Probability - Statistics expected value A box contains $10$ red balls and $6$ blue balls, every time a ball is taken from the box it gets replaced with the same ball, from another unlimited supply of red and blue balls.

What is the expected number of balls you will need to take to have at least $2$ of each colour? (at least $2$ red and at least $2$ blue balls).

Any ideas how to tackle this problem?
So far i have something along the lines of... Let X be the number of balls taken from the bag until 2 of each colours has been taken. a = red ball, b= blue ball.... P(a) = 10/16, P(b) = 6/16. minimum balls needed to be taken is 4, there are 4C2 different ways of choosing these balls
 A: The following diagram depicts the state transition graph of this ball drawing game:

Here, the red/blue arrows denote that we've just drawn a red/blue ball. The numbers in the circles denote the number of balls we have drawn so far. $2$ means that we have drawn at least $2$ balls of the given color. 
There are six different basic pathways to get from state $(0,0)$ to state $(2,2)$. Let $p$ denote that we've just drawn a red ball. Here is the list of the probabilities that we follow the different paths:


*

*The uppermost (black path)


$$p^{2+a+b}(1-p)^2.$$
This is because we have to draw two red balls first then we have to draw $a$  red balls in a row then we have to draw a blue ball then again $b$ red balls in a row and, finally we have to draw a blue ball. In what follows I will not explain the results.


*The next is the brown path:


$$p^{2+c}(1-p)^{2}.$$


*The grey path:


$$p^{2}(1-p)^{2+d}.$$


*The purple path:


$$p^{2+e}(1-p)^{2}.$$


*The yellow path:


$$p^{2}(1-p)^{2+f}.$$


*The pink path:


$$p^{2}(1-p)^{2+g+h}.$$
$a,b,c,d,e,f,g,h$ are the number of steps the process has spent at the loopy states. (The "loopy numbers")
From this point on, I will calculate the probabilities belonging the different numbers of steps needed to get at least two balls of both colors.
$4$ steps. 
We can follow all the paths. However, all the loopy numbers have to be zeros. That is
$$P_1=6p^2(1-p)^2=3(p^2(1-p)^2+p^2(1-p)^2).$$
$5$ steps.
Along the black path, there are two possibilities: $a=1,b=0$ and $a=0, b=1$. Along the brown path, there is only one possibility... Along the pink path there are two possibilities again. As a total, we have 
$$P_5=4(p^3(1-p)^2+p^2(1-p)^3).$$
In general, we have 
$$P_k=(k-1)(p^{k-2}(1-p)^2+p^2(1-p)^{k-2}), \ \ k\geq 4.$$
The expectation is
$$\sum_{k=4}^{\infty}k(k-1)(p^{k-2}(1-p)^2+p^2(1-p)^{k-2}).\tag 1$$
I didn't compute the closed formula.$^*$ But for $p=\frac58$ the result is approximatively $6.0645833.$

$^*$ The key to the closed formula is the observation that $k(k-1)p^{k-2}$ and $k(k-1)(1-p)^{k-2}$ are the second derivatives of $p^k$ and $(1-p)^k$, respectively.
EDIT
(An explanation regarding the remark above.)
In order to calculate the expectation we have to compute two infinite sums. The first is
$$\sum_{k=4}^{\infty}k(k-1)p^{k-2}.$$
This is the second derivative of 
$$\sum_{k=4}^{\infty}p^{k}=p^4\sum_{i=0}^{\infty}p^i=\frac{p^4}{(1-p)}.$$
So, 
$$\sum_{k=4}^{\infty}k(k-1)p^{k-2}=\frac {d^2}{dp^2}\frac{p^4}{1-p}=\frac{12p^2(1-p)+4p^3}{(1-p)^2}+\frac{4p^3(1-p)^2+2p^4(1-p)}{(1-p)^4}.$$
The same applies to the second part of (1). Also, we will have to muliply the first part by $(1-p)^2$ and the second part by $ p^2$. 
