Probability Dealing with Cards Here is the story.
You are dealt 8 cards from a well shuffled deck of 52 cards. What is the probability of getting 4 queens and 4 kings?
I understand the probability is quite small but how do we set up the problem to solve the probability?
 A: Only $1$ of the $\binom{52}{8}$ equally likely hands satisfies our condition. So the required probability is $\dfrac{1}{\binom{52}{8}}$. 
To evaluate numerically, note that
$$\binom{52}{8}=\frac{(52)(51)(50)(49)(48)(47)(46)(45)}{8!}.$$
A: To compute this probability we count:


*

*How many combinations of $8$ cards can be withdrawn.

*How many of these has the four kings and the four queens.


The probability is the quotient of these two numbers.
For this problem, it is important to state how we will count. I think that the simpler method is considering that two groups of eight cards are the same if they have the same cards, no matter the order.
With this in mind the answer to 2 is easy: there is only $1$ combination with $4$ king and $4$ queens.
Now we will answer to 1. If you know combinatorics, it's easy: $\binom{52}8=752538150$. (You have to select $8$ different elements among $52$ of them, no matter the order, so they are combinations).
The probability is roughly of $1$ among $750$ millions. A rare event, indeed ;-)
