Can someone help me find the answer to the probability question?

Question

A gambler mixed a "cheat" die with all sixes into a box of eight normal dice. She chooses one at random, rolls it twice, and gets six both times. What is the probability that she chose the "cheat" die?

Answer 1

$$P(\text{Cheat die|two sixes}) = P(\text{Cheat die and two sixes}) / P(\text{Two sixes}) = P(\text{Two sixes| cheat die}) * P(\text{Cheat die}) / P(\text{Two Sixes}) = 1 * 1/9 * 1/(8/9 * 1/36 + 1/9 * 1) = 9/11$$ using the law of total probability to evaluate the probability of two sixes.

Answer 2

$P(\text{Cheat Die}) = \frac{1}{9}$

$P(\text{Fair Die}) = \frac{8}{9}$

$P(\text{2 6's/Cheat Die}) = 1\times\frac{1}{9}$

$P(\text{2 6's/Fair Die}) = \frac{1}{36}.\frac{8}{9} = \frac{2}{81}$

Apply Bayes Theorem now to find the $P(\text{Cheat Die/ 2 6's}) = \dfrac{\frac{1}{9}}{\frac{1}{9}+\frac{2}{81}} = \frac{9}{11}$