Probability of winning in a die rolling game with six players There are 6 players numbered 1 to 6, 1 Player, Player 2, ..., Player 6.
Player 1 rolls a die , if he gets a 1 wins and the game ends, otherwise the die passes to the player whose number matches the number that presents the die and the player makes a second pitch, if is obtained the number of the player who has thrown, he wins and the game ends, otherwise the given passes to the player whose number matches the number rolled, the player rolls the die, if is obtained the number of the player who has thrown, he wins and the game ends, otherwise the die passes to the player whose number matches the number that presents the die in this third release, and so on.
Calculate the probability that player 1 wins.
 A: I also got to 2/7, but in a different way.
Let $p$ be the probability that player 1 wins the game, either on his current turn or in the future. Let $q$ be the probability that player 1 eventually wins when it is someone elses turn, in particular the other player does not end the game in the current turn.
Then player 1 can win right away with probability $p$, or first pass the turn to someone else and win later on in the game with probability $q$:
$$p = \frac16 + \frac56 q.$$
When it is someone else's turn, and they don't win, then either it will be player 1's turn again or the turn will pass to one of the other 4 players. So if it is now someone else's turn, the probability that player 1 will get the next turn and win is $\frac16 p$. But if the turn goes to yet someone else with probability $\frac46$ (other than player 1 and the current player) player 1 will eventually win is still $q$. Hence,
$$q = \frac16 p + \frac46 q.$$
Solving these two equations for $p$ is easily done by hand, giving $p = \frac27$ (and $q=\frac17$ which, by symmetry, is the probability of each of the other 5 players winning).
A: Let $p$ be the probability Player 1 (ultimately) wins. If Player 1 does not win on her first toss, by symmetry, the other players all have equal probabilities of being "next", so all have equal probabilities of ultimately winning, namely $\frac{1-p}{5}$. 
On the first toss, either P1 tosses a $1$ and wins immediately, or tosses something else and becomes effectively one of the "other" players. Thus
$$p=\frac{1}{6}+\frac{5}{6}\cdot\frac{1-p}{5},$$
and now we can solve for $p$.
