Finding 2013 in Pi
johndcook.com
johndcook.com
The probability any sequence of length d is found in N digits of pi is 1 - 1/exp(N*0.1^d) (Poisson distribution for approximating the binomial). Then the limit as N approaches infinity is 1 for any finite d.
A simple example: 2 binary digits in binary sequences of length 3. There are eight binary sequences of length 3:
000
001
010
011
100
101
110
111
3 of those contain '00' but 4 of them '01'. Reason for the discrepancy is that one of those with '00' has 2 overlapping occurrences, but is counted only once. You get this as soon as overlap can occur, i.e. when the sequence to be found starts with x digits that it also ends with.Of course, none of this matters, especially not when d << N, which it will be if N goes to infinity.
Also, the mathematical term is 'normal number' (http://mathworld.wolfram.com/NormalNumber.html), and we do not know whether pi is normal.
One way to see it is this: The probability of picking any particular point in the interval is 0. But that doesn't mean that picking that point is impossible. _Some_ point has to show up when you pick one at random.
In fact, no one has proved that any particular number is
normal. However, we do know that almost all numbers are
normal. That is, the set of non-normal numbers has
Lebesgue measure zero.