I'm working through one of Gelman's exercises on exchangeability and am stuck on a seemingly simple exercise. We are given a box with N black and white balls but we not know how many of each. Task 1: We pick a ball and observe its color (y1), put it back, and then draw a second and observe its color (y2). We want to know if y1 and y2 are exchangeable. Task 2: we do the same as task 2 but without replacement. I know that observations are exchangeable if the probabilities of the permutations of the observation are all equal. Here is what I have for task 1 and 2 so far. I am going to let $w$ be the number of white balls in the box so the number of black balls will be $N-w$.
Task 1 (with replacement) \begin{align*} P(y_1 = black, y_2 = white) &= \frac{N-w}{N} \times\frac{w}{N} = \frac{Nw-w^2}{N^2}\\ P(y_1 = black, y_2 = black) &= \frac{N-w}{N} \times\frac{N-w}{N} = \frac{(N-w)^2}{N^2}\\ P(y_1 = white, y_2 = white) &= \frac{w}{N} \times\frac{w}{N} = \frac{w^2}{N^2}\\ P(y_1 = white, y_2 = black) &= \frac{w}{N} \times\frac{N-w}{N} = \frac{Nw-w^2}{N^2}\\ \end{align*}
Task 2 (without replacement) \begin{align*} P(y_1 = black, y_2 = white) &= \frac{N-w}{N} \times\frac{w}{N-1} = \frac{Nw-w^2}{N^2-N}\\ P(y_1 = black, y_2 = black) &= \frac{N-w}{N} \times\frac{N-w-1}{N} = \frac{(N-w)(N-w-1)}{N^2-N}\\ P(y_1 = white, y_2 = white) &= \frac{w}{N} \times\frac{w-1}{N-1} = \frac{w^2-w}{N^2-N}\\ P(y_1 = white, y_2 = black) &= \frac{w}{N} \times\frac{N-w}{N} = \frac{Nw-w^2}{N^2-N}\\ \end{align*}
After playing around with some values for N and w, and some intuition, the probabilities for Task 1 (w/ replacement) are all the same when w is exactly half of N (i.e. the number of black balls is equal the white balls), and the probabilities for Task 2 (w/o replacement) are not all the same even with half of ball white and the other half black.
I feel like there is something I am missing having to do with the prior placed on the numbers of each color of ball. According to Gelman et al, "with no information available to distinguish them, we have no logical choice but to model the θj’s exchangeably" [page 107]. Does this mean for Task 1 and 2 we can just state this assumption and assume the number of white and black balls are equal since we have no other "logical choice"?