I claim that no arrangement is possible in the case with cards ace through four.
_A_A_A_A_ (_ is any other card)
must be part of the arrangement. Now consider two cases:
the row starts and ends with aces - this is obviously not
possible, since that arrangement would only hold 7 cards, and we
there are cards to the left and right ...
A similar answer to Quintec's (asserting that it is impossible):
Consider that you only have 12 cards that are not a 4, and that you need 12 total cards between the 4s. So the entire arrangement must look like
4 _ _ _ _ 4 _ _ _ _ 4 _ _ _ _ 4
As per Quintec's observation, somewhere in there we must have
_ A _ A _ A _ A _
But you can see that there ...
As claimed above by other users, the arrangement is imposible with four sets of four cards. The earliest solution occurs for four sets of cards 1 to 24, for which there are three solutions. These are known as Langford Quads and are described at John Miller's excellent blog on Langford's Problem.
These are Richard Noble's three solutions.