# Why does probability recursion not work in this case? “What is the probability that the person who makes the first roll wins the game?”

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?

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.