# Game theory,core and nucleolus [closed]

Let $(N; v)$ be the a TU-game where $N = \{1,2,....,n\}; v(\emptyset) = 0$ and

$$v(S)=\begin{cases} S-1&S\text{ odd}\\ S+1&\text{otherwise} \end{cases}$$

1.Calculate for which $n$ the core of the game is nonempty

2.Find the nucleolus for all $n$

1)I triend and: if $n$ is odd

$\sum_{i\in N/\{i\} }x_i \geq n$ that is in contrast with $\sum_{i\in N }x_i =n-1$ but i don't understand how apply this in the case of $n$ even .

2)The solution of my teacher is that the nucleolus is the same of the shapley value (that it's easy to find) but i can't understand why.

## closed as unclear what you're asking by Batominovski, Daniel W. Farlow, user91500, kingW3, NamasteJul 17 '17 at 13:37

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• I don't think terms like "TU-game" and "nucleolus" are well known. You need to give us some references. – Batominovski Jul 16 '17 at 10:23
• – gdlzzzz Jul 16 '17 at 10:34

## 1 Answer

I assume that your game is defined by $v(\emptyset)=0$, and

\begin{equation*} v(S) = \begin{cases} |S|-1, & \text{if}\; |S| \;\text{odd}\; \\ |S|+1, & \text{otherwise}\; \end{cases} \end{equation*}

Then the game is symmetric, and it is well known that for symmetric games the Shapley value and the nucleolus coincide, which is given by $x_{i}=v(N)/n$ for all $i \in N$. Moreover, if the game is symmetric, then the center $\vec{x}$ must be an element of the core, if it exists. Thus, we have to establish for core existence that $x(S)=|S|\,v(N)/n \ge v(S)$ must hold for all $S \subseteq N$. However, this is not anymore satisfied for $n \ge 3$.