(related question : http://mathoverflow.net/questions/7998/most-general-way-to-generate-pairwise-independent-random-variables)

 Let $X_1,X_2,X_3,X_4$ be four random variables with standard Bernoulli distribution (i.e.  $P(X_i=0)=P(X_i=1)=\frac{1}{2}$ for $1 \leq i \leq 4$) and such that any two of those four variables are independent.

 Then  the fourtuple $\overrightarrow{X}=(X_1,X_2,X_3,X_4)$ takes at least one of the three values
$(0,0,0,0),(0,0,0,1)$ or $(0,0,1,0)$ with positive probability.

  This surprising fact  can be shown by brute force: what we have is essentially a system of linear equalities and inequalities in 16 variables corresponding to the distribution of $\overrightarrow{X}$. The proof is straightforward but not very illuminating. Are there better explanations?


 UPDATE 22:15 Actually, the following stronger property holds : for any $(a,b,c,d)$ in
$\lbrace 0,1 \rbrace ^4$, the event $(X_1,X_2)=(a,b) \Rightarrow (X_3,X_4)=(c,d)$ has probability $<1$ (in this sense, $(X_3,X_4)$ cannot be "dependent" on $(X_1,X_2)$).