Try calculating the $L_2$ norm. It's going to be much bigger for distribution #2. If the histogram is $h_i$, normalized so that the $L_1$ norm is $1$, i.e. $\sum_i h_i = 1$, then the $L_2$ norm (squared) is $$L_2^2 = \sum_i h_i^2.$$ One problem you're going to encounter is that the noise is proportional to $\sqrt{n}$, so you should normalize (divide) the $L_2^2$ norm by $n$ (I think). To make sure you have the correct normalization, sample your input and compare the normalized values; they should be about the same if you take all points or, say, only half of them (try even taking only, say, a tenth).