Randomizing before Shamir secret sharing? As I understand, the security of key-sharing depends on the cleartext actually being random. 

If I split a secret which contains things like ASCII or even fixed strings (think -----BEGIN PGP PRIVATE KEY-----) what kind of preprocessing is recommended?  

My current thinking is either one time Pad:

$ \text{Shamir}_{(N,M)} \left(\left( \text{secret} \otimes \text{random}
\right),\text{random} \right) \rightarrow M \text{ shares} , N \text{ required}$  

or multiple rounds of AES:

$ \text{Shamir}_{(N,M)} \left(\text{AES}_{key}\left(\text{secret}
\right)^{rounds},\text{key} ,\text{rounds}\right)\rightarrow M \text{ shares} , N \text{ required}$
As I understand, the security of key-sharing depends on the cleartext
  actually being random.


Actually not, the premise of the Shamir Secret Sharing (SSS) scheme is that it is information-theoretically secure.

That means if you require $N$ shares for reconstruction and an attacker has $&lt;N$ shares, the following equation holds:
$$\Pr[M=m\mid C=c]=\Pr[M=m]$$

Which basically says that given the ciphertext you know as much about the plaintext as you know without having the ciphertext.

Or, put differently, as tylo said in the comments:


  He can guess, but he can't check if his guess was right.




As for why this holds (as you may ask yourself now): Notice that SSS basically encodes your secret as the point $(0,s)$ and that the shares are points $(x,f(x)), x\neq 0$ of a (nearly-random) polynomial of degree $N-1$ over some field (usually $\mathbb F_p^*$). Now imagine as an example you're given 2 points of a degree-2 polynomial (which would be N-1 shares). For every possible $s$, you can now find a third point such that $(0,s)$ is on the polynomial. Put otherwise: There's no additional constraining condition that would help you narrow specific plaintexts down as more likely.