Understanding what the sigma-field should be in dice-experiment I'm new at measure theory and would like some guidance on how to solve this (probably very easy) task:
Throw a fair dice twice. We have the following information of a given outcome: "Two sixes";"Exactly one six"; "No sixes".
Describe the probability space ($\Omega$,$\mathcal{F}$,$\textbf{P}$) that corresponds to this information. 
Also give all $\mathcal{F}$-measurable functions
$X:\Omega\rightarrow\{-1,1,2,3,...\}$ med $E(X)=0$,
corresponding to a fair game.
My thoughts so far:
The sample space should contain all the "pairs" of dices, where the order is not relevant: $\Omega=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,4),(4,5),(4,6),(5,5),(5,6),(6,6)\}$
I'm having troubles to understand what the $\sigma$-field $\mathcal{F}$ should be. 
I thought that I should include all the three events described, but then it's not a $\sigma$-field. What am I doing wrong?
$\mathcal{F}=\{\{\emptyset\},\{(6,6)\},\{(6,1),...(6,6)\},\{(1,1),...,(5,5)\},\{\Omega\}\}$ 
 A: I suspect that you are used to finite outcome spaces $\Omega$ where the outcomes are equiprobable.
This answer shows you that it can be done differently.
If I understand well then only the number of sixes is involved, and this number takes values in $\{0,1,2\}$.
Then it is actually enough to go for $\Omega=\{0,1,2\}$ equipped with $\sigma$-algebra $\mathcal F=\wp(\Omega)$.
It remains to define the probability measure $\mathbf P:\wp(\Omega)\to\mathbb R$.
Based on the fact that the die is fair we find easily that: 


*

*$\mathbf P(\{0\})=\frac56\frac56=\frac{25}{36}$

*$\mathbf P(\{2\})=\frac16\frac16=\frac{1}{36}$

*$\mathbf P(\{1\})=\frac56\frac16+\frac16\frac56=\frac{10}{36}$


This determines $\mathbf P$ completely for every event $A\in\mathcal F$ by:$$\mathbf P(A)=\sum_{a\in A}\mathbf P(\{a\})$$
Further every function $X:\Omega\to\mathbb R$ is $\mathcal F$-measurable (as is always the case if the chosen $\sigma$-algebra is the powerset of $\Omega$).
Advantage of the constructed probability space is that only relevant information is contained. More concise is not even possible.
