You are simply assuming that what I’ve calculated is irrelevant. But the only way to know absolutely for sure whether it is irrelevant is to actually do the calculation! That is, if you have information X and Y, and you think Y is irrelevant to proposition A, the only way you can justify leaving out Y is if Pr(A | X and Y) = Pr(A | X). We often make informal arguments as to why this is so, but an actual calculation showing that, in fact, Pr(A | X and Y) != Pr(A | X) always trumps an informal argument that they should be equal.

Your “probability of guessing the correct card” presupposes some decision rule for choosing a particular card to guess. Given a particular decision rule, we could compute this probability, but it is something entirely different from “the probability that the card is a king”. If I assume that’s just bad wording, and that you’re actually talking about the frequency of heads when some condition occurs, well now you’re doing frequentist probabilities, and we were talking about *epistemic* probabilities.

The context is *all* applications of probability theory. Look, when I tell you that A or not A is a rule of classical propositional logic, we don’t argue about the context or what assumptions we are relying on. That’s just a universal rule of classical logic. Ditto with conditioning on all the information you have. That’s just one of the rules of epistemic probability theory that *always* applies. The only time you are allowed to NOT condition on some piece of known information is if you would get the same answer whether or not you conditioned on it. When we leave known information Y out and say it is “irrelevant”, what that means is that Pr(A | Y and X) = Pr(A | X), where X is the rest of the information we’re using. If I can show that these probabilities are NOT the same, then I have proven that Y is, in fact, relevant.