Conditional Probability Question, from Casella and Berger

Question

I'm reading the following problem in Casella and Berger Statistical Inference Page 20-21. I'm confused how the event $\{4$ aces in 4 cards $\}$ is a subset of $\{i$ aces in $i$ cards $\}$. Can anyone provide an example of what the set and subset would look like?

My assumption is, for the first event, the sample space is the unordered arrangements of drawing 4 cards from the 52-card deck, which is $52\choose4$. The first event $A_1 = \{\{A_c, A_d, A_h, A_s\}\}$, $|A_1| = 1$. For the second event, given $i = 1$, the sample space changes to $52\choose1$ and the event is now $A_2 = \{A_c, A_d, A_h, A_s\}$, $|A_2|= 4$. Therefore, I'm confused on how $A_1 \subset A_2$. Appreciate any guidance here.

Answer 1

$\{\text{$i$ aces in $i$ cards}\}$ is the set of all possible events that you get $i$ aces in first $i$ cards in a row.

Similarly, $\{\text{$4$ aces in $4$ cards}\}$ is the set of all possible events you get 4 aces in the first $4$ cards in a row.

The second event is possible only when the first event has occurred. Thus, the first event contains the possibility of the second event occurring.

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Okay, let $\omega$ be a function such that $\omega(i)$ is the card you are getting at $i$th order. We call $\omega$ an event. Then, we can represent $$\{\text{$2$ aces in $2$ cards}\}=\{ \omega : \omega(1)=ace, \omega(2)=ace \}$$

Similarly, $$\{\text{$4$ aces in $4$ cards}\}=\{ \omega : \omega(1)=ace, \omega(2)=ace, \omega(3)=ace, \omega(4)=ace \}$$

Then, as you can see, if $\omega \in \{\text{$4$ aces in $4$ cards}\}$, then $\omega\in \{\text{$2$ aces in $2$ cards}\}$.

Answer 2

To see how the event $\{4$ aces in 4 cards $\}$ is a subset of $\{i$ aces in $i$ cards $\}$.

4 aces in 4 cards:

Let $AA_1 = \{x : x$ is one of 52 cards $\}$ $|AA_1| = 52$

Let $AA_2 = \{x : x$ is one of 52 cards $ \land x \ne Ace_0 \}$ $|AA_2| = 51$

Let $AA_3 = \{x : x$ is one of 52 cards $ \land x \ne Ace_0,Ace_1 \}$ $|AA_3| = 50$

Let $AA_4 = \{x : x$ is one of 52 cards $ \land x \ne Ace_0,Ace_1,Ace_2 \}$ $|AA_4| = 49$

Let $S = AA_1 \times AA_2 \times AA_3 \times AA_4$ $|S| = 52*51*50*49$

To construct the event 1 aces in 1 cards: $ACES = {Ace_0, Ace_1, Ace_2, Ace_3}$ $A_1 = AA_1 \cap ACES \times AA_2 \times AA_3 \times AA_4$ $|A_1| = 4*51*50*49$ $P(A_1) = |A_1| / |S| = 4/52$

To construct the event 2 aces in 2 cards: $A_2 = AA_1 \cap ACES \times AA_2 \cap ACES \times AA_3 \times AA_4$ $|A_2| = 4*3*50*49$ $P(A_2) = |A_2| / |S| = 12/(52*51) = 6/1326$

Does 2 aces in 2 cards agree with the formula for Casella and Berger?

$\frac{4 \choose 2}{52 \choose 2} = 6/1326$

Yes.

Now to show that 4 aces in 4 cards is a subset of i aces in i cards for $i = 3,2,1$. Use $i = 3$ because the set will have the least elements to be a subset of: $A_3 = AA_1 \cap ACES \times AA_2 \cap ACES \times AA_3 \cap ACES \times AA_4$ $|A_3| = 4*3*2*49$

For 4 aces in 4 cards: $A_4 = AA_1 \cap ACES \times AA_2 \cap ACES \times AA_3 \cap ACES \times AA_4 \cap ACES$ $|A_4| = 4*3*2*1$

$A_4 \subset A_3$

therefore

$A_4 \subset A_i$ for $i = 1, 2, 3$