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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.

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

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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.

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Forth way: Clearly, the expected number of kings equals the expected number of queens, the expected number of aces and so on. By symmetry, each of these is one thirteenth of the "expected" number of arbitrary cards, which is of course two. – Hagen von Eitzen Feb 27 '13 at 17:56
@HagenvonEitzen: Best! I expect it is expected that OP will calculate the expectation using the first or second way. The third way is an attempt to introduce a useful idea. Exploiting symmetry explicitly, as you did, is cleanest. – André Nicolas Feb 27 '13 at 18:08

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