How to calculate the probability of picking an exact item pair? Let's say we have 4 items in a bag: item A, item B and 2 item X
We extract them in pairs. How to calculate the probability that we end up with an AB pair?
I'm thinking 1/4 to extract item A and then 1/3 to extract item B, for our first pick.
That gives 1/12
Is this correct? What about for N items?
 A: Not quite, because if you want to end up with AB as your first pair, you can also first pick object B, and then object A
So, you have probability of $\frac{1}{2}$ of first picking either A or B, and then a probability of $\frac{1}{3}$ of picking the other one from that pair, giving you a total probability of $\frac{1}{6}$ of picking AB as the first pair.
However, you can also end up with AB if you pick XX as the first pair, and thus pick AB as your third and fourth object. By symmetry, the prbability for this will also be $\frac{1}{6}$, and so the probability of ending up with AB after picking pairs until they are all gone is $\frac{1}{3}$
Another way of thinking about this is that there are ${4\choose2}=6$ possible pairs you can pick for the first pair, and exactly 2 of them (AB and XX) will end up with AB, so the probability is $\frac{1}{3}$.
Now, in general, if you have N objects, and N is even, then let's say that $N =2M$. Thus, in that case you pick M pairs, and you want one of them to be AB.
As with the original case where $N =4$, we can first consider the probability of picking AB as the first pair, and then multiply that probability by $M$ to get the probability of picking AB as any one of the $M$ pairs.
Now, the probability of first picking A and then B is $\frac{1}{N}*\frac{1}{N-1}$, but since you can also first pick B and then A, the probability of picking AB as the first pair is $\frac{2}{N(N-1)}$.
Hence the probability of picking AB as any of the $M$ pairs is $$M*\frac{2}{N(N-1)} = \frac{2M}{N(N-1)} = \frac{N}{N(N-1)} = \frac{1}{N-1}$$
You can also see this as follows. At some point you will pick object A, and in order for that to pair up with B, there is exactly one spot left outof the remaining $N-1$ spots for you to pick B. For example, if you pick A as the 6th object, then you must pick B as the 5th object in order to get AB. If you pick A as the seventh object, you must pick B as the 8th.
So, A can go anywhere, but for B there then is exactly 1 out of$N-1$ spots left to get AB. So the probability is $\frac{1}{N-1}$. 
Note that we could have followed this reasoning for the original case with $N=4$ as well, giving us once again $\frac{1}{3}$
