A radio show conducts a game according to the following rules. A number between $1$ and $100$ is selected at random, and the DJ announces what number is chosen. Call it $n$. The announcer then randomly selects a number between $1$ and $n$ and asks that the first called guess the value of the second number chosen. Assuming that the caller's guess is equally likely to be any number between $1$ and $n$, determine the expected value of the number guessed by the caller.
Do I need to use discrete expected value formula? Or there is a certain method to solve this problem.
Hint: $$\text{If the set }A_i\text{ are mutually exclusive events (}A_i\cap A_j = \emptyset\ \forall i\neq j)$$ $$\text{ such that }P\big(\bigcup_i A_i) = \sum_iP(A) = 1\text{, then}$$ $$E[X] = E[X|\Omega] = E\big[X|\bigcup_i A_i] = \sum_iE[X|A_i]\cdot P(A_i)$$
If you let the event $A_n$ represent that the first draw resulted in the number $n$,
can you find $P(A_n)$, $E[X|A_n]$?
EDIT: (in response to OP comments)
It should be clear that the probability of any $n$ on the first draw should be the same for all $n$. So $P(A_n) = \frac1{100},\ \forall n$.
Now, the question becomes: what is the expected value of the second number, $k$, given that the first number is $n$? Well, if we have fixed $n$, the probability of the second number should be $\frac1n$ for all $k$ values. Using this: $$E[k|A_n] = \sum_{k=1}^{n}k\cdot P(k) = \frac1n\sum_{k=1}^n k$$ Now, taking in the equation from above: $$E[k] = \sum_{n=1}^{100} E[k|A_n]\cdot P(A_n) = \frac1{100}\sum_{n=1}^{100}\frac{n+1}{2} = \ ...$$