This is for the one player game and a initial hand of five cards. I will consider that you have a hand with $k$ cards if after throwing away all groups of four cards of the same value, you have $k$ cards in your hand. For instance, if your initial hand is $\{2,3,3,3,3\}$ I will consider it a one card hand, not a five card hand.
Let $P(k)$ be the probability that at some moment in the game you have a $k$ card hand. The following values are easy: $P(0)=1$ (at the end of the game you have no cards in your hand) and $P(40)=P(41)=\dots=P(52)=0$ (if you have $40$ cards, there must be a group of four cards of the same value). A little thought gives
$$
P(39)=\frac{4^{13}}{\dbinom{52}{13}}=0.000105681
$$
For the rest of the values I have run a simulation in Mathematica of $10^7$ games. These are the results:
k H(k)
0 483
1 25839
2 596131
3 10000000
4 230004
5 10000000
6 10000000
7 10000000
8 10000000
9 10000000
10 10000000
11 10000000
12 10000000
13 10000000
14 10000000
14 9999996
16 9999945
17 9999720
18 9998600
19 9994328
20 9980514
21 9942513
22 9854106
23 9675123
24 9348731
25 8822787
26 8068444
27 7080395
28 5919525
29 4675434
30 3458156
31 2381202
32 1514558
33 876208
34 458693
35 213203
36 84852
37 28160
38 7216
39 1067
For each $k$, $H(k)$ is the number of games in which a hand of $k$ cards has been held before reaching the end of the deck. Observe that the value of $H(39)$ is in accordance with the exact value of $P(39)$. A graph of the results:
It is surprising (at least to me) that for certain values of $k$, like $k=5$, a hand of $k$ cards was held in all $10^7$ games, even if a deck like
1,1,1,1,2,2,2,2,3,3,3,3,...
will give only hands of $1$, $2$, $3$ and $4$ cards.
From the comments, it seems that you are allowing only hands of the $3$-$2$ type, commonly called Full House, and hands of the $2$-$2$-$1$ type, commonly called Two Pairs.
Full House: The kind we have $3$ of can be chosen in $\binom{13}{1}$ ways. For each such way, the actual $3$ cards can be chosen in $\binom{4}{3}$ ways. for every way of doing these things, there are $\binom{12}{1}$ ways of choosing the kind we have $2$ of, and then $\binom{4}{2}$ ways to choose the $2$ cards, for a total of $\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$.
Two Pairs: This one is tricky, it is all too easy to double count. The two kinds we have $2$ each of can be chosen in $\binom{13}{2}$ ways. For each of thse ways, the $2$ cards of the higher-ranking kind can be chosen in $\binom{4}{2}$ ways, and for each of these ways the cards of the other kind can be chosen in $\binom{4}{2}$ ways. Finally, after all this has been done, the "junk" card can be chosen in $\binom{44}{1}$ ways, for a total of $\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{44}{1}$.
Finally, add the two numbers obtained above.
Remark: Your calculation does multiple counting in various ways. For example, consider the hand $\spadesuit$ K, $\heartsuit$ K, $\diamondsuit$ $7$, $\clubsuit$ $7$, $\heartsuit$ Q. This has been double-counted, for you have counted as different the choice $\diamondsuit$ $7$, $\clubsuit$ $7$, $\spadesuit$ K, $\heartsuit$ K, $\heartsuit$ Q. This is because your first choice of kind, among the $\binom{13}{1}$ could have been K, and your second choice among the $\binom{12}{1}$ could have been $7$, or the other way around. But these give the same hand,
Counting in a way that avoids multiple counting can be initially quite tricky. We naturally think in terms of doing things sequentially, "first" then "second." we have to be careful to make sure that different choices of "first" and "second" do not yield the same final hand.
Best Answer
Let's say $AB$ are going to get all the spades. Then $CD$ will have no spades between them, so their 26 cards are going to come from the 39 non-spades. We can choose the pair getting one suit in 6 ways, and the suit they get in 4 ways, so the probability is approximately $$\frac{\binom{39}{26}}{\binom{52}{26}} \cdot 6 \cdot 4 \approx 3.9 \times 10^{-4}.$$ This is only approximate because it's possible for more than one pair to have an entire suit, but that should be a small probability compared to the above. It's even possible for there to be a situation such as: $AB$ own all the clubs, $BC$ own all the diamonds, $CD$ own all the hearts, and $DA$ own all the spades. So the exact calculation will be messy.