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.
What is the probability of the Player 1 to win?
I do this: pn = probability that player 1 wins on the n round. $$p_1 = 1/6$$ $$p_2 = 0$$ $$p_3 = \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}$$ $$p_4 = \frac{5}{6} \times \frac{4}{6} \times \frac{1}{6} \times \frac{1}{6}$$
If I have a general formula for $p_n$ I would sum it for all value of $n$ and I have the probability I want. But I can't find this $p_n$.
Thank you!!!
By conditioned on the first rolling, $P_{1wins}=1/6+(5/6)^5*P_{1 wins}$, then solve the equation.