Probability of picking 4 Aces in 10 Playing cards.

Question

I have 52 Playing cards (no joker). I remove 2 cards randomly and remain with 50 cards. After that, I pick 10 cards of 50 cards. Calculate the probability of there being 4 Aces in 10 cards.

I solve this exercise as following:

Remove $2$ cards randomly, I have two cases:

The first, if $2$ cards have $1$ or $2$ Aces, then the probability which I want to define, is $0$. Then I do not concern about this case.

Second, if 2 cards have no Aces, then I define the probability as follows:

• $\dfrac{C_{48}^2}{C_{52}^{2}}$ is the probability of choosing 2 cards (no Ace) when I pick 2 cards of 52 cards. $C_{48}^2$ is the quantity of choosing 2 cards of 48 cards (no Ace). $C_{52}^2$ is the quantity of choosing 2 cards of 52 cards.
• The probability of choosing 4 Aces when I pick 10 cards of 50 remain cards, \begin{equation} \dfrac{1.1.1.1 C_{46}^6}{C_{50}^{10}} = 9,12.10^{-4} \end{equation} whereas: 1 is the quantiy of choosing every Ace (Diamond or Heart or Spade or Club); $C_{46}^6$ is the quantity of choosing remain 6 cards; $C_{50}^{10}$ is the quantity of choosing 10 cards randomly.

Probability of there being 4 Aces in 10 cards of 50 cards \begin{equation} \dfrac{C_{48}^2}{C_{52}^2} \cdot \dfrac{1.1.1.1 C_{46}^6}{C_{50}^{10}} = 7,76.10^{-4} \end{equation}

I think that my solution may be wrong or have no enough case !

Answer 1

Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\quad\boxed{A}\quad\quad\boxed{A}\quad\quad\quad\boxed{A}\quad\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, and there must be four aces somewhere in the next ten.

Further Simplification

Noting @JMP's comment, a further simplification is possible.

The "strip" of $10$ containing $4$ aces and $6$ non-aces could start anywhere and would necessarily contain two consecutive non-aces on at least one end, thus we can simply write

$Pr = \dfrac{10}{52}\dfrac{9}{51}\dfrac{8}{50}\dfrac{7}{49}$