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Let $A_1, A_2, \ldots, A_n$ be a collection of events in a probability space. There are $2^n - n - 1$ subsets S of $\{1, 2, \ldots, n\}$ for which we may or may not have $P(\bigcap_{j \in S}A_j) = \prod\limits_{j\in S}P(A_j)$ $(*)$. (When $S$ is empty or has a single element, this is trivially true.)

Suppose we know whether or not $(*)$ holds for each $S$, except one, which I shall call $T$. Is there any case in which we can deduce from what we know whether or not the equality holds for $T$?

For instance, suppose $n = 3$, and that we know the equality holds for $\{1, 2\}$, $\{1, 3\}$, $\{2, 3\}$, but we wish to know whether or not $(*)$ holds for $T = \{1, 2, 3\}$. In this specific case, we are just asking whether or not we can tell if three events are mutually independent given that they are pairwise independent. But we know we can find pairwise independent events that are not mutually independent and also pairwise independent events that are mutually independent. Therefore, we cannot tell whether or not $(*)$ holds for $T$ with just the information we were given.

Edit: I'm going to make my question more precise as a first step to answering it. First, I'll need a few definitions:

Definition 1: Let $A_1, A_2, \ldots, A_n$ be a collection of events in a probability space. For each $S \subseteq \{1, 2, \ldots, n\}$, we put $\delta_{S} = 1$ if S satisfies $(*)$ and $\delta_{S} = 0$ otherwise.

Definition 2: Let $A_1, A_2, \ldots, A_n$ be a collection of events in a probability space. Then the configuration of this collection is the hypergraph $H = (\{1, 2, \ldots, n\}, E)$, where $S \in E$ iff $\delta_{S} = 0$.

Now I can reformulate my question thus: is the following conjecture false?

Conjecture: Every hypergraph $H = (\{1, 2, \ldots, n\}, E)$ is the configuration of some collection $A_1, A_2, \ldots, A_n$ of events in a probability space, as long as $\emptyset \notin E$ and $\{1\}, \{2\}, \ldots, \{n\} \notin E$.

Even though I want the conjecture to be false, I expect it to be true. So I guess we must try and prove it. But it is not obvious to me how to.

It might be of interest to note that for each natural number $n$, there are $2^{2^n - n - 1}$ different hypergraphs $H$ (as in the Conjecture)!

(That's quite a lot: http://www.wolframalpha.com/input/?i=2%5E%282%5En+-+n+-+1%29.)

But some of them can be "transformed" into others by re-indexing, like the hypergraph with just one edge linking the vertices $1$ and $2$, which is basically the same as the hypergraph with just one edge linking the vertices $2$ and $3$.

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I don't know the general answer, but here's an example complementary to the one given for $n=3$. Put $8$ cards in a hat, each with a $3$-digit number on it. Two of the cards have the number $121$, two of them have $112$, and there's one card with each of the numbers $211,\,212,\,221,\,222$. Now pick a card uniformly at random, and let $A_i$ be the event that the $i$th digit is a $2$. Then $P(A_i)=1/2$ for each $i$, $P(A_1\cap A_2)=P(A_1\cap A_3)=1/4$, $P(A_1\cap A_2\cap A_3)=1/8$, but $P(A_2\cap A_3)=1/8$.

I suspect something like this can be arranged for any subset of events, for every $n\ge3$.

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Nice. This example begs for a generalization! –  TuringMachine Feb 18 '13 at 1:36
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