>Wouldn't these two questions have different answers?
Yes, though in practice they're close.
To answer the actual question asked. We can say that there is a length n, such that all sequences S, of length n, cannot be contained in our total bits transferred, B. This is a consequence of the pigeonhole principle.
Naively, this would be B = n * S, and S = 2^n. So B = n * 2^n. Note that we can actually compress things more though, if we assume that in the worst case every window in the internet bit-corpus is unique.
That this is possible isn't immediately obvious, but consider
01, 00110, 0001011100, 0000100110101111000, 000001000110010100111010110111110000 (which, interestingly, isn't in OEIS, but related sequences: https://oeis.org/A052944 and https://oeis.org/A242323 are).
These bitstrings contain, perfectly overlapping, every bitstring of length n, for n from 1-5. In general, it takes 2^n + n - 1 bits to convey every possible bitstring, if you manage to overlap them perfectly (if someone can prove this, please do. I thought it was grey code/hamiltonian path over hypercube related, but I unconvinced myself of that).
EDIT: Someone else mentioned De Brujin sequences, which these are. And they are based on hamiltonian paths, although not over hypercubes :(. And my sequence is on OEIS, as https://oeis.org/A166315, just converted to decimal. /EDIT
So the real answer is just B = 2^n + n - 1, but for n > 15 or so, n - 1 is so small we can ignore it. In other words, our length n is just log(B). Given the assumption in the article of 3.4067e22 bits, the base 2 log of that is...74.85. This is exactly 1 more than what the article says is the point where you'll have seen half the messages. This isn't a coincidence.
Which raises an interesting question that I leave as an exercise to the reader:
Why is the length for which you cannot have conveyed all bit strings exactly 1 bit more than the point where you've conveyed about half of them?
You’re giving the range for a 50% to 100% chance of non-transmission.
But in the spirit of the original question, I’d suppose we want to find n such that the probability of non-transmission is 2^-n, and the expected number of such transmissions is 1? Is that number similarly close?
Yes, though in practice they're close.
To answer the actual question asked. We can say that there is a length n, such that all sequences S, of length n, cannot be contained in our total bits transferred, B. This is a consequence of the pigeonhole principle.
Naively, this would be B = n * S, and S = 2^n. So B = n * 2^n. Note that we can actually compress things more though, if we assume that in the worst case every window in the internet bit-corpus is unique.
That this is possible isn't immediately obvious, but consider
01, 00110, 0001011100, 0000100110101111000, 000001000110010100111010110111110000 (which, interestingly, isn't in OEIS, but related sequences: https://oeis.org/A052944 and https://oeis.org/A242323 are).
These bitstrings contain, perfectly overlapping, every bitstring of length n, for n from 1-5. In general, it takes 2^n + n - 1 bits to convey every possible bitstring, if you manage to overlap them perfectly (if someone can prove this, please do. I thought it was grey code/hamiltonian path over hypercube related, but I unconvinced myself of that).
EDIT: Someone else mentioned De Brujin sequences, which these are. And they are based on hamiltonian paths, although not over hypercubes :(. And my sequence is on OEIS, as https://oeis.org/A166315, just converted to decimal. /EDIT
So the real answer is just B = 2^n + n - 1, but for n > 15 or so, n - 1 is so small we can ignore it. In other words, our length n is just log(B). Given the assumption in the article of 3.4067e22 bits, the base 2 log of that is...74.85. This is exactly 1 more than what the article says is the point where you'll have seen half the messages. This isn't a coincidence.
Which raises an interesting question that I leave as an exercise to the reader:
Why is the length for which you cannot have conveyed all bit strings exactly 1 bit more than the point where you've conveyed about half of them?