Let's say there is a deck of 40 cards with 4 aces (just take any 12 non-ace cards of a regular deck for instance)
We deal three cards for each player (so 40-12=24 remain undealt) what is the probability of exactly each player having one ace?
I was thinking that a player has $\frac{4}{40}$ chance of having one ace on the 1st card then the second player has $\frac{3}{39}$ and so on, totalizing $\frac{4!}{39! - 35!}$, but this doesn't work because I just calculated the possibility of each player getting an ace on the 1st card dealt. It could also be that player 1 gets not an ace on the first card dealt but an ace on the second (remember that each player gets 3 cards)
So should I take into account the way the cards are dealt as well? Because for instance concerning the 1st two cards: say the first is an ace for player 1 $\frac{4}{40}$ then the chance of the second being an ace for player 2 is $\frac{3}{39}$. But if the player 1 first card was not an ace, that probability is changed to $\frac{4}{39}$, since there are still 4 aces left on the deck... How could I take all the possibilities in consideration then? Sounds like a pretty boring task since there are $12 \choose 2$ possible configurations, considering the ways the 12 cards are deal to each player... Is there a better way to compute this probability?