You only have half a dice of difference each time.
If you've got 1Ad it's 1 chance out of 2
with 2 Ad its 1 + 1/2 so still "only" half a dice missing
and so on (4Ad would be 2+1+1/2 ...)
So we can assume that it theorically tends toward x hits for x dice
and is on average around x-0.5 hits for x dice
Edit : But I think we all agree on the problem being the dispersion of beam and not its average.
If you've got 1Ad it's 1 chance out of 2
with 2 Ad its 1 + 1/2 so still "only" half a dice missing
and so on (4Ad would be 2+1+1/2 ...)
So we can assume that it theorically tends toward x hits for x dice
and is on average around x-0.5 hits for x dice

Edit : But I think we all agree on the problem being the dispersion of beam and not its average.