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An odd number $N \geq 3$ of players are playing a game - they bet on the outcome of a biased coin that comes up heads $p > \frac{1}{2}$ of the time, where $p$ is known to all of the players in advance.

Each player receives a payoff of $1$ if and only if they guess correctly and a minority of players bet on the outcome that they did. If they guess wrongly, or guess with the majority of players, they receive a payoff of $0$.

Players guess simultaneously and are looking to maximise only their own expected payoff.

Question: What are the set of Nash equilibria in mixed strategies? Is there a unique Nash equilibrium?

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    $\begingroup$ On a side note: never accept to play this game with unknown people, since they could have agreed to bet every time (N-1)/2 of them on head and (N-1)/2 on tail! $\endgroup$ Commented Nov 10 at 8:11
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    $\begingroup$ @PietroMajer Hahah, the effect of coalitions on the game is also pretty interesting.. $\endgroup$
    – Nate River
    Commented Nov 10 at 9:44
  • $\begingroup$ But this means that, if (by some reason or by chance) $(N-1)/2$ players adopt the strategy of betting always Heads, and $(N-1)/2$ always Tails, the last player has no hope to win, whatever is their strategy. $\endgroup$ Commented Nov 10 at 11:29
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    $\begingroup$ Indeed, and that gives a continuum of asymmetric Nash equilibria. In addition there is a symmetric Nash equilibrium, where everyone chooses Heads with the same probability $q$. I think the condition for $q$ is that $pP(X<(N-1)/2)=(1-p)P(X>(N-1)/2)$ where $X\sim\text{Binomial}(N-1,q)$. Such $q$ is unique since the ratio between the LHS and the RHS is monotonic in $q$. Are there further asymmetric ones? I would guess not. $\endgroup$ Commented Nov 10 at 11:55
  • $\begingroup$ @JamesMartin I agree that these all give Nash equilibria, but why continuum? In order for it to be a genuine Nash equilibrium, they would have to deterministically bet heads/tails, and there are only finitely many such configurations of $N-1$ players. $\endgroup$
    – Nate River
    Commented Nov 10 at 12:09

2 Answers 2

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Some thoughts about the effect of coalitions. Assuming that the players are a priori indistinguishable, since at every toss at most $\frac{N-1}2 $ of them can gain, the maximum total average payoff is $p \frac{N-1}2 $. An optimal strategy for a player, in whatever sense, is also optimal for every other one by symmetry, so the optimal rational average payoff one can hope is $p \frac{N-1}{2N} $ (here “rational” means: “assuming rational behaviour of opponents”).

If players can communicate, the evil coalition of players $P_1,\dots, P_{N-1}$ against player $P_N$, as said, is betting $\big( \frac{N-1}2 \text{ Heads}, \frac{N-1}2\text{ Tails}\big)$ every time, so that $P_N$ can do nothing but placing as bet Heads every time by revenge, so that the average total payoff of the others is minimised to $(1-p) \frac{N-1}2$.

However, in a less realistic world, players $P_1,\dots, P_{N-1}$ may agree that each of them bets, cyclically with period $N$, Heads $\frac{N-1}2$ times and Tails $\frac{N+1}2$ times, in such a way that the corresponding total wager placed among them $N-1$, is $\big( \frac{N-1}2 \text{ Heads}, \frac{N-1}2\text{ Tails}\big)$ for $\frac{N+1}2 $ times, and $\big(\frac{N-3}2 \text{ Heads}, \frac{N+1}2 \text{Tails}\big )$ for the remaining $\frac{N-1}2 $ times, and present the scheme to the last player $P_N$. This way $P_N$ can win nothing for $\frac{N+1}2$ times out of $N$, but has the opportunity of betting Heads for the remaining $\frac{N-1}2 $, ensuring the optimal rational average payoff of $p\frac{N-1}{2N}$ to everybody. In exchange, $P_N$ should only agree to bet Tail in the remaining $\frac{N+1}2$ times out of $N$, which in any case are already lost for him/her. If $P_N$ does not agree, the coalition may pass to the evil strategy against him/her in retaliation.

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    $\begingroup$ A further question: always assuming rational behaviour, how many virtuous players are needed for a coalition that forces the other ones to accept a common strategy with total average payoff $p\frac{N-1}{2}$? $\endgroup$ Commented Nov 10 at 14:52
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You may want to consult the following paper: https://pure.uvt.nl/ws/portalfiles/portal/854659/dp2007-61.pdf

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