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Background:

Possibly wrong experiments with the bias of the parity of reputation based on 2012-11 MO dumps show there might be some bias:

 lower bound= 10 all= 11198 even= 1563 odd= 9635 e/o= 0.162221069019 average 416.549651724
 lower bound= 100 all= 3762 even= 1230 odd= 2532 e/o= 0.485781990521 average 1168.58001063
 lower bound= 1000 all= 762 even= 365 odd= 397 e/o= 0.919395465995 average 4597.52099738
 lower bound= 3000 all= 301 even= 153 odd= 148 e/o= 1.03378378378 average 8999.94684385
 lower bound= 10000 all= 84 even= 43 odd= 41 e/o= 1.0487804878 average 18494.6547619
 lower bound= 20000 all= 22 even= 11 odd= 11 e/o= 1.0 average 30738.7727273

Is there bias in the parity of reputation or similar "random" parameter on MO?

For someone with SQL skills probably this will be just two SELECT statements on https://data.stackexchange.com

If there is bias, likely it will be explained by logical reasons.

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    $\begingroup$ For obvious reasons, at the lower end there will be a disproportionately large number of people with reputation = 1 modulo 10. $\endgroup$ Apr 10 '16 at 9:58
  • $\begingroup$ @EmilJeřábek I think yours might be true because at the time question upvote gave 10 points. In case accepting an answer toggles the parity, I am not sure yours is true in a random model. $\endgroup$
    – joro
    Apr 10 '16 at 10:02
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    $\begingroup$ I think Emil was referring to the large number of users whose score is not only 1 modulo 10, but is actually 1 identically. $\endgroup$ Apr 10 '16 at 12:54
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    $\begingroup$ @JoelDavidHamkins Emil might also be including the large group of users with 101 reputation (due to the association bonus). I imagine that those two groups (1 rep and 101 rep) make up at least half of all accounts. $\endgroup$
    – user642796
    Apr 10 '16 at 13:00
  • $\begingroup$ @JoelDavidHamkins Users with reputation 1 are excluded by the lower bound of 10 points reputation. $\endgroup$
    – joro
    Apr 10 '16 at 13:18
  • $\begingroup$ @arjafi In 2012 MO wasn't part of the SE network and I believe 101 users couldn't join for free. $\endgroup$
    – joro
    Apr 10 '16 at 13:19
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Here is a simple SEDE query set up to provide basic data on reputation parity above a given lower bound. Others are of course welcome to use this as a basis for their own queries, though my SQL prowess is not exactly substantial.

Here are some signposts as the data currently stands:

Lower Bound  Total Users  Even Rep  Odd Rep  % Even   % Odd   Even/Odd     Mean
       1        53842       10725    43117   19.919   80.081    0.249     168.654
      10        30344        8414    21930   27.729   72.271    0.384     297.995
     100        18295        3675    14620   20.087   79.913    0.251     472.259
     102         7470        3659     3811   48.983   51.017    0.960    1010.265
     500         1844         930      914   50.434   49.566    1.018    3503.946
    1000         1082         544      538   50.277   49.723    1.011    5476.218
    3000          460         230      230   50.000   50.000    1.000   10543.878
   10000          145          70       75   48.276   51.724    0.933   22167.607
   20000           61          30       31   49.180   50.820    0.968   33077.230

As expected, for a lower bound below 102 there is a definite bias towards odd parity, due to the number of accounts which have either 1 or 101 reputation (31378 of the 53842 total accounts are in this group). The vast majority of these accounts are "unused" in the sense that there are no extant posts associated to them (perhaps they asked a homework question which was later deleted).

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Back in 2012, the only practical ways to change the parity of your score was to have an answer accepted (boosting your score by 15), or to vote -1 on someone else's question or answer (which required a lower bound on your score). People who are not particularly active are unlikely to have even scores for this reason.

Another way to change parity was to obtain an even score and find a way to lower your score to the absolute lower bound of 1. I don't think this happened too often.

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  • $\begingroup$ Thanks, this makes sense. $\endgroup$
    – joro
    Apr 11 '16 at 9:00

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