Following is a partial answer and I am hoping that someone can help me prove that it works.
Let us label the cups from 1 to 13.
Before the game begins, the magician and the apprentice write down all the (13*12)/2 = 78 combinations that I can choose, on a blackboard.
To number 1, they allocate, any arbitrary 4 numbers: a,b,c and d. The only things they make sure are that none of a, b, c or d = 1 and ab, ac, ad, bc, bd and cd are all present on the board.
They, then erase these 6 numbers and then to the number 2, allocate another 4 arbitrary numbers, a', b', c' and d' . Again, the only conditions while choosing these 4 numbers are that none of a', b' , c' and d' = 2 and all their 6 combinations (a'b', a'c', a'd', b'c', b'd' and c'd') are present on the board. They then erase these 6 numbers ( a'b', a'c', a'd', b'c', b'd' and c'd') from the board and move on to to allocating numbers to 3, and so on.
The idea is that if the apprentice calls out 1, then the magician will guess a, b, c and d. If the apprentice calls out 2, then the magician will guess a', b', c' and d', etc.
The thing I am unable to prove though is that the magician and his apprentice will always be able to allocate 4 numbers to all the 13 numbers using this strategy and will never come to a point where they can no longer find 4 numbers to allocate to a number.
For example, let's say that they manage to find 4 numbers for 1 to 6. Then they come to 7 and find that no 4 numbers e,f, g, h exist such that ef, eg, eh, fg, fh and gh exist such that are all present on the board .
Intuitively, I feel strongly that this strategy of theirs will work but am unable to prove it.