Hypergeometric Random Variable Expectation In a binomial experiment we know that every trial is is independent and that the probability of success, $p$ is the same in every trial. This also means that the expected value of any individual trial is $p$. So if we have a sample of size $n$, by the linearity property of the expectation, the expected value of the same is just $n \cdot p$. This is all intuitive.
When the population size is finite and when we don't replace the items after every trial, we can't use the binomial distribution to get the probability of $k$ successes in a sample of size $n$ where the population is of size $N$ and the number of successes is $R$ simply because the probability of obtaining a success after every trial changes as the $R$ or/and $N$ change(s). So far so good. Yet when they calculate the expected value of the hypergeometric random variable, it is $(n \cdot R/N)$. This seems to me as the same as saying the probability of obtaining a success in every trial is the same ($R/N$) which is not intuitive at all because I should at least be expecting to see $N$ reducing by $1$ after every trial. I know that there's a flaw in my thinking. Can someone help point it out ?
Edit: I think I'm going to give up on understanding why the expected value of the hypergeometric random variable (HRV) is at it is. None of the answers have alleviated my confusion. I don't think I've made my confusion clear enough. My problem is I'm going about the process of finding the expected value of the HRV in the same way as that of the binomial random variable (BRV). In the BRV's case, if the sample is of size $n$ and we consider each item in the sample as random variable of its own, then $X = X_1+X_2+\cdots+X_n$. To find the $E[X]$, we simply add the $X_i$. Since an item is returned after it is checked, the probability of success does not change. In the case of the HRV, I should expect the probability of success to change because it is not returned back to the population. However, this doesn't seem to be the case. This is my problem.

As others have pointed out, the probability of a red ball at each of your $n$ draw actually is $R/N$.  They are just correlated.
You can also compute this expectation directly from the identity
$$
\sum_{r=0}^n r\binom{R}{r}\binom{N-R}{n-r} = R\binom{N-1}{n-1}
$$
To see this, the rhs counts the number of ways to pick a red ball and then $n-1$ other balls of any colour.  The lhs counts the number of ways to pick $r$ red balls, $n-r$ white balls, and then pick a red ball.  These are the same.
Since
$$
N\binom{N-1}{n-1} = n\binom{N}{n}
$$
by a similar argument, the expectation you want is
$$
\frac{R\binom{N-1}{n-1}}{\binom{N}{n}} = n\frac{R}{N}
$$
