Expectation of $E(XY^2)$ where $X$ and $Y$ are random events related to drawing balls from a vase. The Question:

a vase has 2 white and 2 black balls in it, we remove balls from the vase randomly one by one without returning them. $X=$ the number of balls that we removed until both white balls were removed (incl the last one), $Y=$ the number of balls we removed until we removed the first black one (including the black one). find $E(XY^2)$

My Process so far: 
I constructed a shared measure chart:
https://imgur.com/a/Vmb5DHG
 which I then used to calculate the expectation directly.
$$\begin{aligned} E(XY^2) &= \sum_{x,y} P_{X,Y}(x,y) \cdot xy^2 \\ &= \tfrac{2}{6}\cdot 3 \cdot 1^2 + \tfrac{1}{6} \cdot 4 \cdot 1^2 + \tfrac{1}{6}\cdot 2 \cdot 3^2 \\ &=1+2/3+2 +8/3+3 = 9\tfrac13\end{aligned}$$
The right answer is $9.5$ - So my question is this: 
where am I wrong? I’m not used to creating charts - is there a different way of solving the problem?

any help is really appreciated!
 A: You appear to have accidentally claimed that for the sequence BWBW that $X = 3$ and $Y=1$ when actually it's $X = 4$ and $Y=1$. Everything else is correct. When you retabulate after this small change you should land on the answer you seek.
A: There are $6$ equiprobable permutations of the set $\{w,w,b,b\}$, and for each one, we compute the values of $X$ and $Y$ and $XY^2$ as follows:
$$\begin{array}{c|c|c|c}
\text{Outcome} & \color{red}{X} & \color{blue}{Y} & XY^2 \\
\hline
(w,\color{red}{w},\color{blue}{b},b) & 2 & 3 & 18 \\
(w,\color{blue}{b},\color{red}{w},b) & 3 & 2 & 12 \\
(w,\color{blue}{b},b,\color{red}{w}) & 4 & 2 & 16 \\
(\color{blue}{b},w,\color{red}{w},b) & 3 & 1 & 3 \\
(\color{blue}{b},w,b,\color{red}{w}) & 4 & 1 & 4 \\
(\color{blue}{b},b,w,\color{red}{w}) & 4 & 1 & 4 \\
\end{array}$$
I have color coded the respective ball draws that correspond to $X$ and $Y$.
Again, since each of these outcomes is equiprobable, the desired expectation is simply
$$\operatorname{E}[XY^2] = \frac{18+12+16+3+4+4}{6} = 9.5.$$

In the general case where there are $n$ balls of each color (so $2n$ in total), there are $\binom{2n}{n}$ equiprobable elementary outcomes.  Let the random variable $X$ denote the number of balls needed to obtain the final white ball, and $Y$ denote the number of balls needed to obtain the first black ball.
Suppose we observe $(X,Y) = (x,y)$.  This means the last occurrence of a white ball is in position $x$ and the first occurrence of a black ball is in position $y$.  This in turn implies that all of the balls prior to position $y$ must be white, and all of the balls following position $x$ must be black.  The order and color of all of the balls in between positions $x$ and $y$ may be freely chosen.  So for example, such an outcome might look like this:
$$(w, \ldots, w, \color{blue}{b}, \ldots, \color{red}{w}, b, \ldots, b)$$
where again I have colored the relevant positions of the white and black balls.  Notice however, there is a unique exception, which is the outcome
$$(w, \ldots, \color{red}{w}, \color{blue}{b}, \ldots, b),$$
corresponding to the outcome in which the first $n$ balls are white and the last $n$ balls are black.  This is the only possible way to have $X < Y$, since if any $b$ precedes the final $w$, then we would have $Y < X$.
So exception aside, we may assume $x > y$ and for each permissible outcome $(X,Y) = (x,y)$, the color of the balls in positions $(1, \ldots, y)$ and $(x, \ldots, 2n)$ are fixed.  The colors in positions $(y+1, \ldots, x-1)$ are free within the constraint that there are $n-y$ white balls to choose from, and $x-n-1$ black balls to choose from.  So this means there are $$\binom{x-y-1}{n-y}$$ outcomes where $X = x$ and $Y = y$.
Hence we have $$\operatorname{E}[XY^2] = \binom{2n}{n}^{-1} \left( n(n+1)^2 + \sum_{y=1}^n \sum_{x=n+1}^{2n} xy^2 \binom{x-y-1}{n-y} \right),$$ where the additional $n(n+1)^2$ term arises from the single exceptional case where $x = n$ and $y = n+1$ described above.
The evaluation of this sum for $n = 2$ yields $9.5$ as explicitly enumerated above.  For $n = 3$, we get $\operatorname{E}[XY^2] = 18.3$, which can also be checked by enumeration.
A closed form for general $n$ is
$$\operatorname{E}[XY^2] = 12n - 28 + \frac{1}{n+1} + \frac{48}{n+2} + \frac{(2n+3)}{2 \binom{2n-1}{n}}.$$  However, as the evaluation of the above double sum is beyond the scope of this question, I have not included it.
