Expected number of cards drawn before drawing a $4$ or $5$ I'm working on the following problem:

Compute the number of expected cards drawn from a standard 52 card
  deck (without replacement) until a $4$ or $5$ is drawn.

I tried to model it using a geometric distribution, but am running into problems since the probability of drawing a $4$ or $5$ increases with each successive card drawn. Could this problem be approached using Markov Chains?
 A: Let's call the 4s and 5s "special" cards.  Add a joker to the deck and pretend it's an additional special card, so that there are now $9$ special cards in a deck of $53$ cards.  Now shuffle all the cards up and then deal them out, face down, in one big circle.  If you think about it, the average distance between consecutive special cards is $53/9$.  Now locate the joker and think of it as identifying the "top" of the deck.  The average distance to the next special card (which is now either a 4 or a 5) is still $53/9$.
A: The number of ways you can draw $n$ cards from $52$ is $P(n,52)=\frac{52!}{(52-n)!}$. Note that order matters here. The number of ways all of those are not $4,5$ will be $P(n,44)=\frac{44!}{(44-n)!}$. Here $P$ stands for permutations.
Now let $P(n)$ (not the same meaning of $P$) denote the probability that we fail the first $n-1$ draws and succeed in the $n$-th. Then
$$
\begin{align}
P(n)&=\frac{P(n-1,44)}{P(n-1,52)}\cdot\frac{8}{(52-(n-1))}\\
&=\frac{(52-(n-1))!44!}{52!(44-(n-1))!}\cdot\frac{8}{(52-(n-1))}\\
&=\frac{8(52-n)!44!}{52!(44-(n-1))!}
\end{align}
$$
and we can "simply" sum
$$
E=\sum_{n=1}^{45}n\cdot P(n)=\frac{53}{9}=5.\overline 8
$$
to get the expected number of cards to draw.
A: For this problem, we can develop a formula for $P(x=k)$.  Where the variable x represents the draw on which we obtain a success.  $$P(x=1)=2/13$$ $$P(x=2) = 44/52*8/51$$ $$P(x=3)=(44/52)*(43/51)*(8/50)$$continuing in this fashion, we obtain the formula: $$P(x=k)=8*(44!/52!) \frac{(52-k)!}{(45-k)!}$$  Thus, we can use the definition of expected value: $E(x)=\sum_{k=1}^{45}k\cdot P(x=k)$  Unfortunately, I know no method to solve this sum. So of course, I resorted to wolfram which spit out $53/9$.  Thus, $$E(x) = 53/9 = 5.888...$$
