Question: "Show to leading order, the probability of (averaged) bit error $p_b$ goes as $9f^2.$
I've been studying David Mackay's solution in the book where he argues something like the bit error is dominated by two flips in the received vector. The optimal decoder will induce a 3rd bit flip when faced with 2 bits flipped by noise. Meaning the decoded vector differs in 3 bits from the transmitted vector. He argues the chance of a bit being flipped in the decoded vector is 3/7 regardless of whether it is a parity bit or a source bit. And the chance of a bit being flipped is 3/7 times the probability of block error. He suggests for our purpose of bit error, we're only interested in the source bits not parity bits being flipped. So far i'm with the Author.
But the next part he loses me when he argues that
Are parity bits or source bits more likely to be among these three flipped bits, or are all seven bits equally likely to be corrupted when the noise vector has weight two? The Hamming code is in fact completely symmetric in the protection it affords to the seven bits (assuming a binary symmetric channel). [This symmetry can be proved by showing that the role of a parity bit can be exchanged with a source bit and the resulting code is still a (7, 4) Hamming code; see below.] The probability that any one bit ends up corrupted is the same for all seven bits. So the probability of bit error (for the source bits) is simply three sevenths of the probability of block error.
How does the symmetry allow us to this conclusion? I understand the symmetry exists but why does it mean it doesn't matter that there are 4 source bits and 3 parity bits and we don't need to take into account that there are more source bits than parity bits, which if the noise acted on only the parity bits we'd still have no bit error in that case? Or a mixture of.