Since all 12 possible choices of die can occur and they all have the same probability, choosing first (at random) gives no advantage, so the probability of A winning is 0.5 and of B winning is 0.5 (a draw is not possible).

Now if A can deliberately choose which die they will use and B make their choice knowing what A has choosen we have a different problem. However whatever A chooses B has a choice that restricts the probability of A winning to 1/3 (assuming I have done the sums correctly).

CB