Let $s=13$ denote the number of cards in each suit and $N$ denote the (random) number of cards drawn when all $k=4$ suits are represented for the first time. Hence, $k\leqslant N\leqslant(k-1)s+1$ with full probability.
For each $n\geqslant1$, the event $[N\gt n]$ depends on the $n$ first cards drawn only. There are ${ks\choose n}$ collections of $n$ cards from a full deck of $ks$ cards and each such collection has the same probability ${ks\choose n}^{-1}$ to be drawn. The event $[N\gt n]$ means that one avoids at least one suit. Using inclusion-exclusion principle, there are $A_n$ ways to do so, where
$$
A_n={k\choose1}{(k-1)s\choose n}-{k\choose2}{(k-2)s\choose n}+\cdots\pm{k\choose k-1}{s\choose n}\mp{k\choose 0}{0\choose n}.
$$
This yields the expectation
$$
\mathbb E(N)=\sum_{n\geqslant0}\mathbb P(N\gt n)=\sum_{n\geqslant0}{ks\choose n}^{-1}A_n.
$$
In the case at hand,
$$
\mathbb E(N)=\sum_{n\geqslant0}{52\choose n}^{-1}\left(4{39\choose n}-6{26\choose n}+4{13\choose n}-{0\choose n}\right),
$$
that is,
$$
\mathbb E(N)=4+\sum_{n=4}^{39}{52\choose n}^{-1}\left(4{39\choose n}-6{26\choose n}+4{13\choose n}\right),
$$
or,
$$
\mathbb E(N)=4+4B_{39}-6B_{26}+4B_{13},\qquad
B_i=\sum_{n=4}^{i}{52\choose n}^{-1}{i\choose n}.
$$
Numerically,
$$
\mathbb E(N)=\frac{4829}{630}=7+\frac23-\frac1{630}\approx7.66508.
$$
I am not familiar with Pinochle decks. So for the count I will assume that for example the two $\heartsuit$ Queens are identical.
Then there are $3$ types of $5$-card hands: (i) all cards distinct; (ii) there is $1$ pair of identicals; (ii) there are $2$ pairs of identicals. For the count, we calculate the number of each type, and add up.
(i) This is easiest. There are $24$ different cards, and there are $\binom{24}{5}$ ways to choose $5$.
(ii) There are $24$ different cards. We choose $1$ of the kinds to have $2$ of, and then $3$ different cards for the rest. That gives total $\binom{24}{1}\binom{23}{3}$.
(iii) This can be tricky. We pick $2$ of the kinds to have $2$ each of, and then an odd card. That gives $\binom{24}{2}\binom{22}{1}$.
Now for poker hand probabilities, we have to forget about the count we just made. For the different hands that we counted above are not equally likely. The Type (i) jands are all equally likely, as are the type (ii) hands, as are the type (iii) hands, but we do not have equal likelihood between different types.
There are two ways to proceed. We can take account of the different probabilities for each type. But my inclination is to imagine that the "repeats" are coloured red and blue, making the $48$ cards distinct. Then there are $\binom{48}{5}$ equally likely hands.
For each type of poker hand, count how many ways there are to produce it, taking colouring into account. I will leave the (unpleasant) details to you. We need to define carefully what we mean by each poker hand. For example, five of a kind is now possible.
Let's do a relatively straightforward one, $1$ pair. The type of card we have $2$ of can be chosen in $6$ ways, $9$ to Ace. Suppose for example it is $2$ Kings. They can be either $2$ of the same suit ($4$ choices) or of different suits. If so, the different suits can be chosen in $\binom{4}{2}$ ways, and then the actual cards in $2^2$ ways (all combinations of red and blue), for a total of $4+(6)(4)=28$ ways. Now we choose $3$ types from the remaining $5$, and for each choose a colour, red or blue. This gives a total of $(6)(28)(80)$. Check it. Now divide by $\binom{48}{5}$ for the probability.
Best Answer
Your counting is a bit off.