There are only very few ways for beer A, say, to have a total rank of 4 or less: either (1,1,1), (2,1,1), (1,2,1) or (1,1,2) (I'm listing the rankings given to beer A by the three drinkers). So that the probability that beer A has a total rank of 4 or less is... (the answer is not 4/(4!)^3...)

And you can notice that it is impossible for two beers to have simultaneously a total rank less than 4 (each of these beers needs to have at least two '1'...). Therefore, we have disjoint events, so that the probability that "some" beer has a total rank less than 4 is four times the previous probability (either beer A, or beer B, or beer C, or beer D has rank less than 4, and these events are disjoint and have same probability).