$P[A > B, A > C]$, where $A$, $B$, and $C$ are i.i.d. random variables Let $A$, $B$, and $C$ be i.i.d. random variables. What is the probability $P[A > B, A > C]$? Does the independence of $A$, $B$, and $C$ imply the independence of the events $A > B$ and $A > C$? In other words, does it hold that $P[A > B, A > C] = P[A > B] P[A > C]$? If not, why? Providing an example would be great.
 A: This is the simplest example I could think of.
If $A$, $B$, and $C$ are independent coin tosses with heads 1 and tails 0 then $$P(A>B, A>C)=1/8$$ since this only happens if A=1 and B and C are both 0, which is one of eight possible outcomes, but $$P(A>B)=1/4=P(A>C)$$ since this only happens if $A=1$ and $B$ or $C=0$, which is only one of four possible outcomes, so $${1\over 8}=P(A>B, A>C)\neq P(A>B)\times P(A>C)={1\over 16}.$$
To look at it another way, $P(A>C)=1/4$, but if we know that $A>B$ already, then we know that $A=1$ and $B=0$, so $P(A>C|A>B),$ the conditional probability that $A>C$ given $A>B$, is just the probability that $C=0$, since we already know that if $A>B$ then $A=1$ and $B=0$, and $C$'s probability is independent, so $$P(A>C|A>B)=P(C=0)=\frac12.$$ So
$$
P(A>B,A>C)=P(A>C|A>B)P(A>B)=\frac12\times\frac14.
$$
A: 
Does the independence of A, B, and C imply the independence of the events A>B and A>C?

No.  The event of $A>B$ is evidence that $A$ is large, thus making it more likely that $A>C$ too.

Rather, the iid nature of the random variables introduces the following symmetries:

*

*consider that these six mutually exclusive events are equally probable to each other: $$\rm A>B>C~,~ A>C>B~,~ B>A>C~,~ B>C>A~,~ C>A>B~,~ C>B>A$$


*consider that these six mutually exclusive events are equally probable to each other: $$\rm A>B=C~,~ A>C=B~,~ B>A=C~,~ B>C=A~,~ C>A=B~,~ C>B=A$$


*consider that these six mutually exclusive events are equally probable to each other: $$\rm A=B>C~,~ A=C>B~,~ B=A>C~,~ B=C>A~,~ C=A>B~,~ C=B>A$$


*Then there is this event $A=B=C$.  This and all the above events are mutually exclusive (aka disjoint).


*Of course, things are much simpler if the random variables are continuous, as that vanishes the chances of ties.  You need some more information about the distribution if the variables are discrete.
