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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.

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    $\begingroup$ I think this title is more "searchable". FYI there is a flag suggesting that the question be closed for not having any. Usually the users raising such flags would like you to share your own thoughts. Now that an elegant answer has been posted, I don't know what to suggest. Try to do better with your next question, and hope you enjoy the site. $\endgroup$ – Jyrki Lahtonen Oct 4 '15 at 16:53
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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 $p$.

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  • $\begingroup$ Is there really such a symmetry here? Probability that Player 1 wins on the first roll is $\frac 16$, but probability that Player 2 wins is $\frac 56\cdot\frac 16$, since Player 2 loses immediately if Player 1 already has won. $\endgroup$ – Ennar Oct 4 '15 at 13:01
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    $\begingroup$ There is symmetry between Players 2 to 6, that is what I used. Of course 1 is special. $\endgroup$ – André Nicolas Oct 4 '15 at 13:06
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    $\begingroup$ And the solution is $p=\frac27$ $\endgroup$ – Alice Ryhl Oct 4 '15 at 16:39
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    $\begingroup$ @KristofferRyhl: Yes, that's right. And the others each have probability $\frac{1}{7}$ of ultimately winning, the first player is twice as likely to win as any of the others. $\endgroup$ – André Nicolas Oct 4 '15 at 16:41
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    $\begingroup$ And, clearly, the way to make the game fair is to roll the die first to choose the first player :-) $\endgroup$ – Mark Hurd Oct 4 '15 at 16:51
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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).

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    $\begingroup$ This is actually a more illuminative way to solve. The second equation gives us $p=2q$ (which explicitly gives us relative advantage of the first player) and we don't need the first equation as $p+(6-1)q=1$. $\endgroup$ – A.S. Oct 17 '15 at 21:04
  • $\begingroup$ "Let q be the probability that player 1 wins when it is someone elses turn." How can player 1 win on someone else's turn ? $\endgroup$ – true blue anil Oct 17 '15 at 22:46
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    $\begingroup$ @trueblueanil $q$ is probability to win eventually/overall if the current turn is not that of a player, not right after this turn. Think of player 2 at the beginning of the game and let $q$ be the probability that he wins. If the first roll is 1, P2 doesn't win. If the roll is 2, his probability to win becomes $p$ as now it's the same game and it's his turn to roll. If the first roll is not 1 or 2, his probability to win stays $q$. Hence average of the first two probabilities must be $q$: $q=(p+0)/2$. $\endgroup$ – A.S. Oct 17 '15 at 22:58
  • $\begingroup$ What's the lacuna here. A can win on 1st turn Pr = 1/6. If she loses, with Pr = 5/6, someone, say X is in the same position as A was, so if A's ultimate winning Pr = p, that of X = 5p/6, and by symmetry ultimate winning Pr of all non-A's = 25p/6, p+25p/6 = 1, p = 6/31 $\endgroup$ – true blue anil Oct 18 '15 at 0:10
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    $\begingroup$ @tr " that of X = 5p/6" is false $\endgroup$ – A.S. Oct 18 '15 at 6:37

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