Having trouble determining the sample space for conditional expectation problems My book states the following theorem

Let $X$ be a random variable with sample space  $\Omega$. If $F_1, F_2, . . . , F_r$
  are events such that $F_i$ and $F_j $ are disjoint (for $i$ not equal to $j$) and $\Omega = \cup_jF_j$ then $E(X) = \sum_jE(X|F_j)P(F_j)$

I understand this to mean that given $X$ which can take on a value / outcome from $\Omega$ and $F_1, F_2, . . . , F_r$ are all pairwise disjoint whose union forms $\Omega$, $E(X)$ can be computed by the given equation. However, I had trouble understanding the example illustrating this theorem:

Let T be the number of rolls in a single play of craps. We can think of a single play as a two-stage process. The first stage consists of a single roll of a pair of dice. The play is over if this roll is a 2, 3, 7,
  11, or 12. Otherwise, the player's point is established, and the second stage begins. This second stage consists of a sequence of rolls which ends when either the player's point or a 7 is rolled. We record the outcomes of this two-stage experiment using the random variables X and S, where X denotes the first roll, and S denotes the number of rolls in the second stage of the experiment (of course, S is sometimes equal to 0). Note that T = S + 1. Then $E(T) = \sum_{j=2}^{12} E(T|X = j)P(X=j)$

Here I think T is analogous to the X in the theorem above and each outcome of a roll of the two dices ($X = j$) is an event analogous to an $F_j$. However, the potential values of T can take on are in the set $\{1, 2, ... \infty\}$ while I think the set formed by the union of events $X = j$ for all $j$ is just $\{2,3..,12\}$. I'm confused about what the sample space $\Omega$ would be in this example since it appears the values of the random variable and event appear to be drawn from different sets.    
 A: This is called the "law of total expectation" and is similar to the "law of total probability."  The sample space $\Omega$ is the set of all outcomes $\omega$ of the form: 
$$ \omega = (\mbox{first roll}, \mbox{sequence of other rolls})=(X, \mbox{sequence of other rolls})$$
This sample space can be partitioned into events where the first roll is 2, the first roll is 3, ..., the first roll is 12.  So the partition is: 
$$ \Omega = \{X=2\} \cup \{X=3\}\cup\{X=4\}\cup...\cup\{X=12\}$$
We know that $X$ can only take values in the set $\{2, ..., 12\}$ and so the events $\{X=2\}, \{X=3\}, ..., \{X=12\}$ are indeed mutually exclusive and collectively exhaustive.  The event $\{X=4\}$ contains all outcomes that start with a first roll of 4. 

The individual events $\{X=i\}$ look like this: 
\begin{align}
\{X=2\} &= \{(2)\}\\
\{X=3\} &= \{(3)\}\\
\{X=4\} &= \{(4, 4), (4, 2, 2, 5, 4), (4, 12, 5, 5, 7), (4, 8, 4), ...\}
\end{align}
and so on.  The event $\{X=4\}$ has a countably infinite number of outcomes, but all of them are sequences that start with $4$ and end with either 4 or 7.  
The events $\{X=2\}, \{X=3\}, \{X=7\}, \{X=11\}, \{X=12\}$ all contain just one outcome each and so we trivially have
\begin{align}
E[T|X=2]&=1\\
E[T|X=3]&=1\\
E[T|X=7]&=1\\
E[T|X=11]&=1\\
E[T|X=12]&=1
\end{align}
On the other hand, $E[T|X=4]$ is equal to 1 plus the expected time to roll either a 4 or a 7.
