We deal 2 cards (from a 52 cards deck) to 4 players, what is the probability that each player has an ace?

Question

I can't find the math solution to this problem. So, if we have a normal 52 cards deck and deal 2 cards to each of the 4 players what is the probability that each player has an ace?

Answer 1

Let's first count the number of possible distributions by giving pairs of cards consecutively to the players. There will be $\binom{52}{2}$ ways to choose the first player's hand, $\binom{50}{2}$ ways to choose the second player's, $\binom{48}{2}$ for the third and $\binom{46}{2}$ for the last, giving a total of $1896396138000$ possible distributions.

Now, if each player has an ace, this must mean that each player has both a card that is an ace, and a card that isn't. We can therefore count the number of distributions in which each player has an ace, again, by choosing the cards of each in order: First their ace, then their other card. For the first player, this can be done in $4\times48$ ways, for the second, in $3\times47$, for the third, in $2\times46$ and for the last, in $1\times45$ ways, giving a total of $112078080$ ways.

Our final answer will be the ratio of these two numbers, which turns out to about $0.005910056%$.