Two regular dice are rolled. Given that one of them is a $6$, what is the probability that the other is also a $6$? I'm having trouble with the following question for and I have an exam in two days:

Two regular dice are rolled. Given that one of them is a $6$, what is the probability that the other is also a $6$? Given that the first is a $6$ what is the probability that the other is also a $6$? Contrast these two situations in terms of Bayes' theorem, with particular attention to the amount of information that is made available in each case.

I think the answer is just $1/6$ for both but I presume that's wrong and I am not sure how Bayes theorem helps as $P(\text{1st is a 6 given 2nd is a 6})$ is not much more helpful than $P(\text{2nd is a 6 given 1st is a 6})$ in this context.
 A: Let $A$ be the value of the first die, and let $B$ be the value of the second die.
a) 
\begin{align*}
P(\text{Two sixes}|\text{One six})&=\frac{P(\text{Two sixes})}{P(A = 6\cup B = 6)}\\
&=\frac{(1/6)^2}{1/6+1/6-1/36}\\
&=\frac{1/36}{11/36}\\
&=\frac{1}{11}
\end{align*}

I think the other/more natural way to state this is
"Given at least one six, what is that chance that you got two sixes". Then this is
\begin{align*}
P(\text{Two sixes | At least one six})&=
\frac{P(\text{Two sixes})}{P(\text{At least one six})}\\
&=\frac{P(\text{Two sixes})}{1-P(\text{No sixes})}\\
&=\frac{(1/6)^2}{1-(5/6)^2}\\
&=\frac{1}{11}
\end{align*}

You could also count.
$$(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2),(6,1)$$
Notice that there are $11$ equally likely outcomes. $1$ of them is desirable. Hence the chance is $1/11$.

b) They tell you the first is a six, hence
$$P(B=6|A=6) = \frac{P(A = 6, B=6)}{P(A = 6)} = \frac{(1/6)(1/6)}{(1/6)} = \frac{1}{6}$$
since the outcome of $A$ is independent of the outcome of $B$.
