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A Nash equilibrium of a multi-person game is a collection of strategies such that each player's strategy is a best response to the others: no unilateral change raises that player's expected payoff. For mixed strategies, every pure action used with positive probability must attain the same maximal payoff. Take $c>0$.

In this costly turnout game, compute the payoff difference between voting and abstaining, holding the others' strategies fixed. The opponent $A$ is pivotal exactly when one of $B,C$ votes. Without $A$ the proposition then passes, while with $A$ the tied ballot fails. The pivotal voting probability is $2\beta (1-\beta )$, soIf $A$ mixes with $0<\alpha <1$, indifference is necessary. Therefore

For supporter $B$, voting is pivotal when $A$ abstains and $C$ abstains, or when $A$ votes and $C$ votes. Thus its voting advantage, and likewise that of $C$, isAt $(\alpha ,\beta )=(1,3/4)$, this equals zero, while ${\Delta }_A=c(16(3/4)(1/4)-3)=0$. Every player is indifferent between voting and abstaining at these opponent strategies; in particular, $A$'s pure vote and both supporters' mixtures are best responses. Hence

It is not unique. At $(\alpha ,\beta )=(0,1/4)$, both supporter advantages are $4c(3/4)-3c=0$, and ${\Delta }_A=0$ again. This gives a second symmetric Nash equilibrium. To see these are the only equilibria with a common supporter probability, note that $\beta =0$ or $1$ makes ${\Delta }_A=-3c$, forcing $\alpha =0$; neither endpoint then satisfies the supporters' best response condition. For $0<\beta <1$, supporter indifference requiresIf $A$ also mixes, its two possible values $\beta =1/4,3/4$ force $\alpha =0,1$ respectively, contradicting strict mixing. Finally $\alpha =0$ gives $\beta =1/4$, and $\alpha =1$ gives $\beta =3/4$. Thus

Allowing $B,C$ to have distinct probabilities $\beta ,\gamma$ gives two additional Nash equilibria. The three voting advantages areIf both supporters mix, subtracting their indifference equations gives $(2\alpha -1)(\gamma -\beta )=0$. At $\alpha =1/2$ a supporter has advantage $-c$, so cannot mix; hence $\beta =\gamma$ and the two symmetric equilibria result. If $B$ always abstains and $C$ mixes, the latter's indifference gives $\alpha =1/4$, and $A$'s indifference gives $\gamma =3/8$. Then ${\Delta }_B=-3c/4<0$, so abstention really is optimal. Interchanging $B,C$ gives the other equilibrium. If $B$ always votes and $C$ mixes, the corresponding equations give $\alpha =3/4$, $\gamma =5/8$, but ${\Delta }_B=-3c/4$, contradicting voting. The four configurations with both supporters pure also fail a best response condition. Consequently the full collection of turnout equilibria with two supporters and one opponent is

Suppose turnout costs $3c>0$, each supporter gains $4c$ on passage, and the opponent loses $8c$. The symmetric-supporter Nash equilibria $(\alpha ,\beta ,\gamma )$ are $(0,1/4,1/4)$ and $(1,3/4,3/4)$. Allowing different supporter probabilities adds $(1/4,0,3/8)$ and $(1/4,3/8,0)$. These follow by comparing the pivotal voting probabilities with the turnout cost; no player benefits by changing their mixed strategy.

// Target: mathematical-optimization.bigb