You're using the two sample chi square correctly so far as I can tell. However, combining the two p-values via Fisher's Method isn't quite right here, because the tests aren't independent. Since all the cards have to end up somewhere in the deck, the location of the first card in the shuffled deck is not independent of the location of the second.

The most obvious solution is to do one test from the beginning by working directly with the simulated joint distribution. So when you simulate/tabulate the data, record the probabilities for all 60 cards ending up in your hand, then calculate the marginal distribution of cards from the first 24 and last 24 by summing over that. This directly gets you a Monte Carlo approximation to the appropriate distribution, so you can calculate a single p-value via the chi square test.


Because the deck is fairly large however, the dependence between the cards will be fairly weak, so this won't change your answer very much. Further, because the dependence is by necessity negative, your current method is very slightly conservative, which, if you're going to be wrong, is the direction to be wrong in.