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(related question : 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 $(X_1,X_2)$).

(related question : 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 $(X_1,X_2)$).

(related question : 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?

(related question : 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 $(X_1,X_2)$).

(related question : 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 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?

(related question : 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?