Probability of drawing in the right order and having the second draw be drawn before a fixed step Suppose I am drawing objects uniformly at random, and I continue drawing without replacement until all objects are listed. So the object I draw at the first step is listed in the first place, the object I draw in the second step is listed in the second place, so on. There are $n$ objects. 
Consider some fixed subset of objects $o_1,o_2,..,o_r$, $r < n$. (these are just their names; we don't know the order in which they are drawn) , and call the times at which they are drawn $t_1,t_2,..,t_r$ (note again that these are just the names of the times; no chronological implication here). 
Suppose I want the following two events to occur:


*

*$t_1 < t_2 <$ any of the $t_i$, $i \in \{3,..,r\}$ 

*$t_2 \leq l$, where $l$ is some fixed positive integer. 


In words, I want it to be the case that object $o_1$ is drawn before object $o_2$, which is drawn before any of the $o_3,..,o_r$. And I would like $o_2$ to be drawn before time $l$.
I want the joint probability of the above two events, but I can't consider them independent, right, since if I know (1) occurs this increases the probability that I drew $o_2$ earlier? But I'm not sure how to compute the desired joint probability. 
 A: I don't think independence matters here...you should be able to do this with counting. Your sample space is the set of $n!$ permutations of your objects. Fix a value of $t_2 \le l$. The number of permutations which meet both of your conditions in this case is:
$(t_2-1) \times \frac{(n-r)!}{(n-r-t_2+2)!} \times (n-t_2)!$
The $(t_2-1)$ is the number of possibilities for $t_1$, which must be less than $t_2$. The middle fraction is the number of ways you can fill in the remaining $t_2-2$ slots of the first $t_2$ with objects that are not the $o_i$, and the $(n-t_2)!$ is the number of ways to fill out the rest of the list.
Add these up as $t_2$ varies from 2 to $l$, and divide the summation by $n!$ to get your probability.
A: Calculating probability of event 1
Given $r$ special objects labeled $o_1,o_2,\dots,o_r$ and $(n-r)$ additional ordinary objects, we ask what is the probability of when considering the special objects specifically that $o_1$ occurs before $o_2$ which occurs before any of the other special objects (with possibly some of the ordinary objects scattered throughout the special objects).
We can use the kind fact that the ordinary objects don't matter at all in this scenario.  Each permutation of the special objects is equally likely to occur.
We see then that $Pr(A_1) = \frac{1}{r}\cdot \frac{1}{r-1} = \frac{1}{r^2-r}$

Calculating probability of event 2
We ask for the probability that the special object $o_2$ occurs within the first $l$ spaces.  To do so, we may find the probability that if we select $l$ objects (special or ordinary) that $o_2$ is among them.
We see then that $Pr(A_2) = \frac{\binom{n-1}{l-1}}{\binom{n}{l}} = \frac{l}{n}$

Calculating intersection of the events
Let $1<t_2\leq l$.  We will cycle over all possible values for $t_2$.
Given a specific $t_2$, we first check the probability that $o_1$ occurs before $t_2$.  This occurs with probability $\frac{t_2-1}{n-1}$.
Given that $o_1$ occurs before $o_2$ and given a specific value for $t_2$, we ask what the probability is that all other special objects occur time greater than $t_2$.  This occurs with probability $\frac{\binom{n-r}{n-t_2}}{\binom{n-2}{n-t_2}}$
The probability that $o_2$ was drawn specifically at time $t_2$ was $\frac{1}{n}$ in the first place.
We have then the probability:
$$\sum\limits_{t_2=2}^l\frac{1}{n}\cdot\frac{t_2-1}{n-1}\cdot\frac{\binom{n-r}{n-t_2}}{\binom{n-2}{n-t_2}}$$
It doesn't appear to simplify nicely from here.  Given that it doesn't simply equal $\frac{1}{r^2-r}\cdot\frac{l}{n}$ (this can be checked given specific values for $r,l,n$) the events are indeed dependent on one another.
