Unconditionally dependent, but conditionally independent

I'm trying to come up with some real life examples of situations where you have unconditional dependence, but conditional independence. Here's one that I came up with, but I'm not sure if it's sound.

Consider $$A, B, C$$ to be the events that you win, lose, or tie in a game.

$$A$$ and $$B$$ are clearly dependent since if $$A$$ happens, $$B$$ can't happen.

Now say you're given $$C$$. Then I think knowing C happens would make $$A$$ and $$B$$ independent. But I am hesitant to believe this is sound because the definition of conditional independence is $$P(A, B | C) = P(A|C) P(B|C)$$. We know that $$P(A, B | C) = 0$$ since $$A$$ and $$B$$ are mutually exclusive regardless if we're conditioning on anything. We also know that $$A,C$$ and $$B,C$$ are mutually exclusive, so $$P(A|C) = P(B|C) = 0$$. But is $$A|C$$ and $$B|C$$ actually conditionally independent here, or is $$P(A, B | C) = P(A|C) P(B|C)$$ naturally satisfied due to the mutual exclusivity?

This is an example of conditional independence, since any probability-zero event (in this case $$A\mid C$$) is independent of any other event. However, it's rather a trivial example.
You can get examples where the independence is non-trivial. For example, roll a standard die, and set $$A$$ to be the event of getting an even number, $$B$$ to be the event of getting a prime number, and $$C$$ to be the event of getting at most $$4$$.
Now $$P(A)=1/2$$, $$P(B)=1/2$$ but these events are not independent since $$P(A\cap B)=1/6$$. However, $$P(A\mid C)=P(B\mid C)=1/2$$ and $$P(A\cap B\mid C)=1/4$$, so the conditional events are independent.
• Is coming up with these kind of examples more art than science? Like what made you think of this die example and having $B$ be the event of getting a prime number. Did you do some guess and checking with the probabilities until you found one that worked? May 4, 2021 at 18:07
• @roulette01 I was looking for conditional events of probability $1/2$. For conditional independence you need their intersection to have probability $1/4$, so have a conditional state space (i.e. event $C$) of size $4$. Then add some other states which correspond to $C^c$ and choose $A\cap C^c$, $B\cap C^c$ any way you like, just making sure that they aren't independent. It's easy to see such events will exist; getting them to have "natural" definitions is a bit more tricky and involves some trial and error. May 4, 2021 at 18:24