Probability of exactly $2$ aces within first $5$ cards?

Question

Every person gets $5$ cards from a deck of cards ($52$). What is the probability that the first $5$ cards will contain exactly $2$ aces?

I have tried to calculate it by $\frac{5}{52} \times \frac{5}{47} = \frac{25}{2444}$.

I know my answer is incorrect, but I dont know how I should approach this.

Answer 1

You need to think about the number of ways you can get two aces, and divide this by the total number of hands you can get.

Firstly, there are $4 \choose 2$ different ace combinations that you can get. And, given that two cards in your hand are aces, there are $48 \choose 3$ different combinations for the remaining $3$ cards in your hand (note we remove all 4 aces to get 48 remaining cards, since you can only have 2 aces). This gives the total number of ways to get 2 aces as

$ 4 \choose 2$ $\times$ $48 \choose 3 $

Get this number in your calculator and divide it by the total number of possible hands, $52 \choose 5$ to get the answer.

Recall that $n \choose x $$= \frac{n!}{x! (n-x)!}$ and $n! = n\times (n-1) \times ... \times 2 \times 1$

Answer 2

If you want exactly two aces, we can compute the chance that the first two cards are aces and the next three are something else. The chance the first card is an ace is $\frac 4{52}=\frac 1{13}$. Given that the first card is an ace, there are $3$ aces left out of $51$ cards, so the chance is $\frac 3{51}=\frac 1{17}$. Given that the first two cards are aces, the chance for each of the next three not to be is $\frac {48}{50}, \frac {47}{49}, \frac {46}{48}$ giving a total probability of $$\frac {48\cdot 47\cdot 46}{13\cdot 17 \cdot 50 \cdot 49 \cdot 48}=\frac {1081}{270725}\approx 0.004$$ There are ${5 \choose 2}=10$ ways to choose which two cards of the five are aces, so the chance of exactly two aces in a five card hand is just about $4\%$

Answer 3

You can do it by considering the various ways of picking two in the first 5. If X is not ace and A is ace then you could have:

AAXXX, AXAXX, AXXAX, AXXXA, XAAXX, XAXAX, XAXXA, XXAAX, XXAXA, XXXAA.

The probability of each of these will be the same:

$\frac{ 48 \times 47 \times 46 \times 4 \times 3}{52 \times 51 \times 50 \times 49 \times 48}$, multiply this by your number of total combinations and that should give you the answer.

Note that if your question is "at least two" then this will change.