# 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.

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Hmm.. If $k>B+R$, then $P(E_k)=0$, no? Else, I didn't believe you, and calculated $P(E_2)$, and it is indeed the same as $P(E_1)$.. :) – Berci Jan 30 '13 at 11:43
Yeah, those details I always forget. Using induction, I've proven it is true for all k. But I can't still catch the idea, the reason is it so. – Temitope.A Jan 30 '13 at 11:45
I'm not sure what sort of explanation your "why" is aiming at. Any permutation of the order in which the balls are drawn is equally likely, so these probabilities are all equal by symmetry. Is that the sort of explanation you were looking for, or do you want a more technical proof? If so, what are you regarding as given? – joriki Jan 30 '13 at 12:12
@joriki Expand your answer. For example, it's not clear to me that "Any permutation of the order in which the balls are drawn is equally likely". I'm not even sure I understood the meaning of the sentence. For the meaning of my "why"...the equalities of those probabilities is not intuitive. I have seen, technically, in my case, that it holds, but I want to acquire it by intuition, grasp the phenomenon behind it, like, maybe, the permutations you were talking about. – Temitope.A Jan 30 '13 at 15:21
Yeah..., let us imagine we are distributing the balls instead of cards. We give n people one ball each, which could be red or black. The event $E_k$ correspond to the $k^{th}$ person having a red ball. Then the probability of the first receiver to have a red ball is obviously equal to the probability of the $k^{th}$ person. I'm starting to understand. – Temitope.A Jan 30 '13 at 16:44

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 $k$.