# The Vote with the Wallet Game: Responsible Consumerism as a Multiplayer Prisoner's Dilemma
## Appendix A  
Proof of Equation (4). : We want to show that if \(\frac{1}{n}b+a>c\) mutual responsible voting is the unique NE of the game, and if \(\frac{1}{n}b+a<c\) then mutual conventional voting (buying the conventional product) is the unique NE of the game (the extensive form game is represented in Figure A1). Without loss of generality, let assume that \(n-k\) responsible voters and \(k\) conventional voters, and we want to analyse whether it is a Nash equilibrium. Now, if \(\frac{1}{n}b+a>c\) (respectively, if \(\frac{1}{n}b+a<c\)), then each responsible voter (respectively, conventional voter) has no profitable deviation while each conventional voter (respectively, responsible voter) has always a profitable deviation, since \(\frac{n-k+1}{n}b+a-c>\frac{n-k}{n}b\) (respectively, \(\frac{n-k+1}{n}b+a-c<\frac{n-k}{n}b\)). Then the unique equilibrium is mutual responsible voting (i.e., \(k=0\)) if and only if \(\frac{1}{n}b+a>c\) and mutual conventional voting (i.e., \(k=n\)) otherwise. Proof of Equation (5). : Solving for \(\delta\), we have that  
\[\frac{n-1}{n}b-(b+a-c) \leq (b+a-c)\sum_{t=1}^{+\infty}{(1-\delta)^{t}}\] \[\frac{n-1}{n}b) \leq (b+a-c)\sum_{t=0}^{+\infty}{(1-\delta)^{t}}\] \[\frac{n-1}{n}b \leq (b+a-c)\frac{1}{\delta}(\text{from the fact that }\sum_{t=0}^{+\infty}(1-\delta)^{t}=\frac{1}{1-(1- \delta)})\] \[\delta \leq \frac{n}{n-1}(\frac{b+a-c}{b})\] \[\delta \leq \frac{n}{n-1}(1-\frac{c-a}{b}).\]  
Proof of Proposition 1.: ([43], Supporting Information, Proposition 4) characterizes all pure memory-one strategies that allow for mutual cooperation in a social dilemma. Hence, for the vote with the wallet game a pure memory-one strategy allowing for mutual responsible voting must satisfy the following conditions: (i) \(p_{VR,n-1}=1\), (ii) \(p_{VR,n-2}=0\), (iii) \(p_{A,1}\leq\frac{b+a-c-\frac{1}{n}b}{\frac{n-1}{n}b-(b+a-c)}\), (iv) \(p_{A,0}\leq\frac{b+a-c}{\frac{n-1}{n}b-(b+a-c)}\). For the [PERSON] strategy, conditions (i) and (ii) always hold, and condition (iii) holds if and only condition (4) holds. Indeed, (4) implies \(0\leq\frac{b+a-c-\frac{1}{n}}{\frac{n-1}{n}b-(b+a-c)}\) since \(b+a-c-\frac{1}{n}\geq 0\) and \(\frac{n-1}{n}b-(b+a-c)\geq 0\). Lastly, for condition (iv) we have \[p_{A,0}=1\leq\frac{b+a-c}{\frac{n-1}{n}b-(b+a-c)} \Longleftrightarrow \frac{n-1}{n}b-(b+a-c)\leq b+a-c\] \[\Longleftrightarrow -b-2(a-c)\leq\frac{1}{n}b\] \[\Longleftrightarrow n(-1-\frac{2(a-c)}{b})\leq 1\] \[\Longleftrightarrow \begin{cases}n\leq\frac{b}{2(c-a)-b}&\text{if }-1-\frac{2(a-c)}{b}\geq 0 \\ n>\frac{b}{2(c-a)-b}&\text{if }-1-\frac{2(a-c)}{b}<0\end{cases} \] \[\Longleftrightarrow \begin{cases}n\leq\frac{b}{2(c-a)-b}&\text{if }c\geq\frac{b}{2}+a \\ n>\frac{b}{2(c-a)-b}&\text{if }c<\frac{b}{2}+a\end{cases} ,\]  
and since \(n\) is a natural number and \(\frac{b}{2(c-a)-b}<0\) if \(c<\frac{b}{2}+a\), then we have that  
\[ \begin{cases}n\leq\frac{b}{2(c-a)-b}&\text{if }c\geq\frac{b}{2}+a\\ n>0&\text{if }c<\frac{b}{2}+a\end{cases} .\]  
\(\Box\)  
**Proof. (pTFT is a ZD strategy). ** From Equation (6), setting \(\phi=c-a\) and \(\sigma=1\), we have \(\mathbf{p}=\mathbf{p}^{R}+\frac{1}{c-a}(\mathbf{g}^{i}-\mathbf{g}^{-i})\) writes  
\[\mathbf{p}_{TFT} = \begin{pmatrix}1\\ 1\\ \vdots\\ 1\\ 0\\ \vdots\\ 0\end{pmatrix} +\frac{1}{c-a}\Big{(} \begin{pmatrix}\frac{1}{n}b+a-c\\ \frac{2}{n}b+a-c\\ \vdots\\ b+a-c\\ 0\\ \frac{1}{n}b\\ \vdots\\ \frac{n-1}{n}b\end{pmatrix} - \begin{pmatrix}\frac{1}{n}b\\ \frac{2}{n}b+a-c+(n-2)\frac{2}{n}b\\ \vdots\\ b+a-c\\ 0\\ \frac{1}{n}b+a-c\end{pmatrix} \Big{)}\] \[= \begin{pmatrix}1\\ 1\\ 1\\ 1\\ 0\\ \vdots\\ 0\end{pmatrix} +\frac{1}{c-a}\cdot \begin{pmatrix}\frac{a-c}{n-1}\\ \vdots\\ 0\\ \frac{c-a}{n-1}\\ \vdots\\ c-a\end{pmatrix} = \begin{pmatrix}0\\ \frac{1}{n-1}\\ \vdots\\ 1\\ 0\\ \frac{1}{n-1}\\ \vdots\\ 1\end{pmatrix} \]  
**Proof of Proposition 2. ** We apply ([43], Supporting information, Proposition 3) to our game, and we assume \(\sigma\geq\frac{n-2}{n-1}\) and \(l=b+a-c\). By contradiction, we also assume that the ZD strategy \((l,\sigma)\) is not a Nash equilibrium.