Probability that at least one player out of two gets 4 cards of the same denomination I came across this question, and I couldn't quite get an answer:

A deck of 52 cards, and two players: Adam and Bart, each gets 5 cards (at random). Calculate the probability that at least one of them has 4 cards of the same denomination.

I started like this:

$\Omega = \{(\square,\square,\square,\square,\square)|\square \in \{cards\}\}.$

$|\Omega| = {52 \choose 5}$
 A: the event that Adam gets 4 of the same denom. 
B: the event that Bart gets 4 of the same denom.
$Pr[A] = \frac{|A|}{|\Omega|} = \frac{13\times48}{{52 \choose 5}}$ 
$Pr[B] = \frac{|B|}{|\Omega|} = \frac{13\times48}{{52 \choose 5}}$ 
(13 options to choose the denomination, then we have 1 options for the 4 cards, then we choose any of the rest 48 for the remaining 5th) 
Pr[at least one] = 1 - Pr[none of them] = $1-Pr[\bar A \cap \bar B]$
$Pr[\bar A \cap \bar B] = Pr[\bar A] Pr[\bar B|\bar A]$ and here lies my problem, calculating $Pr[\bar B|\bar A]$ was too complicated.

How can I calculate that? or maybe a different approach to solution?
A: Let $A$ denote the event that Adam gets $4$ cards of the same denomination and let $B$ denote the event that Bart gets $4$ cards of the same denomination.
Then:
$$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=2P\left(A\right)-P\left(A\right)P\left(B\mid A\right)=$$$$P\left(A\right)\left(2-P\left(B\mid A\right)\right)=$$$$13\times\frac{\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}\left(2-11\times\frac{\binom{4}{4}\binom{43}{1}}{\binom{47}{5}}\right)=\frac{13\cdot48}{\binom{52}{5}}\left(2-\frac{11\cdot43}{\binom{47}{5}}\right)$$
A: $Pr(A)=\dfrac{13\times 48}{\binom{52}{5}}=Pr(B)$, seen by picking the rank used for the four of a kind, then picking the remaining card, this out of all ways of giving a hand of five cards to that respective player.
$Pr(A\cap B) = \dfrac{13\times 12\times 44\times 43}{\binom{52}{5}\binom{47}{5}}$, seen by picking the rank of the four of a kind used by the first player, then the rank used for the four of a kind by the second player, then picking the remaining card for the first, then the remaining card for the second, this out of all ways of giving five cards to the first player and five to the second player.
We get then:
$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = 2\times \dfrac{13\times 48}{\binom{52}{5}} - \dfrac{13\times 12\times 44\times 43}{\binom{52}{5}\binom{47}{5}}$$
