# Probability of orderings

Let $t_1,\ldots,t_n$ be a set of $n$ intervals, in the form $[l_i,u_i]$.

I define the precedence $t_i \succ t_j$ as the event in which a sample drawn from $t_i$ is greater than a sample drawn from $t_j$.

I have this running example: $$\left \{ \begin{array}{l} t_1 = [1, 3] \\ t_2 = [2, 4] \\ t_3 = [5, 9] \\ t_4 = [7, 12] \end{array} \right .$$ and I find that: $$\Pr(t_4 \succ t_3) = \Pr(t_4 \succ t_3 \succ t_1 \succ t_2)+\Pr(t_4 \succ t_3 \succ t_2 \succ t_1)$$ It is always true that $\Pr(t_i \succ t_j) = \Pr(t_i\succ t_j \succ t_k \succ t_m) + \Pr(t_i \succ t_j \succ t_m \succ t_k)$?

• How are you defining $t_i \succ t_j \succ t_k \succ t_m$? Can $t_i \succ t_j$ be interpreted as an interval? – Patrick Oct 9 '13 at 15:10
• It is an ordering: the samples drawn from the four distributions are presented in this order: [i - j - k - m]. – Eleanore Oct 9 '13 at 15:11

In your example, the statement would be true if $t_i = t_2$ and $t_j = t_1$ since:
$$\mathbb{Pr}(t_2 \succ t_1) = \mathbb{Pr}(t_4 \succ t_3 \succ t_2 \succ t_1) + \mathbb{Pr}(t_3 \succ t_4 \succ t_2 \succ t_1)$$