# Most likely error with hamming distance

I have a basic question about hamming distances, something confuse me in the book I’m reading about it.

Let’s assume we have a codeword $$y$$ that received an error $$e$$. Thus we had the following event:

$$y_0 \rightarrow y’=y_0+e$$

In Nielsen & Chuang, page 449, he says:

Provided the probability of a bit flip is less than 1/2, the most likely codeword to have been encoded is the codeword y which minimizes the number of bit flips needed to get from y to y’,that is,which minimizes wt(e)= d(y, y’).

Where $$d(a,b)$$ is the Hamming distance between $$a$$ and $$b$$, and $$wt(e)$$ is the Hamming weight of $$e$$, which is $$d(e,0)$$.

My question is the following:

I agree that if there is more probability of no bit flip than the probability to have one, then the most likely codeword is the closest one from $$y’$$, thus a codeword $$y$$ minimising $$d(y,y’)$$.

However, as he writes $$d(y,y’)=wt(e)$$, he seems to assume that the closest vector from $$y’$$ is necesseraly the one we encoded. Which for me is not true in general, the error could have put us closer to another encoded word than the one we encoded right ?

Where is my misunderstanding here ?