Can anyone help me understand the following conditional probability example from Casella and Berger Statistical Inference Page 20, specifically using set theoretic definitions for the sample space and events? I'm stuck trying to understand how the set theoretic definitions of conditional probability and multiplication theorem work. Appreciate any guidance if there is a better approach to this.
For example, the following question of the probability to draw 4 Aces from a deck of 52-cards is given:
How would this look using sets? For instance this calculation should be according to the definition of the multiplication rule:
$P(A_1 A_2 A_3 A_4) = P(A_1)P(A_2|A_1)P(A_3|A_1A_2)P(A_4|A_1A_2A_3)$
it's my assumption where:
$A_1 = \{A_c, A_s, A_h, A_d\}, |A_1| = 4, S=\{x: x = $ one of 52 cards $\}$, $P(A_1) = |A_1|/|S| = 4/52$.
but substituting $A_1$ into $P(A_2|A_1)$ the results are not what I expected. This is the first confusing issue. I would expect Ace of Clubs to be withdrawn from the condition, essentially creating a new sample space as follows, $B_1 = \{x: x = $ one of 51 cards, $x != A_c\}$ and $A_2 = \{A_s, A_h, A_d\}$
so it becomes: $P(A_2|B_1) = P(A_2 B_1) / P(B_1) = (3/52)/(51/52) = 3/51$
and now things start to make sense, but these results don't agree with the definitions above.
Your misunderstanding is thinking that the $A_i$ are sets of cards; rather, they are events, i.e. $A_2$ is the event that "the 2nd card drawn is an ace."
I think you are overthinking things a little bit. The idea for computing $P(A_2 \mid A_1)$ is just to understand that if $A_1$ occurred, there are $51$ cards remaining in the deck, three of which are $3$, so $P(A_2 \mid A_1) = 3/51$. It's as simple as that.
If you want to go compute it using $P(A_2 \mid A_1) = \frac{P(A_1 \cap A_2)}{P(A_1)}$, you'd just end up with circular reasoning, since the whole point of this discussion is to compute the intersection $P(A_1 \cap A_2) = P(A_1) P(A_2 \mid A_1)$ anyway. But if you are just doing a sanity check, then yes $P(A_1) = 4/52$, $P(A_2 \mid A_1) = 3/51$, and $P(A_1 \cap A_2) = \frac{4}{52} \cdot \frac{3}{51}$.