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!


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.

  • $\begingroup$ Sweet, glad to hear I wasn't too far lost in the work. First, I had left $\Omega$ as that long list because my professor has done that in all the examples. While they were simpler than this one, we weren't introduced to another notation, so I went with it. A couple of questions, was my choice for $\mathscr F$ the best choice for this experiment, or was there a better option I could have found? I went with what seemed to be the most obvious at the time; each possible roll of one die, then two die, then three ($\Omega$). $\endgroup$
    – NoVa
    Jan 12 '18 at 6:04
  • $\begingroup$ An element of the set of observations, is a group of three numbers, showing which number came on which dice. Therefore, "each possible roll of one die" isn't an observation, since when you do the experiment, you do get the information of what has happened to one specific die, however you get far more, namely what has happened to each dice. What you can do, is restrict your attention to one dice, but the elements that are in your $\mathcal F$, like $\{5,6\}$ etc. are not subsets of $\Omega$. $\endgroup$ Jan 12 '18 at 6:10
  • $\begingroup$ Therefore, the $\mathcal F$ as written in your question, is incorrect for the above reason. However, there isn't a "better" $\mathcal F$ in very general application, than the one I have shown you. Do not be disheartened : it will take a short matter of time for you to get to grips with this concept. However, you will have to be a little more careful in presentation, that's all. $\endgroup$ Jan 12 '18 at 6:13
  • $\begingroup$ I think I might be a little confused on what you mean exactly. By "restricting your attention to one dice", do you mean that a correct $\mathcal F$ should be $\{\varnothing, \{1\}, ..., \{1,2\},..., \{1,6\},... \{1,2,...,6\}\}$ because you not only have information on which value was rolled, but you also have information on which values were not rolled? $\endgroup$
    – NoVa
    Jan 12 '18 at 6:20
  • $\begingroup$ Can you clarify what $\{1\}$ means in $\mathcal F$? Note that $\mathcal F$ must contain as it's elements, subsets of $\Omega$, while $\{1\}$ is not a subset of $\Omega$, because $1$ is not an element of $\Omega$ (while $(1,1,1)$ is an element, for example). It is still possible that I am incorrect and have interpreted $\{1\}$ incorrectly, so please reprimand me if that is the case. $\endgroup$ Jan 12 '18 at 6:25

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.

  • $\begingroup$ Thank you for your response. @Actoh's below helped me understand my misunderstanding of $P(A)$, but I'm still a little confused about $\mathscr F$. Maybe it will help me understand if I consider the following example. Consider flipping a coin two times. Here, $\Omega={1,2}$ and $\mathscr F = \{\varnothing, \Omega, \{(1,1)\}, \{(1,2)\}, \{(2,2)\}, \{(1,1),\{(1,2),(2,2)\}\},\{(1,2),\{(1,2),(2,2)\}\}, \{(2,2),\{(1,1),(1,2)\}\} $? $\endgroup$
    – NoVa
    Jan 12 '18 at 6:42
  • 1
    $\begingroup$ Oh, but $\Omega$ is not $1,2$. It is $\Omega = \{(1,1),(1,2),(2,1),(2,2)\}$. That is,each element of $\Omega$ is a possible outcome when two coins are flipped. Most of the subsets you have written are correct, except $\{(1,1),\{(1,2),(2,2)\}\}$ and the ones after that. An element of $\mathcal F$ is a subset of $\Omega$. Therefore, what is correct is $\{(1,1),(1,2),(2,2)\}$, and similarly for the others. I do not know why the inner bracket was used. $\endgroup$ Jan 12 '18 at 7:33

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