How to calculate the probability of a hand in a game where cards are dealt one by one? I wanted to ask a question about finding the chance of a specific hand in a specific game, but then I thought that it would be interesting to have a more generic question related to it. So I will try to explain both cases and hope that someone could help me with it :)
Specific question
There is a Spanish 4-players game called Mus which is played with the Spanish (traditional) deck - 40 cards distributed in 4 colors (colors don't really matter in this game). These are the values of the cards for each color: 
$$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | S | C | R | $$
The most popular variant of this game (ocho reyes) changes some of the original value of the cards: 2's become 1's and 3's become R's. So, simplifying we have this set of cards (values) for each color:
$$ | 1 | 1 | 4 | 5 | 6 | 7 | S | C | R | R | $$
At the beginning of the game a player deals the cards, one by one, until all 4 players have 4 cards. Based on how many cards are left on the deck, the distribution would look like this:
$$ Player A = | 40 | 36 | 32 | 28 | $$
$$ Player B = | 39 | 35 | 31 | 27 | $$
$$ Player C = | 38 | 34 | 30 | 26 | $$
$$ Player D = | 37 | 33 | 29 | 25 | $$
Let's say I want to calculate the probability of a specific (one of the best) hand (let's call it best-hand) which is:
$$ | R | R | R | 1 | $$
My approach
My approach for this scenario was calculating the chance of this hand:
Number of possible best-hand's (there are 8 R's and 8 1's): $$ 8 · 7 · 6 · 8 = 4864 $$
Number of possible scenarios after dealing cards: $$ Player A · Player B · Player C · Player D $$$$ = (40 · 36 · 32 · 28) · (39 · 35 · 31 · 27) · (38 · 34 · 30 · 26) · (37 · 33 · 29 · 25) $$$$ =1,315 · 10^{24} $$
So the probability would be:
$$ \mathbb{P}(\text{best-hand}) = \frac{4864}{1,315 · 10^{24}} = 3,6987 · 10^{-21} $$
Somehow these numbers don't feel right for me. I think I don't have all the knowledge for calculating this, so what would be the best approach and the final result?
Edit
Just to clarify (in case I did not explain the way of dealing cards properly), the dealer gives one card to each player (one to player A, one to player B, one to player C, one to player D. Then starts all over, one to player A...) until they all get 4.
Generic question
Based on the previous question, I was thinking about finding a generic approach for solving this one-by-one dealing in a similar game. Our variables would be:
C: number of cards
P: number of players
D: number of cards for each player once cards have been dealt
H: desired hand (number of possible hands in this game)

Maybe it is too hard to precise, but I would like to know what you think about it.
P.S.: I hope that the question is clear, but maybe some of my notations are not proper for what I wanted to express. Please, feel free to edit it for a better understanding.
 A: The number of hands in total is ${40\choose 4}=91390$. The number of best-hands is ${8\choose 3}{8\choose 1}=448$, so the probability of a specific player (Player B, say) being dealt this hand is the quotient of these numbers, i.e., $\approx 0.0049$.
If we want to compute the probaility that at least one of the four players has a best-hand, things get a bit more complicated. As a rough approximation we may consider the events "Player A has a best-hand", "Player B has a best-hand", etc independent, and as another approximation just multiply the above by four (somewhat justified because two best-hands in one game are rather unlikely) so that in a little less than two percent of all games there will be a best-hand. 
A more precise statement (i.e., not suffering from these assumptions and 
approximations)  would be: The expected number of best-hands in any game is $4\cdot \frac{{8\choose 3}{8\choose 1}}{{40\choose 4}}\approx 0.196$.
A: Out of the 8R's and 81's, only two best hands are possible for any two players. The probability of that would be
$$\frac{ {8\choose3}{8\choose1}{5\choose3}{7\choose1}{4\choose2}{32\choose4}{28\choose4}}{{40\choose4}{36\choose4}{32\choose4}{28\choose4}}$$
Reasoning, any one player gets the best hand in first two terms ways, the second player gets the best hand in the next two terms number of ways, these two players could be chosen in 5th term ways, the third one get the four cards in 6th term ways and last player gets his four cards in last term ways.  The total number of ways is the denominator.  That will be probability.
I hope I understood the problem correctly
