What constitutes an outcome in probability? In probability, I often have trouble determining which situations to take as distinct outcomes for calculation. For instance, if we have a die with its faces numbered $1, 2, 2, 3, 3, 6$ and we roll it twice. 
We get $2$ on the first roll and $3$ on the second. Again rolling it twice we get $2$ and $3$. But the $2$ we got the second time is not the same $2$ as the first one. Its the other $2$ inscribed on the face of the die (the die has two $2$'s). 
So, for the purpose of calculating the probability that the sum of the two rolls in a die will be a certain number $4$, say, will these two situations constitute distinct outcomes?
This is just a simplified, distilled example of a persistent problem I face in probability. Is there any way to think about outcomes that can make this clearer?
 A: Outcomes are members of a sample space, also called a probability space; events are subsets of the sample space, also called the probability space.  Thus if you throw a die twice, then among the outcomes are these three: $(3,1), (2,2),(1,3)$.  Those are three outcomes.  But the set $\{(3,1), (2,2),(1,3)\}$ is just one event, which can be described as the event that the sum is $4$.
A: What you are asking is whether $(2,2)$ is a distinct outcome from $(2,2'), (2',2), \text{or } (2',2')$ where the result of $2$ and $2'$ are each from different faces of the die.
The answer is: it depends.   You can consider them four distinct outcomes, each with a probability mass of $1/36$, or you can consider them a single outcome with probability mass of $1/9$.   The choice of which model to use depends on which makes it easier to calculate the probability of your event.   As long as your model weighs the probability of its outcomes accurately it will not matter.
$$\begin{align}
\mathsf P\big(\{(1,3), (2,2), (3,1)\}\big) & =\; \dfrac{2+4+2}{36}
\\[2ex]
\mathsf P\big(\{(1,3), (1,3'), (2,2), (2,2'), (2',2), (2',2'), (3,1), (3',1)\}\big) & =\; \dfrac{8}{36}
\end{align}$$
The advantage of drilling down to a model of hidden distinct outcomes of equal probability is then you just need to count numbers of outcomes within an event; as you don't need to worry about assigning weights.
A: I introduce terms like "compound experiment" and "compound outcome" to distinguish them from terms like "simple experiment" and "simple outcomes".
The face value of a card, regardless of its color or suite, is a simple outcome of a simple experiment.  So too is the roll of a single dice.
A simple outcome can be expressed as a 1-tuple.  The sample space of a simple experiment consists of such 1-tuples.
A compound outcome is expressed as an n-tuple.  For example, I can express the face value and suite of a card as a 2-tuple.  Similarly, I can express the roll of three dice as a 3-tuple.
The sample space of a compound experiment consists of n-tuples.
Another example of a compound experiment is one which returns the face value of a card and the roll of a dice is also a compound.
Without introducing the concept of a compound outcome, the intersection of two events like the face value of a card equaling 3 and the roll of a dice also equaling 3 makes little sense because the sample spaces of the two simple experiments have nothing in common. However, as a compound experiment, it makes sense to speak of the intersection of two events like the face value of the card equaling 3 (regardless of the roll of the dice) and the roll of a dice (regardless of the face value of the card).
The above considerations imply that Venn diagrams, because they illustrate the intersection of events, represent the intersection of events in a sample space of compound outcomes
Similarly, the conditional probability of one event given another event depends upon expressing both events within the same sample space of compound outcomes
