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It seems like most game theory tutorials focus on 2-player games and often algorithms for finding Nash equilibria break down with 3+ players. So here is a simple question:

Is $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ the only Nash equilibrium in a 3-player game of Rock Paper Scissors? How can we discover this analytically?

Edit: Payoff matrices below, in terms of P1 payoff.


                Rock    Paper   Scissors
    Rock     |   0   |   -1   |    0.5  |
P2  Paper    |  -1   |   -1   |    0    |
    Scissors |  0.5  |    0   |    2    |


                Rock    Paper   Scissors
    Rock     |   2   |   0.5  |    0    |
P2  Paper    |  0.5  |    0   |   -1    |
    Scissors |   0   |   -1   |   -1    |


                Rock    Paper   Scissors
    Rock     |  -1   |    0   |   -1    |
P2  Paper    |   0   |    2   |   0.5   |
    Scissors |  -1   |   0.5  |    0    |
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It's not obvious to me what the payoff of a $3$-player version of the game would be. Wikipedia has a long article on the game but doesn't mention a version with more than $2$ players. Two possible payoff definitions might be that a) a point is awarded for each of the three pairs according to the regular rules or b) a player wins a point if and only if her move beats both of the other moves. – joriki Jan 13 '13 at 20:03
I've added the payoff matrices that I'm thinking about. – Wesley Tansey Jan 13 '13 at 20:27
How on earth did you expect us to guess that that was what you had in mind? – joriki Jan 13 '13 at 21:16
I suppose it was the most intuitive payoff matrix to me. It's similar to poker, where you have a single pot and draws result in split pots. – Wesley Tansey Jan 13 '13 at 21:20
With these payoff matrices, two players can clean out the third player simply by arranging to always play differently from each other. The third player will break even one-third of the time,lose $1$ one-third of the time, and gain $0.5$ one-third of the time. – TonyK Jul 1 '15 at 15:00

The procedure for finding mixed-strategy nash equilibrium should not be different when there are three players than when there are 2.

As in the two players case, the key point is that if it is optimal for you to randomize between different strategies, it must be the case that the expected payoff of each strategy is the same (assuming that agents are expected utility maximizers). If you randomize over say two strategies say Rock and Paper but ${U}((1,0,0), s_{-i}) > U((0,0,1), s_{-i})$ then you are definitely not optimizing.

Let $p_i(s)$ be the probability that player $i = 1,2,3$ plays strategy $s = r,p,s$. So to get a mixed-strategy nash equilibrium it needs to be the case that (assuming the probabilities are independent)

$U_1((1,0,0),s_{-i}) = 0*p_2(r)*p_2(r) + (-1)*p_2(r)*p_3(p) + ...$

be equal to

$U_1((0,1,0),s_{-i}) = 2*p_2(r)*p_3(r) + (0.5)*p_2(r)*p_3(p) + ...$

which should itself be equal to

$U_1((0,0,1),s_{-i}) = (-1)*p_2(r)*p_3(r) + (0)*p_2(r)*p_3(p) + ...$

This should be true for every $i=1,2,3$ which leaves you with a simple system of equation to solve. A profile of strategy is a mixed-strategy nash equilibrium iff it solves this system.

Based on that, you'll see that it is easy to determine whether there are other mixed-strategy nash equilibrium.

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