So this is what I finally get:

$\lim_{n\to \infty k\to \infty} (1- [\frac{k!}{\{k^{n}(k-n)!\}}])$

Can this be solved? Is my formulation even correct? Any advice or help would be highly appreciated!

dorito
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  • Your formulation is not correct. The concept of "a random generator with a range of infinity" is impossible. You analyzed the finite case correctly, but the limit does not exist, and the limit would not correspond to any random process. – Mike Earnest May 09 '23 at 20:00
  • Is there a mathematical reason for the limit to not exist? I mean I get the logical idea for it to not exist, but does it have a mathematical proof or law or something? – dorito May 10 '23 at 21:50
  • See this answer to understand the behavior of $k!/[(k-n)!k^n]$ when $n$ and $k$ are large. It is approximately $e^{-n^2/(2k)}$. Therefore, the limit depends on the rate at which $n,k\to\infty$. If $n$ grows faster than $\sqrt{k}$, then $n^2/(2k)\to +\infty$, so $e^{-n^2/(2k)}\to 0$. If $n$ grows slower than $\sqrt{k}$, then the limit is instead $1$. Since there are multiple limits, the limit does not exist. – Mike Earnest May 10 '23 at 22:44

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