Consider the problem of sampling without replacement. Suppose that we have an urn with $r$ red balls and $b$ black balls, and we choose a sample of $2$ balls without replacement. Let $X_i$ be the random variable that equals to $1$ if the $i$-th ball is red and $0$ otherwise. Obviously $P(X_1=1)=\frac{r}{r+b}$. Also, we have $$P(X_2=1)=P(X_2=1|X_1=1)P(X_1=1)+P(X_2=1|X_1=0)P(X_1=0)=\dots=\frac{r}{r+b}$$.
This seems a bit counter-intuitive to me since this is the result I would expect in the case of sampling with replacement. Can you give me some intuition of why this result is reasonable/intuitive?