Counter-intuitive independent and dependent events

probability

In elementary school, I was taught about "independent" and "dependent" events with examples of taking two balls from a box. Assume there is a box containing 2 blue balls and 3 red balls. Two balls are drawn from the box, one is drawn after the other draw. Let $A$ and $B$ represent the probability of taking a blue ball in the first draw and that of taking a blue ball in the second draw, respectively.

Independent

If the ball taken in the first draw is returned before taking the second ball, find the probability that both balls are blue.

It is clear enough that both events are independent. So the required probability is
\begin{align}
P(A\cap B) &=P(A) \times P(B)\\
&= \frac{2}{5} \times \frac{2}{5}\\
&= \frac{4}{25}
\end{align}

Dependent

If the ball taken in the first draw is not returned before taking the second ball, find the probability that both balls are blue.

It is also clear enough that both events are dependent. So the required probability is
\begin{align}
P(A\cap B) &=P(A) \times P(B)\\
&= \frac{2}{5} \times \frac{1}{4}\\
&= \frac{1}{10}
\end{align}

Here the independence and dependency are extremely obvious.

Today I found a counter-intuitive example as follows.

One card is drawn from a deck of 52 cards. Let $X$ be the event that a club is selected, let $Y$ be the event that a face card (Jack, Queen, or King) is drawn, and let $Z$ be the event that a Jack is drawn.

The solution uses

$$P(X|Y)=P(X)$$

to judge that $X$ and $Y$ are independent. This equation is understandable and I know how it is derived.
The solution also shows that $$P(Z|Y)\not =P(Z)$$ so $Y$ and $Z$ are dependent.

Question

Let's forget about the equations and just use our intuition.

How can we explain that

  • $X$ (taking a club) and $Y$ (taking a face) are independent?
  • $Y$ (taking a face) and $Z$ (taking a Jack) are dependent?

My intuition does work only for the first case about taking two balls because it is so obvious as follows.

  • $A$ (taking a blue) and $B$ (taking a blue) with replacement are independent.
  • $A$ (taking a blue) and $B$ (taking a blue) without replacement are dependent.

Best Answer

A deck of cards has exactly one card for each suit and rank. $X$ specifies the suit and $Y$ the rank, so we should expect those two events to be independent – knowing the card's rank gives us no information about the suit.

Yet $Z$ also specifies the rank like $Y$, so knowing that $Y$ is true gives us some information about the card's rank – it is either jack, queen or king, and the probability that it is a jack should be higher. Thus $Y$ and $Z$ should be dependent.

Indeed, both intuitions are true.

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