I will attempt an answer for the case where:
- There is total independence between each type of names, so that we can ignore the multivariate components of the frequency distributions of first letters. (Meaning that, for example, we can ignore any effects whereby suburbs starting with specific letters will be made of people with predominently other specific letters, compared to others, etc...)
- We are sampling the population with replacement.
So this answer will strictly be made in terms of the four distinct univariate distributions that describes the distribution of the first letter of each types of names.
Since the probability of obtaining a duplicate with replacement is larger than the probability of obtaining a duplicate without replacement, this gives us an upper bound. However, this approximation tends towards the exact answer as the population from which you sample tends towards infinity, or I could even say (although I am not sure how to determine this) that the approximation becomes very good as the population from which you sample becomes sufficiently large compared to $N$, or in other words, $\text{Population}>>N$.
So we have a list of $N$ acronyms. Let $A_i$ be the $i^\text{th}$ acronym, where $1 \leqslant i \leqslant N$. Let $A_{i,j}$ be the $j^\text{th}$ letter of the $i^\text{th}$ acronym, where $1 \leqslant j \leqslant 4$ (in our case anyway) and let $A_{i,j,k}$ be the probability that $A_{i,j}$ is the $k^\text{th}$ letter of the alphabet, where $1 \leqslant k \leqslant 26$.
Now (assuming that we do have replacement during our sampling, as stated above) all acronyms are taken from the same four discrete frequency distributions, so that we can write $A_{i,j,k} = \operatorname{P} \left( j,k \right)$, the common probability that the $j^\text{th}$ type of name has the $k^\text{th}$ letter of the alphabet as its first letter.
The probability that two distinct acronyms ($i_1 \neq i_2$) are the same is:
\begin{equation*}
\begin{split}
\operatorname{P} \left( A_{i_1} = A_{i_2} \right) & = \prod_{j=1}^{4} \operatorname{P} \left( A_{i_1,j} = A_{i_2,j} \right) \text{ (assuming independence, as discussed)} \\
& = \prod_{j=1}^{4} \sum_{k=1}^{26} A_{i_1,j,k} \times A_{i_2,j,k} \\
& = \prod_{j=1}^{4} \sum_{k=1}^{26} \operatorname{P} \left( j,k \right)^2
\end{split}
\end{equation*}
Taking one step further, the probability that $S$ distinct acronyms ($i_1 \neq i_2 \neq \dots \neq i_S$) are all the same is:
\begin{equation*}
\begin{split}
\operatorname{P} \left( A_{i_1} = \dots = A_{i_S} \right) & = \prod_{j=1}^{4} \operatorname{P} \left( A_{i_1,j} = \dots = A_{i_S,j} \right) \\
& = \prod_{j=1}^{4} \sum_{k=1}^{26} A_{i_1,j,k} \times \dots \times A_{i_S,j,k} \\
& = \prod_{j=1}^{4} \sum_{k=1}^{26} \operatorname{P} \left( j,k \right)^S
\end{split}
\end{equation*}
Now, the probability of a duplicate in our sample is the union of the probabilities of any two pairs of acronyms within the sample being equal to each other. These are clearly not mutually exclusive, so the inclusion-exclusion formula applies:
\begin{equation*}
\begin{split}
\operatorname{P} \Big( \bigcup_{1 \leqslant i_1 < i_2 \leqslant N} A_{i_1} = A_{i_2} \Big) & = \sum_{i_1 < i_2} \operatorname{P} \left( A_{i_1} = A_{i_2} \right) - \sum_{i_1 < i_2 < i_3} \operatorname{P} \left( A_{i_1} = A_{i_2} = A_{i_3} \right) \\
& + \dots + \left( -1 \right)^N \sum_{i_1 < \dots < i_N} \operatorname{P} \left( A_{i_1} = \dots = A_{i_N} \right) \\
& = \binom{N}{2} \underset{i_1 \neq i_2}{\operatorname{P}} \left( A_{i_1} = A_{i_2} \right) - \binom{N}{3} \underset{i_1 \neq i_2 \neq i_3}{\operatorname{P}} \left( A_{i_1} = A_{i_2} = A_{i_3} \right) \\
& + \dots + \left( -1 \right)^N \binom{N}{N} \underset{i_1 \neq \dots \neq i_N}{\operatorname{P}} \left( A_{i_1} = \dots = A_{i_N} \right) \\
& = \sum_{S=2}^{N} \left( -1 \right)^S \binom{N}{S} \underset{i_1 \neq \dots \neq i_S}{\operatorname{P}} \left( A_{i_1} = \dots = A_{i_S} \right) \\
& = \sum_{S=2}^{N} \left( -1 \right)^S \binom{N}{S} \prod_{j=1}^{4} \sum_{k=1}^{26} \operatorname{P} \left( j,k \right)^S
\end{split}
\end{equation*}
I have not originally thought about the implications of having replacement or not, so I suppose my question is only partially answered with the following statement:
$\displaystyle \operatorname{P} \Big( \textrm{having a duplicate among $N$ acronyms} \Big) \leqslant \sum_{S=2}^{N} \left( -1 \right)^S \binom{N}{S} \prod_{j=1}^{4} \sum_{k=1}^{26} \operatorname{P} \left( j,k \right)^S$
Does anyone have any comments about this? Did I make any mistakes in my reasoning? What about a lower bound?
