Finding expected value of number of kings drawn before the first Ace and its lower bound? Consider a standard deck of cards. We draw cards from the top until we get either a King or an Ace.
1.Write a lower bound for the expected value of the number of Kings drawn before the first Ace.
The solution given in this problem is 1/2.
I'm not sure how they arrive at this solution. I assumed that both Kings and Aces are equally likely to be drawn so shouldn't the # of kings being drawn before first Ace appear be $n* \frac {1}{52}$ ?
2. Find the expected value of the number of Kings drawn before the first Ace.
Here since the number of kings in a deck is 4 and there are 52 cards in total shouldn't the expected number of kings drawn before the first Ace be $n* \frac {4}{52}$ ?

Part 2. For the purpose of analysis, we may pretend the deck consists only of four kings and four aces.  Let $X$ be the number of kings before the first ace.
Then $P(X>0)$ is the probability that the first card is a king, so $P(X>0) = \frac{4}{8}$.
$P(X > 1)$ is the probability that the first two cards are kings, so $P(X>1) = \frac{4}{8} \cdot \frac{3}{7}$.
Similarly, we see that $P(X>2) = \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}$, and $P(X>3) = \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5}$.
So $$E(X) = P(X>0) + P(X>1)+P(X>2)+P(X>3) = 0.8$$
Here we have made use of the theorem that
$$E(X) = \sum_{n=0}^{\infty} P(X>n)$$
which holds for any random variable $X$ which takes on only values in $\{0, 1, 2, 3, \dots \}$.
