If I consider a poker deck with $13$ ranks of $4$ cards each, what is the probability of obtaining 2 triplets and 1 pair (eg. AAA77733) in the following two scenarios:
(A) when drawing 8 cards?
(B) when drawing 16 cards?
(In the 16-card case, I want two ranks with at least 3 cards each, and a third rank with at least 2 cards.) My approach using combinatorics for (A) is: Choose 3 ranks out of 13 and 3 cards out of 4 each for 2 ranks and 2 cards out of 4 for the third rank, then divide by the total number of combinations of 8 cards:
$$ \frac{{{13}\choose{3}} \cdot {{4}\choose{3}}^2 \cdot {{4}\choose{2}}}{{{52}\choose{8}}} = \frac{27456}{752538150}, $$ or about $1$ in $27409$.
Is this correct?
By what factor do I have to multiply the probability in (A) in order to get the probability in (B)?