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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.

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    $\begingroup$ on (b), if $X,Y,Z$ are independent and identically distributed, then it is not true that they are discrete random variables with the property that their values are distinct with probability $1$ $\endgroup$ – Henry Sep 19 '16 at 14:17
  • $\begingroup$ You can make part (b) consistent if $X,Y,Z$ are continuous variables over $\mathbb R$. (But something must change in the problem statement, because as noted above it is inconsistent as stated.) $\endgroup$ – David K Sep 19 '16 at 14:20
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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.

$$ P(X > Y) = P(Y > Z) = P(Z > X) = \frac59 $$

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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 $c$.

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  • $\begingroup$ We must have: $P(X > Y) + P(Y > X) + P(X=Y) = 1$, but $P(X = Y) = 0$ and $P(X>Y)=P(Y>X)$, hence $a = 1/2$. This is the approach you mention right? The other results are obtained the same way. $\endgroup$ – richitesenpai Sep 19 '16 at 19:32
  • $\begingroup$ That's exactly what I had in mind. $\endgroup$ – David K Sep 19 '16 at 19:44

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