The game: dealer and multiple players, lets say 6 players, 1 dealer. Each participant received 2 cards. Hands: each hand consists of 4 cards, the player's two cards and the dealer's two cards.
Goal: Assuming fresh single deck, Calculate the probability that a given round contains the scope of typical poker and black jack hands (one or more). More specifically, what is the probability of one or more single pairs, two pairs, trips, 4 of a kind and blackjack.
Also, by one or more, I mean 1 exactly, 2 exactly etc as partitioned/separate calculations..
Basically, Im looking for a logic to try to properly frame the calculation, but certainly wont object to a detailed worked out response either.
I'll calculate the distributions for the numbers of hands with four of a kind, triple and two pairs. I'll leave the single pairs and blackjacks to you, since the calculation becomes rather involved.
Let $n$ denote the number of players. Let's start with four of a kind. This is easy – at most one player can have four of a kind, and the probability for a given player to have four of a kind is $\frac{13}{\binom{52}4}=\frac1{20825}$, so the probability of one of $n$ players to have four of a kind is $\frac n{20825}$.
Next, three of a kind. At most $2$ players can have three of a kind. There are two cases: the dealer has a pair, with probability $\frac3{51}=\frac1{17}$, or she doesn't. If she does, then by distributing the two remaining cards of that rank we find that the probability that exactly one player has exactly one matching card is
$$ 2\cdot\frac{50-2n}{50}\cdot\frac{2n}{49}\;, $$
and the probability that two players have exactly one matching card is
$$ \frac{2n}{50}\cdot\frac{2(n-1)}{49}\;. $$
If she doesn't, at most two players can have a matching pair. The probability for a particular player to have a matching pair is
$$ p_1=\frac6{50}\cdot\frac2{49}\;, $$
and the probability for two particular players to have a matching pair is
$$ p_2=\frac6{50}\cdot\frac2{49}\cdot\frac3{48}\cdot\frac2{47}\;, $$
so the probability for exactly one player to have a matching pair is
$$ \binom n1p_1-2\binom n2p_2=\frac{6n}{1225}-\frac{3n(n-1)}{230300} $$
and the probability for exactly two players to have a matching pair is
$$ \binom n2p_2=\frac{3n(n-1)}{460600}\;. $$
Combining the results for the two cases, we find that overall the probability for exactly one player to have three of a kind is
$$ \frac1{17}\left(2\cdot\frac{50-2n}{50}\frac{2n}{49}+16\cdot\left(\frac{6n}{1225}-\frac{3n(n-1)}{230300}\right)\right)=\frac{8n(1153-25n)}{978775}\;, $$
and for exactly two players,
$$ \frac1{17}\left(\frac{2n}{50}\frac{2(n-1)}{49}+16\cdot\frac{3n(n-1)}{460600}\right)=\frac{4n(n-1)}{39151} \;. $$
Finally, two pairs. Now any number of players can have two pairs. Again we can condition on whether the dealer has a pair.
If she does, we want the probability for exactly $m$ players to have a different pair. If $k$ players have such a pair, of which $j$ pairs of players have pairs of the same rank, then $k-2j$ players have pairs of a unique rank. There are $\frac{k!}{(k-2j)!2^j}$ ways to form the pairs of players, $\frac{12!}{(12-(k-j))!}$ ways to select ranks, $(4\cdot3)^{k-2j}$ ways to select the suits for the unique ranks and $(4\cdot3\cdot2)^j$ ways to select the suits for the paired ranks. Thus the probability for $k$ particular players to have a different pair than the dealer is
$$ \frac{(50-2k)!}{50!}\sum_{j=0}^{\left\lfloor\frac k2\right\rfloor}\frac{k!}{(k-2j)!}\frac{12!}{(12-(k-j))!}12^{k-j}\;, $$
so by inclusion-exclusion the probability for exactly $m$ players to have a different pair is
$$ \sum_{k=m}^n(-1)^{k-m}\binom km\binom nk\frac{(50-2k)!}{50!}\sum_{j=0}^{\left\lfloor\frac k2\right\rfloor}\frac{k!}{(k-2j)!}\frac{12!}{(12-(k-j))!}12^{k-j}\;. $$
If the dealer doesn't have a pair, up to $3$ players can match both of the dealer's ranks. The probabilities for $1$, $2$ or $3$ particular players to match both of the dealer's ranks are
\begin{align} q_1&=\frac6{50}\cdot\frac3{49}\;,\\ q_2&=\frac4{48}\cdot\frac2{47}\cdot q_1\;,\\ q_3&=\frac2{46}\cdot\frac1{45}\cdot q_2\;, \end{align}
and the probabilities for exactly $m$ players to match both of the dealer's ranks are
$$ \sum_{k=m}^3(-1)^{k-m}\binom km\binom nkq_k\;. $$
If we define $q_k=0$ for $k\gt3$, we can combine these two results to get the total probability for exactly $m$ players to have two pairs:
$$ \frac1{17}\sum_{k=m}^n(-1)^{k-m}\binom km\binom nk\left(\frac{(50-2k)!}{50!}12!k!12^k\sum_{j=0}^{\left\lfloor\frac k2\right\rfloor}\frac1{(k-2j)!(12-(k-j))!12^j}+16q_k\right)\;. $$
Here's code that I used to check these results against simulations.