Suppose we have a binary symmetric channel with $p=\frac{1}{3}$; that is, a communications channel in which each bit is flipped with independent probability $\frac{1}{3}$. I know that there is a code such that, in the (highly probable) event that no more than $p$ of the bits are corrupted, we can guarantee recovery of any message we put through... so in particular, we can certainly recover a message that is correct in $\frac{1}{4}$-th of its positions, reducing the error introduced by the channel.

Now, let's ask this for a useless channel - a binary symmetric channel $C$ with $p=\frac{3}{4}$. Obviously, this can't carry any information, but bear with me. Is there a code such that, if we feed an encoded message through the channel $C$, we can recover (with high probability) a version of the message with at most two-thirds of the bits flipped? That is, is it possible to reduce the error (without actually recovering any information) even in a capacity-0 channel? Obviously, we can't get our error down under $\frac{1}{2}$, since that would violate the channel capacity theorem... but can we get as close as we like? Or is this generally impossible?

Stating this final question: For fixed $\epsilon>0$ and $\delta\in(\frac{1}{2},1)$, is there a code that can implement a channel with symmetric error no more than $\frac{1}{2}+\epsilon$ over a channel with symmetric error bounded by $\delta$? (For what it's worth, I don't care about the rate, just the error bounds.)

I welcome any suggestions that would clarify the question as explained above - I haven't done much coding theory, so I'm not used to the terms.