Given a random variable $X$ and known (but arbitrary) associated distribution, is it possible to find the probability that it falls between two adjacent order statistics $Y_{(i)}$ and $Y_{(i+1)}$? That is, find $P(Y_{(i)} \le X \le Y_{(i+1)})$. Assume that $X$ and $Y$ are independent.
Finding the probability that $X$ is greater (or smaller) than one of these can easily be found due to independence: $$P(X \ge Y_{(i)})=\int_{-\infty}^{\infty}\int_{-\infty}^{x}f_{X}(x)f_{Y_{(i)}}(y)~dy~dx$$ But I run into the problem that $Y_{(i)}$ and $Y_{(i+1)}$ are not independent. How would you properly account for this in an expression for the probability?
Assuming, as Henry detailed in a comment, that the $Y_{(i)}$ are ordered from an i.i.d. sample of size $n$ of the same distribution as $X$, we have
$$\mathsf P\left(Y_{(i)} \le X \le Y_{(i+1)}\right)=\frac1{n+1}\;,$$
since all $(n+1)!$ orders of the $n+1$ independent variables $Y_1,\ldots,Y_n,X$ are equiprobable and $X$ has each particular rank in a fraction $\frac1{n+1}$ of them.