Why is this conditional probability set up like this?

Question

I am having troubles understanding why it is necessary to add $E_1$,$E_2$, and $E_4$. What is wrong with just using $E_3$?

Answer 1

Because it gives us an expression of terms which are more easy to evaluate.

We do not need to do it this way, but it works.

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We seek the probability for $E_4$, the probability that each ace is in a distinct pile. But we need a method to evaluate this probability, since it is not immediately apparent. There are several methods that may serve; and this is but one among them...

Since $E_4$ is a subset of $E_3$, and $E_3$ a subset of $E_2$, and so on, then $E_4$ is the intersection of all and we may apply the product rule.

$$\small\begin{align}\mathsf P(E_4) &=\mathsf P(E_1\cap E_2\cap E_3\cap E_4)&&\text{since }E_4\subseteq E_3\subseteq E_2\subseteq E_1\\[1ex]&=\mathsf P(E_1)~\mathsf P(E_2\mid E_1)~\mathsf P(E_3\mid E_2\cap E_1)~\mathsf P(E_4\mid E_3\cap E_2\cap E_1)&&\text{via the product rule}\\[1ex]&=\mathsf P(E_1)~\mathsf P(E_2\mid E_1)~\mathsf P(E_3\mid E_2)~\mathsf P(E_4\mid E_3)&&\text{since }E_4\subseteq E_3\subseteq E_2\subseteq E_1\end{align}$$

Now each factor in this product is easy to evaluate from first principles.

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$\mathsf P(E_1)$ is the probability for the ace of spade being in one hand.

$\mathsf P(E_2\mid E_1)$ is the probability for the ace of hearts being in a different hand from the ace of spades, when given that the ace of spades is in one hand.

$\mathsf P(E_3\mid E_2)$ is the probability for the ace of diamonds being in a different hand from both the aces of spades and hearts, when given that the aces of spades and hearts are themselves in two different hands.

$\mathsf P(E_4\mid E_3)$ is the probability for the ace of clubs being in a different hand from the other aces, when given that those aces are themselves in three different hands.