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.