Editing Benford's law (section)

==Generalization to digits beyond the first==
[[File:Benford_law_log_log_graph.svg|thumb|300px|link={{filepath:Benford_law_log_log_graph.svg}}|Log–log graph of the probability that a number starts with the digit(s) ''n'', for a distribution satisfying Benford's law. The points show the exact formula, ''P''(''n'') = log<sub>10</sub>(1&nbsp;+&nbsp;1/''n''). The graph tends towards the dashed asymptote passing through {{nobr|(1, log<sub>10&thinsp;</sub>''e'')}} with slope −1 in log–log scale. The example in yellow shows that the probability of a number starts with 314 is around 0.00138. The dotted lines show the probabilities for a uniform distribution for comparison. (In {{nowrap|[{{filepath:Benford_law_log_log_graph.svg}} the SVG image]}}, hover over a point to show its values.)]]
It is possible to extend the law to digits beyond the first.<ref name=Hill1995sigdig>{{cite journal | last1 = Hill | first1 = Theodore P. | author-link = Theodore P. Hill | year = 1995 | title = The Significant-Digit Phenomenon | url = http://digitalcommons.calpoly.edu/cgi/viewcontent.cgi?article=1041&context=rgp_rsr | journal = The American Mathematical Monthly | volume = 102 | issue = 4| pages = 322–327 | jstor = 2974952 | doi = 10.1080/00029890.1995.11990578}}</ref> In particular, for any given number of digits, the probability of encountering a number starting with the string of digits ''n'' of that length{{snd}} discarding leading zeros{{snd}} is given by

: <math>\log_{10}(n + 1) - \log_{10}(n) = \log_{10}\left(1 + \frac{1}{n}\right).</math>

Thus, the probability that a number starts with the digits 3,&nbsp;1,&nbsp;4  (some examples are 3.14, 3.142, {{pi}}, 314280.7, and 0.00314005) is {{math|log<sub>10</sub>(1&nbsp;+&nbsp;1/314)&nbsp;≈&nbsp;0.00138}}, as in the box with the log-log graph on the right. &nbsp;

This result can be used to find the probability that a particular digit occurs at a given position within a number. For instance, the probability that a "2" is encountered as the second digit is<ref name=Hill1995sigdig />

: <math>\log_{10}\left(1 + \frac{1}{12}\right) + \log_{10}\left(1 + \frac{1}{22}\right) + \cdots + \log_{10}\left(1 + \frac{1}{92}\right) \approx 0.109</math>.

The probability that ''d'' (''d''&nbsp;=&nbsp;0,&nbsp;1,&nbsp;...,&nbsp;9) is encountered as the ''n''-th (''n''&nbsp;>&nbsp;1) digit is

: <math>\sum_{k=10^{n-2}}^{10^{n-1}-1} \log_{10}\left(1 + \frac{1}{10k + d}\right).</math>

The distribution of the ''n''-th digit, as ''n'' increases, rapidly approaches a uniform distribution with 10% for each of the ten digits, as shown below.<ref name=Hill1995sigdig /> Four digits is often enough to assume a uniform distribution of 10% as "0" appears 10.0176% of the time in the fourth digit, while "9" appears 9.9824% of the time.
{| class="wikitable" style="text-align:right"
! Digit
! 0
! 1
! 2
! 3
! 4
! 5
! 6
! 7
! 8
! 9
|-
! 1st
| {{N/A}}
| 30.1%
| 17.6%
| 12.5%
|  9.7%
|  7.9%
|  6.7%
|  5.8%
|  5.1%
|  4.6%
|-
! 2nd
| 12.0%
| 11.4%
| 10.9%
| 10.4%
| 10.0%
|  9.7%
|  9.3%
|  9.0%
|  8.8%
|  8.5%
|-
! 3rd
| 10.2%
| 10.1%
| 10.1%
| 10.1%
| 10.0%
| 10.0%
|  9.9%
|  9.9%
|  9.9%
|  9.8%
|}