Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number of dices that are sixes after n rolls.
$$\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \end{pmatrix}$$
I want to figure out how many turns it takes for me on average to get all sixes.
I'm not even sure where to start with this problem. Is it a right approach to write a program where I calculate $P^n$ and see when the 6th column all equals to 1? 
Any pointers would be greatly appreciated.

An other method
Each turn you set aside any die which shows a six.
Let $F_n$ be the expected number of turns until you set aside at least one of $n$ die.   Let $E_n$ be the expected number of turns until you set aside all $n$ die.   Let $p_n(k)$ be the (conditional)probability of setting aside $k$ die in a turn when given that you set aside at least one.
$$\begin{align}p_n(k) &=\binom nk\dfrac{ 5^{n-k}}{(6^n-5^n)}\mathbf 1_{k\in \{1,..,n\}}
\\ F_n &= 6^n/(6^n-5^n)
\\ E_1 &= F_1\\ &=6
\\ E_2 &= F_2+p_2(1)E_1 \\ &= 36/11+2(5/11)\cdot 6\\ & = 96/11
\\ E_3 &= F_3+p_3(1)E_2+p_3(2)E_1
\\ E_4 &= F_4+p_4(1)E_3+p_4(2)E_2+p_4(3)E_1
\\ E_5 &= F_5+p_5(1)E_4+p_5(2)E_3+p_5(3)E_2+p_5(4)E_1
\\ E_6 &= F_6+p_6(1)E_5+p_6(2)E_4+p_6(3)E_3+p_6(4)E_2+p_6(5)E_1
\end{align}$$
$$$$
