Basic Question about linearity of expectation

Question

I am going through some introductory notes on probability here http://www.stat.berkeley.edu/~aldous/134/gravner.pdf

In Chapter 8, page 89, there is a problem where you get a bag containing 10 Black, 7 red and 5 white balls. What i find surprising is that E[X] where X denotes the number of red balls you get after 5 draws in the following cases

(i) with replacement (ii) w/o replacement

is the same.

I mean its clear that in both cases $E[X] = \sum_{i=1}^{i=5}E[I_i]$ where $I_i$ is an indicator variable which is $1$ if the $i^{th}$ ball you get is red.

And in case(i) the claim that $E[X] = 5*7/22$ is believable as $P(I_i = 1) = 7/22$ is true for each of the draws.

But in case(ii) I am unable to understand why $E[I_i] = 7/22$ for all the $5$ draws. I mean since $I_i$'s are indicator rv's, $E[I_i] = P(I_i = 1)$ and i can see that this quantity is $7/22$ only for the first draw.

I do not understand why all $E[I_i]'s$ are the same as the author claims. This is making me terribly confused. Please help.

Also, I would request if you can shed some more light on the problem. As in, please provide some commentary which kind of makes this result intuitive, easier to digest (and not surprising). That is, I would like to really understand why this result is true at a deeper level.

Hope the question is clear.

Edit: Removed one statement as I had not done calculations as per one of the comment below. The calculations confirm that $Pr(I_j = 1) = 7/22$ for all $0 \leq j \leq 5$. I want to understand this better. I mean, calculations is one thing -- but why are all the numbers coming out to be the same?

Answer 1

Take a very simple case of an urn with one white ball and one black ball.

If you sample with replacement, the chances of getting a white ball on each draw is $\frac{1}{2}$, agreed?

If you sample without replacement, there are only two equally likely outcomes WB and BW, right? and so the probability of getting a white ball on the first draw is $\frac{1}{2}$ (outcome WB occurred) and the probability of getting a white ball on the second draw is also $\frac{1}{2}$ (outcome BW occurred).

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If you draw $n$ distinct objects in succession from an urn, without replacement, there are $n!$ different outcomes. Write the outcomes as forming the rows of a $n!\times n$ matrix. Each column must contain each of the $n$ items an equal number of times, no? So the probability of drawing item #i on the $j$-th draw is $\frac{1}{n}$, same as the probability of drawing item #i on the first draw, or the second draw, or the third draw, or,...