I need to come up with the probability space for rolling two dice, which includes $\Omega$ (the sample space), a $\sigma$-field and a probability function.

Thus far, I have that $\Omega = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2),(4,3), (4,4), (4,5),(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}$

Next, I need to come up with a $\sigma$-field, preferably the $\sigma$-field consisting of all possible outcomes.

I've already done this for rolling one die, where $\Omega = \{ 1,2,3,4,5,6\}$, and the $\sigma$-field $\mathcal{F} =\{\emptyset, \{1\},\{2\},\{3\},\{4\},\{5\}, \{6\},\{1,2 \}, \{ 1,3\}, \{1,4\},\{1,5\}, \{1,6\}, \{2,3\},\{2,4\},\{2,5\},\{2,6\}, \{3,4\},\{3,5\},\{3,6\},\{4,5\},\{4,6\},\{5,6\}, \{1,2,3\}, \{1,2,4\}, \{1,2,5\}, \{1,2,6\}, \{1,3,4\}, \{1,3,5\}, \{1,3,6\}, \{1,4,5\}, \{1,4,6\}, \{1,5,6\}, \{2,3,4\}, \{2,3,5\}, \{2,3,6\}, \{2,4,5\}, \{2,4,6\}, \{2,5,6\}, \{3,4,5\}, \{3,4,6\}, \{3,5,6\}, \{4,5,6\}, \{1,2,3,4\}, \{1,2,3,5\}, \{1,2,3,6\}, \{1,2,4,5\}, \{1,2,4,6\}, \{1,2,5,6\}, \{1,3,4,5\}, \{1,3,4,6\}, \{1,3,5,6\}, \{1,4,5,6\}, \{2,3,4,5\}, \{2,3,4,6\}, \{2,3,5,6\}, \{2,4,5,6\}, \{3,4,5,6\}, \{1,2,3,4,5 \}, \{1,2,3,4,6\}, \{1,2,3,5,6\}, \{1,2,4,5,6\}, \{1,3,4,5,6\}, \{2,3,4,5,6\}, \Omega \}$ consists of the $2^{6}$ possible outcomes of rolling one die. For example, $\{2,3,5,6\}$ corresponds to "rolling a $2$,$3$,$5$, or $6$".

Now, for the case of two dice, I was wondering if the $\sigma$-field, call it $\mathcal{G}$, consists of all possible pairings of the outcomes from $\mathcal{F}$.

So, then I'd expect $\mathcal{G}=\{\emptyset, (\{1\},\,\text{each outcome in}\, \mathcal{F}), (\{2\},\, \text{each outcome in}\,\mathcal{F}),\cdots , (\{2,3,4,5,6\},\,\text{each outcome in}\,\mathcal{F}), \Omega\}$.

Is that correct? If not, how do I figure this out? Also, is it correct to have $\emptyset$ and $\Omega$ as members of $\mathcal{G}$? Or should I have $(\emptyset,\,\text{each outcome in}\,\mathcal{F})$ and $(\Omega , \text{each outcome in}\,\mathcal{F})$?

I think it would help if I really understood what $\emptyset$ and $\Omega$ mean in the context of this problem, or at least in the case of the single die. Does $\emptyset$ correspond to not rolling the die at all, and $\Omega$ correspond to rolling the die and getting either a $1$,$2$,$3$, $4$, $5$, or $6$? So, then would it even make sense to have $(\emptyset,\,\text{each outcome in}\,\mathcal{F})$ in the two dice case? Because then, wouldn't that mean only rolling one of the dice? (So, then for example, $(\{2,6\},\emptyset)$ wouldn't make sense either, because it would correspond to only rolling the first die.) On the other hand, though, it seems $(\{2,3,4,5,6\},\Omega)$ would make sense, because it corresponds to rolling either a $2,3,4,5$, or $6$ for the first die and rolling either a $1,2,3,4,5$, or $6$ for the second one.

As for the probability function, for one die, it would be $P(\{i\})=\frac{1}{6}$, $i = 1,\cdots, 6$. For two dice, we now have $36$ possible rolls, which are all equipositive, but I'm not sure how to write out the probability function in this case. Could somebody help me with this as well?

Thank you for your time and patience.

  • 1
    $\begingroup$ use the definition you have for $\Omega = \{(1,1),(1,2) \dots $, so your points will be pairs; and follow the same logic for one die. You won't have the degenerate points such as $(\{2,6\},\emptyset)$. $\endgroup$
    – karakfa
    Commented Sep 17, 2017 at 3:57
  • $\begingroup$ @karakfa I applied the binomial theorem in that case, and then it was easy to see what taking 6 possible rolls say 2 at a time looked like. Should I perhaps relabel these points $(1,1), (1,2), \cdots$ as $1,2,\cdots, 36$ to make it easier? $\endgroup$
    – user100463
    Commented Sep 17, 2017 at 3:59
  • $\begingroup$ If it makes it easier you can, but these are just labels. Treat (1,1) etc as the label to eliminate renumbering them. $\endgroup$
    – karakfa
    Commented Sep 17, 2017 at 10:02
  • $\begingroup$ "I need to come up with the probability space for rolling two dice" What for? In case you are wondering, this is a serious question, as almost nothing in probability requires to know what the probability space $(\Omega,\mathcal F,P)$ is. In your setting, all that matters is that such a probability space exists, which supports two independent random variables $X$ and $Y$ uniformly distributed on $\{1,2,\ldots,6\}$ (the results of the two rolls). Then one simply works with $(X,Y)$ instead of losing one's time on irrelevant matters. $\endgroup$
    – Did
    Commented Sep 20, 2017 at 13:26
  • 1
    $\begingroup$ A case of sadism? $\endgroup$
    – Did
    Commented Sep 20, 2017 at 17:40


You must log in to answer this question.