No because it's not a random walk. Cards in this problem are not put back in the deck after they are drawn, therefore the probability of going +1/-1 depends on

No because it's not a random walk. Cards in this problem are not put back in the deck after they are drawn, therefore the probability of going +1/-1 depends on how many cards of each color were drawn so far. It's still markovian though, which means it's straightforward to find a solution (unless you want to go crazy with optimizations as OP suggests).