Playing cards: Probability pairs An ordinary deck of $52$ playing cards ($4$ suits, $13$ values) are shuffled. Then player $A$ takes the top two cards. Player $B$ takes the next to top cars. Let $Pr[A]$ denote the probability for player $A$ to have a pair (i.e. two cards with the same value) and $Pr[B]$ the probability for player $B$ to have a pair. Task: Calculate $Pr[A], Pr[B], Pr[A \cap B]$ and $Pr[A|B]$.
My attempt:
$$Pr[A] = 1 \cdot \frac{3}{51} = \frac{3}{51}$$ because $A$ can take any first card but then there are only $3$ out of the remaining $51$ cards left, that match up to a pair.
$$Pr[B] = \frac{3}{49}$$
$$Pr[A \cap B] = Pr[A] \cdot Pr[B]$$ (are they independent?)
$$Pr[A|B] = \frac{Pr[A \cap B]}{Pr[B]}$$
I'm pretty sure that my $Pr[B]$ is wrong (or isn't it??). And because I don't have the exact $Pr[B]$ I can't calculate $Pr[A \cap B]$ and $Pr[A|B]$ just yet.
So, could somebody explain to me how to calculate $Pr[B]$ and if I can then just use the formulas above to calculate the remaining probabilities.
Thank you!
(If anything is unclear, please let me know!)

Here is a perhaps inelegant way to show that $P(B)=P(A)$
Suppose player $A$ gets a pair with probability $\frac{3}{51}$
Then player $B$ gets a pair with probability $$\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}$$
since for his first card, he can either get a different rank than $A$ with probability $\frac{48}{50}$ or get the same rank with probability $\frac{2}{50}$
Suppose player $A$ does not get a pair with probability $\frac{48}{51}$
Then player $B$ gets a pair with probability $$\frac{44}{50}\cdot\frac{3}{49}+\frac{6}{50}\cdot\frac{2}{49}$$
since player $A$ gets one of each rank, and so for his first card, he can either get a different rank than both of $A$'s cards with probability $\frac{44}{50}$ or get the same rank as one of them with probability $\frac{6}{50}$
All together
$$\frac{3}{51}\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)+\frac{48}{51}\left(\frac{44}{50}\cdot\frac{3}{49}+\frac{6}{50}\cdot\frac{2}{49}\right)=\frac{3}{51}$$
From here, the rest is trivial.
We now have
$$P(A\cap B)=\frac{3}{51}\cdot\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)\approx 0.0035$$
and
$$\begin{align*}
P(A \mid B)
&=\frac{P(A\cap B)}{P(B)}\\\\
&=\frac{\frac{3}{51}\cdot\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)}{\frac{3}{51}}\\\\
&\approx0.0596
\end{align*}$$
