Mathematical calculation to determine probability of observing the character space used for a set of string inputs
Well, assuming that the router selects each character of the password in a uniformly and independently distributed fashion, then the probability of an individual character not being a specific character $c$ is:

$$1 - 1/62$$

(where the alphabet size is 62).

Hence, if we generate $n$ characters, the probability $c$ never occurs is:

$$(1 - 1/62)^n$$

Now, we have 62 possible values of $c$, and we want the probability that all of them occur somewhere to be high (say, $&gt;0.99$), and so we have:

$$62 (1 - 1/62)^n &lt; 1 - 0.99$$

Evaluating this, we get $n &gt; 537$, or 67 8-character passwords.

The exact value you obtain will depend on the size of the alphabet (which you might not know precisely when you start), and the desired failure probability.  In addition, this is pessimistic in the sense that if you see every alphabetic character other than n, it's plausible that n can occur (and you just didn't happen to see it).  On the other hand, the router may be programmed to avoid the 1, I, l characters (because they may be easily confused in some fonts), and so that might not be quite so dreadfully pessimistic.

Pedants will note that this analysis isn't precisely correct (as the probabilities aren't independent; each character of the password is some $c$, and so multiplying the nonoccurance probability of each individual character by the number of characters isn't actually justified), however I believe that it's close.