TITLE: Sufficient conditions for establishing a total order on a family of probability distributions?
QUESTION [5 upvotes]: Let $\mathcal{X}$ be some set of independent random variables. Define the ordering on $\mathcal{X}$ by $X_i \prec X_j$ if and only if $\mathcal{P}\left\{X_i \le X_j\right\} \ge \frac{1}{2}$. Are there known weak conditions on $\mathcal{X}$ such that this ordering is a total ordering? A partial ordering? (I am guessing there are multiple answers to this question.)
The counterexample of non-transitive dice illustrates that not all $\mathcal{X}$ have partial orderings, much less total orderings. However, if, for example, $\mathcal{X}$ consisted of Gaussian random variables, a total ordering is established by comparing the means. 
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REPLY [2 votes]: The random variable $X$ stochastically dominates the random variable $Y$, written $X\succeq Y$, if $\Pr(X\ge y)\ge\Pr(Y\ge y)$ for all $y$. This relation is transitive.
Let's write $X\gtrsim Y$ if $\Pr(X\ge Y)\ge 1/2$. This relation is not transitive as the example of intransitive dice shows. However it is "total" in the sense that $X\lesssim Y$ or $Y\lesssim X$ holds.

Theorem. $X\succeq Y$ implies $X\gtrsim Y$.
Proof: In the notation of continuous random variables with pdfs $f_X$, $f_Y$,
\begin{align}
\Pr(X\ge Y)&=\int_{-\infty}^\infty \int_y^\infty f_X(x)f_Y(y)\,dx\, dy
=\int_{-\infty}^\infty \Pr(X\ge y) \,f_Y(y)\, dy\\
&\ge\int_{-\infty}^\infty \Pr(Y\ge y) \,f_Y(y)\, dy=\int_{-\infty}^\infty \int_y^\infty f_Y(x)f_Y(y)\,dx\, dy= \frac12.
\end{align}

By the Theorem, the intransitive dice phenomenon cannot occur when stochastic domination totally orders the elements of $\mathcal X$.