I was reading the problem in this post: Solving for an expected value from discrete random variables:
An urn contains $30$ marbles of which 8 are black, $12$ are red, and $10$ are blue. Randomly, select four marbles without replacement. Let $X$ be the number of black marbles in the sample of four. Compute $E(X)$.
I understand the solution using indicator random variables:
We consider that the drawn marbles can be placed in a line, and on doing so let $X_i$ be the indicator that the $i$-th marble in this line is black. The probability that a particular marble in the line will be black is: $8/30$. Thus $\mathsf E(X_i)=8/30$ for all four marbles.
By the Linearity of Expectation then: $E(X) $ $= \sum\limits_{i=1}^4 E(X_i) = \frac{16}{15}$
By the same logic, if we sampled with replacement, we should get the same answer. To me this is unintuitive. For instance, if we only had two black balls and we ask the same problem, wouldn't the two $E(X)$ still be the same? If we are sampling without replacement shouldn't the fact that sampling $3$ or $4$ black balls being impossible make this value different from where you sample with replacement (and it is possible to sample $3$ or $4$ balls)?
Can someone give me some intuition of why this is reasonable? Much appreciated!