Game probability I hope you could help me with this problem... 
The game goes this way:
There are 6 players, numbered 1 to 6.
Player 1 starts the game, he rolls a die with six faces. If the result (x) of rolling the die is 1 then Player 1 wins. Else the player number x starts his turn. The game goes on and the Player x rolls the die, if the result (y) is equal to x then Player x win, else it's the turn of Player y. And so on.
I thin the probability of the Player 1 to win is $\frac{2}{7}$. But the question is: If player one has won, wich is the expected number of times he thrown the dice?
I try to do this:
$p_n$ = probability that player 1 wins on the n round.
$$p_1 = 1/6$$
$$p_2 = 0$$
$$p_3 = \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}$$
If I have a general formula for $p_n$ I would sum $$n \times p_n$$ for all value of $n$ and I have the expected number of rounds. But I can't find this $p_n$.
Thank you so much!!!
 A: Note: I extended this answer and fixed a missing $z$ in equation $(\star)$ based on the comments.
For each player the probability of win changes depending on whether they are the one rolling. Pick any player. Let $P$ be the probability that the player wins given that they are rolling and $Q$ be the probability that the player wins if someone else is rolling.
If the player is rolling, then they win in that roll with probability $1/6$ and have to pass the dice with probability $5/6$ and become a non-rolling player. So
$$P = 1/6 + 5/6 Q.$$
Similarly,
$$
Q = 4/6 Q + 1/6 P.$$
Solving the equations gives $P = 2/7$ and $Q = 1/7$. So the probability that player 1 wins, as you said, is 2/7.
Now let $f(z)=\sum_{n=1}^\infty p_n z^n$, where $p_n$ is the probability that player 1 wins in $n$ rounds given that she is rolling; and $g(z)=\sum_{n=1}^\infty q_n z^n$, where $q_n$ is the probability that player 1 wins in $n$ rounds given that she is not rolling. Similar to the previous part:
\begin{align}
f(z)&=1/6 z + 5/6 z g(z)\\
g(z)&=1/6 z f(z) + 4/6 z g(z),\qquad(\star)
\end{align}
which results in 
$$f(z)=\frac{2 z (3-2 z)}{(z+6) (6-5 z)}$$ (note that $f(1)=2/7$). The requested expected value is $f'(z)/f(z)$ at $z=1$ so it equals $$27/7.$$ 
If we are only interested in the number of throws of player 1, then the equations change as follows:
\begin{align}
f(z)=&1/6 z + 5/6 z g(z)\\
g(z)=&1/6 f(z) + 4/6 g(z),
\end{align}
Note that $z$ does not count the throws by the other players. So we get
$$
f(z)=\frac{2z}{12-5z}\cdot
$$
Now $f'(1)/f(1)=12/7$ is the expected number of throws of player 1, given that she wins.
A: Edit: As pointed out in a comment, only a part of this calculation is required to answer the question.

Let's analyze the entire game under the condition that player $1$ wins.
If another player rolls, player $1$ would win with probability $\frac27$ if she gets the die and with probability $\frac17$ if another player gets the die, so, given that she wins, she gets the die with probability
$$
\frac{\frac15\cdot\frac27}{\frac15\cdot\frac27+\frac45\cdot\frac17}=\frac13\;.
$$
If she rolls herself, she would win with probability $1$ if she rolls her own number and with probability $\frac17$ if she doesn't, so, given that she wins, she rolls her own number with probability
$$
\frac{\frac16\cdot1}{\frac16\cdot1+\frac56\cdot\frac17}=\frac7{12}\;.
$$
Now we can set up equations for the expected number of rolls $x$ if she's rolling and $y$ if another player is rolling:
\begin{align}
x&=1+\frac5{12}y\;,\\
y&=1+\frac13x+\frac23y\;.
\end{align}
Substituting the first equation into the second yields $y=\frac{48}7$, and thus the desired expected length is $x=\frac{27}7$.
Here's Java code that confirms the result numerically.
