Okay, time for another math post. Here's my question about pi:

We all know, of course, that pi can't repeat because it's irrational (transcendental, even). But that just means that it can't repeat the same string over and over and over forever. There's nothing to prevent any particular string of symbols from repeating *once*, starting right after the string ends. So the question is, for some *n*, do the first *n* digits of pi repeat (once, in positions *n+1* through *2n*)?

Here's the only progress that I've made so far. For a random list of numbers (and I'm pretty sure pi actually *hasn't* been proven to be totally random, but I could be wrong about that...please let me know if I am), the probability that a specific *n* consecutive digits will appear at a given location is clearly 1/(10^*n*). So for that random string, the probability that the first *n* digits repeat as described above is 1/(10^*n*). Call that event *E_n*. I think that the set of events {*not(E_n)* for all positive integers *n*} is a set of independent events—meaning that if one set of *n* digits doesn't repeat, then it has no bearing on whether any other set repeats or not.

Therefore, the probability of *none* of the events *E_n* occuring is the product of the indvidual events:

[1—1/10][1—1/100]...[1—1/(10^*n*)]...

which I'll call P, and which doesn't equal 0, since none of the *factors* (thanks vlorbik) are 0;

and the probability that a string of *n* digits *will* repeat for some *n* must be:

1 —P.

But I sure don't know how to compute that product, and since so many digits of pi are known, I can't decide if this calculation should apply to pi or not.

Any ideas? Am I way off base here?

UPDATE: my spreadsheet tells me that P is surely .9, which makes 1-P=.1, which sounds like a semi-reasonable probability that a random string of digits has that property. but nothing definitive.