Probability of the Same Pair of Balls Drawn from Two Separate Urns This morning, my friends and I discussed following problem.
Problem:
There are two persons named Mr. A and Mr. B. Each person has his own urn containing $N$ different balls. They uniformly randomly draw a ball twice with replacement from their own urns. 
What is the probability that they draw the same pair of balls?
Example: 
Let $N = 3$ and let's label them with integer $i$, where $1 \leq i \leq N$. Let $A_k$ and $B_k$, where $k \in \{1,2\}$,  be the events when Mr. A and Mr. B draw ball $i$ at the $k$ drawing, respectively. 
The pair $((A_1,B_1),(A_2,B_2))$ denotes an outcome from the drawing process. 
Events of interest are, for example, $((1,1),(2,2)), ((1,1),(3,3)), \mathrm{or}~ ((1,2),(2,1)). $
First Answer:
Number of sample space $|\Omega|$ is $\binom{N}{2}\times\binom{N}{2}$. Number of possible outcomes is $\binom{N}{2}$. The probability is $\frac{1}{\binom{N}{2}}$.
Second Answer:
$|\Omega| = N^4$.
Let $X$ be an event where they both draw the same ordered pair of balls, e.g., $((1,1),(2,2)) \mathrm{or}~ ((1,1),(3,3)).$
Let $Y$ be an event where they both draw the same "cross-ordered" pair of balls, e.g., $((1,2),(2,1)) \mathrm{or}~ ((1,3),(3,1)).$
$|X| = N^2$, and $|Y| = N\times(N-1).$
Hence, the probability is $\frac{N^2 + N\times(N-1)}{N^4}$.
Question:
Is either of these answers correct?
 A: OPs second answer is correct. 
We denote with $[N]:=\{1,2,3,\ldots,N\}$ and consider all $N^4$ tuples in
\begin{align*}
  \mathcal{A}=\{((A_1,B_1),(A_2,B_2))|A_j,B_j\in[N],j=1,2\}
  \end{align*}
We denote with $E(A_j=B_k), 1\leq j,k\leq 2$ the event that balls $A_j$ and $B_k$ are drawn and are equal.

Using the inclusion-exclusion principle we can calculate the number of events of drawing equal pairs. We obtain
  \begin{align*}
  &\#\left(E(A_1=B_1)\cap E(A_2=B_2)\right)\\
&\qquad+\#\left(E(A_1=B_2)\cap E(A_2=B_1))\right)\\
&\qquad-\#\left(E(A_1=B_1)\cap E(A_2=B_2)\cap E(A_1=B_2) \cap E(A_2=B_1)\right)\\
  &=|\{(A_1,A_1),(A_2,A_2)|A_1,A_2\in[N]\}|\\
&\qquad+|\{(A_1,A_2),(A_2,A_1)|A_1,A_2\in[N]\}|\\
  &\qquad -|\{(A_1,A_1),(A_1,A_1)|A_1\in[N]\}|\\
  &=N^2+N^2-N\\
  &=2N^2-N
  \end{align*}

The number of all events according to OPs rule is $|[N]|^4=N^4$. We conclude the probability $P(N)$ that Mr. A and Mr. B draw equal pairs of balls (with replacement) from urns containing $N$ balls is

\begin{align*}
  P(N)=\frac{2N-1}{N^3}
  \end{align*}

A: There are $\binom{N}{2}$ pairs. There is 1 way to order the match. The probability of getting a match for a particular pair is $(1/\binom{N}{2})^2$. Thus,
$$P(\text{Match}) = 1\cdot\binom{N}{2}\frac{1}{\binom{N}{2}}\frac{1}{\binom{N}{2}} = \frac{1}{\binom{N}{2}}.$$
A: I think the answer should be $\frac{2}{N^2}$ because the probability that they draw balls in the same order is $\frac{1}{N^2}$ and I do not see why it'd be different for the cross-ordered case as we can swap the order of drawing without any affect to the outcome. And as they're independent cases, we can say that the chance that one of them occurs is $\frac{2}{N^2}$.
