Conditional probability with Vigenere Cipher given specific plaintexts and key sizes
The answer to this (thanks to help from fkraiem) was literally included in the first sentence of the question.

I went wrong by assuming that the probability of the occurrence of a 1 or 2 character key didn't weigh into the calculation. My (rather questionable) reasoning was that since the key did in fact have to be either 1 character or 2 characters (i.e. it couldn't be 0 or 3 or more characters), that the 50% probability didn't weigh into it. Yes, that reasoning is in fact both as silly and as incorrect as it sounds.

So, the corrected components of the calculations above are:


  Therefore:
  
  
  Pr[C=ff | M=dd] x Pr[M=dd] = [1/26 x 0.5 + 1/(26^2) x 0.5] x 0.2
  Pr[C=ff | M=de] x Pr[M=de] = [0 x 0.5 + 1/(26^2) x 0.5] x 0.8
  
  
  Ergo:
  
  
  Pr[C=ff] = [1/26 + 1/(26^2)] x 0.5 x 0.2 + [0 + 1/(26^2)] x 0.5 x 0.8
  Pr[C=ff] = 0.004585799