I am trying to get to know probability a little better since it's a weak point for me and I was wondering what other ways there were to intuitively think about the problem of finding the probability that 4 bridge players each have exactly one ace after being dealt a deck of 52 cards.
My solution started with that there are $4!$ ways of distributing the aces initially, and then subsequently there are $48 \choose 12$ ways of distributing the rest of the first hand, $36 \choose 12$ ways of distributing the rest of the second hand, and $24 \choose 12$ ways of distributing the third hand, and one way to distribute the last hand. Thus there are $4! *$ $48 \choose 12$ $*$ $36 \choose 12$ $*$ $24 \choose 12$ ways of the players having exactly one ace.
The total sample space, however, is counted for by first giving the first player his hand ($52 \choose 13$), the second player has hand from the remaining deck, ($39 \choose 13$), etc.; so the sample space is $ 52\choose 13$ $*$ $39 \choose 13$ $*$ $26 \choose 13$.
The final answer after simplifying is that the probability that each player receives exactly one ace is $\frac{4!48!13^{3}}{52!}$
I think this is correct (correct me if my solution is wrong), but what are the other ways of thinking of this problem?