A continuous function vanishes on $(-\infty,a]$ and on $[c,\infty),$ its graph is a straight line on the interval $[a,b]$ and another straight line on $[b,c],$ and its integral is $1.$


The mean of the probability distribution whose density this is, is
$$
\frac{a+b+c} 3
$$
and the variance is
$$
\frac{a^2+b^2+c^2-ab-ac-bc} {18}.
$$
Both of these are symmetric functions of $a,b,c,$ despite the fact that the role of $b$ in the first paragraph above is different from those of $a$ and $c.$


Is this symmetry somehow surreptitiously present in the characterization of this distribution in the first paragraph above?



Symmetry in the triangular distribution
The distribution function equals
$$
p(x)=\frac{|x-a|}{(a-b)(a-c)}+\frac{|x-b|}{(b-a)(b-c)}+\frac{|x-c|}{(c-b)(c-a)}.
$$
This is pretty symmetric. If you need a $k$-th moment, it equals 
$\frac2{(k+1)(k+2)}h\_k(a,b,c)$, where $h\_k$ is complete homogeneous polynomial (sum of all monomials of degree $k$.)


