Is my understanding of Expected Value of a Random Variable correct?

Question

I have come up with the following understanding about Expectation of RV.

Please, correct me if I am wrong.

Definition: Expected value of a RV says that, if a random experiment is repeated n number of times, what would be the value that you expect to see most of the times. Expected value of a Random Variable is a value. It is not a probability.

Example: In case of a dice, if the dice is rolled $n$ number of times, the value we would see most of the times is 3.5.

Calculation: We don't have to repeat the same experiment $n$ number of times to find the expected value of a random variable. It can be done using the random variable and its associated probability distribution. In case of a dice,

$ \sum_{x=1}^6 x(1/6) = 1*(1/6) + 2*(1/6) + 3*(1/6) + 4*(1/6) + 5*(1/6) + 6*(1/6) = 3.5 $

Continuous Case

If $X$ is a continuous random variable with density function $f_X(x),$ then the expectation of $X$ is $\displaystyle E(X)=\int_{-\infty}^\infty x\,f_X(x)\,dx$

If $Y=g(X)$ is a function of $X$(i.e. a density function)$,$ then $\displaystyle E(Y)=\int_{-\infty}^\infty g(x)\,f_X(x)\,dx$

Edit:

After reading comments and answers,

Definition:

Expected value of a RV says that, if a random experiment is repeated n number of times, what would be the average value of the outcomes.

Expected value of a Random Variable is a value. It is not a probability.

Example:

In case of a dice, if the dice is rolled $n$ number of times, the average value of the outcomes would be 3.5.

Answer 1

The expected value can be thought of as the long-run average. If you roll your die many, many times, and look at the average of the rolls, you will end up close to 3.5.

This is not the value that you expect to see most often. This would be the mode of the distribution (in the discrete case - for a continuous distribution, any particular value has the same probability of zero). After all, you will never see a roll of 3.5 with your standard six-sided die.

Answer 2

I would like to help - you put $n$ in your definition, but this is exactly what we don't want. Since $n$ is never referenced elsewhere, its presence serves to inform the reader of your definition that the experiment is repeated a finite number of times. So, we can either remove $n$ altogether and state that the experiment is repeated infinitely many times (which is a problem logically) or we can leave $n$ right where it is and state that the expectation is the limit of the average of the values as $n\to\infty$. That is, in your dice example, let's take the average of $n$ values. As we increase $n$, the average gets closer and closer to $3.5$.