One sufficient condition comes from stochastic domination.

Suppose the $X_i\in\mathcal X$ are continuously distributed with distributions depending on a parameter $\theta=\theta(X_i)$ so that for $X,Y\in\mathcal X$,

$$\theta(X)\ge\theta(Y)\qquad\Longrightarrow\qquad\Pr(X\ge y)\le\Pr(Y\ge y)\quad\forall y.$$

Then $\theta(X)\ge\theta(Y)$ also implies \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} Hence the $X_i$ inherit (a homomorphic image of) the order of the numbers $\theta(X_i)$, and so are totally ordered.

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 $\mathcal X$.

Here is an easy and common scenario.

Suppose the $X_i\in\mathcal X$ are continuously distributed with distributions depending on a parameter $\theta=\theta(X_i)$ so that

$$\theta(X)\ge\theta(Y)\qquad\Longrightarrow\qquad\Pr(X\ge y)\le\Pr(Y\ge y)\quad\forall y.$$

Then for $X,Y\in\mathcal X$, $\theta(X)\ge\theta(Y)$ if and only if $$ \Pr(X\ge Y)=\int_{-\infty}^\infty \int_y^\infty f_X(x)f_Y(y)\,dx\, dy =\int_{-\infty}^\infty \Pr(X\ge y) \,f_Y(y)\, dy$$ $$ \ge\int_{-\infty}^\infty \Pr(Y\ge y) \,f_Y(y)\, dy=\int_{-\infty}^\infty \int_y^\infty f_Y(x)f_Y(y)\,dx\, dy= \frac12.$$ Hence the $X_i$ inherit the order of the numbers $\theta(X_i)$, and so are totally ordered.

One sufficient condition comes from stochastic domination.

Suppose the $X_i\in\mathcal X$ are continuously distributed with distributions depending on a parameter $\theta=\theta(X_i)$ so that for $X,Y\in\mathcal X$,

$$\theta(X)\ge\theta(Y)\qquad\Longrightarrow\qquad\Pr(X\ge y)\le\Pr(Y\ge y)\quad\forall y.$$

Then $\theta(X)\ge\theta(Y)$ also implies \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} Hence the $X_i$ inherit (a homomorphic image of) the order of the numbers $\theta(X_i)$, and so are totally ordered.