Non-independence of three events, given their intersection is independent

Question

Suppose you flip a fair coin three times, find three events $A$, $B$, and $C$, such that no two of them are independent, but $P(A \cap B \cap C) = P(A)P(B)P(C)$

I am given the question above. From what I can tell, this is a trick question since:

$P(A \cap B \cap C) = P(A)P(B|A)P(C|B,A) = P(A)P(B)P(C)$

then

$P(B|A)$ must equal $P(B)$ (and the same for $P(C|B,A) = P(C)$), which means

$P(A \cap B) = P(A)P(B)$, that is, they are independent, which goes against the premise.

Am I missing something here or is this correct? (I feel like I'm missing something)

EDIT: If someone could answer this visually (using a venn diagram or something) I'd really appreciate it

Answer 1

Consider rolling a fair die.
Let $A = \{1,2,3\}$, $B=A$ and $C = \{2,4,5,6 \}$ where the sets stand for the outcome of the die.

Then we have $\mathbb{P}(A) = \mathbb{P}(B) = \frac{1}{2}$ and $\mathbb{P}(C)=\frac{2}{3}$. However, we have $\mathbb{P}(A \cap B ) = \mathbb{P}(A) \neq \mathbb{P}(A) \mathbb{P}(B)$, so $A$ and $B$ are dependent. Furthermore we have $\mathbb{P}(A \cap C) = \mathbb{P}( \{2\} ) = \frac{1}{6} \neq \frac{1}{3} = \mathbb{P}(A) \mathbb{P}(C)$, likewise for $B$ and $C$. So $A$,$B$ and $C$ are mutually dependent.

However, we have $\mathbb{P}(A \cap B \cap C) = \mathbb{P}(\{2\}) = \frac{1}{6} = \frac{1}{2} \frac{1}{2} \frac{2}{3} = \mathbb{P}(A) \mathbb{P}(B) \mathbb{P}(C)$.

Answer 2

Since you're supposed to flip a fair coin three times, you have 8 different equally probable outcomes to work with. Call them $\{1,2,3,\ldots,8\}$.

The simplest numbers that seem likely to work here are $P(A)=P(B)=P(C)=\frac12$, in which case we need $A\cap B\cap C$ to contain a single outcome only.

One strategy would be to pick $A$ and $B$ to be as blatantly dependent as they can be: $A=B=\{1,2,3,4\}$, and then adjust $C$ to get the right value of $P(A\cap B\cap C)$. Whether this $C$ is independent of $A$ and/or $B$ will then not matter for the validity of our counterexample.

Can you find a 4-element $C$ such that $\{1,2,3,4\}\cap\{1,2,3,4\}\cap C$ has one element?

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The problem with your reasoning in the question is that although you can conclude that $$P(B)\cdot P(C)=P(B\mid A)\cdot P(C\mid A,B)$$ this does not allow you to conclude that the two factors on each side are the same pair. A conditional probability can be either smaller or larger than the probability without the condition, so just because the products are the same doesn't mean that the factors have to be. In the above answer, the equation works out as $\frac12\cdot\frac12=1\cdot\frac14$.