Probability Space of Rolling a Fair Die Three Times I'm a little confused about what this problem is asking exactly and would simply like some advice on my solution. The problem is:

We roll a die three times.  Give a probability space $(\Omega, \mathcal F, P)$ for this experiment. 

The following is my written solution. 

Assuming the die is fair and six-sided, the sample space for one roll is $\Omega_1 = \{1, 2, 3, 4, 5, 6 \}$. Which, for all three rolls gives:
$$ \Omega = \{ (1,1,1), ..., (1,6,6), (2,1,1), ..., (2,6,6),(3,1,1),...,(5,6,6),...,(6,6,6) \} $$
Our sigma field is then
$$ \mathcal F = \{ \varnothing, \Omega, \{1\}, ...,\{6\},\{1,1\},\{1,2\},...,\{5,6\},\{6,6\}\} $$
Now, let $A\subset \Omega$ such that $ A=A_{111}\cup A_{112} \cup ... \cup A_{665} \cup A_{666}= \bigcup_{i,j,k=1}^6 A_{ijk}$. 
Then, $$ P(A)=\frac{A}{\Omega} $$
Our probability space is then defined by $(\Omega, \mathcal F, P(A))$, whose values are listed above. 

Here are my questions regarding this problem: 
1) Are my sample space and sigma field correct for this experiment? 
2) The initial question was a little vague and am unsure of what $P$ I'm looking for here, so I took a guess at that solution. I'm fairly certain my answer for that part is incorrect. From examples I've seen online, a specific event is typically given, and you're required to find the probability of that event. So, I tried to expand that to include any possible rolling combinations. 

To clarify some notation just in case, $A_{111}$ is the event that you roll a 1 three times. Similarly, $A_{352}$ would be the event that you roll a 3, then a 5, then a 2. 
Thank you!
 A: I think you have a good hang of the concept. However, things can always be written better.
For example, the sample space for three independent dice, rather than the suggestive $\{(1,1,1),...,(6,6,6)\}$(which is correct, so credit for that) can be written succinctly as $\Xi \times \Xi \times \Xi$, where $\Xi = \{1,2,3,4,5,6\}$. This manages to express every element in the sample space crisply, since we know what elements of cartesian products look like.
The sigma field is a $\sigma$-algebra of subsets of $\Omega$. That is, it is a set of subsets of $\Omega$, which is closed under infinite union and complement. 
Ideally, the sigma field corresponding to a probability space, is the set of events which can be "measured" relative to the experiment being performed i.e. it is possible to assign a probability to this event, with respect to the experiment being performed. What this specifically means, is that based on your experiment, your $\sigma$-field can possibly be a wise choice.
In our case, we have something nice : every subset of $\Omega$ can be "measured" since $\Omega$ is a finite set (it has $6^3 = 216$ elements) hence the obvious candidate, namely the cardinality of a set can serve as its measure(note : this choice is not unique! You can come up with many such $\mathcal F$).
Therefore, $\mathcal F = \mathcal P(\Omega)$, where $\mathcal P(S)$ for a given set $S$ is the power set of $S$, or the set of all subsets of $S$. This is logical since this contains every subset of $\Omega$, and is obviously a $\sigma$-algebra. 
Now, $P(A)$, for $A \subset \mathcal F$(equivalently, $A$ any subset of $\Omega$) is, for the reasons I mentioned above, simply the ratio between its cardinality, and the cardinality of $\Omega$. Therefore, $P(A) = \frac{|A|}{216}$. That is exactly what you wrote, except well, division by sets isn't quite defined.
So there you have it, an answer, along with what you've done right and wrong.
NOTE : $\mathcal P$ for the power set, and $P$ for the probability of a set is my notation here. I still think the letter $P$ is used for both, but it's causing confusion here, hence the change.
A: Good news and bad news: your $\Omega$ is correct, but your $\mathcal{F}$ and $P$ are not.

Let's see what's wrong with your proposed
$$\mathcal{F}=\{\varnothing,\Omega,\color{red}{\{1\},\ldots,\{6\},\{1,1\},\{1,2\},\ldots,\{5,6\},\{6,6\}}\}.$$
By definition, a $\sigma$-algebra $\mathcal{F}$ must be a set of subsets of the sample space $\Omega$, i.e. $\mathcal{F}\subseteq\mathcal{P}(\Omega)$, subject to some conditions. (Here $\mathcal{P}(\Omega)$ stands for the power set of $\Omega$, which is the set of all its subsets.) To reiterate: each element of $\mathcal{F}$ must be a subset of $\Omega$.
Let's check:


*

*$\varnothing\subseteq\Omega$? Certainly, yes.

*$\Omega\subseteq\Omega$? Certainly, yes.

*$\{1\}\subseteq\Omega$? That means that $1\in\Omega$, but there's no such element in $\Omega$. Each element of $\Omega$ is an ordered triple, not a single number.

*$\cdots$

*$\{6\}\subseteq\Omega$? No, for the same reason as above.

*$\cdots$

*$\{5,6\}\subseteq\Omega$? That means that $5\in\Omega$ and $6\in\Omega$, but there are no such elements in $\Omega$, as explained above.

*$\{6,6\}\subseteq\Omega$? Note that $\{6,6\}=\{6\}$, already considered above.


What is the correct $\mathcal{F}$? For finite sample spaces, we can take the entire power set $\mathcal{P}(\Omega)$ to be the $\sigma$-algebra of measurable events. So
$$\mathcal{F}=\mathcal{P}(\Omega)=\{\varnothing,\Omega,\{(1,1,1)\},\ldots,\{(6,6,6)\},\{(1,1,1),\{(1,1,2)\}\ldots\ldots\ldots\}.$$
Shown above, besides the obvious $\varnothing$ and $\Omega$, are all six singletons and an example of a two-element subset. In fact, trying to write it explicitly is kind of a bad idea, because the set is really huge:
$$|\mathcal{F}|=|\mathcal{P}(\Omega)|=2^{|\Omega|}=2^{216}.$$

Now, the problem with you definition of the probability function $P$ is that you didn't actually define it. You only gave one value of this function. From what you wrote about your notation — "$A_{111}$ is the event that you roll a $1$ three times. Similarly, $A_{352}$ would be the event that you roll a $3$, then a $5$, then a $2$" and "$A=A_{111}\cup A_{112}\cup\ldots\cup A_{665}\cup A_{666}$" — we see that in fact this $A$ includes all possible outcomes, so $A=\Omega$. Thus you've only defined $P(\Omega)$ and nothing else!
Moreover, your expression
$$P(A)=\color{red}{\frac{A}{\Omega}}$$
isn't quite meaningful. $A$ and $\Omega$ are sets and can't be divided by each other. You meant to divide their cardinalities:
$$P(A)=\frac{|A|}{|\Omega|}.$$
That would be the correct definition, if we let $A$ be an arbitrary subset of $\Omega$, not just the one that you defined.
