Choosing 10 cards randomly from a 52 card deck Given a 52-card deck, if we pick 10 cards, what's the probability of having all four aces among the 10 cards we picked?
My attempt was defining $\Omega = \{(1,2,...,52)^{10}\}$. Now $|\Omega|=\binom{52}{10}$
Let $A$ be the event that we're looking for. $A=\{\{1,2,3,4\}\cup(5,6,...,52)^{6}\}$. $|A|=\binom{48}{6}$
Now $Pr(A)=\dfrac{|A|}{|\Omega|}=\dfrac{\binom{48}{6}}{\binom{52}{10}}$
I don't have the answer but I noticed I got a really small number ($0.07\%$), so I don't think I solved it right.
 A: For any 4 cards from 10 there are $^{10}C_4 = 210$ combinations.
For any 4 cards from 52 there are $^{52}C_4 = 270725$ combinations.
$P(4A) = \frac{^{10}C_4}{^{52}C_4} = \frac{210}{270725} = .0007757$
In other words, your $10$ cards only comprises of $210$ four card combinations from a possible $270725.$
A: Comment: You have several correct answers already--including your own! I just wanted to show
the connection with the hypergeometric distribution.
The number $X$ of Aces among ten cards chosen at random without replacement
from a 52 card deck has a hypergeometric distribution. There are four favorable cards (Aces) and 48 unfavorable cards (non-Aces).
In R statistical software, one can compute the probabilities $P(X = k)$ for
$k = 0, 1, 2, 3, 4.$
cbind(k, pdf)
     k          pdf
     0 0.4134453782
     1 0.4240465417
     2 0.1431157078
     3 0.0186166774
     4 0.0007756949

You seek 
$$P(X=4) = \frac{{4 \choose 4}{48 \choose 6}}{{52 \choose 10}} = 0.0007756949.$$
You are correct that getting all four Aces, even among ten cards, has a very
small probability. The figure below shows a bar chart of the hypergeometric
distribution of $X.$ Your probability corresponds to the very short bar at the far right.

Notes: (1) As an experiment, I tried doing ten million draws of 10 cards and counting
the Aces each time. The expected number of aces is $10^7 \times 0.0007756949 \approx 7757.$ In my simulation I got 7591, which is fewer than 7757, but
within the margin of simulation error.
(2) In many versions of poker each player's hand consists of five supposedly randomly chosen cards. A poker player would be delighted to find all four Aces in his/her
hand. But the probability of that is even smaller than your probability:
$1.847 \times 10^{-5}.$ 
dhyper(4, 4, 48, 5)
## 1.846893e-05

(3) If you sample the ten cards with replacement, then the number $Y$ of Aces in ten independent draws has $Y \sim \mathsf{Binom}(10, 1/13).$
Then $P(Y = 4) = 0.004548553.$ This is larger than your probability because
the Aces don't get "used up" as they get chosen. (Each Ace could possibly
be chosen more than once.)
dbinom(4, 10, 1/13)
## 0.004548553

A: Yes, the probability for selecting four specific cards (and six others) when selecting ten cards from a standard 52 card deck (without replacement) is $${\left.{\dbinom 44}\dbinom{48}6\middle/\dbinom{52}{10}\right.}$$
The probability that those four cards shall be placed among the top ten places in the deck is $${\left.\dbinom{10}4\middle/\dbinom{52}{4}\right.}$$
Which is the same small value $6/7735$.
A: The chance that A♣️ is in your hand is $10/52$. Given that, the chance the A♦️ also is is then $9/51$, etc, so the answer is $\dfrac{10\cdot9\cdot8\cdot7}{52\cdot51\cdot50\cdot49}$.
