Benford's Law
To examine the distribution of the mantissas of a dataset, we can
examine the fractional parts of the common logarithms of the data.
That's because the fractional part of the common logarithm is the
common logarithm of the mantissa.
For example, consider numbers with mantissa $2.5$:
$$
\begin{align}
\log(.025) &= .39794000867203760957 - 2\\
\log(.25) &= .39794000867203760957 - 1\\
\log(2.5) &= .39794000867203760957 + 0\\
\log(25) &= .39794000867203760957 + 1\\
\log(250) &= .39794000867203760957 + 2\\
\log(2500) &= .39794000867203760957 + 3
\end{align}
$$
Thus, if the mantissa of a number is $2.5$, the fractional part of its
common logarithm is $.39794000867203760957$.
If the data spans several decades (powers of $10$, not years; see
Decade (log scale),
when we combine the data from all of the
decades, it tends to even out. Thus, the fractional parts of the
common logarithms of the data should be evenly distributed.
For example, suppose the common logarithm of the data is distributed
across $4$ decades as shown below:
$\hspace{7mm}$
Summing the distributions of the fractional parts of the logarithms,
that is, moving the decades on top of each other and adding curves,
we get the black curve at the top of the image below, which is close
to evenly distributed:
$\hspace{64mm}$
Thus, we arrive at the principal assumption of Benford's Law: the
logarithm of the mantissa of data which spans several decades is
typically distributed evenly.
Note the hash marks on the bottoms of the graphs above. These marks
separate where the different leading digits of the mantissa live.
On the line segment below, we expand these hash marks and align the
leading digit of the mantissa with the fractional part of the common
logarithm. The leading digit of the mantissa is $1$ if the fractional
part of the common logarithm is between $0$ and $.30103$; the leading
digit is $2$ if the fractional part is between $.30103$ and $.47712$; and
so on.
$\hspace{7mm}$
If the principal assumption of Benford's Law holds, the fractional
part of the common logarithm is evenly distributed. In view of the
previous diagram, it is obvious that the probability of $1$ being the
leading digit is greater than that of $2$ being the leading digit; the
probability of $2$ is greater than that of $3$; and so on. This is made
precise below.
Data that has a mantissa starting with the digit $1$ has a common
logarithm whose fractional part ranges from $\log(1)$ to $\log(2)$. If
the fractional part of the common logarithm of the data is evenly
distributed, then the portion of the data that starts with $1$ would
be
$$
\frac{\log(2)-\log(1)}{\log(10)-\log(1)}=.30102999566398119521
$$
Similarly, data that has a mantissa starting with the digit $2$ has
a common logarithm whose fractional part ranges from $\log(2)$ to
$\log(3)$. Thus, the portion of the data starting with $2$ would be
$$
\frac{\log(3)-\log(2)}{\log(10)-\log(1)}=.17609125905568124208
$$
In the same manner, data that has a mantissa starting with the
digit $d$ has a common logarithm whose fractional part ranges from
$\log(d)$ to $\log(d+1)$. Thus, the portion of the data starting with $d$
would be
$$
\frac{\log(d+1)-\log(d)}{\log(10)-\log(1)}\tag{1}
$$
Using $(1)$, we can compute the probability that such data will start
with the digit $d$:
$$
\begin{array}{}
d&P(d)\\
\hline\\
1&.30102999566398119521\\
2&.17609125905568124208\\
3&.12493873660829995313\\
4&.096910013008056414359\\
5&.079181246047624827723\\
6&.066946789630613198203\\
7&.057991946977686754929\\
8&.051152522447381288949\\
9&.045757490560675125410
\end{array}
$$
This distribution of leading digits is called Benford's Law.
Further Digits
The probability that the first two digits are $10$ is
$$
\frac{\log(11)-\log(10)}{\log(100)-\log(10)}=.041392685158225040750
$$
The probability that the first two digits are $20$ is
$$
\frac{\log(21)-\log(20)}{\log(100)-\log(10)}=.021189299069938072794
$$
Adding the probabilities for all first digits, we can compute the
probability that the second digit is $0$ to be $.11967926859688076667$.
In this manner, we can compute the probability that the second digit
is $d$:
$$
\begin{array}{}
d&P(d)\\
\hline\\
0&.11967926859688076667\\
1&.11389010340755643889\\
2&.10882149900550836859\\
3&.10432956023095946693\\
4&.10030820226757934031\\
5&.096677235802322528359\\
6&.093374735783036121570\\
7&.090351989269603369600\\
8&.087570053578861399175\\
9&.084997352057692199898
\end{array}
$$
For reference, here are the probabilities that the third digit is $d$:
$$
\begin{array}{}
d&P(d)\\
\hline\\
0&.10178436464421710175\\
1&.10137597744780144287\\
2&.10097219813704129959\\
3&.10057293211092617495\\
4&.10017808762794737592\\
5&.099787575692177452606\\
6&.099401309944962084127\\
7&.099019206561896092170\\
8&.098641184154777437875\\
9&.098267163678253538152
\end{array}
$$
Here are the probabilities that the fourth digit is $d$:
$$
\begin{array}{}
d&P(d)\\
\hline\\
0&.10017614693993552632\\
1&.10013688811757926504\\
2&.10009767259461432585\\
3&.10005850028348653742\\
4&.10001937109690488020\\
5&.099980284947840433784\\
6&.099941241749525329518\\
7&.099902241415451708313\\
8&.099863283859370683672\\
9&.099824368995291309873
\end{array}
$$