Simulate a sock drawer
Jelly, 8 bytes

ḤX€Ṛ&lt;RTḢ


Try it online! or verify the distribution for N = 7.

Background

Let n be the number of pairs; there are 2n individual socks.

For the first draw, there are 2n socks and 0 of them would result in a matching pair. Therefore, the probability of success is 0 / 2n = 0.

Since the first draw wasn't successful, there are 2n - 1 socks on the pile and 1 of them would result in a matching pair. Therefore, the probability of success is 1 / (2n - 1).

If the second draw wasn't successful, there are 2n - 2 socks on the pile and 2 of them would result in a matching pair. Therefore, the probability of success is 2 / (2n - 2).

In general, if the first k draws were unsuccessful, there are 2n - k socks on the pile and 2 of them would result in a matching pair. Therefore, the probability of success is k / (2n - k).

Finally, if none of the first n draws was successful, there are 2n - k socks on the pile and all of them would result in a matching pair. Therefore, the probability of success is n / (2n - n) = 1.

How it works

ḤX€Ṛ&lt;RTḢ  Main link. Argument: n

Ḥ         Unhalve; yield 2n.
 X€       Map `random draw' over [1, ..., 2n], pseudo-randomly choosing an integer
          from [1, ..., k] for each k in [1, ..., 2n].
   Ṛ      Reverse the resulting array.
     R    Range; yield [1, ..., n].
    &lt;     Perform vectorized comparison.
          Comparing k with the integer chosen from [1, ..., 2n - (k - 1)] yields 1
          with probability (k - 1) / (2n - (k - 1)), as desired.
          The latter half of elements of the left argument do not have a counter-
          part in the right argument, so they are left untouched and thus truthy.
      T   Truth; yield all indices of non-zero integers.
       Ḣ  Head; extract the first one.