Spread of a benign virus in a population over time
To many, many decimal places, the percentage of II in the population is 100%.

To answer this you need to make some assumptions about mating structure. I assume mating is panmictic here for simplicity. Ditto sex ratio, but I'm assuming equal number M/F (and equal risk of environmental infection). Finally, I’m assuming NI and II are at equal risk of soil-borne infection (and infections to II are simply wasted), rather than the environmental reservoir somehow seeking out a fixed proportion of NIs, which I think is probably what you meant. I’m also applying the soil-acquired infection to generation 0 here (so they have a chance to become infected before they mate). Finally, II and NI are more traditionally called $I$ (for infectious) and $S$ (for susceptible) in this type of model, so I've followed that convention below.

Unless I’m missing something the population size makes no difference: the chance of a child acquiring infection does not depend on the number of children, and the background rate of environmental infection is proportional to population size, so it all cancels out. I'll therefore stick to proportions from here on, which means $I$ and $S$ are proportions below, not absolute numbers.

Each generation, the proportion of matings between two $I$s is $I^2$, between two $S$s is $I^2$, and between an S and an I is $2·S·I$. So the proportion of Is in the next generation is $I^2 + 2·I·S$. Or, more simply, $1-S^2$.

So:


$S(0) = \frac{3,999,850}{4,000,000} = 0.9999625$
$S(1) = S(0)^2$
$S(2) = S(1)^2$, which is ${S(0)^2}^2$


More generally,
$S(n) = {S(0)^2}^n$

So at generation 600:

$S(600) = {S(0)^2}^{600}$

$2^{600}$ is a very large number. So it shouldn’t be too surprising that everyone is infected by this point.

We can mess around (technical term) with the equation:

$S(n) = {S(0)^2}^n$

...to solve for (for example) how many generations it takes for 50% of the population to be infected:

$$\begin{alignat}{1}
0.5 &amp;= {0.9999625^2}^n \\
\log(0.5) &amp;= \log(0.9999625)\times2^n\\
\frac{\log(0.5)}{\log(0.9999625)} &amp;= 2^n\\
\log_2\left(\frac{\log(0.5)}{\log(0.9999625)}\right) &amp;= n
\end{alignat}$$

So 50% of your global population is infected by generation 15, only 300 years into the simulation.


  however a few NI persons will still pair with NI mates--for a while


The probability of anyone ($S$, or $I$) mating with an $I$ individual is directly proportional to the $I$ proportion of the population, so the chance of this happening just keeps going down faster and faster.