voter paradox. Show that $\min\{a,b,c\} \leq 2/3$. 
Let $X,Y,Z$ to be discrete random variables with the property that their values are distinct with probability $1$. Let $a = P(X > Y), \ b = P(Y > Z), c = P (Z>X)$.

(a) Show that $\min\{a,b,c \} \leq 2/3.$
I do not know if this is the correct reasoning, but this is what i think of the problem so far. The sum of $a,b,c$ is 
$$ P(X > Y) + \  P(Y > Z) +  P (Z>X) = E(I_{X > Y} + I_{Y >Z} + I_{Z > X}).$$
I think this sum is two because if such $t$ happens to be in $I_{X > Y}, \ I_{Y >Z}$, then it cannot be in $I_{Z < X}$. Then two out of three are satisfied and $\min\{a,b,c \} \leq 2/3$ is achieved.
(b) If $X,Y,Z$ are independent and identically distributed, then $a = b = c = 1/2$. 
Since there are identically distributed, we have for example $P(X > Y) = P(Y > X)$. Also, $P(Z > X) = P(X > Z)$ and $P(Y >Z) = P(Z > Y).$ Does this means that we can use independence to get
$$P(X > Y) + P(Z > X) = 1,$$
so $P(X > Y) = 1/2 = P(Z > X)$, but $P(X > Y) = P(Y > Z)$, so $a =b=c=1/2$.
I can feel that i am doing something wrong, but maybe is that i am not understanding the problem at all.
 A: I think you're almost there, notice that
$$ I_{X > Y} + I_{Y >Z} + I_{Z > X} \le 2 $$
from where you can write
$$ a+b+c \le 2 $$
and follows from $\min \{\dots\} \le \text{avg} \{ \dots \}$
$$\min\{a,b,c\} \le \frac{a+b+c}3  $$
Note that you can apply this to the non-transitive dice game.  3 dice with faces

Die X has sides 2, 2, 4, 4, 9, 9.
Die Y has sides 1, 1, 6, 6, 8, 8.
Die Z has sides 3, 3, 5, 5, 7, 7.

$$ P(X > Y) = P(Y > Z) = P(Z > X) = \frac59 $$
A: The use of $E(I_{X > Y} + I_{Y >Z} + I_{Z > X})$ is a good idea.
As you observed, this value cannot be greater than $2$
(because $I_{X > Y} + I_{Y >Z} + I_{Z > X} \leq 2$).
If $X,Y,Z$ are independent and identically distributed
discrete variables, then $a < \frac12$.
You can make sense of the question if you drop the word "discrete",
allowing continuous variables to be used.
Then you have
$$P(X>Y)=P(Y>X)$$
as you observed, but also
$$P(X=Y) = 0.$$
Now apply total probability to the events $X>Y$, $X<Y$, and $X=Y$
and you should easily be able to show that $a = P(X>Y)=\frac12$.
Notice that you don't need to use variable $Z$ at all to get the value
of $a$, but you can repeat the argument twice more
(or simply invoke the symmetry of the problem statement)
to solve for $b$ and $c$.
