[bigpooch]

Say AA4 is on board. Now there are only 52-3 = 49 cards
left in the deck. For a two card holding there are now
C(49,2) = 49 x 48 / 2 combinations. The holding of 7s 2c
is exactly the same as 2c 7s so the combinations just
count the possible hands exactly once. Each of these
C(49,2) = 1176 combinations are equally likely.

The number of combinations for trips or better is:

Quads: 1
An ace+ other: 2 x 47 = 94
(two choices of which remaining ace x 47 choices of
the remaining non-ace: one 4 is gone on the flop)
pocket 44: 3

so that makes 98/1176 = 1/12.

There is another way of determining this instead of
enumerating the above possibilities (there may be several
ways of getting the correct answer): consider the
chance of not getting an ace in two cards. For a hand
not to have an ace, the probability is 47/49 x 46/48 or
the probability of not having an ace 190/(49x48) and the
same result follows (add 3x2/(49x48) for 44).