It's certainly a "very long random string" without context but as people have pointed out above, it's actually not a very good password if people adopted this pattern widely (and you said the attacker knows this).
2000^4 = 16000000000000 possible passwords = 1.6E13 = ([A-Z] + [a-z] + [0-9] + [!@#$%^&()])^7.1ish. So, your four words from the 2000 word list are equal to a 7ish character password that looks like "Av#12GH". I'm not sure if you meant that seven characters was "very long" but I wouldn't say it is. Still a very strong password but maybe not as random as it appears to be when the pattern is known.
My point was that adding a character to something like "valve tangle hastens accept" is like adding a couple bits to something like "Av#12GH", and yet people feel like it's accomplishing something valuable. They feel that way because "Av#12GH" looks random, but "valve tangle hastens accept" doesn't.
Obviously the literal length of the string is not strictly relevant, and it was probably inarticulate of me to include that.
Knowledge of the pattern has nothing to do with it. That 2048^4 figure is what I mean when I say such a password is strong, and such a figure presumes the attacker knows what system I am using.
Recall that since the passphrase is randomly generated, that 2048^4 is the true probability of guessing it--all the elements of the set are live possibilities. To compete on equal footing, a seven character password must also be randomly generated.
A password is not necessarily strong simply because it spans a large character set. "Sp1d3r!", for example, may as well be a dictionary word. Raw length "spiderspiderspider" is not necessarily helpful either. Randomness is what you need.
I have a password file here with several hundred passwords just like that (actually, they're all 12 chars with upper/lower case, digits, and "special chars", as chosen and stored by 1Password...)
Joe Public is unlikely to use passwords like that, but I'm 100% sure I'm not the only hackernews reader who does.
2000^4 = 16000000000000 possible passwords = 1.6E13 = ([A-Z] + [a-z] + [0-9] + [!@#$%^&()])^7.1ish. So, your four words from the 2000 word list are equal to a 7ish character password that looks like "Av#12GH". I'm not sure if you meant that seven characters was "very long" but I wouldn't say it is. Still a very strong password but maybe not as random as it appears to be when the pattern is known.