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.
 A: Just a quick correction to your transition matrix...
$$P =\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} 
\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}
$$
and it only represents your transition probabilities on a single round of the game, not after n rolls. 
One thing you can try doing is calculating 
$$X_n = [1 \ 0 \ 0 \ 0 \ 0 \ 0] P^n = [p_{n,1} \ p_{n,2} \ p_{n,3} \ p_{n,4} \ p_{n,5} \ p_{n,6}]$$
The expected number of rounds played in this game would be given by
$$
S=1+\sum_{n=1}^\infty (p_{n,1}+p_{n,2}+p_{n,3}+p_{n,4}+p_{n,5})
$$
i.e. You're adding up all the elements in the first five columns for $n=1,2,3...$, since those represent the probability of still playing the game after n rounds. You can easily approximate $S$ using a spreadsheet. Calculating it analytically using this method might be tedious, but I would assume is definitely doable.
A: If $E_n$ is the expected number required when $n$ out of $5$ are already a six then clearly $E_5=0$ and you can write (dropping your final column) 
$$\begin{pmatrix}E_0 \\ E_1 \\ E_2  \\ E_3 \\ E_4  \end{pmatrix} =\begin{pmatrix}1 \\ 1 \\ 1  \\ 1 \\ 1  \end{pmatrix} + \begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} \\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} \\ 0& 0 & \frac{125}{6^3} &\frac{75}{6^3}& \frac{15}{6^3}  \\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2} \\ 0 & 0 & 0 & 0 & \frac{5}{6}  \end{pmatrix} \begin{pmatrix}E_0 \\ E_1 \\ E_2  \\ E_3 \\ E_4  \end{pmatrix}$$
which you can either solve by hand or, if $M$ is that square matrix,  find the row sums of $(I-M)^{-1}$ 
I think it should give you $E_5 \approx 13.023662$
A: Yet another way:
Let's call rolling all five dice a "turn".  Let $X$ be the number of the first turn on which all five dice have rolled at least one six.  If $X>n$ then we have at least one die which has not rolled a six by turn $n$.  So
$$\begin{align}
E(X) &= \sum_{n>0} P(X > n) \tag{1} \\
&= \sum_{n=0}^{\infty} \{1 - [1-(5/6)^n)]^5 \} \\
&= \sum_{n=0}^{\infty} \left( 1 - \sum_{i=0}^5 (-1)^i \binom{5}{i} (5/6)^{ni} \right) \tag{2}\\
&= \sum_{i=1}^5 (-1)^{i+1} \binom{5}{i} \sum_{n=0}^{\infty} (5/6)^{ni} \\
&= \sum_{i=1}^5 (-1)^{i+1} \binom{5}{i} \frac{1}{1-(5/6)^i} \tag{3}\\
&= 13.02366
\end{align}$$
$(1)$ is true for any discrete random variable which only takes on non-negative values.
$(2)$ is by the binomial theorem.
$(3)$ is by the formula for the sum of an infinite geometric series.
A: 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.
$$\begin{align}p_n(k) &=\binom nk\dfrac{ 5^{n-k}}{(6^n-5^n)}\mathbf 1_{k\in \{1,..,n\}}
\\ F_n &= 6^n/(6^n-5^n)
\\ E_1 &= F_1\\ &=6
\\ E_2 &= F_2+p_2(1)E_1 \\ &= 36/11+2(5/11)\cdot 6\\ & = 96/11
\\ E_3 &= F_3+p_3(1)E_2+p_3(2)E_1
\\ E_4 &= F_4+p_4(1)E_3+p_4(2)E_2+p_4(3)E_1
\\ E_5 &= F_5+p_5(1)E_4+p_5(2)E_3+p_5(3)E_2+p_5(4)E_1
\\ E_6 &= F_6+p_6(1)E_5+p_6(2)E_4+p_6(3)E_3+p_6(4)E_2+p_6(5)E_1
\end{align}$$
$$$$
