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I attemped to solve this problem using a method that is slightly different from the solutions provided (but results in a correct answer), and I'd like to check if the logic is correct (it seems incorrect).

I created two opposing "players" - player 1, and then a group-player(player 2) that is players 2-6. I labeled the probability player 1 wins (eventually) as $p_1$ and the probability player 2 wins (eventually) as $p_2$.

Then, I wrote $p_1$ as the below, since player 1 hits 1 with 1/6 probability, and the game moves to player 2 with probability 5/6. When player 2 is rolling, player 2 must lose for player 1 to win, so I multiply by $P(\text{player 2 loses}) = 1-p_2$.

$p_1 = \frac{1}{6} + \frac{5}{6}(1-p_2)$

I wrote $p_2$ as the below, since player 2 wins with probability 1/6, gets to roll again with probability 4/6, and gives the dice back to player 1 with probability 1/6.

$p_2 = \frac{1}{6} + \frac{4}{6}p_2 + \frac{1}{6}(1-p_1)$.

Solving this system of equations yields $p_1 = 2/7$ and $p_2 = 6/7$. Clearly this doesn't make sense since by my construction, $p_1$ and $p_2$ are disjoint and should add to 1. Where did I go wrong in my setup?

mk0219
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  • This is hard to follow. Given that it is player $2's$ turn, the probability that they lose (on that round) is $\frac 56$, not what you wrote. I suggest: study the answer posted by Andre Nicholas. I think that he does (correctly) what you try to do here. – lulu Mar 19 '24 at 18:10
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    How would your calculations change if "player(s) 2" were the beginning player rather than player 1? Since they didn't I should think it has to do with that. – JMoravitz Mar 19 '24 at 18:43
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    To correct this, I believe if we interpret $p_2$ instead as the probability that player(s) 2 eventually win given that player 1 did not win on their first turn that this fixes all of the problems. Indeed, if $P_2$ is the probability player(s) 2 win overall, we would have $P_2 = \Pr(\text{player1 didn't win on first turn}\cap \text{player2 wins eventually}) = \frac{5}{6}\cdot \frac{6}{7}=\frac{5}{7}$ as expected. – JMoravitz Mar 19 '24 at 18:48
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    $p_1 = 1/6 + 5/6(1-p_2)$ and $p_1 + p_2 = 1$ gives an incorrect answer (it gives $p_1 = 1$), so either you are not defining your variables correctly or the first equation is wrong. – DanielV Mar 19 '24 at 18:52
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    Your setup is fine. $p_1$ is the probability that player $1$ eventually wins given that it's currently his roll. $p_2$ is the probability that any other player eventually wins given that it's currently any other player's roll. There's no reason these should add up to $1$. – mjqxxxx Mar 19 '24 at 19:58
  • Thank you for the comments - my question is answered. – mk0219 Mar 19 '24 at 20:57

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