Expected value of number of draws We have $5$ number in a bag: $(1,3,5,7,9)$. We draw one from the bag and then put it back. We do this until the sum of the numbers can be divided by $3$. Whats the expected value of the number of draws?
My idea was to solve it with Markov-chains:
States: $0,1,2$. So numbers$\mod 3$.
The matrix will be:
$\begin{pmatrix}
2/5 & 2/5 & 1/5 \\
1/5 & 2/5 & 2/5 \\
2/5 & 1/5 & 2/5
\end{pmatrix}$
Then we have an equation system: 
$k_1=1+2/5k_1+2/5k_2$, $\ k_2=1+1/5k_1+2/5k_2$ and $k_3=0$.
Even if I solve this, I'm not sure how to continue. Thanks for help.

We use your idea, but partly for ease of typing we use different notation. Say that we are in State $1$ if the sum so far is congruent to $1$ modulo $3$, and that we are in State $2$ if the sum is congruent to $2$. Let $m$ be the mean waiting time.  Let $a_1$ be the additional mean waiting time, given we are in State $1$, and $a_2$ the additional mean waiting time given that we are in State $2$.
On the first pick, we either get something divisible by $3$, in which case we have spent $1$ pick, and our additional waiting time is $0$. Or else we get a $1$ or a $7$, and our mean additional waiting time is $a_1$. Or else we get a $5$, and our mean additional waiting time is $a_2$. Thus
$$m=1+\frac{2}{5}a_1+\frac{1}{5}a_2.\tag{1}$$
Suppose we are in State $1$, and do another trial.  If we get a $3$ or a $9$, we stay in State $1$, and our additional expected time remains at $a_1$. If we get a $1$ or a $7$, our additional expected waiting time becomes $a_2$. And if we get $5$, we are finished. Thus
$$a_1=1+\frac{2}{5}a_1+\frac{2}{5}a_2.\tag{2}$$
Similarly, 
$$a_2=1+\frac{1}{5}a_1+\frac{2}{5}a_2.\tag{3}$$
Solve the last two equations for $a_1$ and $a_2$, and substitute in Equation (1).
