Probability Question with Deck of Cards - three players, 5 cards each, P(at least one person has exactly two aces) Three players are each dealt, in a random manner, five cards from a deck containing 52 cards. Four of the 52 cards are aces. Find the probability that at least one person receives exactly two aces in their five cards.
Effort: I think the probability of one person getting two aces would be $\dfrac{3*(4\ _nC_r\ 2)*(48 \ _nC_r\  3)}{(52 \ _nC_r\  5)}$, but assuming that the first person gets two aces, to find the probability of the second person, I think this might require some sort of multinomial coefficient, I am just unsure of what that is!

Here's another way, taking the aces to be  prize winning #s in a lottery.
Imagine the 52 numbers are divided into 4 groups $A-B-C-D\;$ of $\;5-5-5-37$,
and we want to count either A alone has exactly $2$ aces, or $A-B$ have 2 aces each.
There aren't too many patterns to count, we can just add up.
$2-1-0-1\text{ and } 2-0-1-1 : 2{5\choose2}{5\choose 1}{37\choose1}$
$2-1-1-0 : {5\choose 2}{5\choose1}{5\choose1}$
$2-0-0-2 : {5\choose 2}{37\choose 2}$
$2-2-0-0 : {5\choose 2}{5\choose 2}$
We counted only $A$ and $AB$, so we need to multiply by $3$
$Pr = \dfrac{3\left[2\dbinom{5}{2}\dbinom{5}{1}\dbinom{37}{1}+ \dbinom{5}{2}\dbinom{5}{1}\dbinom{5}{1}+\dbinom{5}{2}\dbinom{37}{2}+\dbinom{5}{ 2}\dbinom{5}{2}\right]}{\dbinom{52}{4}}$
You might find this more straightforward.
