In the process of computing Shapley values, I observed an interesting combinatorial constant. I am not exactly sure where such behavior comes. And here is the conjecture.

**Notations**

For any finite non-empty sequence of number $Q \subset \mathbb{R}$, I can construct a vector $C = (c_0, c_1, c_2, \cdots, c_{|Q|})$ using such a mapping $C(Q)$: $$ \sum_{i=0}^{|Q|} c_i x^ i = \prod_{q_i \in Q} (1+q_ix) $$

Given a sequence $A = a_1, a_2, \cdots, a_m $, denote $A_{\neg i}$ as $A \setminus \{a_i\}$. $|A| = m$ denotes size/length of $A$ is $m$. With size $m \in \mathbb{N}$, denote a constant vector $N(m)$ as $(1/{m-1 \choose 0}, 1/{m-1 \choose 1}, \cdots, 1/{m-1 \choose m-1 } )$. And $\langle\cdot, \cdot \rangle$ is the inner product operation.

**Conjecture**

For any finite non-empty sequence $A \subset \mathbb{R}$ with $|A| = m$, i conjecture if $0 \in A$ then:

$$ \sum_{i \in A} (1 - i) \langle C(A_{\neg i}), N(m) \rangle = m $$

I'm able to obtain consistent results from simulations, but I don't know how to prove it. Any help on a proof/dis-proof would be appreciated!

**Example**

For $A = 1, 2, 3$, the size $m$ is clearly 3.

$A_{\neg 1} = 2, 3$, $A_{\neg 2} = 1, 3$, $A_{\neg 3} = 1, 2$

$C(A_{\neg 1}) = (1, 5, 6)$, $C(A_{\neg 2}) = (1, 4, 3)$, $C(A_{\neg 3}) = (1, 3, 2)$.

$N(3) = (1, 0.5, 1)$

$$ (1-1)* \langle (1, 5, 6), (1, 0.5, 1) \rangle \\ + (1-2)* \langle (1, 4, 3), (1, 0.5, 1) \rangle \\ + (1-3)* \langle (1, 3, 2), (1, 0.5, 1) \rangle \\ = -15 $$ which is not $m$.

On the other hand: For $A=0, 2, 3$, the size $m$ is still 3.

$A_{\neg 0} = 2, 3$, $A_{\neg 2} = 0, 3$, $A_{\neg 3} = 0, 2$

$C(A_{\neg 0}) = (1, 5, 6)$, $C(A_{\neg 2}) = (1, 3, 0)$, $C(A_{\neg 3}) = (1, 2, 0)$.

$N(3) = (1, 0.5, 1)$

$$ (1-0)* \langle (1, 5, 6), (1, 0.5, 1) \rangle \\ + (1-2)* \langle (1, 3, 0), (1, 0.5, 1) \rangle \\ + (1-3)* \langle (1, 2, 0), (1, 0.5, 1) \rangle \\ = 3 $$ which is $m$.

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