Exchangeability in probability schemes - a clarification Let A be a box with B black balls and R red balls. We extract balls from the box without putting them back. Let define the event $E_k:=$ The $k^{th}$ ball extracted is red, $1\leq k\leq B+R$. Then P($E_1$)=R/(B+R), and I've manually shown that P($E_3$)=P($E_2$)=P($E_1$). I know this is true for all $E_k$, but why? I know there is some results of De Finetti about this, but can you explain this particular situation? The events $\{E_n\}_{n\geq1}$ are not independent, so I don't know how to treat them in a recurrence relation.
 A: So, we have $B+R$ balls, let's name them $a_1,\dots,a_B,a_{B+1},\dots,a_{B+R}$ with the first $B$ being black, and let $A:=\{a_1,\dots,a_{B+R}\}$.
Then consider that any such experiment continues until the end, i.e., we pick out all the $B+R$ balls. Then, this experiment corresponds to a permutation of $A$, and each such permutation is equally possible (hence with probability $1/(B+R)!$).
Then the event $E_k$ can also be written as 
$$(E_1\lor \lnot E_1) \land \dots \land (E_{k-1}\lor \lnot E_{k-1}) \land E_k \land (E_{k+1}\lor \lnot E_{k+1}) \land \dots \land (E_{B+R}\lor \lnot E_{B+R})$$
expanding that, we get $E_k = \displaystyle\bigvee_\epsilon \left(E_k\land \bigwedge_{i\ne k}\epsilon_iE_i\right) $ where $\epsilon=(\epsilon_i)_i$ is an array of symbols 'empty' or '$\lnot$', that is, determines whether the $i$th ball is red or not. Since these events are disjoint, we get
$$P(E_k)=\sum_\epsilon P\left(E_k\land \bigwedge_{i\ne k}\epsilon_iE_i\right). $$
In other words, we divide our events to the union of (kind of) elementary events, which now are the full experiments.
Now, consider the bijection between the permutations which always exchanges the $1$st and $k$th ball. By this, it could be clearer now that the above sum is independent from $k$.
