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.

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 $p$.