Probability of getting red ball at ith step 
An urn has r red and w white balls that are randomly removed one at a
  time. Let $R_i$ be the event that the $i$th ball removed is red. Find
  $P(R_i)$

I started with calculating $P(R_1) = \frac{r}{r+w}$. Next, 
$$P(R_2) = P(R_2|R_1)\cdot P(R_1) + P(R_2|W_1)\cdot P(W_1) = $$
$$ = \frac{r-1}{r-1+w}\cdot\frac{r}{r+w}+\frac{r}{r+w-1}\cdot\frac{w}{r+w} = \frac{\left(r-1+w\right)r}{\left(r+w-1\right)\left(r+w\right)} = \frac{r}{r+w}$$
And so on. The answer is $P(R_i) = \frac{r}{r+w}$.
The textbook's answer shows that the calculations weren't necessary:

$\frac{r}{r+w}$ because each of the $r + w$ balls is equally likely to
  be the $i$th ball removed.

I don't see why each ball is equally likely. At any ith step we don't have $(i-1)$ balls! Could someone help me to understand this?

It is important to understand this. Try to think in a way of: I draw all balls at the same time and then I give them at random the numbers $1$, $2$ etcetera. The one with number $1$ on it is considered to be the ball drawn first, etc.. Maybe that helps.
Edit
Person A and person B both are ordered to pick a ball from the urn. Habit of A: he takes the first ball touched by him. Habit of B: he touches $i-1$ distinct balls (let's say without looking at them) and after that he picks a ball distinct from the balls allready touched by him. Will B have a larger or smaller probability to pick a red ball? No. If B would have taken the touched balls out of the urn then it would be the $i$-th ball taken out that we are talking about here.
