5e doesn't have a rules-based cutline below which you can re-roll, then?
The problem with balance in this case is that the supposedly-balanced methods eliminate a great many possible outcomes, which IMO is a cost rather than a feature.
That would certainly serve to put a lot of stats into the mushy middle; I'm no mathmologist but I'd guess the odds of getting anything outside the 8-13 range would be very very low.
You're right. Here are 10 sets of 15d2-12, where the average is 10.5 and the standard deviation is 1.94. (That's tight!) This is comparable to the Classic (3d6) method where the average is 10.5 and the standard deviation is 2.96.
(10, 13, 11, 7, 10, 8) = Unevaluatable.
(10, 10, 10, 12, 9, 9) = 12 points
(9, 12, 14, 13, 8, 6) = Unevaluatable.
(8, 12, 9, 10, 10, 12) = 13 points.
(6, 11, 9, 8, 10, 13) = Unevaluatable.
(11, 10, 10, 14, 9, 13) = 20 points.
(9, 9, 11, 11, 11, 14) = 18 points.
(14, 8, 10, 12, 13, 11) = 21 points.
(12, 11, 9, 13, 9, 9) = 15 points.
(7, 12, 7, 11, 11, 10) = Unevaluatable.
Here are 10 sets of 18d2k15-12, where the average is 12 and the standard deviation is 2.12. (That's less tight, for whaterver math reason.) This is comparable to the Standard (4d6k3) method, where the average is 12.24 and the standard deviation is 2.85.
(8, 10, 8, 12, 12, 9) = 11 points.
(15, 8, 15, 13, 15, 12) = 36 points.
(13, 12, 11, 17, 7, 11) = Unevaluatable.
(12, 7, 15, 13, 13, 9) = Unevaluatable.
(13, 9, 11, 13, 15, 13) = 28 points.
(13, 9, 12, 13, 12, 14) = 26 points.
(15, 11, 12, 8, 10, 12) = 22 points.
(10, 9, 12, 14, 17, 10) = Unevaluatable.
(9, 11, 13, 14, 15, 10) =
27 points.
(15, 14, 15, 12, 12, 12) = 37 points.