The correct answer is $\boxed{\frac{{{}^{\text{n}}{{\text{C}}_{\text{d}}}}}{{{2^{\text{n}}}}}}$.
To calculate the probability of a Hamming distance of $d$, we need to count the number of ways to choose $d$ bit positions where the two strings differ, and then divide this by the total number of possible strings. The total number of possible strings is $2^n$, since each bit can be either 0 or 1. The number of ways to choose $d$ bit positions where the two strings differ is ${{n \choose d}}$, which is the binomial coefficient. The binomial coefficient is the number of ways to choose $d$ objects from a set of $n$ objects.
Therefore, the probability of a Hamming distance of $d$ is $$\frac{{{}^{\text{n}}{{\text{C}}_{\text{d}}}}}{{{2^{\text{n}}}}}$$
Option A is incorrect because it divides by $2^d$ instead of $2^n$. Option B is incorrect because it divides by $2^d$ instead of $2^n$ and also includes the case where $d=0$, which is not a possible Hamming distance. Option C is incorrect because it does not take into account the number of possible ways to choose the bit positions where the two strings differ. Option D is incorrect because it is the probability that the two strings are identical, which is not the same as the probability that the Hamming distance is 0.