probability

Yeah, like in the title, how do Regular Conditional Probabilities assign values to $P(A|B)$ when $P(B) = 0$? Because I am not trained in advanced probability, please explain it as simply as possible. It'll be great if you can demonstrate with an example. Let $X$ be a random variable between $[0, 1]$. What is the $P(X=\frac{2}{3}|X=\frac{2}{3})$?

Also, can it also assign values to $P(A|B)$ in finite spaces? For example, imagine you throw a die, and you observe that the die landed on an odd number at 12 noon. So letting $O$ be the event that the die landed odd, $P(O)$ changes from $\frac{1}{2}$ to $1$ after noon. Hence, after noon, letting $E$ be the event that the die landed even, $P(E)$ changes to $0$. But even after 12 noon, I still intuitively want to say that $P(E|E)$ is still $1$. It seems to me that for any non-empty set $A$ in the sigma-algebra, $P(A|A) = 1$. Can regular conditional probabilities assign $P(E|E)$ a value of $1$? Is it possible? Or do I need to give that intuition up? Thanks for your help.

Edit: I think my question is different from this: Probability, conditional on a zero probability event

Because I'm asking about finite spaces too. Also, I'm asking for a demonstration of how regular conditional probabilities assign value to $P(X=\frac{2}{3}|X=\frac{2}{3})$.

Edit 2: Let the sample space in the dice example be {1, 2, 3, 4, 5, 6}.

Edit 3: The sample space for the dice example is {1, 2, 3, 4, 5, 6}. The initial probability function before noon is $P$. After noon, the probability function becomes $P'$. So $P(O)$ = $P(E)$ = 0.5. But $P'(O) = 1$ and $P'(E) = 0$. Still, I want to say that $P'(E|E) = 1$.