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.