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?

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.
