You should have started by making that clear.
There are \(\displaystyle \dfrac{256!}{(256-16)!}\) (or permutations of 256 taken 16 at a time) strings of length sixteen from two hundred and fifty six possible choices in which no two are the same value.
There are \(\displaystyle 256^{16}-\dfrac{256!}{(256-16)!}\) strings of length sixteen from two hundred and fifty six possible choices in which at least two are the same value.
Or, to put it another way: If we want at least 2 cards to be the same, that means 2, 3, 4, ...., 16 cards the same which is everything except 1 card the same and so we can solve our problem by getting the probability of 1 card the same, i.e. no duplicates, and subtracting that from one.
Numerator: Well if we need every thing different, the choices are 256 on the first round, 255 on the second, ..., 241 on the sixteenth round or the
256P
16=\(\displaystyle \dfrac{256!}{(256-16)!}\)
given by pka.
Denominator: For the total number of possibilities we have a choice of 256 on the first round, 256 on the second, ..., 256 on the sixteenth round or 256
16 possibilities.
So the probability of having all different is
\(\displaystyle \dfrac{_{256}P_{16}}{256^{16}}\)
and the probability of having at least two matching is
1 - \(\displaystyle \dfrac{_{256}P_{16}}{256^{16}}\) = \(\displaystyle \dfrac{256^{16}\, -\, _{256}P_{16}}{256^{16}}\)
BTW: To avoid those large numbers, the calculation proceeds as
1 - \(\displaystyle \dfrac{256}{256}\, \dfrac{255}{256}\, \dfrac{254}{256}\, ...\, \dfrac{241}{256}\) ~ 0.5696
