Probability of winning in a die rolling game with six players

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 $p$.

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).