If partial information is revealed, are the events independent? Let's say that A, B, C takes on any integer values between 1 to 10.
Event I: A is greater than C.
Event II: B is greater than C.
Why are Event I and Event II independent? The intuitive definition of independence is that one event gives NO information about the other (as in two independent coin flips). But, in this case, Event I prevents C from taking on 10 (since A is guaranteed to be greater than C) which impacts our thinking of Event II.
My question, in general, is how should be think about independence when "some information" is revealed.
Thanks!!
 A: Let $E_1$ and $E_2$ be the events that $A > C$ and $B > C$, respectively.  Then $$\Pr[E_1] = \sum_{c=1}^{9} \Pr[A > C \mid C = c]\Pr[C = c] = \sum_{c=1}^9 \frac{10-c}{10} \cdot \frac{1}{10} = \frac{9}{20}.$$  Similarly, $\Pr[E_2] = 9/20$.
What is $\Pr[E_2 \mid E_1]$?  That is to say, given that $A > C$, what is the probability that $B > C$ also?  We have by definition of conditional probability,
$$\Pr[E_2 \mid E_1] = \frac{\Pr[E_1 \cap E_2]}{\Pr[E_1]},$$ and $$\Pr[E_1 \cap E_2] = \sum_{c=1}^9 \Pr[(A > C) \cap (B > C) \mid C = c]\Pr[C = c].$$  Now, because $A$ and $B$ are independent and identically distributed (note we are not saying $E_1$ and $E_2$ are independent!), we have for each $c$,
$$\Pr[(A > c) \cap (B > c)] \overset{\text{ind}}{=} \Pr[A > c]\Pr[B > c] \overset{\text{id}}{=} \left( \frac{10 - c}{10} \right)^2.$$  Hence $$\Pr[E_1 \cap E_2] = \sum_{c=1}^9 \left(1 - \frac{c}{10}\right)^2 \frac{1}{10} = \frac{57}{200},$$ therefore $$\Pr[E_2 \mid E_1] = \frac{57/200}{9/20} = \frac{19}{30}.$$  But if $E_1$ and $E_2$ are independent events, then we should have $\Pr[E_2 \mid E_1] = \Pr[E_2]$.  Since $19/30 \ne 9/20$, this shows they are not independent.
