When 100 systems have been discovered the chance of connecting to an existing system is still 100-1
But the chance of a loop existing somewhere in that 100 systems is O(1). Intuitively, this can be understood by thinking about the next 100 systems. You'll be rolling against odds of at least 100-1 100 times, so the odds are that at least one of those rolls will come up a winner. For the first 100 systems, the odds will be lower than 100-1, but the same general principle applies - systems 51-100 will give you at least roughly 1-in-4. This is very similar to the "compare birthdays" cocktail game
http://en.wikipedia.org/wiki/Birthday_problem - the odds of two people in a room having the same birthday approaches unity for a surprisingly small number of people, because it doesn't matter which birthday gets a match. The reason that this maps to the birthday problem is that the "new" end of a link can be thought of as that link's "birthday".
As discussed in the article, the true probability of no matches occurring in m links with N systems is roughly m!*(N choose m)/N^m (up to shifts by 1 or two in the value of m). Also as discussed in the article, the easy way to see when the probability of a match approaches unity is to compare N to (m choose 2) = m*(m-1)/2. When m ~ sqrt(2N), these will be roughly equal and the odds are high of a match somewhere.
So if you want to be able to generate roughly M systems before the odds are good of getting a match, you need roughly M^2 total systems. I would choose N at either 1,000,000 or even 1,000,000,000 (assuming Steve didn't use 16-bit ints for system index). I wouldn't go much above 1,000,000,000, since that would overflow a 32-bit integer.
John