Deck of cards random variable

Question

How would I compute the expected value of the number of kings randomly drawn pair of cards from a single deck? I tried

$0: (\frac{48}{52})\cdot(\frac{47}{52})$

$1: (\frac{4}{52})\cdot (\frac{48}{52})$

$2: (\frac{4}{52})\cdot (\frac{3}{52}) $

Just to get a broad idea, but it doesn't add up to $1$. Should I add in $n\choose k$? I'm not sure how to approach the problem since drawing one card leaves one less in the deck. Any help would be appreciated.

Answer 1

In the deck of 52 cards there are 4 kings. The probability to get no kings is thus $$ \mathbb{P}\left(\text{no. kings} = 0\right) = \underbrace{ \frac{48}{52}}_\text{1st card is not a king} \cdot \underbrace{\frac{47}{51}}_\text{2nd card is not a king, given that the 1st one is not} $$ $$ \mathbb{P}\left(\text{no. kings} = 2\right) = \underbrace{ \frac{4}{52}}_\text{1st card is a king} \cdot \underbrace{\frac{3}{51}}_\text{2nd card is a king, given that the 1st one is as well} $$

Answer 2

We do the problem is several ways. The reason we are writing out a solution is because of the third approach we will take.

First way: Imagine that we draw a card at random, and then draw another card. Record the cards we got, in the order we got them.

The probability that the first card is a non-King is $\frac{48}{52}$. Given that the first card was a non-King, the probability the second is a non-King is $\frac{47}{51}$. So with probability $p_0=\frac{48}{52}\cdot\frac{47}{51}$ the number of Kings is $0$.

Similarly, the probability the first card is a King and the second is a non-King is $\frac{4}{52}\cdot \frac{48}{51}$. Similarly, the probability of a non-King followed by a King is $\frac{48}{52}\cdot \frac{4}{51}$. These two probabilities are equal. So the probability of $1$ King is given by $p_1=2\cdot\frac{4}{52}\cdot\frac{48}{51}$.

similarly, the probability $p_2$ of $2$ Kings is given by $p_2=\frac{4}{52}\cdot \frac{3}{51}$.

So the expected number of Kings (the mean number of Kings) is $(0)(p_0)+(1)(p_1)+(2)(p_2)$.

Add up. Leave the denominators alone for a while. Then simplify the fraction as much as you can. You should get that the mean number of Kings is $\frac{2}{13}$.

Second way: There are $\binom{52}{2}$ equally likely ways to choose a hand of $2$ cards. The number of $0$-King hands is $\binom{48}{2}$. So the probability the hand has $0$ Kings is $\frac{\binom{48}{2}}{\binom{52}{2}}$.

To count the number of $1$-King hands, note that the King can be chosen in $\binom{4}{1}$ ways, and for each of these the non_king can be chosen in $\binom{48}{1}$ ways. For the probability, divide by $\binom{52}{2}$.

Similarly, the number of $2$-King hands is $\binom{4}{2}$. Divide by $\binom{52}{2}$ for the probability.

Now that we have the probabilities, we can find the mean as in the first way.

Third way: Let random variable $X_1$ be $1$ if we got a King on the first pick, and $0$ otherwise. Let $X_2=1$ if we got a King on the second pick, and $0$ otherwise. Then random variable $K$, the number of Kings, is given by $K=X_1+X_2$. By the linearity of expectation, we have $$E(K)=E(X_1+X_2)=E(X_1)+E(X_2).\tag{$1$}$$ We need to calculate $E(X_1)$. Since $X_1=1$ with probability $\frac{4}{52}$, and $0$ otherwise, we have $E(X_1)=\frac{4}{52}=\frac{1}{13}$.

Similarly, $E(X_2)=\frac{1}{13}$. Thus by Formula $(1)$, we have $E(K)=\frac{2}{13}$.

Remark: The third way is more sophisticated than the other two, but in a sense it is quite a bit simpler. The ideas of the third way may be very useful to you later.