Any predetermined sequence in the decimal expansion of an irrational number I came up with this question in a random math discussion with my friend. I am wondering if one can always find a predetermined sequence of numbers, such as 123456, 33333, in the decimal expansion of a given irrational number, say, pi. Since the decimal expansion goes on forever, it seemsvery likely that we can hit this sequence eventually. But my friend and I did not have a single idea of how to even start to prove/disprove this proposition.
(And since I study philosophy as well, so (to digress a bit) I feel this question is somewhat similar to the problem of monkeys and shakespeare, that, imagine you have infinitely many monkeys, each of which has a typing machine. Suppose they are all typing and can type on forever, is it possible that they can write a sonnet by shakespeare at some time. ) 
 A: The answer is very negative. Consider for instance the number $\sum_{n=1}^\infty \frac{1}{10^{n!}}$ is irrational (easily shown, but in fact is even transcendental (a bit more difficult to show)), yet clearly it fails to contain lots and lots of predefined sequences. 
This example can be altered slightly to show that for any predetermined sequence of digits (practically in any base) there exists infinitely many (even of cardinality $c$) real numbers that fail to contain that sequence in their expansion. 
A related property is that of normality. A real number is normal if the probability for digit occurrences in its expansion is equal over all digits. For such a number any finite sequence of digits appears with probability $1$ in its decimal expansion (but that does not mean it surely appears in it). Normal numbers are very common, and in fact the measure of the non-normal ones is $0$. 
The analogy with the typing monkeys is a bit weak since the typing monkeys define a stochastic process whereas a real number is a static object. 
