I know the title is confusing. I was just thinking about this question in my head and didn't know how to word it correctly.
Let's say I have a 6-sided die, a coin, and an ace of spades (not the whole deck, just the one card). And I will either roll the die, flip the coin, or give you the ace of spades for free. But you don't know which of these actions I will do.
(a) What is the chance you will get a 2?
(b) What is the chance you will get a head?
(c) What is the chance you will get the ace of spades?
I was thinking the answer could be one of two possibilities: The first is $\frac{1}{9}$ for all three because there are 9 possible outcomes (1, 2, 3, 4, 5, 6, H, T, Ace) and each one comes up exactly once.
The second is $(\frac{1}{18}, \frac{1}{6}, \frac{1}{3})$ for (a, b, c), respectively.
Because there is an equal chance of doing each action (so $\frac{1}{3}$ for each), then we apply a $\frac{1}{3}$ probability to each action.
So we would have
(a): $(\frac{1}{3})(\frac{1}{6})+(\frac{1}{3})(\frac{0}{2})+(\frac{1}{3})(\frac{0}{1}) = \frac{1}{18}$
(b): $(\frac{1}{3})(\frac{0}{6})+(\frac{1}{3})(\frac{1}{2})+(\frac{1}{3})(\frac{0}{1}) = \frac{1}{6}$
(c): $(\frac{1}{3})(\frac{0}{6})+(\frac{1}{3})(\frac{0}{2})+(\frac{1}{3})(\frac{1}{1}) = \frac{1}{3}$
I have never taken a probability class before, so I don't have much background knowledge. Are either of these right? If so, why?
Edit: Suppose each action (roll die, flip coin, grab ace) is equally likely. Sorry, as I said, I know very little about probability.