Marginal distribution of $n$-th trial in a sequence of experiments with stopping rule Imagine I'm running a series of independent trials where the outcome $X_{i}$ can be either 1 or 0 (a bernoulli variable) with probability $p$.
I will run the trials as in a negative binomial way: I will run as many trials as I need until I get, say, $3$ success.
Question: what is the marginal probability of $X_4$ (the result of the fourth trial)? By marginal, I mean not conditioning on $X_1, X_2, X_3$.
 A: For context.
$X_4$ can be expressed as the triple $(\Omega, \mathcal{F}, P)$ where $\Omega$ is a sample space, $\mathcal{F}$ is a set of events, sets of zero or more elements of $\Omega$, and $P$ is a measure over events such that $P(\mathcal{\Omega})=1$. For a randomly chosen outcome $\omega \in \Omega$, any event $A\in \mathcal{F}$ such that $\omega \in A$ is said to occur.  Clearly, $\Omega = \{0,1\}$ and $\mathcal{F}=\{\varnothing, \{0\},\{1\}, \{0,1\}\}$. We see then by definition of the probability space, the empty set (non-existence) cannot occur; it is outside the scope of the probability space. Therefore all probability spaces require the actual realization of outcomes and $X_4$ is only defined when it is realized. 
There are other formulations of probability, but this is the one that dominates statistical theory.
A: You could use the following sample space, events and probabilities:
Event X1 X2 X3 X4 X5 X6  Prob
3     1  1  1           p^3
4     0  1  1  1        (1-p)*p^3  
4     1  0  1  1        (1-p)*p^3
4     1  1  0  1        (1-p)*p^3
5     1  1  0  0  1     (1-p)^2*p^3
5     1  0  1  0  1     (1-p)^2*p^3
5     1  0  0  1  1     (1-p)^2*p^3
5     0  1  1  0  1     (1-p)^2*p^3
5     0  1  0  1  1     (1-p)^2*p^3
5     0  0  1  1  1     (1-p)^2*p^3
6 .... etc

The marginal probabilities for $X_n$ can then be found by summing the probabilities of the sample space, and for $n>3$ this will be 
$$P(x_n) = \begin{cases} P(E<n) &\quad \text{ if } x_n=\emptyset \\
(1-P(E<n))\cdot(1-p) &\quad \text{ if } x_n=0  \\
(1-P(E<n))\cdot p  &\quad \text{ if } x_n=1 
\end{cases} $$ 
where $P(E<n)$ is the probability for the event that the number of trials until stopping is smaller than $n$ and can be expressed with the negative binomial.
You could also use:
$$P(x_n|x_n \neq \emptyset) = \begin{cases} 
1-p &\quad \text{ if } x_n=0  \\
p  &\quad \text{ if } x_n=1 
\end{cases} $$ 
Whichever these two you use is, I guess, up to personal taste which depends on the context and on how you regard the marginal probability and the sample space.
