dice - 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!!!