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The experiment is the following:

take a pool of paperclips for which in each round we follow these steps:

1) choose two paperclips randomly

2) if they are not linked with each other then link them to each other and follow with step 3); if they are already linked with each other go back to step 1)

3) add a new paperclip in the pool

4) go and do step 1)

[ you can see a video of a similar process here: https://www.youtube.com/watch?v=F-I-BVqMiNI&t=6s

I have modified that process by adding the step 3) ]

In the limit, this process apparently follows a Pareto distribution, so a certain x% of the paperclips-chains that are formed, hold a percentage y% of the paperclips where x% < y% and x%+y% = 100%. (That is the Pareto principle)

Now, how can I calculate the two parameters of the Pareto distribution (tail index or exponent, and the scale parameter) with the information I have of the process as above?

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    $\begingroup$ You might make your Question clearer if you simulate the paper-clip game or discuss it--in order to show the connection to a Pareto distribution. Supposing that the procedure you propose leads to a Pareto distribution, you might start by looking at Wikipedia about estimation of what is denoted $\alpha$ there. $\endgroup$ – BruceET Jul 10 at 5:16
  • $\begingroup$ Thank you. It still puzzles me that there’s no reference in Wikipedia or elsewhere as to the derivation of the alpha corresponding to the famous 80/20. That’s actually more relevant to me that the particular clip-paper game I mentioned. $\endgroup$ – Kronecker Jul 10 at 15:21
  • $\begingroup$ If you still have an unresolved question, suggest you start a new well-focused question with descriptive title. You may get useful answer or link. $\endgroup$ – BruceET Jul 10 at 18:31
  • $\begingroup$ I have made clearer and more specific the question and provided a link to a video explaining the experiment, can you remove the put on hold? Thank you $\endgroup$ – Kronecker Jul 11 at 20:33
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    $\begingroup$ I edited that part to address your observation, than you $\endgroup$ – Kronecker Jul 12 at 13:27

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