Let $X_1,\dots,X_n\geq 0$ be independent, continuous random variables, and assume that $X_i$ stochastically dominates $X_j$ if $i<j$. This means that for any $x>0$, \begin{align} \mathbb{P}(X_i\geq x) \geq \mathbb{P}(X_j\geq x). \end{align} In this case we clearly have $\mathbb{P}(X_i=X_{(1)})\geq\mathbb{P}(X_j=X_{(1)})$, where $X_{(1)}=max(X_1,\dots,X_n)$. In other words, $X_i$ is more likely than $X_j$ to be the largest random variable.
Can we show that for a given $\alpha>0$, \begin{align} \mathbb{P}(X_i=X_{(1)}|X_{(2)}\geq \alpha) \geq \mathbb{P}(X_j=X_{(1)}|X_{(2)}\geq \alpha). \end{align} Note that the event $X_i=X_{(1)}|X_{(2)}\geq\alpha$ means that given at least two of the random variables exceed the level $\alpha$, the random variable $X_i$ is the largest one.
P.S. $X_i$ stochastically dominating $X_j$ also means that $X_i$ has the same distribution as $X_j+Y$, where $Y\geq 0$ is a random variable independent of $X_j$.
