You don't know it's alphabetic, you don't know it's an actual English phrase, you don't know how long it is, you don't basically know jack shit to be able to effectively use a dictionary attack against a password like that. You don't know it's a passphrase, basically.
The fact that there's google results for that phrase means it's a useless passphrase.
http://www.banjohangout.org/archive/251990/14
"You always find something in the very last place you look."
Bamn.
Now show me the actual algorithm that generated the passphrase without knowing the cleartext beforehand.
If the attackers knows that your password is constructed in this fashion, then it is trivial to track the password, as we've restricted the search space to a multiple of the number of common English words. The entropy argument only makes sense if the human readable strings are just as likely to be chosen as passwords as other random strings, which is not at all the case.
For example, my /usr/share/dict/american-english contains just shy of 100,000 words. A random word chosen from that set has 16.6 bits of entropy, and four randomly chosen words has over 66 bits of entropy. If anything, XKCD's comic is understating the entropy involved.
This is why some websites assign passwords to users and do not allow users to pick their own custom passwords. The only safe passwords are those generated by machines.
This does not mean that picking words to form a pass-phrase is less secure than picking letters to form a password.
Does less entropy not mean less secure? Or am I just reasoning about the entropy all wrong?
Then all bets are off, but they don't, so we're sorted.
Mind you, my /usr/share/dict has ~ 100,000 words in it. 100,000 5 is around the same order of magnitude as 62 12, which is the number of 12 character passwords of upper and lower letters + digits.
This is probably quite close to what a brute force passphrase cracking software would do as well, and he's not even adding bits for common alterations, such as capitalisation of first letter(s), spaces between words, common substitutions, etc. So the 44 bits estimate is for a software matching exactly this pattern, using exactly this common English dictionary.
Also, I suspect throwing in a single word from another language would greatly increase overall strength, especially if it's an uncommon word.
2048^4 = 17592186044416
2^44 = 17592186044416[0] The original http://www.princeton.edu/~wbialek/rome/refs/shannon_51.pdf
[1] and some evidence that it's still correct http://en.wikipedia.org/wiki/Hutter_Prize
Diceware uses a set of 7776 words. You select words from the list using 5 dice. 5 words, picked using 5 rolls of the set of 5 dice, gives you about 64 bits of entropy.
> A five-word Diceware passphrase has an entropy of at least 64.6 bits; six words have 77.5 bits, seven words 90.4 bits, eight words 103 bits
Because our attacker knows that we've used Diceware, and knows what diceware wordlist we used, and knows that we've used a 5 word passphrase, there are 7776^5 phrases to try. That's 28,430,288,029,929,701,376.
http://world.std.com/~reinhold/dicewarefaq.html
I'd be interested if you think Diceware is broken.
Crackers are using phrases from literature and the bible. Trawling the rest of the internet is not far behind.