Construction of binary random variable

We throw two coins in a row and thus get the event space $\{ZZ, WW, ZW, WZ\}$.
Each of the 4 elementary events has a probability $1/4$.
how can I construct 3 binary random variable $x_1$, $x_2$, $x_3$ about this event space, which are 2-fold independent, but not independent.

Let $E = \{ZZ,WW,ZW,WZ\}$. A binary random variable is a mapping $X\colon E \rightarrow \{0,1\}$. The probability of an event, say $X=0$, is defined as
$$ \Pr[X = 0] =\frac{|\{e \in E : X(e) = 0\}|}{4}. $$
For example, let $C_1(ZZ)=C_1(ZW) = 0$, $C_1(WZ)=C_1(WW) = 1$. Then $\Pr[C_1=0]=1/2$.

Two (binary) random variables $X_1,X_2$ are independent if for every $b_1,b_2 \in \{0,1\}$,
$$\Pr[X_1 = b_1 \text{ and } X_2 = b_2] = \Pr[X_1 = b_1] \Pr[X_2 = b_2]. $$
For example, if $C_2(ZZ) = C_2(WZ) = 0$ and $C_2(ZW) = C_2(WW) = 1$, then you can check that $C_1$ and $C_2$ are independent, by checking that $\Pr[C_1 = b] = \Pr[C_2 = b] = 1/2$ and $\Pr[C_1 = b_1 \text{ and } C_2 = b_2] = 1/4$. On the other hand, $C_1$ and $C_1$ are not independent since $\Pr[C_1 = 0 \text{ and } C_1 = 0] = 1/2$.

Similarly you can define when more than two random variables are independent. It should be clear that there are no three binary random variables $X_1,X_2,X_3$ with $\Pr[X_1 = 0] = \Pr[X_2 = 0] = \Pr[X_3 = 0] = 1/2$ which are independent, since that would imply $\Pr[X_1 = X_2 = X_3 = 0] = 1/8$, which is impossible (the only values that $\Pr$ can obtain are $0,1/4,1/2,3/4,1$). On the other hand, any three constant random variables are independent.

A set of random variables is $k$-wise independent if any $k$ of them are independent. The question asks you to find three random variables $X_1,X_2,X_3$ so that any two are independent, but all three aren't. My hint is that there is a solution with $X_1 = C_1$ and $X_2 = C_2$.

StackExchange Computer Science Q#6386, answer score: 4