I'll start with a couple of questions:
- Is there a systematic way to construct the sample space for a given problem in naive probability theory? Or is it more of an art than a science?
- Is there a unique correct sample space for a problem or there can be multiple ones (maybe depending on what we condition on in the background)?
I'll illustrate my conundrum with exercise 32 from Blitzstein-Hwang's textbook, page 38 (2nd edition);
Four cards are face down on the table. You are told that two are red and two black, and you need to guess which two are red and which two are black. You do this by pointing to the two cards you're guessing are red (and then implicitly you're guessing that the other two are black). Assume that all configurations are equally likely, and that you do not have psychic powers. Find the probability that exactly 4 of your guesses are correct.
I approach. Suppose we lay cards on the table left to right, indicating the position on the table by numbers 1 to 4. It seems plausible that we could take as the elementary outcome the information about how the red cards were dealt, and then how we choose the red cards. We can encode such an elementary outcome as, e.g $$(\{1,3\}, \{1,4\}),$$meaning that the red cards were dealt at positions 1 and 3, and then we chose 1 and 4 as our guess for where the red cards are. This way, if we assume that dealing the cards is independent from guessing, there are $${4 \choose 2} \cdot {4 \choose 2} =36$$elementary outcomes in the sample space. For our problem, we select all the outcomes of the form $$(\{x, y\}, \{x,y\}),$$ and since there are 6 of those, the probability is $$\frac{6}{36}=\frac{1}{6}.$$
II approach. Another approach would be to simply assume that the cards have already been dealt (say the red ones are $\{1,2\}$, but we do not know it when guessing), and there are 6 ways we can choose 2 out 4 cards with equal probability. Ergo, the sample space consists of 6 equiprobable elementary outcomes, and since only one leads to the event we are interested in, the probability is $$\frac{1}{6}.$$
I see the second approach used more often to solve problems. However, I find the second approach less satisfactory than the first one, as it seems that it is collapsing the random nature of how we deal the cards. In that sense, it feels like the second approach is computing the conditional probability $$\mathbb{P}(\text{we guess both red cards correctly} \ | \ \text{red cards were dealt in slots} \{1,3\}),$$but since we do not know the position in advance and still are guessing randomly, the events are independent and the answer is the same. Is that a good explanation for why the answer is the same but sample spaces are different?