What is the probability of the $n^\mathrm{th}$ biggest $x$? Suppose I have $n^2$ exponential random variables $x_1,\ldots,x_{n^2}$ that are independent and identically distributed (i.i.d) and have common parameter $\lambda$. I also have a positive number $\alpha$. (Maybe, we can assume that $\alpha$ is less than $1$.)
I sort in descending order these random variables to obtain $x_{(1)}\ge x_{(2)}\ge\ldots \ge x_{(n^2)}$. 
I would like to find the probability that $n^\mathrm{th}$ biggest $x$ times $\alpha$ minus $1$ is greater than the $(n+1)^\mathrm{th}$ biggest $x$. That is,
$$\Pr\left[\alpha x_{(n)}-1\ge x_{(n+1)}\right],$$
or
$$\Pr\left[x_{(n)}\ge \dfrac{x_{(n+1)}+1}{\alpha}\right].$$
 A: This is not an easy problem, but does appear to be amenable to solution.
Notation: decreasing vs increasing order
The OP poses the problem:  


*

*In a sample of size $n^2$, consider the sampled sorted in descending order with square braces:  $$x_{[1]}> x_{[2]} > \dots > x_{[n]} > x_{[n+1]} > \dots > x_{[n^2]}$$

*By contrast, to apply conventional order statistic notation, define the sample in increasing order as:
$$x_{(1)} < x_{(2)} < \dots < x_{(r)} < x_{(s)} < \dots <  x_{(n^2)}$$
As an example, if $n = 3$, the OP considers a sample of size 9, and is interested in the $3^{rd}$ and $4^{th}$ largest values, namely $$x_{[3]} \text{ and } x_{[4]}  \quad \quad \text{(sorted in decreasing order)}$$
or equivalently
$$x_{(6)} \text{ and } x_{(7)}  \quad \quad \text{(sorted in increasing order)}$$

1 2 3 4 5 6 7 8 9
9 8 7 6 5 4 3 2 1

Note that: if $n = 3$, then $x_{[3]} = x_{(7)} \quad \text{ and } \quad x_{[4]} = x_{(6)}$. 
More generally, we have:  
$$x_{[n]} = x_{(n^2-n+1)} \quad \text{ and } \quad x_{[n+1]} = x_{(n^2-n)}$$
Solution
Let $X \sim \text{Exponential}(\lambda)$ with pdf $f(x)$:

For a size $n^2$ random sample drawn on $X$, the joint pdf of the order statistics $(X_{(r)}, X_{(s)})$, for $r < s$, as say $g(x_r, x_s)$:

where I am using the OrderStat function from the mathStatica package for Mathematica to automate the calculation. 
Substituting $r = n^2-n$ and $s = n^2-n+1$ yields the joint pdf of $(X_{[n]}, X_{[n+1]})$ :

Finally, we seek:  
$$\Pr\left[X_{[n]}\ge \dfrac{X_{[n+1]}+1}{\alpha}\right] \quad = \quad \Pr\left[X_{(s)}\ge \dfrac{X_{(r)}+1}{\alpha}\right]$$

The Good News ... is that an exact closed-form solution is obtained for $0 < \alpha \leq 1$, and that it appears to be correct. For instance, when $n = 4$, $\lambda  = 1$ and $\alpha = 1$, the exact probability simplifies to:
$$\frac{1}{e^4} \approx 0.0183156$$
which is consistent with a quick Monte Carlo check (which generated 0.0183).
The not so good news ... in the case when $\alpha > 1$ , I am unconvinced that Mathematica's assertion that the integral is indeterminate is, in fact, correct. In particular, if one tries specific numerical values for say $n = 4$, then exact and correct symbolic probabilities are obtained. Either way, exact closed-form solutions are certainly obtainable for given parameter values. 

Notes


*

*As disclosure, I should add that I am one of the authors of the software used above.

*Monte Carlo check of probability:
Here is some quick Mathematica code to check the probability calculation, when $n = 4$, assuming a standard Exponential parent:
...
data = ParallelTable[Sort[RandomVariate[ExponentialDistribution[1], 16], Greater][[4;;5]], 10^6]; 

{ndata, nplus1data} = Transpose[data];

Count[ndata - (nplus1data+1)/alpha, x_ /; x>0]/Length[ndata] //N

... where one simply has to replace alpha with the chosen numerical value.
