There is a difference in my computed expected number of throws by hand and by simulation. The difference is about 1/4, but my question is, which one is wrong?

**Clarification of the rules of this game:**
We have $6$-sided dice, with colours Red, Green, Blue, Cyan, Magenta and Yellow. Based on the throw and the eligible paths you move a pawn. 
The starting state in the game is position $0$. When throwing the right colour you can move your pawn to the next state, otherwise the pawn remains in the same position. The eligible paths are (with the numbers denoting the column):

* 0 -> 1R, 1B
* 1B -> 2C, 2G, 2Y
* 1R -> 2G, 2M, 2Y
* 2C -> 3B, 3M
* 2G -> 3B, 3R
* 2M -> 3R
* 2Y -> 3B, 3M
* 3B -> 4C, 4G, 4M, 4Y
* 3R -> 4M, 4R, 4Y
* 3M -> 4B, 4C, 4G, 4Y

**My interest:**
$E(X\_i)$, with $X\_i$ the number of throws to get in column $i$. $E(X\_0) = 0$, since it is the begin state.

**My computation by hand:**
Let $Z\_j$ the number of throws to get in one of the eligible next states from state $j$,
$$
\begin{align\*}
E(X\_1) &= 3 + E(Z\_{0}), \text{since the propability of throwing one of those is 2/6.}\\
E(X\_2) &= (E(Z\_{1R}) + E(Z\_{1B})/2 + E(X\_1) = 2 + 3, \text{assuming independence between throws.}\\
E(X\_3) &= (E(Z\_{2C}) + E(Z\_{2G}) + E(Z\_{2M}) + E(Z\_{2Y}))/4 + E(X\_2) = (3 + 3 + 6 + 3)/4 + 5\\
E(X\_4) &= ... = 1\frac{2}{3} + 8\frac{3}{4} = 10\frac{5}{12}
\end{align\*}
$$

The difference starts occurring at $E(X\_3)$ in my simulation I get about $8.5$ a difference of about $0.25$, but when I compute $E(X\_4)$ the difference is still about the same.
So I feel like the manual computation of the expected number of throws is wrong, what do you think?