Probability conditional on entry of exchangeable random vector: does the index matter? I have a random varbiable $Y\in\{0,1\}$, which is dependent on a random vector $X \in \{0,1\}^n$. Therebey, the entries in $X$ are exchangeable. That is, $P(X_i=1)=P(X_j=1)$. 
I want to compute/approximate $P(Y=1 | \exists j: X_j=1)$. 
I know $P(Y=1 | X_1=1)$ and $P(Y=1)$. I have the intuition that $$|P(Y=1) - P(Y=1 | \exists j: X_j=1)| \leq |P(Y=1) - P(Y=1 | X_1=1)|,$$ because knowing the index of the known component of $X$ is more information than just knowing that $X_j=1$ for some index. Conditioning on more information, in turn, should change the probability more than conditioning on less information.
Is that correct? (Does it even hold with equality?) Why? How could I check that?
 A: It is true that the event $B_1 = \{X_1 = 1\}$ is a subset of the event $B_2 = \{\exists j: X_j = 1\}$ so you do "know more information", but how does that affect the probability $P(Y=1)$?  You are essentially evaluating $P(Y=1 \mid B_k)$ for three events: $B_1, B_2$, and $B_3 = \Omega$ (i.e. the entire space, i.e. unconditioned).  While it is true that $B_1 \subset B_2 \subset B_3$, the prob $P(Y=1 \mid B_k) = {P(B_k \cap Y=1) \over P(B_k)}$ depends on how the event $\{Y=1\}$ intersects with each of them, and there is no reason to believe there will be a trend.  To be more explicit, as you go from $B_1$ to $B_2$ to $B_3$, both the numerator and denominator increase (or: do not decrease), so there is no telling what can happen to the ratio which is $P(Y=1 \mid B_k)$.
Here is a specific counter-example.  I found myself using this as a counter-example to many different MSE questions, BTW, so this seems to be a good one to test any conjecture about conditional probabilities. :)


*

*$X_1, X_2 =$ two different independent fair coin flips, where $X_i = 1$ means Heads

*$Y=$ the two flips show the same result

*$P(Y=1) = {HH,TT \over HH, HT, TH, TT} = \frac12$

*$P(Y=1\mid X_1= 1) = {HH \over HH, HT} = \frac12$

*$P(Y=1 \mid X_1 =1 \cup X_2 = 1) = {HH \over HH, HT, TH} = \frac13$
