How do I compute the probability in this exercise? I need to define the probability space $\Omega$  and it's probability function $\Bbb{P}$ for the following exercise:

For a fair die which we toss two times, compute the probability that the parity of the two numbers that show up matches.

My Idea was the following. We define $\Omega=\{1,...,6\}^2$, then $|\Omega|=36$. We have the following partition into even and odd numbers:$$\Omega=\{1,3,5\}~\dot\cup~\{2,4,6\}=:A~\dot \cup ~B$$Now let us define $\Lambda\subset \Omega$ such that it contains all pairs $(u,v)$ where $u,v\in A$ or $u,v\in B$.

*

*Case $1$: $u,v\in A$. Then we have $3^2=9$ possibilities.

*Case $2$: $u,v\in B$. Then we have $3^2=9$ possibilities.

Thus we have $18$ possibilities and $$\Bbb{P}(\Lambda)=\frac{|\Lambda|}{|\Omega|}=\frac{18}{36}=\frac{1}{2}$$ Is this correct so?
 A: Looks fine overall!
One suggestion: you might be a bit more explicit about what the measure $\mathbb P$ is actually defined to be. It's not hard, of course -- it just assigns a measure of $1/36$ to each of the $36$ atomic elements (i.e. single pairs). I think it'd be good to specify that for completeness' sake, since in principle you could construct an alternative probability measure that might correspond to, say, rolling weighted dice. Additionally, you were apparently asked to construct the probability measure, and your posted solution sees to just use it without ever stopping to say how it's defined.
That said, that comment is a bit nitpicky. The probability measure you have implicitly chosen is, in any reasonable sense, the default one.
A: Alternative approach:
When you define the sample space, it is very helpful if each element in the space is equally likely.  Within this parameter, you could simplify the problem by defining the sample space to have only $4$ elements, instead of $36$ elements.
Let $E$ be the event that the die roll is even. 
Let $O$ be the event that the die roll is odd.
Then, you could define the sample space to be
$\{E,O\}^2~$ (four elements)
Here, the favorable outcomes would be the union of $\{E\}^2$ (one element) with $\{O\}^2$ (one element).
So, the probability would be computed as
$$\frac{1 + 1}{4}.$$
The underlying idea behind this alternative approach is that the problem asks you to focus (only) on the die's odd/even parity on each of the  two rolls.  This can be (validly) done without focusing on the (exact) number of spots face-up on the die rolls.
