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 expan

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 https://en.m.wikipedia.org/wiki/Normal_number