Probability of rolling two dice with either showing a six We roll two dice, one red and one green. Under each assumption below, what is the probability that the roll is double sixes?
a) The red die shows six.
b) At least one of the dice shows a six.
I think this is pretty straight forward:
For a) it is $\dfrac{1}{6}$ 
For b) it is $\dfrac{6+6+6-2}{36} \quad$ ($6$ for only red, $6$ for only green, $6$ for both green and red have sixes, $2$ redundant cases) 
Please correct me if I'm wrong.

I am suspicious of your second computation.  I think it would be easier to think in terms of inclusion-exclusion and independence:
\begin{align}
P(\text{at least one 6})
&= P(\color{red}{\text{red 6}} \lor \color{green}{\text{green 6}})\\
&= P(\color{red}{\text{red 6}}) + P(\color{green}{\text{green 6}}) - P(\color{red}{\text{red 6}} \land \color{green}{\text{green 6}}) && (\text{inclusion-exclusion}) \\
&= P(\color{red}{\text{red 6}}) + P(\color{green}{\text{green 6}}) - P(\color{red}{\text{red 6}})\cdot P(\color{green}{\text{green 6}}) && (\text{independence}) \\
&= \frac{1}{6} + \frac{1}{6} - \frac{1}{6}\cdot\frac{1}{6} \\
&= \frac{1}{6} + \frac{1}{6} - \frac{1}{36} \\
&= \frac{6 + 6 - 1}{36} \\
&= \frac{11}{36}.
\end{align}

Alternatively, you can just write out the entire sample space (albeit in an organized manner):
\begin{array}{r|rrrrrr}
 & \color{green}{1} & \color{green}{2} & \color{green}{3} & \color{green}{4} & \color{green}{5} & \color{green}{6} \\\hline
\color{red}{1} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{2} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{3} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{4} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{5} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{6} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare& \blacksquare \\
\end{array}
Here, the black squares represent "good" outcomes (i.e. at least one 6), while the white squares represent "bad" outcomes.  There are a total of 36 outcomes, 11 of which are good, thus
$$ P(\text{at least one 6}) = \frac{\text{good outcomes}}{\text{all outcomes}} = \frac{11}{36},$$
as above.
$\color{white}{\text{Boy, I hope no one is red/green colorblind...}}$
