If I randomize time points $$T=\{266.14646, 107.28526, 108.89631, 593.03129,\\ 118.10284, 425.60470, 39.02817, 291.90210\}$$ into two groups of equal size, then what is the probability that $T_i$ (the $i$-th time point) will be the 3rd order statistic of the second group?
Given that $T_i$ follows an exponential with rate $\lambda_i$.
I am getting confused to calculate this probability. Especially with what should be the probability that $T_i$ is the 3rd largest value of the 2nd group.
Let's generalize a little: you have $n=8$ data in sorted order, $x_1 \lt x_2 \lt \cdots \lt x_n$, which you wish to divide randomly into groups of size $\alpha=4$ and $\beta=4$. Denote the division by the indicator of $\beta$: this is, in effect, an $n$-digit binary number having exactly $\beta$ ones. (Examples appear below.) In order for $x_i$ to be the $k=3$rd smallest in the second group, we need three things to happen. Binomial coefficients count the number of ways they can happen:
Digit $i$ of the indicator is $1$. This happens in $1 = \binom{1}{1}$ ways.
There are exactly $k-1$ $1$'s among digits $1$ through $i-1$. This happens in $\color{red}{\binom{i-1}{k-1}}$ ways.
There are exactly $\beta-k$ $1$'s among digits $i+1$ through $n$. This happens in $\color{blue}{\binom{n-i}{\beta-k}}$ ways.
These three events are independent because they describe non-overlapping positions in the indicator, so their product is the number of ways of performing the split.
The total number of ways in which the data can be split is given by the binomial coefficient $\binom{n}{\beta}$, each of which is equally likely, whence the chance that $x_i$ is $k$th smallest in the second group is
$$\frac{\color{red}{\binom{i-1}{k-1}} \color{blue}{\binom{n-i}{\beta-k}}}{\binom{n}{\beta}}.$$
(Here and later, red objects denote or count numbers ranked ahead of $x_i$ and blue objects denote or count numbers ranked after $x_i$.)
For example, let $n=8$, $\alpha=4$, $\beta=n-\alpha=4$, and $k=3$ (which is the specific instance in the question). Let's tabulate $i$, the corresponding binomial coefficients, and their product:
i Choose(i-1,2) Choose(8-i,4-3) Product
1 0 7 0
2 0 6 0
3 1 5 5
4 3 4 12
5 6 3 18
6 10 2 20
7 15 1 15
8 21 0 0
The total, $0+0+5+12+\cdots+15+0$, is $70$, which is precisely $\binom{8}{4}$, confirming the law of total probability. The interpretations are:
There is no chance that either $x_1$ or $x_2$ could be the third smallest elements in the second group.
There are $1\times 5=5$ ways out of $70$ that $x_3$ could be third smallest in the second group, whence the answer to the question about the $T_i$ is $5/70 = 1/14 \approx 7.1$%. In terms of the binary indicators, these five ways can be written
$$\color{red}{11}\ 1\ \color{blue}{10000},\quad \color{red}{11}\ 1\ \color{blue}{01000},\quad \color{red}{11}\ 1\ \color{blue}{00100},\quad \color{red}{11}\ 1\ \color{blue}{00010},\quad \color{red}{11}\ 1\ \color{blue}{00001}.$$
For instance, the fifth indicator $\color{red}{11}\ 1\ \color{blue}{00001}$ identifies $\{\color{red}{x_1}, \color{red}{x_2}, x_3, \color{blue}{x_8}\}$ as the second group.
$$\color{red}{110}\ 1\ \color{blue}{1000},\quad \color{red}{110}\ 1\ \color{blue}{0100},\quad \color{red}{110}\ 1\ \color{blue}{0010},\quad \color{red}{110}\ 1\ \color{blue}{0001} \\ \color{red}{101}\ 1\ \color{blue}{1000},\quad \color{red}{101}\ 1\ \color{blue}{0100},\quad \color{red}{101}\ 1\ \color{blue}{0010},\quad \color{red}{101}\ 1\ \color{blue}{0001} \\ \color{red}{011}\ 1\ \color{blue}{1000},\quad \color{red}{011}\ 1\ \color{blue}{0100},\quad \color{red}{011}\ 1\ \color{blue}{0010},\quad \color{red}{011}\ 1\ \color{blue}{0001}.$$