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So the first example (where recursion works) the author provided is

You play a dice game with a friend. You roll a fair 6-sided die and your friend rolls a fair 8-sided die. You add $2$ to your roll, and then compare the results. Whoever has the higher result wins the game. If the results are the same, then this process is repeated until there is a winner. What is the probability that you win this game?

His solution is

$$p = \frac{9}{16} + \frac{1}{8}p \implies p = \frac{9}{14}$$

Now the problem that I am interested in is

You and a friend take turns rolling a fair six-sided die, and the first person to roll a 6 wins. What is the probability that the person who makes the first roll wins the game?

I've tried doing this

$$p = \frac{1}{6} + \frac{5}{6}p$$

Which gives me $p = 1$, but the correct answer is $p = \frac{6}{11}$

Am I doing something wrong, or the recursion is not applicable in the second example? If it is not, why?

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Say the first person to roll has probability $p$ of winning. Then that person has a $\frac16$ chance of winning on the first roll. If that does not happen ($\frac56$ of the time), he will only have another chance if the second person doesn't win on their turn, which has a $\frac56$ chance of occurring. Thus the $\frac56$ should be squared and the equation becomes $$p=\frac16+\frac{25}{36}p$$ and $p=\frac6{11}$ as desired.

The erroneous quation $p=\frac16+\frac56p$ would describe a game where a sole player keeps on rolling until getting a six. The result of $p=1$ is correct here, since a six will be rolled almost surely.

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