## 3 System Model
### Expected Payoff  
To reach the equilibrium, the nodes play mixed strategies. This means any node has permission to declare itself as cluster head (CH) with probability \(p\) and non-cluster head (NCH) with probability \(p=1-q\). Theorem 1: _The mixed strategies Nash equilibrium exists for the symmetrical clustering game, and the probability \(p\) for the player as cluster head (CH) in the equilibrium is given as_  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]  
Proof: To find out the Nash equilibrium in mixed strategies, which corresponds to the probability \(p\) of a node as cluster head (CH) by using the methodology as presented in the [36]. First of all, we need to find out the expected payoff for every available choice. When the node act as a cluster head (CH), the expected payoff is \(U_{CH}=z-c\), which is independent of other node strategies. When the node is playing non-cluster head (NCH) then the expected payoff is  
\[U_{NCH}=Pr\{\text{no player declare as CH}\}\cdot 0+Pr\{\text{at least one declare as CH}\}\cdot z\]  
\[=z\cdot(1-q^{N-1})\]  
\[=z\cdot(1-(1-p)^{N-1})\]  
The payoffs are equal in the equilibrium. Therefore, no player has the incentive to change its strategy. Thus,  
\[z-c=z\cdot(1-(1-p)^{N-1}).\] \begin{table}
\begin{tabular}{c c c} \hline \hline  & & **Player 2** \\ \hline
**Player 1** & **Cluster Head (CH)** & **Non-Cluster Head (NCH)** \\ \hline CH & \(z-c_{j},z-c_{j}\) & \(z,z-c_{j}\) \\ NCH & \(z-c_{j},z\) & _0, 0_ \\ \hline \hline \end{tabular}
\end{table} 
Table 2: The (2 \(\times\) 2) players payoff. We can compute the probability \(p\) by solving the above equation, which corresponds to the equilibrium. \[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]  
Let us denote \(\omega=\frac{c}{z}<1\). Figure 4 shows the probability values as the nodes increases for different \(\omega\) parameter values (i.e., 0.05, 0.1, 0.3, 0.5, 0.7 and 0.9). As the number of players increases, the probability \"\(p\)\" increases. When the attributes of the nodes (i.e., mobility speed, energy level and distance) are similar in a network, then the equilibrium condition is given by the probability as:  
\[ForN=1,p=1-(\frac{c}{z})^{\frac{1}{N-1}}=1. \tag{9}\]  
The probability of being cluster head (CH) is always 1 if \(N=1\), i.e., that at least one node must play as cluster head (CH). \[For^{\prime}N^{\prime}nodes\Rightarrow p=1-(\frac{c}{z})^{\frac{N}{N-1}}. \tag{10}\]  
The average payoff of an arbitrary node \"\(i\)\" is specified as  
\[P=(z-c)\cdot Pr\{S_{i}=CH\}+z\cdot Pr\Big{\{}S_{i}=NCH\bigcap\exists jis.tS_{j} =CH,j\
eq i\Big{\}}\]  
Figure 4: Probability \(p\) of a player for being a cluster head (CH) versus number of players for different ‘\(\omega\)’ values. \[=(z-c)\cdot Pr\{S_{i}=CH\}+z\cdot Pr\{S_{i}=NCH\}\cdot Pr\{\exists js.tS_{j}=CH,j \
eq i\}\]  
\[=(z-c)\cdot p+z\cdot(1-p)\cdot(1-Pr\{S_{j}=CH,j\
i N,j\
eq i\})\]  
\[=(z-c)p+z(1-p)(1-(1-p)^{N-1})\]  
\[P=z-c_{j}p-z(1-p)^{N}. \tag{11}\]  
For the equilibrium strategy, the average payoff \(P_{NE}\) is given as  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]0  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]1  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]2  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]3  
To understand the scenario with nodes consisting of different attributes (i.e., mobility speed, energy level and distance), consider two players' games as shown in \[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]4 The probability of a node being a cluster head (CH) for two nodes \((n_{1},n_{2})\) with mixed strategy equilibrium is calculated as follows: let us suppose \(P_{i}\) is the probability of every node in the network with the cost \(c_{i}\), for \(i=1,\,2,\,3,\,4,\ldots n\). The probability of \(n_{1}\) node being a CH with cost \(c_{1}\) is derived as follows on the basis of Equation (11) :  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]5  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]6  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]7  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]8  
\[p=1-(\frac{c}{z})^{\frac{1}{N-1}}.\]9  
where \(c_{1}\) and \(c_{2}\) are the costs to be a cluster head (CH).