Suit Up!

Each of the four probabilities will have the same denominator so we really just need the numerators. In other words the number of possible hands of each type.

Let $A'(n)$ be the number of possible full house hands from the deck with $n$ suits, etc. These enumerations are modified from this Wikipedia entry

$A'(n)=\binom{13}{1} \binom{n}{3} \binom{12}{2} \binom{n}{2}$

$B'(n)=\binom{13}{1} \binom{n}{4} \binom{12}{1} \binom{n}{1}$

$C'(n)=\binom{13}{1} \binom{n}{3} \binom{12}{2} \binom{n}{1}^{2}$

$D'(n)=\binom{13}{2} \binom{n}{2}^{2} \binom{11}{1} \binom{n}{1}$

Setting up the big fraction and canceling out every term that appears in the numerator and denominator leaves

$\large \frac{A(n)D(n)}{B(n)C(n)}=\frac{A'(n)D'(n)}{B'(n)C'(n)}=\frac{\binom{n}{2}^{3} \binom{13}{2} \binom{11}{1}}{\binom{n}{4} \binom{13}{1} \binom{12}{2} \binom{n}{1}^{2}} = \frac{(\frac{n(n-1)}{2})^{3}\cdot \frac{13 \cdot 12}{2} \cdot 11}{\frac{n(n-1)(n-2)(n-3)}{24}\cdot 13 \cdot \frac{12 \cdot 11}{2} \cdot n^{2}} = \frac{3(n-1)(n-1)}{(n-2)(n-3)}$

${\displaystyle \lim_{n\to\infty}} \frac{3(n-1)(n-1)}{(n-2)(n-3)} = \boxed{3}$

This is a bit sketchy, but it gives the right idea.

First, $A(n) = 13 \cdot 12 \cdot P(AAAKK),$ where $AAAKK$ is any hand with three aces and two kings. And $P(AAAKK) = \binom{5}{2} P(A/A/A/K/K),$ where $A/A/A/K/K$ is the hand in order: first three aces, then two kings. We'll use this notation in the following parts as well. Now the probability of any ordered hand where we're given the ranks of each card is dependent on $n,$ but in the limit as $n\to\infty$ the cards are essentially being chosen with replacement, i.e. drawing one ace doesn't decrease the probability of drawing another one. (This is the sketchy part--it's correct but it probably could use a more rigorous argument.) That is, $P(A/A/A/K/K) \rightarrow \frac1{13^5}$ as $n\to \infty.$

So $A(n) \rightarrow 13 \cdot 12 \cdot 10 \cdot \frac1{13^5}.$

Now $B(n) = 13 \cdot 12 \cdot P(AAAAK) = 13 \cdot 12 \cdot 5 \cdot P(A/A/A/A/K) \rightarrow 13 \cdot 12 \cdot 5 \cdot \frac1{13^5}.$

Now $C(n) = 13 \cdot \binom{12}{2} \cdot P(AAAKQ)$ (the binomial coefficient is because K and Q are interchangeable), so $C(n) = 13 \cdot \binom{12}{2} \cdot 20 \cdot P(A/A/A/K/Q) \rightarrow 13 \cdot \binom{12}{2} \cdot 20 \cdot \frac1{13^5}.$ The $20$ comes from the product of five choices for the king's position and then four choices for the queen's position, or if you like, it's the multinomial coefficient $\binom{5}{3,1,1}.$

Finally, $D(n) = \binom{13}2 \cdot 11 \cdot P(AAKKQ) \rightarrow \binom{13}2 \cdot 11 \cdot 30 \cdot \frac1{13^5}.$ The $30$ comes from five choices for the queen and then $\binom{4}{2}$ choices for the position of the king pair, or if you like, it's the multinomial coefficient $\binom{5}{2,2,1}.$

Putting it together (and noticing that the $\frac1{13^5}$ factors cancel), we get $\frac{A(n)D(n)}{B(n)C(n)} \rightarrow \frac{(13 \cdot 12 \cdot 10)\left(\binom{13}2 \cdot 11 \cdot 30\right)}{(13 \cdot 12 \cdot 5)\left(13 \cdot \binom{12}2 \cdot 20\right)} = \fbox{3}.$