[tzahola]

Assuming that pi is a normal number [1], its digits are uniformly distributed. Therefore the density of a given binary sequence of length N in pi’s binary expansion is 2^-N.

So on average, the offset for a given binary sequence will be on the order of 2^N (I’ve used some hand-waving here). You need N bits for that, you you’ve gained basically nothing.

[1] https://en.m.wikipedia.org/wiki/Normal_number

[jimktrains2]

A friend and I tried to do this in highschool and realized it wasn't so useful except for a cool factor.