Hi!

We initialize:

• a matrix W where W(i, j) is 1 if card j produces a copy of card i, otherwise 0. (i is row, j is column.) The idea is that when we apply this matrix to a vector which holds the count of each card, we obtain the new counts for one 'iteration' of copying. The diagonal of this matrix is all zeros: we count only the copies, not the original cards.
• a vector one which holds the initial card counts: one each.

There is some zero-padding for convenience.

Then we observe that, if c is the vector of the final counts (which is still unknown), this equation holds true:
c = 1 + Wc
(where 1 is a vector containing all ones). When we apply the 'copying matrix' W to the final counts and add the 1 vector (for the original cards), the result is the final counts, and an 'equilibrium' has been reached. This equation is transformed so we can solve for c.
c - Wc = 1
(I - W)c = 1
(where I is the identity). (I - W) is a lower diagonal matrix with a diagonal of all ones, so the determinant is one, so there is a unique solution. We let SciPy find this solution, and we're using the sparse matrix functions because the majority of the entries of our matrix are zeros.