There isn't a single correct answer to this. It actually depends on the way that the
get(position) method is implemented.
If the implentation always traverses the list in one direction (either forward or backward) then the worst case is N steps.
If the implentation traverses the list from the one end or the other depending on the value of the position, then the worst case is N/2 steps.
But there is a catch. In order to do it the "smart" way, the list implementation has to keep a record of the list's size. Without this, it must calculate the size ... which involves traversing the entire list.
(Oracle / OpenJDK
LinkedList implementations are specified as traversing from either end in all Java versions that I looked at. However, you have not said if you are talking about the standard implementations.)
The average case is also not quite as straight forward as you would think. It actually depends on the distribution function for the positions you look up. If the distribution function is "flat"; i.e. all positions are equally probable, then the average number of steps is half of the worst case. Otherwise ... not.