Theoretical uncertainty of a circuit's total resistance when made entirely of resistors My question in short(ish) is: Will the fractional uncertainty of a circuit made entirely of resistors with equal fractional uncertainties be the same as the fractional uncertainty of those resistors. 
For example, suppose we have three resistors with a $5\%$ tolerance in parallel. The values of the resistors is completely arbitrary. Will the fractional uncertainty of $R_{total}$ be $5\%$ as well?
Suppose instead of three resistors we have $n$ resistors instead. Does it still hold up? Theoretically and in practice?
What drives this question is the thought that even if mathematically the fractional uncertainty should be constant I can't imagine that a circuit made of 100,000 resistors will behave even remotely close to how one would expect it to. It just seems that at a certain number of variables in your circuit you can no longer accurately describe the situation with simple circuit rules. Something more complicated must have to happen right?
 A: If you slap together a large number of resistors in a random fashion, then you will not be able to analyze it with simple circuit rules, but this happens even with no uncertainty.
The short answer is that the series rule and the parallel rule are not sufficient for complicated circuits and if someone told you otherwise, they were wrong.

I can't imagine that a circuit made of 100,000 resistors will behave even remotely close to how one would expect it to. It just seems that at a certain number of variables in your circuit you can no longer accurately describe the situation with simple circuit rules. Something more complicated must have to happen right?

Your imagination is on track, but it is not a function of the uncertainty, rather it is a function of how they are arranged.
Specifically, even will no uncertainty, it is not the case that you can analyze the situation with simple circuit rules.
In particular the parallel and series rules do not allow you to model an arbitrarily complicated circuit. Even with no uncertainty.
You can find the effective resistance by applying an arbitrary voltage and then using Kirchoff's circuit analysis to find the current. If it is all resistors, the current will be proportional to the voltage so you have an effective resistance.
Now we can address uncertainty. If you write each resistance as variable then you might be able to solve Kirchoff's laws algebraically as an expression of the individual resistances. Then if you use interval arithmetic to compute the effective resistance then you should get a range.
I see no reason to think the range will be symmetric about the point you would have gotten if you'd put the rated resistance in, which I think is the true heart of concern related to your original imagination. However the fractional uncertainty is probably less than or equal to the largest fractional uncertainty of your components.
Another concern is that no quality control is perfect so if you wired up trillions of resistors one of them might be outright broken above and beyond the rated uncertainty.
A: Let's say we have a bunch of resistors.  Let's call the "ith" resistor $R_i$ with fractional tolerance $t_i$.  In that case, each resistor's actual value is somewhere in
$$R_i \pm R_i t_i$$
If we connect N resistors in series, then the total resistance is somewhere in
$$R_{total} = \sum_1^N (R_i \pm R_i t_i)$$
$$R_{total} = \sum_1^N R_i \pm \sum_1^N R_i t_i$$
Now, the fractional tolerance of our overall resistor must be the variation divided by the mean, or
$$t_{total} = \frac{\sum_1^N R_i t_i}{\sum_1^N R_i}$$
If they are all the same resistors, so $R_i = R$ and $t_i = t$, then
$$t_{total} = \frac{N R t}{N R}$$
$$t_{total} = t$$
Ergo, if you connect a bunch of 5% resistors in series, the combination of them acts like a 5% resistor.
If they're in series, it's a different story.
$$R_{total} = \frac{1}{\sum_1^N \frac{1}{R_i \pm R_i t_i}}$$
I'm still thinking about how to find the overall tolerance in this case.  If it turns out that the parallel tolerance is also the same as the individual tolerance, then I think we can conclude that an arbitrary combination of resistors will have the same tolerance as the individual ones.
A: No. As long as there is not a biased error, the deviations from the ideal value will tend to cancel each other by simply adding when in series (The ideal - and + some error) or by the average when in parallel. The more resistors the better it gets because of the larger sample size for averaging or adding. A mental picture of the string of resistors or the bundle in parallel should be convincing.
Using @Brionius notation, for series resistors (I leave out the +- and assume the t's can be positive or negative)
$R_{total\; =\; \sum_{1}^{n}{\left( R_{i}\; +\; R_{i}t_{i}\;  \right)}=\; R\sum_{1}^{n}{\left( 1\; +\; t_{i} \right)}}=\; nR\; +\; R\left( 0 \right)$  
Also, it is not hard to try with 2, 4, 9, 16 resistors and see if it is converging. I chose those numbers because you might also see a relation to the square root of the number of resistors. I don't see it in the equations written so far, but who knows?
For the parallel case, you can simplify with an approximation for large n and t<<1.  $\frac{1}{1+t}=1-t$  where I don't have a symbol for "approximately" so used an = sign. As you can see in some other comments, you get R total is R/n and again t goes to zero, or I would say the error decreases as n increases.
Unfortunately, resistors do not tend to be like this. Today a batch is pretty consistent in the deviation from ideal. There is probably someone here who does quality control for a living, we just need to get them to read this question.
A: 
I can't imagine that a circuit made of 100,000 resistors will behave even remotely close to how one would expect it to.

A two-terminal network of 100,000 resistors will behave as a single resistor.
If you knew the exact values of the 100,000 components, you could calculate the exact value of the network as a whole.  If you only knew a probability distribution for each of the component's values, then you could compute a probability distribution for the network's value.
I'm not saying it would be a simple calculation, but it definitely would be possible.  You would do it by applying either Norton's Theorem or Thévenin's theorem

But you don't need to think about a hundred thousand resistors to answer you question.  Just think about a one Ohm resistor in series with a one million Ohm resistor.
If the values are exact, then the total resistance of the network should be 1,000,001 Ohms.
But what's the percent change in the network's value if the value of the 1,000,000Ω changes by 5%?  And what's the percent change in the network's value if the 1Ω resistor's value changes by 5%?
They're different!  There is no answer to your question that does not take into account all of the the different values and they way in which they are all connected.
