In a deck of 52 cards, you draw two cards without replacement. Are the events conditionally independent given that the first card drawn is a heart?

Question

In a deck of 52 cards, you draw two cards consecutively without replacement. Are the events "drawing a heart on the first draw" and "drawing a black card on the second draw" conditionally independent given that the first card drawn is a heart?

I solved this problem as follows and I would like to know if the approach is correct:

$A: \text{Drawing a heart on the first draw} \\ B: \text{Drawing a black card on the second draw}\\ C: \text{First card drawn is a heart} \\ P(A): \frac{13}{52} = 0.25\\ P(B): \frac{26}{51} \approx 0.5098 \\$

If events $A$ and $B$ are independent then: $P(A\cap B) = P(A)P(B)$.

$P(A\cap B) = \frac{_{13}C_1 \times _{26}C_1}{_{52}C_2} \approx 0.2549\\ P(A)P(B) \approx 0.1275$

Therefore these events are not independent because $P(A)P(B) \neq P(A\cap B)$. Now, we check for conditional independence on $C$.

$P(A|C) = 1$: Because if we know the first card drawn is a heart then it changes the probability of A. Probability of drawing a heart on the first draw given the first draw is a heart is 1.

$P(B|C) = \frac{P(C|B)P(B)}{P(C)}\\ P(C) = P(C|B)P(B) + P(C|B^c)P(B^c) = P(C\cap B) + P(C \cap B^c) = \frac{_{13}C_1 \times _{26}C_1}{_{52}C_2} + \frac{_{13}C_1 \times _{25}C_1}{_{52}C_2} \approx 0.2549 + 0.2451 \approx 0.50\\ P(B|C) = \frac{0.2549}{0.50} \approx 0.5098 \\ P(A \cap B | C) = P(A|C)P(B|C) = 1 \times 0.5098 = 0.5098\text{ must hold for conditional independence.}\\ P(A \cap B|C) = \frac{P(C|A \cap B)P(A \cap B)}{P(C)} = \frac{P(A \cap B \cap C)}{P(C)} = P(A \cap B) = 0.2549.$

Therefore, these events are not independent and, not conditionally independent since, $P(A \cap B|C) \neq P(A|C)P(B|C).$

Is this approach correct. Are there gaps in my understanding that jump out to you? Is there a better way to solve this question?

Answer 1

Some of the issues:

• $A$ and $C$ are the same events so, conditioned on $C$, every event is independent of $A$, including $B$: $P(A \cap B \mid C)= P(B \mid C) =P(A \mid C)P(B \mid C)$
• $P(A \cap B) = \frac{13 \times 26}{52 \times 51}$ which is half what you say
• $P(A)P(B)=\frac{13}{52}\times \frac{26}{52}$, close to but not equal to what you say
• $P(B \mid C) = \frac{26}{51}$ as you correctly say but $P(A \cap B \mid C)$ is the same as this rather than half of it.

You can say there is not (unconditional) independence between $A$ and $B$ because $\frac{13 \times 26}{52 \times 51} \approx 0.12745$ is not the same as $\frac{13}{52}\times \frac{26}{52}=0.125$, but this is not the calculation you did.

Intuitively, drawing a heart first makes it more likely the second card is black than not knowing the first card would (you have changed your knowledge about the red cards remaining), so you would not expect independence.

Conditioned on the first card being a heart, drawing a heart on the first draw would be inevitable so trivially everything else would be independent of that.