Vtyjkyuk

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 ($Omega$,$mathcalF$,$textbfP$) that corresponds to this information.
Also give all $mathcalF$-measurable functions
$X:Omegarightarrow-1,1,2,3,...$ 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 $mathcalF$ 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?
$mathcalF=emptyset,(6,6),(6,1),...(6,6),(1,1),...,(5,5),Omega$

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 $mathbf P:wp(Omega)tomathbb R$.

Based on the fact that the die is fair we find easily that:

• $mathbf P(0)=frac56frac56=frac2536$


• $mathbf P(2)=frac16frac16=frac136$


• $mathbf P(1)=frac56frac16+frac16frac56=frac1036$

This determines $mathbf P$ completely for every event $Ainmathcal F$ by:$$mathbf P(A)=sum_ain Amathbf P(a)$$

Further every function $X:Omegatomathbb R$ 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.