Such a password scheme provides much less than 44 "bits" of entropy. Considering the use of 4 randomly chosen words from the c.170000 english words in general use, means we can guess the paraphrase in around 2^22 tries - even less than "Tr0ub4d0r3&".
EDIT: I'm totally wrong, it's more like 2*10^22 ... oops!
EDIT: I'm totally wrong, it's more like 2*10^22 ... oops!