What is the probability of rolling 6 on two cubes when we only count the higher of the two numbers

Question

We roll a six-sided die twice in a row and count the larger of the two different numbers. How likely is it to get a 6 this way? (This problem was translated and there isn't any additional information).

I am very confused - firstly, I tried counting all of the different possibilities which would fit the criteria such as rolling:

1. 1,6
2. 2,6
3. 3,6
4. 4,6
5. 5,6
6. 6,1
7. 6,2
8. 6,3
9. 6,4
10. 6,5

There are 10 different possibilities where the higher of the 2 numbers would be 6, so I originally wrote the answer such as 10/36 But then I came with more possibilities and I do not know if they are correct such as:

1. Counting an additional 11th possibility of rolling 6,6 (I don't know if it should be included due to the two numbers being the same value, so I don't know if it counts as a six, because otherwise I would assert that 6 is bigger than 6)

2. Reducing the total number of possibilities from 36 to 21 (so that I would remove the repeating ones but I am not sure if it's correct)

Answer 1

Let $X$ and $Y$ represent two dice, then

$$ \Pr [\max\{X,Y\}\leqslant k]=\Pr [X\leqslant k\text{ and }Y\leqslant k]=\Pr [X\leqslant k]\Pr [Y\leqslant k]\tag1 $$

where in the last equality we used the assumption that $X$ and $Y$ are independent (the result on any of them doesn't influence the result in the other). Now, for a (fair) die we have that $\Pr [X=k]=\frac1{6}$ so $\Pr [X\leqslant k]=\frac{k}{6}$, then from (1) we have that $\Pr [\max\{X,Y\}\leqslant k]=\frac{k^2}{36}$, and consequently

$$ \begin{align*} \Pr [\max\{X,Y\}=k]&=\Pr [\max\{X,Y\}\leqslant k]-\Pr [\max\{X,Y\}\leqslant k-1]\\ &=\frac{k^2-(k-1)^2}{36}\\ &=\frac{2k-1}{36} \end{align*} $$