EXAMPLE:
A 7-segment display shows any number from 0 to 9 at random (equal
probabilities).

Let $X$ be the indicator random variable of
whether the blue segment is on. Similarly, $Y$ is the
indicator for the red segment. Find the conditional distribution
of $Y$ given $X.$
SOLUTION:
Here $X,Y$ both take values in $\{0,1\}.$
We need to find $P(Y=y | X=x)$ for $x,y\in\{0,1\}.$
Now $P(Y=1|X=1) = P(X=1,Y=1)/P(X=1).$
Both the blue and the red segments are on in only the numbers
3,4,5,6,8,9. So $P(X=1,Y=1) = \frac{6}{10}.$
The blue segment is on in the numbers 2,3,4,5,6,8,9. So $P(X=1) =\frac{7}{10}.$
Hence $P(Y=1|X=1) = P(X=1,Y=1)/P(X=1) = \frac 67.$
You should now be able to work out the other three conditional
probabilities similarly.
We can define conditional CDF or conditional PMF in the obvious
way.
It is important to understand that the conditional
expectation/variance is a random variable, which is a function of
the conditioning random variable.

Remember the throm of total probability:
$$
P(A) = P(B) P(A|B) + P(B^c)P(A|B^c),
$$
where combined the two conditional probabilities of $A$ to
arrive at the (unconditional) probability of $A?$
Well, we can do similar things with conditional
expectation/variance also.

Proof:
Let $X$ take values $x_1,x_2,...$ and $Y$ take
values $y_1,y_2,...$. Let the joint PMF of $(X,Y)$ be
$$
P(X=x_i~\&~Y=y_j) = p_{ij}.
$$
Then $P(Y=y_j | X=x_i) = \frac{p_{ij}}{p_{i\bullet}}.$

So $E(Y|X=x_i) = \sum_j y_j \frac{p_{ij}}{p_{i\bullet}}.$
Expectation of this is
$$
\sum_i E(Y|X=x_i) p_{i\bullet} = \sum_i \sum_j y_j
\frac{p_{ij}}{p_{i\bullet}}p_{i\bullet} = \sum_i \sum_j y_j p_{ij} =
\sum_j y_j \sum_i p_{ij} = \sum_j y_j p_{\bullet j} = E(Y),
$$
as required.
[QED]
Many expectation problems can be handled step-bystep using this
result. Here are some examples.

EXAMPLE:
A casino has two gambling games:

Roll a fair die, and win Rs. $D$ if $D$ is the
outcome.

Roll two fair dice, and win Rs 5 if both show the same
number, but lose Rs 5 otherwise.

You throw a coin with $P(Head)=\frac 13$ and decide to play game
1 if $Head,$ and game 2 if $Tail.$ What is your
expected gain?
SOLUTION:
Let $X$ be your gain (in Rs), and let $Y$ be the outcome of the
toss.
Then $E(X|Y=Head) = 3.5$ and $E(X|Y=Tail) = 5\times\frac{6-30}{36}=-\frac{10}{3}.$
So, by the tower property, $E(X) = P(X|Y=Head)\times P(Y=Head)+P(X|Y=Tail)\times P(Y=Tail) = \cdots.$
The tower property is very useful for computing expectations
involving a random number of random variables. Here is an
example.

EXAMPLE:
A random number $N$ of customers enter a shop in a
day, where $N$ takes values in $\{1,...,100\}$ with
equal probabilities. The $i$-th customer pays a random amount $X_i$,
where $X_i$ takes values in $\{1,2,...,10+i\}$
ith equal probabilities. Assuming that $N,X_1,...,X_N$ are
all independent, find the total expected payments by the
customers on that day.

SOLUTION:
We have $E(X_i) = \frac{11+i}{2}.$
So $E\left(\sum_1^N X_i|N\right) = \sum_1^N E(X_i|N) = \sum_1^N E(X_i) = \sum_1^N \frac{11+i}{2} = 5.5N+\frac{N(N+1)}{4}.$
By tower property, the required answer is $E\left(5.5N+\frac{N(N+1)}{4}\right)=\cdots.$

EXAMPLE:
10 holes, numbered 1 to 10, in a row. 5 balls are dropped
randomly in them (a hole may contain any number of balls). Call a
ball "lonely" if there is no other ball in its hole or the
adjacent holes. Find the expected number of lonely balls.

SOLUTION:
Define the indicators $I_1,...,I_5$ as
$$
I_i = \left\{\begin{array}{ll}1&\text{if }i\mbox{-th ball is lonely}\\0&\text{otherwise.}\end{array}\right.
$$
Then the total number of lonely balls is $X = \sum I_i.$
So we are to find $E(X) = \sum E(I_i).$
Let $Y_i = $ the hole where the $i$-th ball has fallen.
Then $E(I_i|Y_i=1)$ is the conditional probability that
all the balls except the $i$-th one has landed in
holes $2,...,10$ given that the $i$-th ball has landed
in hole 1.
You should be able to compute this easily. Similarly, you can
compute $E(I_i|Y_i=k)$ for $k=1,...,10.$
Notice that $Y_i$ can take values $1,...,10$ with equal probabilities.
So tower property should provide the answer as
$$
E(X) = \sum E(E(I_i|Y_i)) = \cdots.
$$

Proof:
This follows directly from the tower property.

If $X,Y,Z$ are jointly distributed random variables, then we
can talk about conditional distribution of $Z$
given $(X,Y)$ or $X$ given $Z$ or $(X,Z)$
given $Y,$ etc. We can even condition step by step. For
example, we can talk about $E(E(Z|X,Y)|X).$ This is a
function of $X$ alone.

Let $I_j$ be the indicator variable for whether there is a
record at position $j.$ Then $P(I_j=1)$ may be computed
by total probability:
$$
P(I_j=1) = \sum_{k=j}^n P(X_j=k)P(I_j=1|X_j=k).
$$
Similarly for $P(I_jI_k=1).$

The problem is basically optimising $\sum P_i^2$ subject
to $\sum P_i$ being fixed. Cauchy-Scwartz might help.