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→Simple First Steps
Proof using the Shapley value for a single stage of negotiation:
<math>v(\{\empty\}) = 0, \;v(\{I\}) = 0, \; v(\{E\}) = 0, \; v(\{I,E\}) = 2\cdot\frac{1}{2}^\frac{1}{2}\,</math>
:<math>\therefore \phi_I(v)= \frac{1!0!}{2!}(2\cdot\frac{1}{2}^\frac{1}{2} - 0) + \frac{0!1!}{2!}(0 - 0) = \frac{1}{2}^\frac{1}{2} \approx 0.707\,</math>
For the second stage, the characteristic function is:
:<math>v(\{\empty\}) = 0, \;v(\{I\}) = \frac{1}{2}^\frac{1}{2}, \; v(\{E\}) = \frac{1}{2}^\frac{1}{2}, \; v(\{I,E\}) = 2\cdot\frac{1}{2}^\frac{1}{2}\,</math>
This gives:
:<math>\phi_{I(2)}(v)= \frac{1!0!}{2!}(2\cdot\frac{1}{2}^\frac{1}{2}-\frac{1}{2}^\frac{1}{2}) + \frac{0!1!}{2!}(\frac{1}{2}^\frac{1}{2} - 0) = \frac{1}{2}^\frac{1}{2} \approx 0.707\,</math>
