Contract bridge probabilities ten I love playing contract bridge and am trying to understand the game better. In a standard game of contract bridge, there are 4 players that are dealt 13 poker cards each. There is also a point system also known as High Card Point (HCP) that can be used to gauge the strength of your given hand.
The HCP system is as follows:
Ace: 4 point/hcp
King: 3 point/hcp
Queen: 2 point/hcp
Jack: 1 point/hcp
All other cards: 0 point/hcp
So in a deck of 52 poker cards, each suite has a total of 10 HCP, and the entire deck has a total of 40 HCP
I am interested in knowing how I can go about calculating the probability that a single hand has a total of 0 HCP, 1 HCP, 2 HCP... 40 HCP
Thanks !
 A: Personally, I think that the OP (i.e. original poster) provided reasonable context in his posting.  There is no reason to doubt that the OP is totally untrained in Math, and is merely posing the question out of intellectual curiosity.  Therefore, it does not seem reasonable to require the OP to show work.

Assume that a player is dealt $(13)$ cards (without replacement) from the standard $(52)$ card deck.  I will illustrate how to compute the probability that the person's hand has exactly $(10)$ HCP's.  This will introduce the OP to concepts that he will then have to explore on his own.
First, you need to know how to interpret the expression
$$\binom{n}{k} ~: ~n \in \Bbb{Z^+}, ~k \in \{0,1,2,\cdots,n\},$$
and what it's signficance is.
$$\binom{n}{k} = \frac{n!}{k! [(n-k)!]}.$$
$0!$ is arbitrarily denoted to equal $(1)$.
$\displaystyle \binom{n}{k}$ refers to the number of distinct ways of selecting $(k)$ items, from a group of $(n)$ items, where the selection is done without replacement, and the order that the items are selected is regarded as not relevant.

I will express the probability as
$$\frac{N}{D} ~: ~D = \binom{52}{13}. \tag1 $$
So, the problem has been reduced to enumerating $N$.

The first thing to do is identify the explicit mutually exclusive cases that need to be enumerated.  They are shown in the following table:
\begin{array}{| r | r | r | r| r |}
  \hline                       
  \text{Counter} & \text{# Aces} & \text{# Kings} & \text{# Queens} & \text{# Jacks}\\
  \hline                       
  N_1 & 2 & 0 & 1 & 0 \\ \hline
  N_2 & 2 & 0 & 0 & 2 \\ \hline
  N_3 & 1 & 2 & 0 & 0 \\ \hline
  N_4 & 1 & 1 & 1 & 1 \\ \hline
  N_5 & 1 & 1 & 0 & 3 \\ \hline
  N_6 & 1 & 0 & 3 & 0 \\ \hline
  N_7 & 1 & 0 & 2 & 2 \\ \hline
  N_8 & 1 & 0 & 1 & 4 \\ \hline
  N_9 & 0 & 3 & 0 & 1 \\ \hline
  N_{10} & 0 & 2 & 2 & 0 \\ \hline
  N_{11} & 0 & 2 & 1 & 2 \\ \hline
  N_{12} & 0 & 2 & 0 & 4 \\ \hline
  N_{13} & 0 & 1 & 3 & 1 \\ \hline
  N_{14} & 0 & 1 & 2 & 3 \\ \hline
  N_{15} & 0 & 0 & 4 & 2 \\ \hline
  N_{16} & 0 & 0 & 3 & 4 \\
  \hline  
\end{array}
So, all that remains is instructions for how to compute each of 
$N_1, N_2, \cdots, N_{16}.$
Then, you have that 
$N = N_1 + N_2 + \cdots + N_{16}.$
Then, you take this computation for $N$, and feed it into the expression in (1) above.

Rather than providing separate instructions for each of 
$N_1, N_2, \cdots, N_{16},$ 
I can provide one set of instructions that may be followed for each of the $(16)$ partial enumerations.
For any such partial enumeration, assume that it represents 
$(a)$ Aces, $(b)$ Kings, $(c)$ Queens, $(d)$ Jacks.
Here, it is to be understood that

*

*$a,b,c,d$ are each elements in $\{0,1,2,3,4\}$, as indicated by the specific variable $N_r$ that is being enumerated, where $r \in \{1,2,\cdots,16\}.$


*Let $s = 13 - (a + b + c + d)$. 
This signifies that the bridge hand will have contain exactly $s$ cards from the $(36)$ cards represented by the ranks $(2)$ through $(10)$ inclusive.
Then, the computation for the specific variable $N_r$ is
$$\binom{4}{a} \times \binom{4}{b} \times \binom{4}{c} \times \binom{4}{d} \times \binom{36}{s}.$$
