Probability question: $100$ balls with $r$ red balls.

Denote $N=100$ (number of balls), $K=r$ (number of red balls);

$P_{N,K}(n,k) = \Pr$ ($k$ balls are red, when $n$ draws are made), where $0\le k \le n$.

Looking at Hypergeometric distribution, one can see that

$$ P_{N,K}(n,k) = \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}. $$

Then the probability that the $(n+1)$th ball (where $n=1,2,...,N-1$) drawn will be red, is: $$ P_{n+1} = \sum\limits_{k=0}^{n} P_{N,K}(n,k)\cdot \dfrac{K-k}{N-n} $$ $$ = \sum\limits_{k=0}^{n} \dfrac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}\cdot \dfrac{K-k}{N-n} $$ $$ = \sum\limits_{k=0}^{n} \frac{ \frac{K\cdot(K-1)!}{k!(K-1-k)!} \binom{N-K}{n-k}}{ \frac{N\cdot(N-1)!}{n!(N-1-n)!}} $$ $$ =\dfrac{K}{N} \sum\limits_{k=0}^{n} \dfrac{ \binom{K-1}{k} \binom{N-K}{n-k}}{\binom{N-1}{n}} =\dfrac{K}{N} \sum\limits_{k=0}^{n} P_{N-1,K-1}(n,k)= \dfrac{K}{N} \cdot 1 = \dfrac{K}{N}, $$ where $$\sum\limits_{k=0}^{n} \dfrac{ \binom{K-1}{k} \binom{N-K}{n-k}}{\binom{N-1}{n}}=1$$ follows from Vandermonde's identity. So, $P_{2}=P_{50} = P_{100} = \dfrac{K}{N} = \dfrac{r}{100}$.

Consider all the possible results of drawing the balls. Each result can be represented by a binary number having 1 when we draw a red ball, so that the number of 1s in the number is $r$. Your question is then equivalent to asking what is the probability of 1 occurring at any particular place of the number and this is easily seen to be $r/100$ independently of the position.