If we draw the contingency table of x (vertical) against y (horiz.), we have a square.
For n=4, we have (legend: < : x<y = : x=y > : x>y
y 1 2 3 4
x
1 = < < <
2 > = < <
3 > > = <
4 > > > =
We see that there are n(n-1)/2 cases of x<y out of n^2.
Therefore,
p(x<y)=n(n-1)/(2n^2)=(n-1)/(2n)
However, if the sample space is continuous, it will be simply p(x<y)=1/2.
