Strictly, is length(key) >= length(message) wrong for one time pads? It's a common adage that for a perfectly secret one time pad, length(key) >= length(message) must hold.  But that's wrong isn't it?  At least very sloppy maths. Isn't the strict mathematical requirement that:-

entropy(key) >= entropy(message)?

An example.  Consider a OTP used for a message in the English language.  From various studies we guesstimate that the information entropy rate of English is ~1.6 bits/character.  Lets not quibble about the exact value as anything &lt;8 will suffice for this example. Now take a standard OTP key of random 0-255 value bytes.  That's an entropy of 8 bits/byte.  Therefore for long English OTP messages, the key can be one 1/5 of the message length and still guarantee perfect secrecy. Essentially the message could be compressed five fold and then XORed with the much shorter binary key.

And the equation is reversible.  If you encrypt binary data (say reasonably in-compressible compiled code) with a basic English alphabet, you'd need five times the length of key material compared to the message.

Clearly as the message alphabet increases, then the mathematical requirement tends towards the original adage.  But isn't my general case more accurate?

Am I wrong/confused/hungry?
Your confusion stems from a misinterpretation of a formal statement about the limitations of perfect secrecy. Let us first recall the definition

Perfect Secrecy.
An encryption scheme $(\mathsf{Gen}, \mathsf{Enc}, \mathsf{Dec})$ with message space $\mathcal{M}$ is perfectly secret if for every probability distribution over $\mathcal{M}$, every message $m \in \mathcal{M}$, and every ciphertext $c \in \mathcal{C}$ for which $\Pr[C = c] &gt; 0$:
$$Pr[M = m | C = c] = Pr[M = m].$$

Note that there is no mention of bits or bitstrings here and "length of the message" is not really well defined for a general message space.
The statement that is often informally paraphrased as 


  The key must be at least as long as the message.


therefore doesn't really make any sense here. This is where you seem to be confused. The formal statement that can be proven about perfectly secret encryption schemes is the following:

Limitation of Perfect Secrecy.
If $(\mathsf{Gen}, \mathsf{Enc}, \mathsf{Dec})$ is a perfectly secret encryption scheme with message space $\mathcal{M}$ and key space $\mathcal{K}$, then $|\mathcal{K}| \ge |\mathcal{M}|$.

Note that this is only a statement about the size of the key space! It makes no mention of any length.
Now, the most well known perfectly secret encryption scheme is the one-time pad, which is most commonly defined over bitstrings¹ with a message space $\mathcal{M}=\{0,1\}^n$ for some message length $n$ as well as a keyspace $\mathcal{M}=\{0,1\}^\ell$ for some key length $\ell$. In this specific case case $|\mathcal{K}| \ge |\mathcal{M}|$ indeed implies that $\ell \ge n$.

In your specific case however you used a different message space, in particular you defined $\mathcal{M}$ to be the set of "messages in the English language" which is something we would need to define more formally. We can define this for example as "any concatenation of words in the English language with total length at most $\ell$ characters".

In this case, clearly $|\mathcal{M}| &lt; 2^{8\ell}$ and therefore obviously you do not need $|\mathcal{K}|\geq 2^{8\ell}$ but the definition never claimed otherwise. What you are defining here is simply not the one-time-pad, but a different encryption scheme with a different message space. And the same condition on the key space still holds. It is just that this does not necessarily correspond to an immediate lower bound on the length of the keys when encoded as bitstrings.



¹Technically the one-time pad can be defined over strings over any finite alphabet $\Sigma$ as long as we can define an order over $\Sigma$.