Fun question/"brainteaser"

[CRGreathouse]

I don't think this problem really counts as a brainteaser, because I don't know the answer.

Consider a sequence like A097614 (http://www.research.att.com/~njas/sequences/A097614) which works as follows:
Given a constant (pi in this case), find the first position in the constant with a decimal "0". This is a1. Then find the first position in the constant with a decimal a1; this is a2, and so on.

If the constant were 0.11777777770... instead, the sequence would be cyclic:
0, 11, 1, 1, 1, 1, ...

What is the probability that such a base-b sequence is eventually cyclic on a random constant? Here, "random constant" means that each decimal place to the right of the decimal point has a 1/b chance of taking each value in 0, 1, ..., b-1.

[Dodo]

My wild guess would be that the probability for the sequence to be infinitely long is 0. That it will either terminate abruptly (due to some string not existing in the decimals of pi), or that at some point a string will be found at one of the positions 4, 41, 415, 4159, 41592, ..., bringing you back to position 2.

Edit:
On second thought, I'm not so sure. The list of positions I mentioned is countable, while the set of all possible search strings could, for all we know, be close to the set of all possible finite strings of digits, which is awfully close to the power set of N, thus way bigger. No? Maybe?

[CRGreathouse]

If the chance of finding a given n-digit sequence at any position in the random constant is b^-n, you'd expect to find n somewhere around position n. But it's much easier to slip up a digit then down one, and that quickly moves you many places forward.

I'm not sure of my intuition on this one either. It seems so likely that eventually you'd become cyclic, and yet the chance drops every time the number of digits increases; I'm not sure which would outpace the other.