Conditional probability of min and max of two dice Consider the following problem, from Tijms's Understanding Probability:

Two dice are rolled. Let the random variable $X$ be the smallest of the two outcomes and let $Y$ be the largest of the two outcomes. What are the conditional mass functions $P (X = x | Y = y)$ and $P (Y = y | X = x)$?

My attempt:
$$
P(X=x, Y=y) = \begin{cases}
  \frac{1}{36} & \text{if }x=y \\
  0            & \text{if }x>y \\
  \frac{2}{36} & \text{if }x<y \\
\end{cases}.
$$
For the individual probabilities, we have that one of the two outcomes is fixed and has to be equal to the minimum/maximum, the other dice can roll any number between the minimum and 6, or between 1 and the maximum. The order does not count, so I multiply by 2:
$$
P(X=x) = \frac16 \frac{6-x+1}{6}\cdot 2,
$$
and
$$
P(Y=y) = \frac16 \frac{y}{6}\cdot 2.
$$
Putting everything together, we have:
$$
P(X=x|Y=y) = \frac{P(X=x,Y=y)}{P(Y=y)} = \begin{cases}
  \frac{1}{2y} & \text{if }x=y \\
  0           & \text{if }x > y \\
  \frac{1}{y} & \text{if }x < y \\
\end{cases}.
$$
and
$$
P(Y=y|X=x) = \frac{P(X=x,Y=y)}{P(X=x)} = \begin{cases}
  \frac{1}{2(6-x+1)} & \text{if }x=y \\
  0           & \text{if }x > y \\
  \frac{1}{6-x+1} & \text{if }x<y
\end{cases}.
$$
Does it sound right?

Partial answer due lack of time
For $P(Y=y)$ I have something slightly different. I´ve worked with the table below:
$P(Y=y|X=x)$
$$ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|}  \hline \text{ X/Y }  & 1 &2 &3 &4 &5 &6  \\ \hline \hline \hline 1 &\frac{1}{36} &\frac{2}{36} &\frac{2}{36}& \frac{2}{36}& \frac{2}{36}& \frac{2}{36}\\  \hline  2& &\frac{1}{36} &\frac{2}{36} &\frac{2}{36} &\frac{2}{36} &\frac{2}{36} \\ \hline 3& & &\frac{1}{36} &\frac{2}{36} &\frac{2}{36} &\frac{2}{36}\\ \hline 4 & &  & &\frac{1}{36} &\frac{2}{36} &\frac{2}{36} \\ \hline  5 & & & & &\frac{1}{36} &\frac{2}{36} \\ \hline  6& & & & & &\frac{1}{36}  \\ \hline \end{array}$$
Now we can sum up the cells in column $y$ to obtain $P(Y=y)=\frac{2\cdot y-1}{36}$.
We can check if this is plausible by calculating the sum ( sanity check).
$\sum\limits_{y=1}^6 P(Y=y)=\sum\limits_{y=1}^6\frac{2\cdot y-1}{36}=\frac1{36}\cdot \left(  2\cdot \sum\limits_{y=1}^6  y-\sum\limits_{y=1}^61 \right)=\frac1{36}\cdot \left(  2\cdot \frac{6\cdot 7}{2}-6 \right)=1$
This is the result we have expected.
