Parallel resistors deviation
If you are just looking for the mathematics, then it's just a sensitivity analysis. Suppose:
$$\begin{align*}
R_{parallel} &amp;= \frac{R_1\,R_2}{R_1+R_2} &amp; R_{series} &amp;= R_1+R_2
\end{align*}$$
Then, for parallel:
$$\begin{align*}
\% R_{parallel} &amp;= \frac{\text{d}\,R_{parallel}}{R_{parallel}}\\\\
&amp;= \frac1{R_{parallel}}\left[\frac{R_2^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_1 + \frac{R_1^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_2\right]\\\\
&amp;=\frac{R_1+R_2}{R_1\,R_2}\left[\frac{R_2^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_1 + \frac{R_1^{\:2}}{\left(R_1+R_2\right)^2}\,\text{d}R_2\right]\\\\
&amp;=\frac{R_2}{R_1+R_2}\,\frac{\text{d}R_1}{R_1} + \frac{R_1}{R_1+R_2}\,\frac{\text{d}R_2}{R_2}\\\\
&amp;=\frac{\%R_1}{1+\frac{R_1}{R_2}}+\frac{\%R_2}{1+\frac{R_2}{R_1}}
\end{align*}$$
It's pretty easy to see from the above that the smaller resistor's %-deviation dominates the %-deviation for parallel combinations. And that just makes sense, of course.
And, for series:
$$\begin{align*}
\% R_{series} &amp;= \frac{\text{d}\,R_{series}}{R_{series}}\\\\
&amp;= \frac1{R_{series}}\bigg[\text{d}R_1 + \text{d}R_2\bigg]\\\\
&amp;=\frac1{R_1+R_2}\bigg[\text{d}R_1 + \text{d}R_2\bigg]\\\\
&amp;=\frac{R_1}{R_1+R_2}\,\frac{\text{d}R_1}{R_1} + \frac{R_2}{R_1+R_2}\,\frac{\text{d}R_2}{R_2}\\\\
&amp;=\frac{\%R_1}{1+\frac{R_2}{R_1}}+\frac{\%R_2}{1+\frac{R_1}{R_2}}
\end{align*}$$
It's pretty easy to see from this case above that the larger resistor's %-deviation dominates the %-deviation for series combinations. And that just makes sense, too, of course.
That's the mathematics, sans Beyes and probabilities. And I've no idea if that's where you wanted to go. So I'm just putting it out there for your consideration.
Notes
I left the above with the following two statements:
$$\begin{align*}
\% R_{parallel} &amp;=\frac{\%R_1}{1+\frac{R_1}{R_2}}+\frac{\%R_2}{1+\frac{R_2}{R_1}}\\\\
\% R_{series} &amp;= \frac{\%R_1}{1+\frac{R_2}{R_1}}+\frac{\%R_2}{1+\frac{R_1}{R_2}}
\end{align*}$$
Let's set \$k_1^{'}=\frac1{1+\frac{R_2}{R_1}}\$ and \$k_2^{'}=\frac1{1+\frac{R_1}{R_2}}\$. But note also that \$1 = k_1^{'}+k_2^{'}\$! So let's use \$k=k_1^{'}\$ and therefore \$\left(1-k\right)=k_2^{'}\$:
Then:
$$\begin{align*}
\% R_{parallel} &amp;= k_2^{'}\cdot \%R_1+k_1^{'}\cdot \%R_2\\\\&amp;= \left(1-k\right)\cdot \%R_1+k\cdot \%R_2\\\\
\% R_{series} &amp;= k_1^{'}\cdot \%R_1+k_2^{'}\cdot \%R_2\\\\&amp;= k\cdot \%R_1+\left(1-k\right)\cdot \%R_2
\end{align*}$$
The above is just algebraic manipulation. But there's a deeper meaning.
You were asking about resistor values that are near to each other versus resistor values that are far from each other. Your tables show entries for \$199:\$1 and \$39:1\$ ratios, for example. How does that apply here?
A ratio expressed as \$a:b\$ is no different from saying \$\frac{a}{a+b}:\frac{b}{a+b}\$ and as it turns out \$1=\frac{a}{a+b}+\frac{b}{a+b}\$. But recall that \$1 = k_1^{'}+k_2^{'}\$. So we can just as well write: \$ k_1^{'}:k_2^{'}\$ or \$k:1{-k}\$, in equivalent form.
If we choose things so that \$R_2\ge R_1\$ (and we can always do that), then it follows that \$\frac{R_2}{R_1}:1\$ is the equivalent relationship. Set \$r=\frac{R_2}{R_1}\$, where \$R_2\ge R_1\$ and \$\therefore r\ge 1\$, so that \$r:1\$ applies. (As \$r\to \infty\$, \$k\to 0\$ and as \$r\to 0\$, \$k\to 1\$.) Then:
$$\begin{align*}
\% R_{parallel} &amp;= \frac1{r+1}\bigg[r\cdot \%R_1+ \%R_2\bigg]\\\\
\% R_{series} &amp;= \frac1{r+1}\bigg[ \%R_1+ r\cdot\%R_2\bigg]
\end{align*}$$
Since \$r\ge 1\$, and keeping in mind \$R_2\ge R_1\$, the above makes it pretty clear in each case whether the larger or smaller valued resistor is dominant.
Now, if the % specification is considered Gaussian (integral of Poisson events) then you can use that fact in your computations:
$$\begin{align*}
\% R_{parallel} &amp;= \frac1{r+1}\sqrt{r^{\,2}\cdot \%R_1^{\:2}+ \%R_2^{\:2}}\\\\
\% R_{series} &amp;= \frac1{r+1}\sqrt{ \%R_1^{\:2}+ r^{\,2}\cdot\%R_2^{\:2}}
\end{align*}$$
Let's take the case of a \$39:1\$ and where both are 1% resistors placed in parallel or series. Then we'd compute \$\frac{1}{39+1}\sqrt{\left[ 1\%\right]^2+\left[39\cdot 1\%\right]^2}\approx 0.97532\%\$. (Doesn't matter whether we use parallel or series.) Not much of an improvement.
To relate this to your table, where the author proposes the sum of two \$10\:\Omega\$ resistor with 1% tolerance to make a single \$20\:\Omega\$ resistor, you'd now expect \$20\:\Omega\cdot \pm 0.97532\%\approx \pm 0.1951 \:\Omega\$, which is what you'll see in your table for the case of \$39:1\$.
(However, Self calls this value the &quot;standard deviation,&quot; which absent any clarification implies to me \$1\,\sigma\$. If so, I disagree. More on that in the Summary below.)
For \$4:1\$, find \$20\:\Omega\cdot \frac{1}{4+1}\sqrt{\left[1\%\right]^2+\left[4\cdot 1\%\right]^2}\approx \pm 0.1649\:\Omega\$ and this also matches the table entry you have in your question.
Summary
The only problem I have with Self's discussion is that he appears to assume that a 1% resistor has a single \$\sigma\$ of 1%. I frankly doubt this and instead earlier believed that it is a \$3\,\sigma\$ specification. But I'm no expert on the subject. Perhaps someone will add a reference to an official document from a resistor manufacturer on this topic.
However, that said, the above shows that the blind application of statistical rules (absent any real empirical knowledge of the statistics of modern resistors made by modern manufacturers and sold &quot;unbinned&quot;) combines nicely together with common sensitivity analysis on parallel and series equations to achieve similar results to the use of Excel and hundreds of monte carlo runs in a grand numerical exercise by Douglas Self.
Douglas Self didn't really need to resort to numerical, monte carlo methods to get to the same place. It could have been similarly achieved merely by adding sensitivity analysis. That analysis can also be applied to far more complex combinations and more interesting questions, allowing wide-ranging exploration, if needed.
Best wishes!