Explanation
If Carol wins in the first round, then she must have rolled a six after two non-sixes have occurred. This happens with the probability
$$\left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{1}{6}\right)=\left(\dfrac{1}{6}\right)\left(\dfrac{5}{6}\right)^2$$
If Carol wins in the second round, then five non-sixes preceded her six. This happens with probability
$$\left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{1}{6}\right) =\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{5}$$
This pattern continues. Carol wins in the third round with probability $$\left(\dfrac{1}{6}\right)\left(\dfrac{5}{6}\right)^{8}$$ and in the fourth with probability $$\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{11}$$.
It is possible (though unlikely) that the game could continue forever. The probability that Carol wins the game is equal to the sum of the probabilities that she wins in any given round:
$$=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}+\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{5}+\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{8}+....=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}\left[1+\left(\dfrac{5}{6}\right)^{3}+\left(\dfrac{5}{6}\right)^{6}+.....\right]$$
$$=\displaystyle \left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}\sum_{n=0}^{\infty} \left(\dfrac{5}{6}\right)^{3n}=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}\sum_{n=0}^{\infty} \left(\dfrac{5^3}{6^3}\right)^{n}$$
$$=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2} \left[\dfrac{1}{1-\dfrac{5^3}{6^3}}\right]$$
$$=\dfrac{25}{91}$$
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